Ardakov - Noetherian Algebras

8/3/2019 Ardakov - Noetherian Algebras

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Noetherian Algebras

1. Basic Theory

1.1. R ring with a 1, not necessarily commutative.

Definition. A left R-module is an abelian group M (under +) together with a

multiplication

R × M → M

(r, m) → r.m

satisfying

r(m + n) = rm + rn

(r + s)m = rm + sm

(rs)m = r(sm)

1m = m

for all m, n ∈ M and r, s ∈ R.

Right R-modules are defined similarly.

Examples. (1) R itself is both a left R-module and a right R-module.

(2) Any abelian group M is a left (and right) Z-module.

(3) If R = k is a field, an R-module is just a k-vector space.

1.2. Definition. The left R-module M is said to be cyclic if it can be generated

by a single element: M = Rx for some x ∈ M . M is finitely generated if it can be

written as a finite sum of cyclic submodules M = Rx1 + Rx2 + . . . + Rxn.

Lemma. Any cyclic left R-module M = Rx is isomorphic to a quotient of R.

Proof. The map R → M given by r → rx is a left R-module homomorphism which

is onto and has kernel ann(x) = {r ∈ R : rx = 0}. This is a left ideal of R, known

as the annihilator of x. Hence M ∼= R/ ann(x).

1.3. Lemma. Let M be a left R-module. The following are equivalent:

(i) Every submodule of M is finitely generated

(ii) Ascending chain condition: There does not exist an infinite strictly as-

cending chain of submodules of M

(iii) Maximum condition: Every non-empty subset of submodules of M contains

at least one maximal element. (If S is a set of submodules, then N ∈ S is a

maximal element if and only if N ∈ S , N ≤ N implies N = N ).

Proof. (i) ⇒ (ii). Suppose M 1 M 2 . . .. Let N = ∪M n. Then N is a submodule

of M so N is finitely generated by m1, . . . , mr say. If mi ∈ M ni , then it follows

that N = M n where n = max ni, a contradiction.

(ii) ⇒ (iii) If S is a nonempty subset with no maximal element, pick M 1 ∈S . Since S has no maximal element, we can find M 2 ∈ S such that M 1 M 2.

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Continuing like this gives a strictly ascending infinite chain M 1 M 2 . . ., a

contradiction.

(iii) ⇒ (i) Let N be a submodule of M and let S be the set of submodules of

N which are finitely generated. Since 0 ∈ S , S has a maximal element L, say. Let

x ∈ N . Since L + Rx is a finitely generated submodule of N and L is maximal in

S , L + Rx = L so x ∈ L. Hence N = L is itself finitely generated.

Dually, we have the descending chain condition and the minimum condition ;

these are equivalent to each other.

1.4. Definition. An R-module satisfying (i), (ii), (iii) of Lemma 1.3 is Noetherian .

The ring R is left Noetherian if it is Noetherian as a left R-module.

An R-module satisfying the descending chain condition is said to be Artinian .

The ring R is left Artinian if it is Artinian as a left R-module.We have similar definitions ”on the right hand side”. Note that if the ring is

commutative, there is no difference between ”left” and ”right”.

Examples. (1) Any division ring D is right and left Noetherian (as well as

right and left Artinian)

(2) Any principal ideal domain R is Noetherian - for example, R = Z or R =

Q[x].

1.5. Proposition. Let N be a submodule of M , a left R-module. Then M is

Noetherian if and only if both N and M/N are Noetherian.

Proof. (⇒) Since any submodule of N is a submodule of M , N is left Noetherian.

If Q1/N < Q2/N < .. . is a strictly ascending chain in M/N then Q1 < Q2 < . . . is

a strictly ascending chain in M , so M/N is left Noetherian.

(⇐) Suppose N and M/N are Noetherian and let L1 ≤ L2 ≤ . . . be an ascending

chain in M . Then (L1 + N )/N ≤ (L2 + N )/N ≤ . . . is an ascending chain in

M/N which terminates since M/N is Noetherian. Hence there exists k such that

Li + N = Lk + N for all i ≥ k. Since N is Noetherian, there exists m such that

Li ∩ N = Lm ∩ N for all i ≥ m.

We claim that Li = Ln for all i ≥ n := max(k, m). Clearly Ln ⊆ Li; let x ∈ Li.

Then x ∈ Li ⊆ Ln + N so there exists y ∈ Ln and z ∈ N such that x = y + z. Now

z = x − y ∈ Li ∩ N = Ln ∩ N so x ∈ Ln also, as required.

1.6. Lemma. Suppose M can be expressed as a sum M = M 1 + M 2 + . . . + M n.

Then M is Noetherian if and only if each M i is Noetherian.

Proof. Suppose each M i is Noetherian. By an easy induction using (1.5), the di-

rect sum N = M 1 ⊕ M 2 ⊕ · · · ⊕ M n is Noetherian. Now there is an R-module

homomorphism f : N → M such that f (m1, . . . , mn) = m1 + . . . + mn. Since

M = M 1 + M 2 + . . . + M n, f is onto so M ∼= N/ ker f is Noetherian by (1.5).



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Conversely, since each M i is a submodule of M , if M is Noetherian then each

M i is Noetherian, again by (1.5).

Corollary. Let R be a left Noetherian ring. Then any finitely generated left R-

module is left Noetherian.

Proof. Any finitely generated module is a sum of cyclic modules. The result follows

from (1.2) and (1.6).

1.7. Proposition. Let R be a left Noetherian ring and let S be a ring containing

R such that S is a finitely generated left R-module under left multiplication by R.

Then S is Noetherian.

Proof. By Corollary 1.6, S is Noetherian as a left R module. Since any left ideal

of S is necessarily an R-submodule, S satisfies the ACC on left ideals. Hence S isNoetherian.

1.8. Examples of Noetherian rings.

(1) Rings which are finite dimensional vector spaces over a field k, for example

M n(k) and group algebras kG of finite groups (see (1.12)). Use Proposition

1.7 with R = k.

(2) Rings of integers in algebraic number field, M n(Z). These are both finitely

generated Z-modules.

(3) k[[X ]], the power series ring in one variable over a field k (this is a PID).

(4) Let p be a fixed prime and let Z( p) = {q = m/n ∈ Q : (m, n) = 1 and p n}.

This is an example of a localisation of Z. The localisation of a Noetherianring is Noetherian - see later.

(5) R = k[X 1, . . . , X d], k a field. This is Hilbert’s Basis Theorem - see

Theorem 1.14.

1.9. Free algebras. Let k be a field. A k-algebra is a ring R containing k as a

central subfield.

Definition. The free associative algebra on n generators kx1, . . . , xn is the k-

vector space with basis given by all possible products y1 · · · ym where y1, . . . , ym ∈{x1, . . . , xn}. Multiplication is given by concatenation on basis elements and is

extended by k-linearity to the whole of kx1, . . . , xn

.

Note that kx1, . . . , xn is not finite dimensional over k.

For example, if n = 1 then kx has {1, x , x2, . . .} as a basis. In fact kx ∼= k[x],

the polynomial algebra.

Similarly, kx, y has as a k-basis the set {1, x , y , x2,xy,yx,y2, x3, x2y , . . .}. It

can be shown (Exercise) that kx, y is not Noetherian. For n ≥ 3 there is a ring

homomorphism kx1, . . . , xn → kx1, x2 which sends xi to xi if i = 1 or 2 and to

0 if i ≥ 3. This map is surjective, so kx1, . . . , xn is not Noetherian for n ≥ 2.



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Let V be the vector space spanned by {x1, x2, . . . , xn}. Then

T (V ) := k⊕

V ⊕

(V ⊗

V )⊕

(V ⊗

V ⊗

V )⊕ · · ·

is the tensor algebra on V . Multiplication is given on homogeneous components

by concatenation and is extended by linearity to T (V ). Check that T (V ) ∼=kx1, . . . , xn where n = dim V , as k-algebras.

1.10. Enveloping algebras.

Definition. Let k be a field. A Lie algebra over k is a k-vector space g, equipped

with a Lie bracket [.] : g × g → g satisfying

(1) (x, y) → [x, y] is bilinear

(2) [x, x] = 0 for all x ∈ g and hence [y, z] = −[z, y] for all y, z ∈ g

(3) [x, [y, z]] + [y, [z, x]] + [z, [y, x]] = 0 for all x,y,z∈g.

Note that this bracket is not associative.

Examples. (1) Any (associative) k-algebra R becomes a Lie algebra under the

commutator bracket [x, y] = xy − yx.

(2) gln(k), the set of all n × n matrices over k with the commutator bracket.

(3) sln(k), the set of traceless n × n matrices over k with commutator bracket.

(4) If V is any vector space, we can define the trivial bracket [x, y] = 0 for all

x, y ∈ V . This is the abelian Lie algebra.

Definition. The universal enveloping algebra U (g) of the Lie algebra g is defined

to be

U (g) := kx1, . . . xn/I

where {x1, x2, . . . , xn} is a basis for g and I is the ideal of kx1, . . . xn generated

by the set {xixj − xjxi − [xi, xj ], 1 ≤ i, j ≤ n}.

For example, if g is abelian, then U (g) is just the polynomial algebra k[x1, . . . , xn].

Proposition. U (g) is left (and right) Noetherian for any finite dimensional Lie

algebra g.

Proof. See Corollary 7.8.

1.11. Weyl algebras. We begin with a motivational result.

Lemma. Let A = k[x] and consider the k-linear maps ∂ ∂x : A → A and x : A → A,where x is multiplication by x. Then

[∂

∂x, x] = 1.

Proof. By the product rule,

[∂

∂x, x](f ) = (xf ) − xf = f

for all f ∈ A.



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Now consider the polynomial algebra A = k[x1, . . . , xn] and the k-linear maps

xi : A → A and

∂

∂xi : A → Af → xif f → ∂f

∂xi,

for 1 ≤ i ≤ n. These maps are examples of differential operators on A. It can be

verified that all these operators commute, except for ∂ ∂xi

and xi, which satisfy the

relation

[∂

∂xi, xi] = 1.

Definition. Let k be a field. The n-th Weyl algebra An(k) over k is defined to be

An(k) := kx1, . . . , xn, y1, . . . yn/I

where I is the ideal of kx1, . . . , xn, y1, . . . yn generated by

xixj − xjxi 1 ≤ i, j ≤ n,

yiyj − yjyi 1 ≤ i, j ≤ n,

yixi − xiyi − 1 1 ≤ i ≤ n,

xiyj − yjxi i = j.

When n = 1, we obtain

A1(k) = kx, y/yx − xy − 1.

There is a surjection An(k) S where S is the k-subalgebra of Endk(A) gen-

erated by

{ x1, . . . , xn, ∂f ∂x1

, . . . , ∂f ∂xn

}, mapping xi to xi and yi to ∂f

∂xi. In fact,

An(k) ∼= S if and only if the characteristic of k is zero.

Proposition. The Weyl algebras are left (and right) Noetherian for all n ≥ 1.

Proof. See Corollary 7.8.

1.12. Group Algebras.

Definition. Let G be a group and let R be a ring. The group algebra RG consists

of formal linear combinations g∈G

rgg,

where rg ∈ R for all g ∈ G and all but finitely many rg are zero. Addition and

multiplication is given by

(g∈G

rgg) + (g∈G

sgg) =g∈G

(rg + sg)g

(h∈G

rhh)(k∈G

skk) =g∈G

(h,k∈G

hk=g

rhsk)g.



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When k is a field and V is a k-vector space, recall that a representation of G on

V is a group homomorphism

ϕ : G → Autk(V ).

Such a homomorphism extends uniquely to a k-algebra homomorphism

ϕ : kG → Endk(V )

and V may then be regarded as a left kG-module, via

x.v = ϕ(x)(v)

for all x ∈ kG.

Conversely, if V is a left kG-module, there is a representation ϕ : G → Autk(V )

given by

ϕ(g)(v) = g.v

for all v ∈ V .

Thus there is a 1-1 correspondence between representations of G and left kG-

modules.

1.13. Group algebras of poly-(cyclic or finite) groups.

Definition. The group G is said to be poly-(cyclic or finite) if there is a chain

1 = G0 G1 . . . Gn−1 Gn = G

of subgroups of G such that each Gi/Gi−1 is cyclic or finite for each i = 1, . . . , n.

Examples. (1) Finite groups G.(2) Infinite cyclic G = x ∼= Z. In this case, kG ∼= k[x−1, x], the ring of

Laurent polynomials.

(3) Free abelian G = x1, . . . , xn ∼= Zn: kG ∼= k[x−11 , x1, . . . , x−1n , xn].

(4) G =

1 Z Z

0 1 Z0 0 1

. Here we have the chain

1 G1 G2 G3 = G

where G1 =

1 0 Z0 1 0

0 0 1

and G2 =

1 Z Z0 1 0

0 0 1

.

Proposition. Let R be a left(right) Noetherian ring and let G be a poly-(cyclic or

finite) group. Then RG is left(right) Noetherian.

Proof. After the proof of Hilbert’s Basis Theorem.

Question. Suppose ZG is left Noetherian. Must G be poly-(cyclic or finite)?

Theorem (Auslander-Swan). Let G be a poly-(cyclic or finite) group. Then



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(1) G is isomorphic to a subgroup of GLn(Z) for some n.

(2) G has a subgroup H of finite index isomorphic to a subgroup of Tr n(

O),

the group of upper-triangular matrices in GLn(O), where O is the ring of

integers of an algebraic number field.

Proof. Omitted.

1.14. Hilbert’s Basis Theorem. We will prove the following noncommutative

version of Hilbert’s Basis Theorem:

Theorem (McConnell, 1968). Let S be a ring, R a left Noetherian subring and let

x ∈ S .

(1) If R + xR = R + Rx and S = R, x, then S is left Noetherian.

(2) Suppose ϕ

∈Aut(R) is such that rx = xrϕ for all r

∈R. If S =

R, x

,

then S is left Noetherian.

(3) Suppose x is a unit in S such that x−1Rx = R. If S = R,x,x−1, then S

is left Noetherian.

Proof. (1). R + Rx + . . . + Rxn = R + xR + . . . + xnR:

This follows by from R + xR = R + Rx. To see this, use induction to show that

xnR ⊆ R + Rx + . . . + Rxn and Rxn ⊆ R + xR + . . . xnR for all n ≥ 1.

Consequences:

(a) The set of all elements of S of the form

r0 + xr1 + . . . + xnrn, n ≥ 0 (∗)

forms a subring of S . Since it contains both R and x and S = R, x, we seethat S is the ring of all such ’polynomials’. Note that elements of S need not

be uniquely expressible in the form (∗).

(b) The set of polynomials of degree ≤ n, namely R + Rx + . . . + Rxn, is both a

left and a right R-submodule of S .

(c) For each r ∈ R and n ≥ 0 there exists r ∈ R such that rxn = xnr + s where

deg s < n.

Now, let I be a left ideal in S . We will show that I is finitely generated. Let

I n = {rn ∈ R : there exists s ∈ I such that s = r0 + xr1 + . . . + xnrn}.

It’s clear that I n is closed under addition. Let r ∈ R. By part (c) above, we can

find r ∈ R such that rxn −xnr has degree < n. Since I is a left ideal, rs ∈ I , and

rs ≡ rxnrn ≡ xn(rrn)

modulo terms of degree < n. Hence rrn ∈ I n so I n is a left ideal of R.

Next, if s =n

i=0 xiri ∈ I , then xs =n+1

i=1 xiri−1 ∈ I so rn ∈ I n+1. Hence

I n ≤ I n+1 for all n ≥ 0.

Since R is left Noetherian, the increasing chain

I 0 ≤ I 1 ≤ . . . ≤ I n ≤ . . .



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must terminate. Say I m = I m+1 = . . .. For i = 0, . . . , m let {rij} be finitely many

elements of R generating I i as a left ideal of R. Choose sij = xirij+ lower degree

terms ∈ I .

Claim: X = {sij : 0 ≤ i ≤ m, all j} generates I as a left ideal.

Let s = r0 + xr1 + . . . + xnrn ∈ I , so that rn ∈ I n; we’ll show that s ∈ RX .

Proceed by induction on n, the case n = 0 being trivial.

If n ≥ m then rn ∈ I m so rn =

ajrmj for some aj ∈ R. Choose aj ∈ R such

that ajxn = xnaj+ lower degree terms. Then s − ajxn−msmj ∈ I and modulo

terms of degree < n,

s −

ajxn−msmj ≡ xnrn −

ajxnrmj ≡ xnrn −

xnajrmj = 0.

So s − ajxn−msmj has smaller degree than s and we can apply induction.

If n≤

m then rn = ajrnj for some aj

∈R, so for suitable a

j ∈R, s

− ajsnj ∈ I also has smaller degree than s. By induction, these smaller degree

elements of I lie RX , as required.

(2) If rx = xrϕ for all r ∈ R, then Rx = xR so R + xR = R + Rx. Apply (1).

(3) Let T = R, x. Then T is left Noetherian by (2). Let I be a left ideal of S .

Now, I ∩ T is a left ideal of T and is hence finitely generated: I ∩ T =n

i=1 T si,

say. If s ∈ S , then xms ∈ I ∩ T for some m ≥ 0, so s =n

i=1 x−maisi for some

ai ∈ T . Hence the si’s generate I as a left ideal of S .

Proof of Proposition 1.13. Choosing a chain of subgroups

1 = G0 G1 . . . Gn−1 Gn = G

we see that it’s sufficient to show that if RGi−1 is left Noetherian then so is RGi

for all i = 1, . . . , n.

Suppose first that Gi/Gi−1 is finite. Choose a finite transversal X for Gi−1 in

Gi, so that Gi = ∪x∈XGi−1x. Then

RGi =x∈X

RGi−1x

is a finitely generated left RGi−1-module, so RGi is left Noetherian by Proposition

1.7. Now, if Gi/Gi−1 is infinite cyclic, choose a generator xGi−1; then RGi is gen-

erated by RGi−1,x and x−1. Since Gi−1 Gi, RGi−1 is invariant under conjugation

by x, so RGi is left Noetherian by Theorem 1.14(3).



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2. Ideal structure

Throughout this chapter, R denotes an arbitrary ring, unless stated otherwise.2.1. Simple modules.

Definition. An R-module is M is simple or irreducible if M = 0 and the only

submodules of M are 0 and M .

Suppose M is simple. Choose 0 = x ∈ M ; then M = Rx so M ∼= R/I where

I = ann(x) is the point annihilator of x, by Lemma 1.2. Note that ann(x) need

not be equal to ann(y) if x, y are distinct nonzero elements of M , unless R is

commutative.

Note that M = Rx is simple if and only if ann(x) is a maximal left ideal of R.

2.2. Zorn’s Lemma.

Definition. A poset is a set equipped with a binary relation ≤ which is reflexive,

transitive and antisymmetric. A chain in a poset S is a countable subset C =

{x1, x2, . . .} of S such that

x1 ≤ x2 ≤ . . . .

An upper bound for a subset C of S is an element u ∈ S such that x ≤ u for all

x ∈ C. We say that x ∈ S is a maximal element if x ≤ y with y ∈ S forces x = y.

Theorem (Zorn’s Lemma). Let S be a nonempty poset. Suppose every chain in S has an upper bound. Then S has a maximal element.

This is equivalent to the Axiom of Choice, which we will always assume.

2.3. Lemma. Suppose L is a proper left ideal of R. Then L is contained in a

maximal ideal I of R. Equivalently, every cyclic module has a simple quotient.

Proof. Since L is proper, 1 /∈ L. Let S = {K l R : L ⊆ K, 1 /∈ K }. Since L ∈ S ,this set is nonempty. S is partially ordered by inclusion. If K 1 ⊆ K 2 ⊆ . . . is a chain

in S , then ∪K n also contains L and doesn’t contain 1, i.e. ∪K n ∈ S . Hence every

chain in S has an upper bound in S . By Zorn’s Lemma, S has a maximal element

I . It’s clear that I is now a maximal left ideal of R containing L as required.

2.4. Primitive ideals. Recall that by an ideal of R we mean a two-sided ideal.

Definition. Let I be a two-sided ideal of R. Then I is primitive if I is the anni-

hilator of a simple left R-module M :

I = AnnR(M ) = {x ∈ R : xM = 0} =

x∈M

ann(x).

This definition is not symmetrical, and it is known that left primitivity does not

imply right primitivity. Nonetheless, the attribute ”left” is usually omitted. Note

that the annihilator I of any module M is always an ideal of R.



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Lemma. Let M = Rx be a simple left R-module. Then I = AnnR(M ) is the

largest two-sided ideal contained in L = ann(x).

Proof. Note that this largest two-sided ideal K exists, since the sum of all two-sided

ideals contained in L is itself a two-sided ideal contained in L. Certainly I ⊆ L, so

I ⊆ K . Now KM = KRx ⊆ Kx ⊆ Lx = 0 since K is two-sided, so K ⊆ I .

Corollary. Every maximal ideal of R is primitive. Moreover, if R is commutative,

every primitive ideal is maximal.

2.5. Jacobson radical.

Definition. The Jacobson radical J (R) of R is defined to be the intersection of all

primitive ideals of R. The ring R is said to be semiprimitive if J (R) = 0.

It will be shown in Corollary 2.7 that this definition is left-right symmetric. Note

that J (R) is the set of elements of R which annihilate every simple left R-module.

Lemma. J (R) is equal to the intersection K of all maximal left ideals of R.

Proof. Let I be a maximal left ideal. Then P = AnnR(R/I ) is primitive, so J (R) ⊆P ⊆ I by Lemma 2.4. Hence J (R) ⊆ K .

Now let P = AnnR(M ) be a primitive ideal, where M is a simple R-module.

Note that P = ∩0=x∈M ann(x) is an intersection of maximal left ideals, so K ⊆ P .

It follows that K ⊆ J (R) as required.

2.6. Nakayama’s Lemma.

Lemma. Let M be a finitely generated nonzero left R-module and let J = J (R).

Then JM is strictly contained in M .

Proof. Since M is finitely generated, we can choose a cyclic quotient N of M , which

has a simple quotient M/K by Lemma 2.3. Then J.(M/K ) = 0 so JM ⊆ K which

is strictly contained in M .

Note that the condition that M is finitely generated is necessary here: if p is a

prime, R = Z( p) and M = Q then J = pR and JM = pQ = M .

2.7. Recall that an element x ∈ R is a unit if there exists y ∈ R such that

xy = yx = 1.Proposition.

J (R) = {x ∈ R : 1 − axb is a unit for all a, b ∈ R} =: K.

Proof. Let x ∈ K , let I be a maximal left ideal of R and suppose that x /∈ I . Since

I is maximal, I + Rx = R, so 1 − ax ∈ I for some a ∈ R. Since x ∈ K , 1 − ax

is a unit, a contradicting the fact that I is proper. Hence x ∈ I so K ⊆ I for all

maximal left ideals I of R. By Lemma 2.5, K ⊆ J (R).



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Now let x ∈ J (R). Since J (R) is a two-sided ideal, to show that x ∈ K it’s

sufficient to show 1

−x is a unit. Now, if R(1

−x) is a proper left ideal, we can find

a maximal left ideal L containing it by Lemma 2.3. By Lemma 2.5, x ∈ J (R) ⊆ L

and 1 − x ∈ L so 1 ∈ L, a contradiction. Hence there exists y ∈ R such that

y(1 − x) = 1.

Now, 1 − y = −yx ∈ J (R), so by the above argument applied to 1 − y, we can find

z ∈ R such that

z(1 − (1 − y)) = zy = 1.

Hence zy(1 − x) = 1 − x = z so zy = 1 and yz = 1, meaning that z = 1 − x is a

unit. as required.

This result shows that J (R) is the largest ideal A of R such that 1

−A consists

entirely of units of R.

Corollary. The Jacobson radical is left-right symmetric. It follows that the inter-

section of all maximal left ideals of R is equal to the intersection of all maximal

right ideals.

2.8. Prime ideals.

Definition. The ideal P of R is said to be prime if P = R and whenever A, B are

ideals of R such that AB ⊆ P , either A or B is contained in P . The ring R is

said to be prime if 0 is a prime ideal. The ideal N of R is nilpotent if there exists

n

≥0 such that N n = 0.

Note that if R is commutative, then P is prime if and only if R/P is an integral

domain, which agrees with the old definition.

Lemma. (1) Let P be prime and let N be nilpotent. Then N ⊆ P .

(2) Any primitive ideal P = AnnR(M ) is prime.

Proof. (1) Since N n = 0 ∈ P , either N ⊆ P or N n−1 ⊆ P . Continue.

(2) Suppose A, B R are such that AB ⊆ P . Then ABM = 0. If BM = 0 then

B ⊆ P . Otherwise, BM = M since M is simple, so AM = 0 and A ⊆ P .

2.9. Minimal primes.

Definition. Let I R. A prime P of R is a minimal prime over I if P ⊇ I and

I ⊆ Q ⊆ P with Q prime forces Q = P . P is a minimal prime of R if it is a

minimal prime over 0.

Proposition. Let R be a left Noetherian ring and let I R be a proper ideal. Then

(1) There exist primes P 1, . . . , P n containing I such that P 1 · · · P n ⊆ I .

(2) The set of minimal primes over I is equal to the set of minimal primes in

{P 1, . . . , P n}.



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Proof. Suppose that (1) is false. Since R is left Noetherian, we can choose a maximal

counterexample I . Thus I contains no finite product of prime ideals containing I ,

and I is maximal with respect to this property.

Claim: I is prime.

If I is not prime, we can find A, B R such that AB ⊆ I but A ⊆ I and B ⊆ I .

By maximality of I , I + A contains the product of primes P 1, . . . , P n containing

I + A, and similarly Q1 · · · Qm ⊆ I + B for some primes Q1, . . . , Qm containing

I + B. Hence

P 1 · · · P nQ1 · · · Qm ⊆ (I + A)(I + B) ⊆ I 2 + AI + IB + AB ⊆ I,

so I itself contains a finite product of primes containing it. This contradicts the

definition of I , so in fact I is prime.

Thus we have a contradiction, and (1) follows.

Hence we have a finite set of primes P 1, . . . , P n containing I such that P 1 · · · P n ⊆I . Let {X 1, . . . X m} be the distinct minimal primes of {P 1, . . . , P n}. Thus each P j

contains some X ij so I contains some product of the X k’s, possibly with repetition:

X i1 · · · X in ⊆ P 1 · · · P n ⊆ I.

Now, suppose Q is any prime containing I . Then X i1X i2 · · · X in ⊆ I ⊆ Q which

forces X ij ⊆ Q for some j. If Q is a minimal prime over I , Q must equal X ij .

Finally, we show that each X k is a minimal prime over I . If I ⊆ Q ⊆ X k then

X j ⊆ Q ⊆ X k for some j by the above. But the X ’s are minimal in {P 1, . . . , P n},

so X j = Q = X k and (2) follows.

2.10. Prime radical.

Definition. The prime radical N (R) of R is the intersection of all prime ideals of

R. R is semiprime if N (R) = 0. An ideal I R is semiprime if it’s an intersection

of some collection of prime ideals, or equivalently, if N (R/I ) = 0.

Note that by Lemma 2.8(1), N (R) contains every nilpotent ideal of R.

Proposition. Let R be a left Noetherian ring with nilradical N = N (R) and let

P 1, . . . , P n be the minimal primes of R. Then

(1) N = P 1 ∩ . . . ∩ P n.

(2) N is nilpotent.

Proof. Part (1) is clear. By the proof of Proposition 2.9, 0 contains a product of k

of the P i’s, possibly with repetition, so N k = 0, as required for (2).

2.11. When R is commutative, there is a nice characterisation of the nilradical.

Recall that an element x is nilpotent if xn = 0 for some n ≥ 0.

Proposition. Let R be a commutative ring. Then the set K of all nilpotent ele-

ments of R is an ideal and equals N (R).



13

Proof. Let x, y ∈ K , so that xn = ym = 0 for some integers n, m ≥ 0. Clearly

(xy)n+m = 0, so xy

∈K . Now, each term in the binomial expansion of (x + y)n+m

is a multiple of either xn or ym so (x + y)n+m = 0 and x + y ∈ K . Hence K is an

ideal.

Now, if x ∈ K , xR is nilpotent so xR ⊆ N (R). Hence K ⊆ N (R). Suppose

therefore that x /∈ K . Then x is not nilpotent, so the set C = {1, x , x2, . . .} doesn’t

contain 0. Let S = {I R : I ∩ C = ∅}. Since 0 ∈ S , S is nonempty and is clearly

closed under unions. By Zorn’s Lemma, S has a maximal element P . We will show

that P is prime.

Suppose A,B R are such that AB ⊆ P but A, B are not contained in P . Then

xn ∈ P +A and xm ∈ P +B for some integers n, m, so xn+m ∈ (P +A)(P +B) ⊆ P ,

contradicting P ∈ S .Hence x /∈ K implies x /∈ P for some prime ideal P , whence x /∈ N (R).

There is a similar characterisation for N (R) when R is arbitrary in terms of

strongly nilpotent elements - see McConnell and Robson.



14

3. Artinian rings

3.1. Recall that a right Artinian ring R is one which is satisfies the DCC as aright module.

Proposition. Suppose S be a ring containing a right Artinian ring R such that S

is a finitely generated right R-module under right multiplication by R. Then S is

right Artinian.

Proof. This can be proved in exactly the same way as Proposition 1.7. In fact, the

Artinian analogues of Propositions 1.5 and 1.6 are also valid.

Examples. The following rings are all right Artinian:

(1) The group algebra kG of a finite group G over a field k.

(2) The matrix ring M n(D), where D is a division ring.

(3) Any finite ring, for example Z/mZ, m = 0.

(4) k[x]/(x2), k a field.

The structure of a right Artinian ring is quite well understood. The following is

a summary of what’s known:

Theorem. Let J = J (R) be the Jacobson radical of the right Artinian ring R.

(1) (Artin-Wedderburn) R/J is isomorphic to a finite direct sum of matrix

rings over division rings D1, . . . , Dk:

R/J ∼= M n1(D1) ⊕ M n2(D2) ⊕ . . . ⊕ M nk(Dk)

(2) J is nilpotent and J = N (R).(3) (Hopkins) R is right Noetherian.

3.2. Semisimple modules.

Definition. A right R-module is said to be semisimple or completely reducible if

M is a direct sum of simple submodules.

Lemma. Let M be a semisimple R-module. The following are equivalent:

(1) M is Noetherian.

(2) M is Artinian.

(3) M is a direct sum of finitely many simple modules.

Proof. By Proposition 1.6 and its Artinian analogue, (3) implies (1) and (2). If M is not a finite direct sum, we can find a submodule N of M which is a countably

infinite direct sum N = M 1 ⊕ M 2 ⊕ . . . of nonzero submodules M i. Now the chains

M 1 < M 1 ⊕ M 2 < M 1 ⊕ M 2 ⊕ M 3 < . . .

and

M 1 ⊕ M 2 ⊕ M 3 ⊕ · · · > M 2 ⊕ M 3 ⊕ · · · > . . .

show that M is not Noetherian nor Artinian.



15

3.3. Lemma. Any submodule N of a semisimple Artinian module M is itself

semisimple Artinian.

Proof. Let M = M 1 ⊕ M 2 ⊕ . . . ⊕ M n be a direct sum of simple submodules M i.

Proceed by induction on n, the case n = 1 being obvious.

If N ∩ M i = 0 for some i, then N → M 1 ⊕ . . . ⊕ M i−1 ⊕ M i+1 ⊕ . . . ⊕ M n, so N

is semisimple by induction. Hence we may assume that N ∩ M i = 0 for all i. But

M i is simple, so M i ⊆ N for all i, whence N = M is semisimple.

3.4. Proposition. Suppose R is semiprimitive and right Artinian. Then the right

R-module RR is semisimple.

Proof. Let S consist of all finite intersections I 1 ∩ . . . ∩ I n of maximal right ideals

I j . Since R is right Artinian, S has a minimal element L = I 1 ∩ . . . ∩ I n, say. Nowif I is a maximal right ideal, L ∩ I ∈ S , so L ∩ I = L by minimality of L. Hence

L ⊆ I for all maximal I r R, whence L ⊆ J (R) = 0 by Lemma 2.5.

Now, the natural map R −→ ⊕nj=1(R/I j) has kernel I 1 ∩ I 2 ∩ . . . ∩ I n = 0, so RR

is a submodule of the semisimple Artinian module ⊕nj=1(R/I j). The result follows

from Lemma 3.3.

Because of this result, a semiprimitive right Artinian ring is more commonly

known as semisimple Artinian . This is left-right symmetric by Wedderburn’s the-

orem - see comments in (3.7).

3.5. Some linear algebra.

Proposition. Let M = M 1 ⊕ M 2 ⊕ . . . ⊕ M n be a direct sum of right R-modules.

Then EndR(M ) = HomR(M, M ) is isomorphic to the matrix ring

S =

Hom(M 1, M 1) Hom(M 2, M 1) · · · Hom(M n, M 1)

Hom(M 1, M 2) Hom(M 2, M 2) . . . Hom(M n, M 2)

· · · · · ·· · · · · ·

Hom(M 1, M n) Hom(M 2, M n) · · · Hom(M n, M n)

with the obvious multiplication.

Proof. This is best seen by writing elements of M as column vectors and thinking

of R-module endomorphisms acting by matrix multiplication on the left of these

column vectors.

Let σj : M j → M and πi : M M i be the canonical injections and projections.

Formally, we can define a map α : EndR(M ) → S by setting the (i, j) element of

α(f ) to be the composition

M jσj→ M

f → M πi→ M i;



16

thus α(f )ij = πif σj . We can also define β : S → EndR(M ) by

β (T ) =i,j σjT jiπi.

Check that α and β are mutually inverse ring homomorphisms.

3.6. Schur’s Lemma. Let M and N be simple right R-modules. Then

(1) EndR(M ) is a division ring.

(2) If M N then HomR(M, N ) = 0.

Proof. Let ϕ : M → N be a nonzero R-module homomorphism. Then ker(ϕ) < M

and Im(ϕ) > 0. The simplicity of M and N forces ker(ϕ) = 0 and Im(ϕ) = N , so

ϕ is an isomorphism.

Part (2) follows immediately, whereas if N = M , the above argument shows that

every nonzero element of EndR(M ) is left and right invertible. Therefore EndR(M )is a division ring, proving (1).

3.7. Artin-Wedderburn.

Theorem. Let R be a semiprimitive right Artinian ring. Then R is isomorphic to

a direct sum of matrix rings over some division rings D1, . . . , Dk:

R ∼= M n1(D1) ⊕ M n2(D2) ⊕ . . . ⊕ M nk(Dk)

Proof. Consider θ : R → HomR(RR, RR) given by θ(x)(y) = xy. This is a ring

homomorphism. If θ(x) = 0, θ(x)(1) = x = 0, so θ is an injection. Also, if f : R →R is a right module map, then f (r) = f (1)r = θ(f (1))(r) so that f = θ(f (1)) and

θ is an isomorphism.By Proposition 3.4, we can write R as a direct sum of simple right R-modules.

Grouping these together, we may write

RR∼= An1

1 ⊕ An22 ⊕ . . . ⊕ Ank

k

where the Ai are pairwise nonisomorphic simple modules. Applying Proposition

3.5, we obtain

R ∼= EndR(RR) ∼=

EndR(An1

1 ) 0 · · · 0

0 EndR(An2

2 ) · · · 0

· · · · · ·

· · · · · ·0 0 · · · EndR(Ank

k )

∼=

M n1(D1) 0 · · · 0

0 M n2(D2) · · · 0

· · · · · ·· · · · · ·0 0 · · · M nk(Dk)

where each Di = EndR(Ai) is a division ring, by Schur’s Lemma (3.6).



17

As an exercise, the reader should check directly that a ring of the form

M n1

(D1)⊕

M n2

(D2)⊕

. . .⊕

M nk

(Dk)

is semisimple Artinian.

Note that if we were dealing with left R-modules, we would obtain that the op-

posite ring Rop is isomorphic to the direct sum of matrix rings. Since M n(Dop)op ∼=M n(D) and the opposite ring of a division ring is also a division ring, the theorem

holds with ’right’ replaced by ’left’. This justifies the term semisimple Artinian

ring , without reference to side.

3.8. Before we can prove Hopkins’ Theorem, we will need some results on semisim-

ple modules.

Proposition. Let R be an arbitrary ring and let M be an R-module. Suppose M

is a sum of simple submodules. Then M is semisimple.

Proof. Write M =

α∈A M α for some simple submodules M α of M . Let

S = {B ⊆ A : the sumα∈B

M α is direct}.

Since S contains singletons {α} for all α ∈ A, S is nonempty. It’s easy to check that

unions of chains in S are again in S , so by Zorn’s Lemma, S contains a maximal

element A0.

Now, if X = ⊕α∈A0M α = M , we can find M β such that M β X . Hence

M β ∩ X = 0 and the sum X + M is direct. Therefore A0 ∪ {β } ∈ S , contradicting

the maximality of A0. Hence X = M is a direct sum of simple modules, so M issemisimple.

3.9. Lemma. Suppose the ring R is semisimple Artinian. Then every R-module

is semisimple.

Proof. R is a sum of simple modules by Proposition 3.4. Any cyclic module is a

homomorphic image of R and hence is a sum of simple modules. Now, an arbitrary

module is a sum of cyclic submodules, and hence also sum of simple modules. The

result follows, again from Proposition 3.8.

3.10. Proposition. The Jacobson radical J of a right Artinian ring R is nilpotent.

Proof. The descending chain J ⊇ J 2 ⊇ J 3 ⊇ . . . must terminate since R is rightArtinian. Hence J n = J n+1 = . . . for some n ≥ 0. Let X = lann(J n) = {x ∈ R :

xJ n = 0}, this is a two-sided ideal of R.

Suppose for a contradiction that X < R. Then R/X has a minimal nonzero sub-

module Y /X , being Artinian. It’s easy to see that Y /X is simple. Now (Y /X ).J = 0

so Y J ⊆ X . It follows that Y J n = Y J n+1 ⊆ XJ n = 0, so Y ⊆ lann(J n) = X ,

contradicting Y /X = 0.

Hence X = R so J n = RJ n = XJ n = 0 as required.



18

Corollary. If R is right Artinian, then J (R) = N (R).

Proof. Since every primitive ideal is prime by Lemma 2.8(2), N (R) ⊆ J (R) inany ring. On the other hand if R is right Artinian then J (R) is nilpotent by

Proposition 3.10, so J (R) ⊆ P for any prime ideal P by Lemma 2.8(1). Hence

J (R) ⊆ N (R).

3.11. Hopkins’ Theorem. Let R be a right Artinian ring. Then R is right Noe-

therian.

Proof. By Proposition 3.10, J n = 0 where J is the Jacobson radical of R. Now,

each J m/J m+1 is an R/J -module and R/J is semisimple Artinian. By Lemma 3.9,

each J m/J m+1 is a semisimple Artinian R/J -module, and as such is Noetherian by

Lemma 3.2. The result follows from Proposition 1.5.

It’s tempting to think that the left-right symmetry is so strong that every right

Artinian ring is left Artinian. This is not the case, however.



19

4. Commutative Rings

R is commutative with 1 throughout this chapter.

4.1. The Nullstellensatz. We saw in Corollary 3.10 that if R is right Artinian,

then N (R) = J (R).

Theorem. Let k be a field and R be a finitely generated k−algebra (equivalently,

a quotient of the polynomial ring k[X 1, . . . , X n]). Then

N (R) = J (R).

This is an unusual way of stating the Nullstellensatz, see (4.3).

4.2. Weak Nullstellensatz.

Theorem. Suppose k is a field and R is a finitely generated k−algebra which is

also a field. Then

(1) R is algebraic over k.

(2) If k is algebraically closed, then R ∼= k.

Corollary. If k = k then the maximal ideals of k[X 1, . . . , X n] are of the form

ma = (X 1 − a1, . . . , X n − an) where a = (a1, . . . , an) ∈ kn.

Proof. It’s clear that each ma is a maximal ideal, being of codimension 1. Now if I

is maximal, then the R/I ∼= k so X i +I → ai for some ai ∈ k. Hence X i−ai ∈ I for

all i = 1, . . . , n so ma

⊆I . Since the former ideal is maximal, the result follows.

4.3. Geometric viewpoint. Recall that by Proposition 2.11, I R is semiprime

if and only if f n ∈ I ⇒ f ∈ I .

Definition. If R is commutative, an ideal is radical if it is semiprime.

In the first lecture, we defined an algebraic set to be a subset of kn of the form

Z (A) = {(a1, . . . , an) ∈ kn : f (a1, . . . , an) = 0 for all f ∈ A}where A is an ideal of k[X 1, . . . , X n]. We also defined, for X ⊆ kn

I (X ) = {f ∈ k[X 1, . . . , X n] : f (a1, . . . , an) = 0 for all (a1, . . . , an) ∈ X }this is an ideal in k[X 1, . . . , X n].

It’s easy to check that Z (I (X )) = X for any algebraic set X . The Nullstellensatz

is usually stated as follows:

Theorem. Suppose k = k. If A is a radical ideal of k[X 1, . . . , X n], then

I (Z (A)) = A.

So we have a bijection between radical ideals and algebraic sets.



20

Proof. f ∈ I (Z (A)) ⇒ f (a) = 0 for all a ∈ Z (A) ⇒ f ∈ ma for all a ∈ Z (A) ⇒f

∈ a∈Z(A)

ma = ma⊇A

ma

since a ∈ Z (A) iff A(a) = 0 iff A ⊆ ma.

By Corollary 4.2, all maximal ideals of R = k[X 1, . . . , X n]/A have the form

ma/A for some a ∈ kn, so f + A ∈ J (R) = 0 by Theorem 4.1 and f ∈ A.

Finally, if f ∈ A then f (Z (A)) = 0 so f ∈ I (Z (A)).

4.4. Proof of Theorem 4.2. We will only deal with the case when k is an un-

countable field in this course. For the general case, see Atiyah and Macdonald, (7.8)

and (7.9).

Suppose R is not algebraic. Then we can find α ∈ R such that k(α) ∼= k(t),

the field of fractions of the polynomial ring k[t]. Choose an uncountable subset{ci : i ∈ I } of k and suppose λ1, . . . , λm ∈ k are such that

λ1α − ci1

+λ2

α − ci2

+ . . . +λm

α − cim

= 0.

Clearing denominators gives

λ1(α − ci2)(α − ci3) · · · (α − cim) + . . . + λm(α − ci1)(α − ci2) · · · (α − cim−1) = 0.

Substituting α1 = ci1 into this equation gives λ1(ci1−ci2)(ci1−ci3) · · · (ci1−cim) = 0

so λ1 = 0 and similarly λj = 0 for all j = 1, . . . , m.

Hence { 1α−ci

∈ k(α) : i ∈ I } is an uncountable linearly independent subset of

R. But R is a quotient of the polynomial ring k[X 1, . . . , X n] and as such has a

countable spanning set, consisting of the images of the monomials in R. This is acontradiction, proving (1).

Now, a finitely generated field extension of k which is algebraic must necessarily

be finite dimensional. If k = k then R is a finite field extension of k so R ∼= k as

required for (2).

4.5. Lemma. Let k be a field and let R be a finite dimensional k−algebra which

is a domain. Then R is a field.

Proof. If 0 = x ∈ R then multiplication by x is an injective k−linear map from R

to itself. Since dimk R < ∞ this map must be surjective. Hence there exists y ∈ R

such that xy = 1 and the result follows.

4.6. Proof of Theorem 4.1. As we have observed in the proof of Corollary 3.10,

N (R) ⊆ J (R) for any ring R. Suppose a /∈ N (R) so a is not nilpotent. By Sheet 1

Exercise 6(2), 1 − ax is not a unit in the polynomial ring R[x]. By Lemma 2.3 we

can find a maximal ideal M of R[x] containing 1 − ax.

Since R is a finitely generated k−algebra, so is R[x]/M . By Theorem 4.2, R[x]/M

is algebraic over k and is hence finite dimensional over k. Now, we have a k−linear

injection R/(R ∩ M ) → R[x]/M , so R/(R ∩ M ) is a finite dimensional k−algebra



21

which is a domain. Hence R ∩ M is a maximal ideal of R by Lemma 4.5 and

a /

∈R

∩M as otherwise 1

∈M . Hence a /

∈J (R) as required.

4.7. Associated primes. From now on we’ll be dealing with a commutative Noe-

therian ring R and finitely generated R−modules M .

Definition. A prime P of R is said to be an associated prime of M if there exists

x ∈ M such that ann(x) = P , or equivalently, if R/P → M . The set of all

associated primes of M is denoted by Ass(M ).

Example. Let R = Z and let M be a finite abelian group. Then M has an element

of order p for all primes p dividing |M |, so

Ass(M ) = { pZ : p | |M |}.

4.8. Lemma. If M = 0 then Ass(M ) = ∅.

Proof. Let S = {ann(x) R : 0 = x ∈ M }. Since R is Noetherian, we can pick a

maximal element P = ann(x) ∈ S . Suppose a /∈ P and b /∈ P but ab ∈ P . Then

bx = 0 so ann(bx) ∈ S . Since P ⊆ ann(bx), P = ann(bx) by maximality of P . Now

abx = 0 so a ∈ P , a contradiction. Hence P is prime and so P ∈ Ass(M ).

4.9. Proposition. Suppose M = 0.

(1) If P R is a prime then Ass(R/P ) = {P }.

(2) Let N be a submodule of M . Then

Ass(N ) ⊆ Ass(M ) ⊆ Ass(N ) ∪ Ass(M/N ).

Proof. Let L = R/P and 0 = y ∈ L. Clearly, P ⊆ ann(y). Now if ry = 0 then

r ∈ P since R/P is a domain. Hence P = ann(y) and (1) follows.

Next, the first inclusion of (2) is obvious. Let P = ann(x) ∈ Ass(M ). If

Rx ∩N = 0 then R/P ∼= Rx → M/N and P ∈ Ass(M/N ), so suppose Rx ∩N = 0.

Choose y ∈ Rx ∩ N ; then by the above ann(y) = P so P ∈ Ass(N ) as required.

4.10. Theorem. Suppose M = 0. Then

(1) There exists a chain of submodules

0 = M 0 < M 1 < .. . < M n = M

of M such that M i/M i−1 ∼= R/P i for some prime ideals P 1, . . . , P n of R.

(2) Ass(M ) is finite.

Proof. (1) By Lemma 4.8, Ass(M ) = ∅. Let M 1 = x1R for some x1 ∈ M such

that ann(x1) = P 1 ∈ Ass(M ). Then M 1/M 0 ∼= R/P 1. If M 1 = M we are done.

Otherwise, M/M 1 is nonzero so we can find P 2 = ann(x2) ∈ Ass(M/M 1) for some

x2 ∈ M/M 1. Let M 2 be defined by M 2/M 1 = x2R ∼= R/P 2. This process must

stop since M is Noetherian.



22

(2) By Proposition 4.9,

Ass(M ) ⊆ Ass(M 1) ∪ Ass(M 2/M 1) ∪ . . . ∪ Ass(M n/M n−1) = {P 1, . . . , P n}so Ass(M ) is finite.

This result shows that any finitely generated R−module M can be thought of

as being built up from finitely many modules of the form R/P i for some primes P i.

4.11. Proposition. Every minimal prime of R is in Ass(R).

Proof. Let {P 1, . . . , P n} be the distinct minimal primes of R. By Proposition 2.9,

we can find integers s1, . . . , sn ≥ 1 such that

P s11 P s22 · · · P snn = 0.

It’s enough to show that P 1 ∈ Ass(R). Let M = P s22 · · · P snn if n > 1 and M = Rotherwise.

By Theorem 4.10(1) we can choose a chain

0 = M 0 < M 1 < .. . < M t = M

where M i/M i−1 ∼= Qi for some primes Q1, . . . , Qt. Note that P s11 kills each

M i/M i−1 so P s11 ⊆ Qi and hence P 1 ⊆ Qi for all i = 1, . . . , t by the primality

of Qi.

Also, Q1Q2 · · · QtM = 0 implies Q1Q2 · · · QtP s22 · · · P snn ⊆ P 1. Since P 1 is prime

and P i P 1 for all i ≥ 2, Qk ⊆ P 1 for some k, so in fact Qk = P 1 for some k.

Pick k ≥ 1 least such that Qk ⊆ P 1. Hence if k > 1 then Q1 · · · Qk−1 P 1. If k = 1 let r = 1 and if k > 1 choose r ∈ Q1 · · · Qk−1\P 1. Since Ann(M k/M k−1) =

Qk = P 1 and r /∈ P 1 we can find x ∈ M k such that rx /∈ M k−1.

Claim: P 1 = ann(rx). Note that rP 1x ⊆ Q1Q2 · · · Qk−1Qkx = 0 s o P 1 ⊆ann(rx). If srx = 0 then s(rx + M k−1) = 0. We chose x so that rx /∈ M k−1, and

hence s ∈ P 1 because M k/M k−1 ∼= R/Qk∼= R/P 1 is a domain. Hence ann(rx) =

P 1 ∈ Ass(M ) as required.

4.12. Primary ideals.

Definition. Let P be a prime. An ideal Q R is P −primary if Ass(R/Q) = {P },

and primary if it’s P -primary for some prime P .

One should think of primary ideals as generalisations of prime powers in Z.

Lemma. (1) If Q is P −primary, then P =√

Q.

(2) If P is maximal and Q R has P =√

Q, then Q is P −primary.

Proof. Since any minimal prime over Q must be in Ass(R/Q) by Proposition 4.11,

we see that R/Q has precisely one minimal prime, namely P/Q. Hence N (R/Q) =√Q/Q = P/Q so P =

√Q.



23

For (2), note that P/Q is the prime radical of R/Q and hence is the only prime

ideal of R/Q, being maximal. Hence Ass(R/Q) must be

{P

}, being nonempty by

Lemma 4.8, so Q is P −primary.

Examples. (1) The primary ideals of Z are the prime powers and 0.

(2) Let R = k[x, y], Q = (x, y2). Then √

Q = (x, y) =: P is maximal so Q is

primary. However, Q is not a prime power since P 2 Q P .

Warning: not all prime powers are primary!

4.13. Primary decomposition.

Definition. A primary decomposition of an ideal I R is an expression of the form

I = Q1 ∩ Q2 ∩ . . . ∩ Qk

for some primary ideals Qi. The decomposition is minimal if

• The primes P i’s are pairwise distinct, where Qi is P i−primary

• For all i, Qi

j=i Qj .

Example. In Z, a nonzero ideal I has a minimal primary decomposition of the

form

I = ( ps11 ) ∩ ( ps2

2 ) ∩ · · · ( psnn )

where I = (x) and x = ps11 · · · psn

n is a factorization of x into prime powers.

4.14. Theorem.

(1) (Existence) Every proper ideal has a minimal primary decomposition.

(2) (Uniqueness) Let I = Q1 ∩ . . . ∩ Qk be a minimal primary decomposition

of I , with Qi a P i−primary ideal. Then

Ass(R/I ) = {P 1, . . . , P n}.

Proof. Omitted.



24

5. Commutative localisation

R is a commutative ring throughout this chapter.5.1. Motivation. Let S be a subset of R which is multiplicatively closed (m.c. for

short). We will stick to the convention that any such S necessarily contains 1.

We will construct a ring RS, called the localisation of R at S by ”inverting” the

elements of S - thus the elements of S are units in RS .

This generalises the process of constructing the field of fractions of a commutative

integral domain R - there the set S = R\{0} of nonzero elements is m.c., S consists

of units in the field of fractions F of R, and moreover every element of F can be

written in the form rs−1 for some r ∈ R and s ∈ S .

What if S contains zero divisors? If we want to ”invert” S , then the elements

of R which kill an element of S must be zero in RS because ab = 0 and bc = 1together imply a = abc = 0.

Definition. The assassinator of S is defined to be

ass(S ) = {x ∈ R : xs = 0 for some s ∈ S } =s∈S

ann(s).

Lemma. ass(S ) is an ideal of R.

Proof. Let x, y ∈ ass(S ). So there exist s, t ∈ S such that xs = yt = 0. Then

(x + y)st = 0 and st ∈ S so x + y ∈ ass(S ). Also, if z ∈ R then (zx)s = 0 so

zx ∈ ass(S ).

5.2. Commutative localisation.Definition. Let S ⊆ R be a m.c. subset. A localisation of R at S is a ring RS

together with a homomorphism ϕ : R → RS such that

(a) ϕ(s) is a unit in RS for all s ∈ S .

(b) Every element of RS can be written in the form ϕ(r)ϕ(s)−1 for some r ∈ R

and s ∈ S .

(c) ker ϕ = ass(S ).

Note that ϕ is not in general injective.

Lemma. If RS exists, it’s unique upto isomorphism.

Proof. Suppose θ : R → T is a ring homomorphism satisfying properties (a),(b),(c).Define α : RS → T by α(ϕ(r)ϕ(s)−1) = θ(r)θ(s)−1. If ϕ(r)ϕ(s)−1 = ϕ(u)ϕ(v)−1

then rv −us ∈ ker(ϕ) = ass(S ) so there exists t ∈ S such that (rv −us)t = 0. Hence

rvt = ust so θ(r)θ(v)θ(t) = θ(u)θ(s)θ(t) and θ(r)θ(s)−1 = θ(u)θ(v)−1. Therefore

α is well defined and is easily checked to be a ring homomorphism.

The conditions on θ force α to be onto, whereas if θ(r)θ(s)−1 = 0 then r ∈ker(θ) = ass(S ) = ker(ϕ) so ϕ(r)ϕ(s)−1 = 0 and α is injective. Hence α is a ring

isomorphism.



25

5.3. Construction.

Definition. An element x ∈ R is regular if ann(x) = 0. Equivalently, x is not a zero divisor.

Assume first that S consists of regular elements, so that ass(S ) = 0.

Define a relation on R × S by setting

(r, s) ∼ (u, v)

if and only if

rv = us.

This is clearly reflexive and symmetric. If (r, s) ∼ (u, v) and (u, v) ∼ (a, b) then

rv = us and ub = av, so (rb

−as)v = rvb

−asv = ubs

−avs = 0 so rb

−as

∈ass(S ) = 0 and (r, s) ∼ (a, b). Hence ∼ is an equivalence relation.

Let RS = R × S/ ∼ as a set. Write r/s for the equivalence class of (r, s) in RS

and define addition and multiplication on RS by

r/s + u/v = (rv + us)/(sv)

r/s · u/v = (ru)/(sv)

Exercise. Check the following:

(1) Addition is well defined.

(2) Multiplication is well defined.

(3) RS forms a commutative ring with zero element 0/1 and identity 1/1.

(4) The map ϕ : R → RS given by ϕ(r) = r/1 is a ring homomorphism.(5) Conditions (a), (b) and (c) in Definition 5.2 hold.

We check that addition is well-defined: Suppose r/s = r/s so rs = rs. Now,

r/s + u/v = (rv + us)/(sv) and (r/s) + (u/v) = (rv + us)/(sv),

but (rv + us)(sv) = (rs)v2 + ussv = (rs)v2 + ussv = (rv + us)(sv), so

r/s + u/v = r/s + u/v = r/s + u/v

if u/v = u/v. Hence addition is well-defined.

Hence, the localisation at S exists and is unique.

5.4. The general case. If ass(S ) is not necessarily zero, let R = R/ ass(S ) and

let S denote the image of S in R. Then ass(S ) = 0 so we can form the localisation

RS , together with a ring homomorphism θ : R → RS as constructed above. Let

ϕ : R → RS

be defined by ϕ = θ ◦ π where π is the natural projection of R onto R. So we have

Theorem. The localisation RS exists.



26

Proof. Since ass(S ) = 0, ker(θ) = 0 so ker(ϕ) = ass(S ). Thus ϕ : R → RS satisfies

all the conditions of Definition 5.2.

Note that ϕ(r)ϕ(s)−1 = ϕ(u)ϕ(v)−1 iff ϕ(rv − us) = 0 iff rvt = ust for some

t ∈ S . So we could have constructed RS by imposing the equivalence relation ∼ on

R × S instead, where

(r, s) ∼ (u, v) if and only if rvt = ust for some t ∈ S.

5.5. Examples.

(1) If R is an integral domain then S = R\{0} is a m.s. set. Then RS is just

the field of fractions of R. This is a special case of (2).

(2) If R is arbitrary and P is a prime ideal, then S = R\P is m.c. The

localisation RS is usually denoted by RP and is called

the localisation at the prime ideal P .

The ring Z( p) from Example 1.8 (4) is a special case of this construction.

(3) For any x ∈ R, the set {1, x , x2, . . .} is m.c., and the localisation is usually

denoted by Rx.

(4) If 0 ∈ S then ass(S ) = R so RS = 0.

(5) If I R then S = 1 + I is m.c.

Of these, Example (2) is the most important.

5.6. Localisation of modules.

Definition. Let S be a m.c. subset of R and let M be an R−module. The locali-sation of M at S is defined to be the set of equivalence classes

M S = {m/s : m ∈ M, s ∈ S }in M × S under the equivalence relation ∼ given by

(m, s) ∼ (n, t) if and only if mtu = nsu for some u ∈ S.

This is an RS−module, with the addition and RS−action given by

m/s + n/t = (mt + ns)/(st)

m/s · r/u = (mr)/(su)

for m/s, n/t ∈ M S and r/u ∈ RS. The S -torsion submodule of M is defined to be

assM (S ) = {m ∈ M : ms = 0 for some s ∈ S }.

Exercise. Check that

(1) The operations are well defined and turn M S into an RS−module.

(2) N = assM (S ) is a submodule of M .

(3) assM/N (S ) = 0.

(4) The map ϕ : M → M S given by ϕ(m) = m/1 has kernel precisely assM (S ).



27

(5) M S = 0 if and only if assM (S ) = M .

In fact, M S is naturally isomorphic to M ⊗R RS .

5.7. Exact sequences.

Definition. Let A, B, C be R−modules and suppose we have R−module maps

f : A → B and g : B → C . The sequence

Af → B

g→ C

is said to be exact at B if Im(f ) = ker(g). The sequence

00→ A

f → Bg→ C

0→ 0

is called a short exact sequence if it is exact at A, B and C , or equivalently, if f is injective, Im(f ) = ker(g) and g is surjective.

Note that whenever N is a submodule of N , we have a natural short exact

sequence

0 → N ι→ M

π→ M/N → 0

where ι is the inclusion of N into M and π is the quotient map.

5.8. Exactness of localisation. One of the most important properties of locali-

sation is that it is exact , meaning that it takes exact sequences to exact sequences.

Suppose S is a m.c. set in R and f : M → N is a map of R−modules. Then we

can define a map f S : M S → N S by setting

f S(m/s) = f (m)/s for all m ∈ M, s ∈ S.

Exercise. Check that

(1) f S is well defined.

(2) f S is an RS−module homomorphism.

Proposition. Localisation is exact. In other words, whenever

Af → B

g→ C

is a short exact sequence of R

−modules,

ASf S→ BS

gS→ C S

is a short exact sequence RS−modules.

Proof. Since Im(f ) = ker(g), g ◦ f = 0. Hence

gS(f S(a/s)) = gS(f (a)/s) = g(f (a))/s = 0

for any a ∈ A and s ∈ S , so gS ◦ f S = 0 and Im(f S) ⊆ ker(gS).



28

Suppose b/s ∈ ker(gS). Then gS(b/s) = g(b)/s = 0 so there exists t ∈ S such

that g(b)t = g(bt) = 0. Hence bt

∈ker(g) = Im(f ) since the first sequence is exact,

so we can find a ∈ A such that bt = f (a). Now

f S(a/(st)) = f (a)/(st) = bt/(st) = b/s

so b/s ∈ Im(f S) and Im(f S) = ker(gS) as required.

5.9. Localisation at a prime ideal. Let P be a prime ideal of R and set S =

R\P . A commutative ring is called local if it has a unique maximal ideal.

Proposition. (1) ass(S ) ⊆ P .

(2) RP is a local ring with maximal ideal P P .

Proof. If s /∈ P and xs = 0 then xs ∈ P which is prime so x ∈ P , proving (1).

The natural short exact sequence0 → P → R → R/P → 0

gives rise to the short exact sequence of RP −modules

0 → P P → RP → (R/P )P → 0

by Proposition 5.8. This means that P P is an ideal of RP with quotient isomorphic

to (R/P )P .

If x/s ∈ RP \P P then x /∈ P so x ∈ S and x/s is a unit in RP . Conversely, if x/s

is a unit then x/1 is a unit also, so x/1.y/t = 1 for some y/t ∈ RP . Hence xyu = t

for some t ∈ S , meaning that x /∈ P .

Hence P P is the set of all nonunits of RP . If y + P P ∈ RP /P P is nonzero theny /∈ P P so y is invertible in RP and hence in RP /P P . Hence RP /P P is a field so

P P is maximal. Moreover, any maximal ideal consists of nonunits and must hence

be contained in P P , proving (2).

Exercise. Let F denote the field of fractions of R = R/P . Check that there is

a well defined ring isomorphism α : (R/P )P → F given by α(r/s) = r/s. Hence

RP /P P ∼= F .

5.10. Theorem. Suppose R is Noetherian and S is a m.c. subset. Then RS is

Noetherian.

Proof. Since R = R/ ass(S ) is Noetherian by Proposition 1.5 and RS

∼= RS, we

may assume that ass(S ) = 0. Hence R → RS .

Now, if I RS then (I ∩ R)RS ⊆ I . If x/s ∈ I then x = x/s.s ∈ I ∩ R so

x/s = x.(1/s) ∈ (I ∩ R)RS . Hence I = (I ∩ R)RS for any I RS .

Suppose I 1 ⊆ I 2 ⊆ . . . is an increasing chain of ideals in RS . Since R is Noe-

therian, the chain I 1 ∩ R ⊆ I 2 ∩ R ⊆ . . . terminates, so there exists k such that

I k ∩ R = I m ∩ R for all m ≥ k. But then I k = (I k ∩ R)RS = (I m ∩ R)RS = I m for

all m ≥ k, so RS is Noetherian as required.



30

If z ∈ R then (zx)s = 0 so zx ∈ ass S . Also the Ore condition gives z ∈ S

and s

∈S such that zs = sz. Hence (xz)s = (xs)z = 0 and s

∈S so

xz ∈ ass S .

6.3. Ore’s Theorem. We must first refine Definition 5.3.

Definition. An element x ∈ R is said to be left regular if lann(x) = 0 and right

regular if rann(x) = 0. x ∈ R is regular if it’s both right and left regular, and thus

not a zero-divisor.

Assuming S is a right Ore set, we may factor out by the two-sided ideal ass( S )

and obtain a ring R such that the image S of S in R consists of left regular elements.

If the right localisation RS exists, then R is a subring of RS and ϕ(S ) consists of

units. Hence every element of S must actually be regular (and not just left regular).

This is another necessary condition on the set S for RS to exist.

Theorem (Ore, 1930). Let S be a m.c. subset of the ring R. Then the right

localisation RS exists if and only if

(1) S is a right Ore set, and

(2) S consists of regular elements in R = R/ ass(S ).

Proof. We have shown in the above discussions the necessity of these conditions.

Assuming (1) and (2), it remains to construct the ring RS satisfying the conditions

of Definition 6.1.

By Lemma 6.2, ass(S ) is a two-sided ideal in R. It’s easy to check that the

image S also satisfies the right Ore condition. As before, by passing to the quotientR = R/ ass S and using condition (2), we may assume that S is a right Ore set in

R consisting of regular elements.

We can now construct RS as a set of equivalence classes in R × S under a

certain equivalence relation, in a very similar spirit to the construction of RS in the

commutative case (5.3). The construction is very tedious and is non-examinable.

See handout.

Definition. A m.c. set S is said to be a right divisor set if conditions (1), (2) are

satisfied.

Note that if R is commutative, or slightly more generally, if S consists of central

elements (i.e. sr = rs for all s ∈ S, r ∈ R), then the two conditions are auto-matically satisfied and the localisation RS exists. One can view this Theorem as a

generalisation of Theorem 5.4.

6.4. Ore domains.

Definition. A ring R is a domain if all nonzero elements are regular, or equiva-

lently if it has no zero divisors. R is a right Ore domain if S = R\0 is a right Ore

set in R.



31

By Theorem 6.3, it’s clear that if R is a right Ore domain, then the localisation

RS exists. Moreover, every nonzero element r/s

∈RS is invertible, so RS is a

division ring, known as the division ring of fractions of R.

Theorem. Let R be a right Noetherian domain. Then R is an Ore domain and

hence R has a division ring of fractions.

Proof. Let x, y ∈ S and suppose that xR ∩ yR = 0. Then xR ⊕ yR → R. Now,

x ∈ S is regular so xR ∼= R as right R−modules. Hence

x(xR ⊕ yR) = x2R ⊕ xyR → xR,

so x2R ⊕ xyR ⊕ yR → R. Repeating this trick, we see that

xnR ⊕ xn−1yR ⊕ xn−2yR ⊕ · · · ⊕ xyR ⊕ yR → R

for any n ≥ 1. Hence the sum yR + xyR + x2yR + . . . is direct, contradicting the

fact that R is right Noetherian.

Hence we can find a, b ∈ R such that xa = yb = 0. Since x, y are regular, a, b

are nonzero so S is a right Ore set.

This proof might be reminiscent of Sheet 1, Exercise 12.

Examples. Let k be a field. The following rings are all right Ore domains and

hence have division rings of fractions:

(1) The Weyl algebras An(k).

(2) kG, where G is a torsionfree polycyclic group.

(3) U (g) where g is a f.d. Lie algebra over k.(4) The Iwasawa algebra ΛG if G is a torsionfree compact p−adic Lie group.

6.5. Modules.

Definition. Let S be a right divisor set in R and let M be a right R−module. The

localisation of M at S is defined to be the set of equivalence classes

M S = {m/s : m ∈ M, s ∈ S }in M × S under the equivalence relation ∼ given by

(m, s) ∼ (n, t) if and only if mtu = nsu for some u ∈ S,

where s, t ∈ R are such that st = ts ∈ S . The S -torsion submodule of M isdefined to be

assM (S ) = {m ∈ M : ms = 0 for some s ∈ S }.

Proposition. (1) M S has the structure of a right RS−module.

(2) N = assM (S ) is an R−submodule of M .

(3) assM/N (S ) = 0.

(4) The map ϕ : M → M S given by ϕ(m) = m/1 has kernel precisely assM (S ).



32

(5) M S = 0 if and only if assM (S ) = M .

(6) Localisation is exact.

Proof. Exercise - mimic proofs in (5.6), (5.8) and the handout.

Again, an alternative way of constructing M S is given by the isomorphism

M S ∼= M ⊗R RS.

6.6. Goldie’s Theorem.

Definition. The classical right ring of quotients Q(R) is the localisation of R at

the m.c. subset S consisting of all regular elements of R, if it exists.

Theorem (Goldie, 1958, 1960). The ring R has a classical right ring of quotients

which is semisimple Artinian if and only if

(1) N (R) = 0,

(2) R contains no infinite direct sum of right ideals, and

(3) R satisfies the ACC on right annihilators.

Corollary. If R is semiprime Noetherian, then Q(R) exists and is isomorphic to

M n1(D1) ⊕ M n2(D2) ⊕ . . . ⊕ M nk(Dk)

for some division rings D1, . . . , Dk.

Proof. Follows directly from Theorem 3.7 and Theorem 6.6.

We will only prove (

⇐), the other direction being easier and less interesting. For

the remainder of this chapter, assume that R is a ring satisfying conditions (1), (2)

and (3) of Theorem 6.6. Let S denote the set of all regular elements of R.

6.7. Essential right ideals.

Definition. Let M be an R−module. Then N ≤ M is essential if N ∩ K = 0 for

any nonzero K ≤ M .

There is a connection between right regular elements and essential right ideals.

Let a∗ = rann(a) for any a ∈ R. Note that x∗ = 0 iff x is right regular and x∗ = R

iff x = 0.

Proposition. T r R is essential if and only if T ∩ S = ∅.

Proof. (⇐) Suppose x ∈ T is right regular. It’s sufficient to prove that xR is

essential since xR ⊆ T .

Suppose 0 = I r R and I ∩ xR = 0. Consider the sum

L = I + xI + x2I + . . .

If k

i=0 xiri ∈ L, then r0 = −xr1 − . . . − xkrk ∈ I ∩ xR = 0 so r0 = 0 and

xr1 + x2r2 + . . . + xkrk = 0. Since x∗ = 0, r1 + xr2 + . . . + xk−1rk = 0. Continuing



33

like this, r0 = r1 = . . . = rk = 0 so the sum is direct, contradicting (2). Hence xR

is essential.

(⇒) See (6.11).

6.8. Proof of Goldie’s Theorem - Step 1: RS exists.

Proof. By Theorem 6.3 it’s sufficient to show that S is a right Ore set. Let r ∈ R

and s ∈ S ; since s is regular, sR is essential in R by Proposition 6.7(⇐). Let

T = {a ∈ R : ra ∈ sR}.

T is a right ideal in R; we’ll show that T is essential. Suppose I r R is nonzero. If

rI = 0 then I ⊆ T so I ∩ T = 0. Otherwise sR ∩ rI = 0 since sR is essential. If

0 = rx ∈ sR ∩ rI then 0 = x ∈ I ∩ T so I ∩ T = 0 as claimed.

By Proposition 6.7(

⇒), we can find a

∈T

∩S . Then ra = sb for some b

∈R

and S is a right Ore set.

6.9. Lemma. Let x ∈ R be right regular. Then x is regular.

Proof. Suppose yx = 0 but y = 0. Then y∗ = R. Now xR ⊆ y∗ so y∗ is essential

by Proposition 6.7(⇐). Let z∗ R be a maximal annihilator containing y∗ - this

exists by assumption (3).

Now, if zRz = 0 then (RzR)2 = 0, so z = 0 by assumption (1) and z∗ = R, a

contradiction. Hence there exists b ∈ R such that zbz = 0.

Hence z∗ ⊆ (zbz)∗ = R so z∗ = (zbz)∗ by maximality of z∗. If u ∈ bzR ∩ z∗ then

u = bzv and zu = 0 so zbzv = 0. Hence v ∈ z∗ and u = bzv = 0. This shows that

bzR ∩ z∗

= 0 and bz = 0, so z∗

is not essential, a contradiction.Hence y = 0 and x is regular.

6.10. Nil right ideals.

Definition. X ⊆ R is nil if x is nilpotent for all x ∈ X .

Lemma. Let J r R.

(a) If J is nil then J = 0.

(b) If J = 0 then there exists y ∈ J such that yR ∩ y∗ = 0.

Proof. (a) Suppose for a contradiction that there exists nonzero a ∈ J . Then

aR ⊆ J is nil, and hence Ra is also nil, because (ax)

n

= 0 ⇒ (xa)

n+1

= 0.Let S = {b∗ : 0 = b ∈ Ra}. By (3), we can pick 0 = b ∈ Ra so that b∗ is maximal

in S . Since N (R) = 0 by (1), bRb = 0, so we can find x ∈ R such that bxb = 0.

Since Ra is nil and 0 = xb ∈ Ra, there exists m ≥ 2 such that (xb)m−1 = 0 but

(xb)m = 0. Now ((xb)m−1)∗ ∈ S and b∗ ⊆ ((xb)m−1)∗ as by = 0 ⇒ (xb)m−1y = 0.

By maximality of b∗, xb ∈ ((xb)m−1)∗ = b∗ so bxb = 0, a contradiction.

(b) By (a), J is not nil. Choose z ∈ J which is not nilpotent. Now, the chain

z∗ ⊆ (z2)∗ ⊆ (z3)∗ ⊆



34

stops by (1), so (zn)∗ = (zn+1)∗ = . . . for some n. Let y = zn so that (y2)∗ = y∗.

Note that y

= 0 since z is not nilpotent.

Suppose that u = yv ∈ yR ∩ y∗. Then yu = y2v = 0 so v ∈ (y2)∗ = y∗ and

u = yv = 0. Hence yR ∩ y∗ = 0 as required.

6.11. Proof of Proposition 6.7(⇒).

Proof. Suppose for a contradiction that T ∩ S = ∅. We will construct an infinite

direct sum inside T . More precisely, we will construct a sequence x1, x2, . . . ∈ T

such that for all m ≥ 1

• I m = 0 where I m := {x1, . . . , xm}∗.

• The sum x1R + x2R + . . . + xmR + (I m ∩ T ) is direct.

Proceed by induction on m.

Base case m = 1: Pick 0 = x1 ∈ T such that x1R∩ x∗1 = 0 using Lemma 6.10(b).Since x1 is not regular, it’s not right regular by Lemma 6.9, so I 1 = x∗1 = 0. The

sum x1R + x∗1 is direct by construction, so x1R + (I 1 ∩ T ) is also direct.

Inductive step: Since I m = 0 by induction and T is essential, I m∩T = 0. Choose

xm+1 ∈ I m ∩ T such that x∗m+1 ∩ xm+1R = 0 using Lemma 6.10(b).

Note that since xm+1 ∈ I m ∩ T , the sum

x1R + x2R + . . . + xm+1R

is direct, being contained in x1R + x2R . . . + (I m ∩ T ) which is direct by induction.

Hence I m+1 = {x1, . . . , xm+1}∗ = (x1 + . . . + xm+1)∗. Since x1 + . . . + xm+1 ∈ T

and T contains no regular elements, I m+1

= 0.

It remains to show that x1R + x2R + . . . + xm+1R + (I m+1 ∩ T ) is direct. Now,

xm+1R ⊆ I m ∩ T

by construction, so if Z = x1R + . . . + xmR then

(Z + xm+1R) ∩ (I m ∩ T ) = (Z ∩ I m ∩ T ) + xm+1R = xm+1R

by the modular law, because Z ∩ I m = 0 by induction. Since I m+1 = I m ∩ x∗m+1,

(Z + xm+1R) ∩ (I m+1 ∩ T ) = (Z + xm+1R) ∩ (I m ∩ T ) ∩ x∗m+1 ⊆⊆ xm+1R ∩ x∗m+1 = 0,

as required. Thus the sum x1R + x2R + . . . is direct contradicting (2), so T must

contain a regular element.

6.12. Closed right ideals.

Definition. J r R is closed if for any I r R such that I J there exists 0 = Lr R

with L ⊆ I and L ∩ J = 0.

Draw a picture!

Lemma. R has the minimum condition on closed right ideals.



35

Proof. Suppose J 1 J 2 J 3 . . . is a strictly decreasing chain of closed right

ideals. Since J 1 J 2, there exists 0

= L1 r R such that L1

⊆J 1 and L1

∩J 2 = 0.

Similarly, since J n J n+1, there exists 0 = Ln r R such that Ln ⊆ J n and

Ln ∩ J n+1 = 0. Then we get an infinite direct sum

L1 ⊕ L2 ⊕ L3 ⊕ . . . ,

contradicting (2).

6.13. Proof of Goldie’s Theorem - Step 2: RS is semisimple Artinian.

Proof. Let K r RS . By adapting the proof of Theorem 5.10, K = (K ∩ R).RS . We

claim that J = K ∩ R is closed.

For, suppose I J . Pick z ∈ I \J and let T = {r ∈ R : zr ∈ J }, a right ideal.

If s

∈T

∩S then zs

∈J so z

∈J.RS = K and z

∈K

∩R = J , a contradiction.

Hence T ∩ S = ∅, so T is not essential by Proposition 6.7 (⇒). Hence there exists

0 = V r R such that V ∩ T = 0.

It follows that zV ⊆ I and zV ∩ J = 0 as required.

Now, if K 1 K 2 K 3 . . . is a strictly descending chain of right ideals in RS ,

then K 1 ∩ R K 2 ∩ R K 3 ∩ R . . . is a strictly descending chain of closed right

ideals in R, contradicting Lemma 6.12. Hence RS is right Artinian.

Finally, Let J be the Jacobson radical of RS . By Proposition 3.10, J is nilpotent

and hence so is J ∩ R. As R is semiprime, J ∩ R = 0. But now J = (J ∩ R).RS = 0

so RS is semisimple Artinian, as required.



36

7. Filtrations, associated graded rings and completions

Throughout this chapter, k denotes a field.7.1. Filtered rings.

Definition. A (Z−)filtration on a ring R is a set of additive subgroups (Ri)i∈Z

such that

• Ri ⊆ Ri+1 for all i ∈ Z,

• Ri.Rj ⊆ Ri+j for all i, j ∈ Z,

• 1 ∈ R0, and

• ∪i∈ZRi = R.

If R has a filtration, R is a filtered ring. The filtration on R is said to be

• positive if Ri = 0 for all i < 0,• negative if Ri = R for all i ≥ 0,

• separated if ∩i∈ZRi = 0.

Note that the axioms imply that R0 is a subring of R and that each Ri is a left

and right R0−module. Note also that ∩i∈ZRi is always an ideal in R.

Examples. (1) R = {q = m/pk ∈ Q, m ∈ Z( p), k ≥ 0}, Ri = p−iZ( p) where p

is a fixed prime.

(2) R = k[x, x−1], Ri = xik[x−1] = k{xi, xi−1, xi−2, . . .}.

(3) Suppose R is a finitely generated k−algebra with generating set {x1, . . . , xn}.

If i

≥0 let Ri be the k

−subspace of R spanned by words in the xj ’s of length

at most i, let R0 = k and let Ri = 0 whenever i < 0. This is a positive

filtration on R.

(4) R = Z, Ri = p−iZ for i ≤ 0 and Ri = R otherwise. This is a negative

filtration. More generally,

(5) Let R be any ring and let I be an ideal of R. Then Ri = I −i is a negative

filtration, called the I −adic filtration.

7.2. Graded rings.

Definition. A (Z−)graded ring is a ring S which can be written as

S =i∈Z S i

for some additive subgroups S i ⊆ S , satisfying S i.S j ⊆ S i+j for all i, j ∈ Z and

1 ∈ S 0. S i is called the i−th homogeneous component of S , and an element s ∈ S

is homogeneous iff it lies in some S i.

A graded right ideal J r S is a right ideal of the form ⊕i∈ZJ i with J i ≤ S i.

Lemma. If J is a finitely generated graded right ideal of a graded ring S , then J

has a finite generating set consisting of homogeneous elements.



37

Proof. Since J is graded, note that J ∩ S i = J i for each i ∈ Z. Let {x1, . . . , xn} be

a generating set for J . If xj =

i∈Z aij for some aij

∈J i, we see that each aij lies

in J and is homogeneous. Clearly, {aij : aij = 0} is now a finite generating set for

J consisting of homogeneous elements.

7.3. Associated graded ring.

Definition. Let R be a filtered ring with filtration (Ri)i∈Z. Define the abelian group

gr R =i∈Z

Ri/Ri−1.

Equip gr R with multiplication, which is given on homogeneous components by

Ri/Ri−1 × Rj/Rj−1 −→ Ri+j/Ri+j−1

r + Ri−1 , s + Rj−1 → rs + Ri+j−1

and on the whole of gr R by bilinear extension. Then gr R becomes a ring called the

associated graded ring of R.

Note that the multiplication is well-defined because RiRj ⊆ Ri+j−1, Ri−1Rj ⊆Ri+j−1 and RiRj−1 ⊆ Ri+j−1. One should think of gr R as an approximation

to the ring R which is often easier to understand but nonetheless contains useful

information about the ring R itself.

Notation. If r ∈ Ri\Ri−1 then the symbol of r in gr R is the homogeneous element

σ(r) = r + Ri−1

∈gr R.

7.4. Almost commutative algebras.

Definition. An almost commutative k−algebra is a finitely generated k−algebra

with generators {x1, . . . , xn} such that

xixj − xjxi ∈ k{1, x1, . . . , xn} for all i, j = 1, . . . , n.

Examples. (1) The commutative polynomial k−algebra k[x1, . . . , xn].

(2) The Weyl algebra An(k).

(3) The universal enveloping algebra

U (g) of a k

−Lie algebra g, dimk g <

∞.

(4) Any homomorphic image of U (g).

Proposition. Let R be an almost commutative algebra generated by {x1, . . . , xn}.

Equip R with the positive filtration as in Example 7.1(3). Then there exists a

surjective homomorphism of k−algebras

ϕ : k[X 1, . . . , X n] gr R

given by ϕ(X i) = σ(xi), i = 1, . . . , n.



38

Proof. Note that xi ∈ R1 for all i. If xi, xj /∈ k, we have

σ(xi) = xi + R0 ∈ R1/R0 and σ(xj) = xj + R0 ∈ R1/R0.

Since R is almost commutative,

σ(xi)σ(xj) = xixj + R1 = xjxi + R1 = σ(xj)σ(xi),

meaning that σ(xi) and σ(xj) commute. If one of xi, xj lies in k then this is also

true. Hence the k−algebra map ϕ exists.

To show that ϕ is surjective, it’s sufficient to show that u + Rt−1 lies in Im ϕ for

any u ∈ Rt\Rt−1. Write u as a sum of words of length at most t in the generators

{x1, . . . , xn}. Since we’re interested in u + Rt−1, we can assume that each word

actually has length t.

Writing u =mj=1 λjxi(j)1 xi(j)2 · · · xi(j)t for some i

(j)1 , . . . , i

(j)t ∈ {1, . . . , n}, we have

u + Rt−1 =

mj=1

λjϕ(X i(j)1

)ϕ(X i(j)2

) · · · ϕ(X i(j)t

) ∈ Im ϕ.

7.5. Proposition. Any almost commutative k−algebra R is a quotient of U (g) for

some finite dimensional k−Lie algebra g.

Proof. Let {x1, . . . , xn} be the generating subset of R and let g = k{1, x1, . . . , xn}.

R is a k−Lie algebra under the commutator bracket [a, b] = ab − ba, and the

condition that R is almost commutative says precisely that g is a Lie subalgebra of R and hence a finite dimensional Lie algebra itself.

Now we get a ring homomorphism θ : U (g) → R extending the natural injection

g → R of k−Lie algebras. Since {x1, . . . , xn} generates R as a k−algebra, θ is onto,

as required.

Example. Let h2n+1 be the Heisenberg Lie algebra of dimension 2n + 1 with gen-

erators {x1, . . . , xn, y1, . . . , yn, z} and relations [yi, xi] = z for i = 1, . . . , n, all the

other brackets being zero. Then there is a surjective map of k−algebras

θ : U (h2n+1) An(k)

given by θ(xi) = xi, θ(yi) = yi and θ(z) = 1.

Note that one can also represent h2n+1 by (n + 2) × (n + 2) matrices:

h2n+1 ∼=

0 k · · · k k

0 0 · · · 0 k

· · · · · · ·0 0 · · · 0 k

0 0 · · · 0 0

≤ gln+2(k).



39

7.6. Filtered and graded modules.

Definition. Let R be a filtered ring with filtration (Ri)i∈Z and let M be a right R−module. A filtration on M is a set (M i)i∈Z of additive subgroups of M satisfying

• M i ⊆ M i+1 for all i ∈ Z,

• M i.Rj ⊆ M i+j for all i, j ∈ Z,

• ∪i∈ZM i = M .

Positive, negative and separated filtrations are defined analogously to (7.1). Fil-

tered left modules are defined similarly.

Example. Let M be a finitely generated right R−module with generating set A.

Then M i = ARi for all i ∈ Z gives a filtration of M , known as the standard

filtration.

Definition. Let S = ⊕i∈ZS i be a graded ring. A graded right S −module is a right

S −module V of the form

V =i∈Z

V i

such that V iS j ⊆ V i+j for all i, j ∈ Z.

7.7. Associated graded modules.

Definition. Let R be a filtered ring and let M be a filtered right R−module with

filtration (M i)i∈Z. Define the abelian group

gr M =i∈Z M i/M i−1.

Equip gr M with a gr R−action, which is given on homogeneous components by

M i/M i−1 × Rj/Rj−1 −→ M i+j/M i+j−1

m + M i−1 , r + Rj−1 → mr + M i+j−1

and on the whole of gr M by bilinear extension. Then gr M becomes a graded

gr R−module, called the associated graded module of M .

If N is a submodule of M , define the subspace filtration (N i)i∈Z on N by N i =

N ∩ M i. Also, define the quotient filtration ((M/N )i)i∈Z on M/N by (M/N )i =

(M i + N )/N .

Proposition. Let R be a filtered ring, let M be a filtered right R−module with filtration (M i)i∈Z and let N be a submodule of M . Equip N with the subspace

filtration and M/N with the quotient filtration. Then

(1) There exists an injection α : gr N → gr M of right gr R−submodules.

(2) Identifying gr N with its image in gr M , we have

gr(M/N ) ∼= gr M/ gr N

as right gr R−modules.



40

Proof. The natural composition of maps N i → M i and M i M i/M i−1 has kernel

N i

∩M i−1 = N

∩M i−1 = N i−1. So we have an injection of abelian groups

αi : N i/N i−1 → M i/M i−1

for all i ∈ Z. Putting these together we get an injection

α = ⊕αi : gr N → gr M.

Exercise: check that α is a right gr R−module homomorphism.

This proves (1).

We have the quotient filtration ((M i + N )/N )i∈Z of M/N . Consider the compo-

sition

β i : M i/M i−1ui→ M i + N

M i−1 + N

vi→ (M i + N )/N

(M i−1 + N )/N

where ui(m + M i−1) = m + M i−1 + N and vi is the natural isomorphism.Note that ui is onto, whereas

ker(ui) =M i ∩ (M i−1 + N )

M i−1=

M i−1 + (M i ∩ N )

M i−1=

M i−1 + N iM i−1

= Im(αi)

by the modular law. Since vi is an isomorphism, β i is onto and ker(β i) = Im(αi)

for all i ∈ Z. Letting

β = ⊕β i : gr M → gr(M/N )

we see that β is onto and ker(β ) = Im(α).

Exercise: check that β is a right gr R−module homomorphism.

Hence β induces the required isomorphism of right gr R−modules

gr M/α(gr N ) = gr M/ ker(β ) ∼= Im(β ) = gr(M/N ).

7.8. Lifting the Noetherian property - I.

Theorem. Suppose R is a positively filtered ring such that gr R is right Noetherian.

Then R is right Noetherian.

Proof. If I ⊆ J are right ideals of R, equip I and J with the subspace filtrations

and J/I with the quotient filtration. Proposition 7.7 shows that gr I and gr J are

right ideals of gr R, and moreover that gr I ⊆ gr J and gr(J/I ) ∼= gr J/ gr I .

SupposeI 1 ≤ I 2 ≤ I 3 ≤ . . .

is an increasing chain of right ideals of R. Applying gr we obtain an increasing

chain of right ideals of gr R

gr I 1 ≤ gr I 2 ≤ gr I 3 ≤ . . .

which stops because gr R is right Noetherian. So it’s sufficient to prove that if

gr I = gr J for right ideals I ⊆ J then I = J .



41

Let (Ri)i∈Z be the filtration on R. If I < J then I ∩ Ri < J ∩ Ri for some i.

Since the filtration on R is positive, we can choose a least such i. Choose some

x ∈ (J ∩ Ri)\I . Then x /∈ Ri−1 as I ∩ Ri−1 = J ∩ Ri−1 by minimality of i. Now

σ(x) = x + Ri−1 ∈ gr J = gr I

so there exists y ∈ I such that σ(x) = σ(y) = y + Ri−1. Hence x − y ∈ Ri−1. But

x − y ∈ J since y ∈ I ⊆ J , so x − y ∈ J ∩ Ri−1 = I ∩ Ri−1. Hence x − y ∈ I so

x ∈ I , a contradiction. The result follows.

Of course, the same result holds with ”right” replaced by ”left”.

Corollary. Any almost commutative algebra R is right and left Noetherian. This

includes An(k) and U (g).

Proof. By Proposition 7.4, gr R is a quotient of a polynomial algebra k[X 1, . . . , xn]for some n, which is right and left Noetherian by Theorem 1.14. Hence R is right

and left Noetherian by Theorem 7.8.

7.9. Topologies. Assume until the end of this chapter that R is a negatively fil-

tered ring and M is a negatively filtered right R−module with filtration (M i).

We can define a topology on M called the filtration topology , by choosing the open

subsets to be unions of sets of the form m + M i. Then as required, the collection

of open sets is closed under arbitrary unions, and also under finite intersections -

since the intersection of m1 + M i(1), . . . , mn + M i(n) is a union of sets of the form

m + M i where i = min

{i(1), . . . , i(n)

}. If the filtration is separated, we can define

a metric on M : fix a real number c > 1 and set

d(x, y) = ck where k = min{i : x − y ∈ M i}.

Note that we need ∩M i = 0 to ensure that d(x, y) = 0 ⇒ x = y. It can be checked

that the topology we get on M from the metric is the filtration topology, so the

choice of the constant c is irrelevant.

Two filtrations (M i) and (M i) give the same filtration topology on M if and

only if for all i there exist s(i) such that M s(i) ⊆ M i and t(i) such that M t(i) ⊆ M i .

The filtrations are then said to be topologically equivalent .

7.10. Cauchy sequences and completions.

Definition. A Cauchy sequence on M is a sequence (xn)∞n=0 such that for any

open set U containing 0, there exists c(U ) ∈ N such that xn − xm ∈ U whenever

n, m ≥ c(U ). Two Cauchy sequences (xn) and (yn) are said to be equivalent if

xn−yn → 0. Equivalently, given any open set U containing 0, there exists N (U ) ∈ Zsuch that xn − yn ∈ U whenever n ≥ N (U ).

Exercise. Check that this gives an equivalence relation on the set of all Cauchy

sequences on M .



42

Let [xn] denote the equivalence class of the Cauchy sequence (xn) under this

equivalence relation, and let

M denote the set of all such equivalence classes. The

operations

[xn] + [yn] = [xn + yn]

[xn] . [yn] = [xnyn] [xn], [yn] ∈ Rturn R into an ring, and

[xn] + [yn] = [xn + yn] [xn], [yn] ∈ M

[xn] . [rn] = [xnrn] [xn] ∈ M , [rn] ∈ Rturn M into a right R−module. There is a ring homomorphism

R → Rr

→[r]

where [r] is the equivalence class of the constant sequence with value r. The kernel

of this map is ∩Ri. This also enables us to view M as a right R−module, and we

have an analogous map of R−modules

M → M

m → [m]

Definition. We say that R ( M ) is the completion of R ( M , respectively). We call

R ( M ) complete if the natural map R → R ( M → M ) is an isomorphism.

Note also that we can define a filtration on

M by setting

M i = {[xn] : ∃ci ∈ N such that xn ∈ M i whenever n ≥ ci}the set of equivalence classes of Cauchy sequences ”which end up in M i eventually”.

Thus R and M have naturally defined topologies, and the natural maps M → M

and R → R are continuous with respect to these topologies.

Note that a complete module M is necessarily separated, since the kernel of the

natural map M → M is ∩M i.

Exercise. Check all the details above!

7.11. Inverse limits. There is another way of approaching completions. Let r < s

and let πrs : M/M r M/M s be the natural projection. Define

lim←−

M/M i = {(yn+M n)n∈Z ∈i∈Z

M/M n : πrs(yr+M r) = ys+M s whenever r < s.}

This becomes a right R−module with pointwise addition and scalar multiplica-

tion. When M = R, these operations turn lim←−

R/Ri into a ring.

Lemma. M is isomorphic to lim←−

M/M i as R−modules.



43

Proof. If (xn) is a Cauchy sequence in M , then for each i ∈ Z there exists c(M i) ∈ Nsuch that xn

≡xm mod M i for all n, m

≥c(M i). We can choose the c(M i) to

satisfy

c(M 0) ≤ c(M −1) ≤ c(M −2) ≤ . . . .

Now if r < s then πrs(xc(M r) + M r) = xc(M r) + M s = xc(M s) + M s since c(M r) ≥c(M s), which means that (xc(M i) + M i)i∈Z ∈ lim

←−M/M i.

Define

θ : M → lim←−

M/M i

[xn] → (xc(M i) + M i)i∈Z.

Note that this doesn’t depend on the choice of the numbers c(M i): if d(M i) are

such that xn ≡ xm mod M i whenever n, m ≥ d(M i), then

xc(M i) ≡ xmax(c(M i),d(M i)) ≡ xd(M i) mod M i

for all i ∈ Z.

Now, if [xn] = [yn] ∈ M , then xn − yn → 0 as n → ∞. Hence for all i ∈ Z there

exists t(M i) such that xn − yn ∈ M i whenever n ≥ t(M i). Let c(M i), d(M i) be the

integers corresponding to [xn] and [yn]. Letting e(M i) = max(c(M i), d(M i), t(M i)),

we have

xc(M i) ≡ xe(M i) ≡ ye(M i) ≡ yd(M i) mod M i

for all i ∈ Z. Hence θ([xn]) doesn’t depend on the choice of Cauchy sequence inside

the equivalence class [xn], so θ is well defined.

Exercise. Check that θ is a map of right R−modules.

It remains to show that θ is an isomorphism. If [xn] ∈ ker(θ), then xc(M i) ∈ M i

and hence xn ∈ M i for all n ≥ c(M i). Hence xn → 0 and [xn] = 0, so θ is injective.

If y = (yi + M i)i∈Z ∈ lim←−

M/M i, let xn = y−n ∈ M for all n ∈ N. Then (xn)∞n=0

is a Cauchy sequence and that θ([xn]) = y. Hence θ is surjective.

Equip each M/M i with the discrete topology and

i∈Z M/M i with the (Ty-

chonov) product topology. Then the map θ constructed above is actually a home-

omorphism, where we give

M the filtration topology and lim

←−M/M i ⊆

i∈Z M/M i

the subspace topology.Also, in the case when M = R, θ is a ring isomorphism.

The advantage of inverse limits is that the elements are easy to manipulate; the

disadvantage is that everything is dependent on the filtration.

7.12. Examples.

(1) Z p, the ring of p−adic integers is the completion of R = Z with respect to

the p−adic filtration Ri = p−iZ for i ≤ 0. Thus Z p = lim←−

Z/piZ.



44

(2) More generally, if R is any ring and I R is a two-sided ideal, we can

complete R with respect to the I

−adic filtration:

R = lim←−

R/I i.

(3) If R = k[X 1, . . . , X n] and I = (X 1, . . . , X n) R then R ∼= k[[X 1, . . . , X n]],

the ring of power series. Here we identify a power series with the sequence

of partial sums. For example, if n = 1,

a0 + a1X + a2X 2 + . . . ↔ (a0 + I, a0 + a1X + I 2, . . .)

This works similarly for arbitrary n - the notation is cumbersome.

(4) Consider the group of all n × n invertible matrices with coefficients in Z p.

This is GLn(Z p). Let

K = {M ∈ GLn(Z p) : M ≡ I n×n mod p} = ker(GLn(Z p) → GLn(F p)).

K is an example of a torsionfree compact p−adic Lie group.

Let R = Z p[K ] be the group algebra of K over Z p. Let I be the mod p

augmentation ideal of R: I = ker(ϕ) where

ϕ : R → F pg∈K λgg →

g∈K λg

where λ ∈ F p is the reduction of λ ∈ Z p modulo p. Then R, the completionof R with respect to the I −adic filtration is the Iwasawa algebra of K ,

written ΛK .

7.13. Proposition. Let R be a ring and I R an ideal such that R/I is a division

ring. Then the completion R of R with respect to the I −adic filtration is a local

ring , that is, it has a unique maximal left and right ideal.

Proof. Let x ∈ I = {(xn + I n) ∈ lim←−

R/I n : x1 ∈ I }. Then

y = 1 + x + x2 + x3 + . . . = (1 + I, 1 + x2 + I 2, 1 + x3 + x23 + I 3, . . .)

∈ Rand y(1 − x) = (1 − x)y = 1, so 1 − x is invertible in R whenever x ∈ I . Now if

u = (un + I n) /∈ I , then u1 /∈ I . Since R/I is a division ring, we can find v ∈ R

such that u1v ≡ vu1 ≡ 1 mod I . Hence uv − 1 ∈ I so uv and vu are invertible by

the above. Hence u is invertible whenever u /∈ I so I is the unique maximal left

and right ideal of R as required.

This shows that Z( p) → Z p for any prime p.



45

7.14. Lifting the Noetherian property - II.

Theorem. Let R be a complete negatively filtered ring such that gr R is right Noe-therian. Then R is right Noetherian.

Proof. Let I r R be a right ideal. Then gr I is a graded right ideal of gr R by

Proposition 7.7. Since gr R is right Noetherian, gr I is finitely generated. We can

choose a homogeneous generating set, by Lemma 7.2:

gr I =

mi=1

si gr R

where s1, . . . , sm are all homogeneous. Each si is the symbol of some xi ∈ I :

si = σ(xi) ∈ Rn(i)/Rn(i)−1, say.

We will show that

I =

mi=1

xiR.

Since each xi ∈ I and I is a right ideal, the reverse inclusion is obvious. Let y ∈ I

and suppose that y ∈ Rj\Rj−1 (if y ∈ ∩Rj then y = 0 ∈ mi=1 xiR since the

filtration on R is separated.)

Now σ(y) = y + Rj−1 ∈ gr I =m

i=1 si gr R so we can find ti ∈ gr R such that

σ(y) =

mi=1

siti.

Since σ(y) and the si are homogeneous, we may assume that ti ∈ Rj−n(i)/Rj−n(i)−1,i = 1, . . . , m. If ti = 0 set ri1 = 0, otherwise choose ri1 ∈ Rj−n(i)\Rj−n(i)−1 such

that ti = ri1 + Rj−n(i)−1. Hence

y ≡m

i=1

xiri1 mod Rj−1.

Now let y1 = y −mi=1 siri1 ∈ I ∩ Rj−1. If y1 = 0, stop; if not, repeat the

above with y1 in place of y. This gives a sequence (ri1, ri2, ri3, . . .) such that rik ∈Rj−n(i)−k+1 and

yk = y

−

m

i=1xi

k

l=1ril

∈I

∩Rj−k

for every k ≥ 1. But now (k

l=1 ril)∞k=1 is a Cauchy sequence in R which converges

to some ri ∈ R because R is complete. Also, yk → 0 since yk ∈ Rj−k for all k.

Hence, letting ri =∞

l=1 ril, we obtain

y =

mi=1

xiri ∈m

i=1

xiR

as required.



46

Examples. (1) Giving Z p the p−adic filtration, gr Z p∼= F p[t] which is Noe-

therian, and of course Z p is complete with respect to this filtration. Hence

Z p is Noetherian.

(2) Suppose K is a compact p−adic Lie group of a special type, namely a uni-

form pro- p group, of dimension d. One example is the K appearing in

Example 7.12(4), it has dimension n2. It can be shown that there is a fil-

tration of ΛK which is topologically equivalent to the I −adic filtration of

ΛK , such that

gr ΛK∼= F p[X 0, X 1, . . . , X d],

the polynomial algebra over F p in d + 1 variables. Hence ΛK is right and

left Noetherian.



47

8. Weyl algebras

Throughout this chapter, k denotes a field of characteristic 0, and all moduleswill be left modules.

8.1. Recall. The n-th Weyl algebra An has generators {x1, . . . , xn, y1, . . . , yn} as

a k−algebra and relations [yi, xi] = 1 for i = 1, . . . n.

Let A = k[X 1, . . . , X n] be the polynomial algebra. From (1.11), we know that

there is a natural k−algebra homomorphism

ρ : An → Endk(A)

xi → (f → X if )

yi → (f → ∂f ∂Xi

).

We can therefore think of A as a left An−module, viar.f = ρ(r)(f ) for all r ∈ An, f ∈ A.

8.2. Notation.

• When α = (α1, . . . , αn) ∈ Nn, write xα = xα1

1 · · · xαnn and yα = yα1

1 · · · yαnn .

• The degree of the monomial xα is defined to be |α| = α1 + . . . + αn. We

also write α! = α1! · · · αn!.

• ((An)m) is the positive filtration on An defined in Example 7.1(3): (An)m

consists of the span of all words in the generators {x1, . . . , xn, y1, . . . , yn}of length at most m. This is also known as the Bernstein filtration on An.

• The degree of a nonzero element r ∈ An is the integer m = deg r such thatr ∈ (An)m\(An)m−1.

8.3. Proposition. The set {xαyβ : α, β ∈ Nn} is a basis for An.

Proof. Any word in the generators can be brought to a linear combination of the

xαyβ ’s by ”pushing x’s to the left”, using the relations yixi = xiyi + 1. Hence

{xαyβ} spans An.

Next, suppose r =

α,β∈Nn λαβxαyβ ∈ An is zero but some coefficient λαβ is

nonzero. Choose δ ∈ Nn such that λγδ = 0 for some γ ∈ Nn, but λαβ = 0 whenever

|β | < |δ|.Now, if

|β | ≥ |

δ|

then either β = δ or β i > δi for some i. But in the latter case,

yβii .X δii = 0 so yβ .X δ = 0, whereas yδ.X δ = δ!. Hence

r.X δ =|β|<|δ| λαβxαyβ .X δ +

|β|≥|δ|,β=δ λαβxαyβ .X δ +

α∈Nn δ!λαδX α

= δ!

α∈Nn λαδX α = 0

since char(k) = 0 and λγδ = 0. This contradicts r = 0 and the result follows.

In fact, this is true even when char(k) > 0.



48

Corollary. Equip An with the positive filtration (An)m given in Example 7.1(3).

Then

gr An ∼= k[Z 1, . . . , Z 2n].

Proof. Let ϕ : k[Z 1, . . . , Z 2n] gr An be the surjective k−algebra homomorphism

given by ϕ(Z i) = σ(xi), ϕ(Z i+n) = σ(yi), i = 1, . . . , n, as in Proposition 7.4.

Note that ϕ sends homogeneous polynomials of degree m into the m−th graded

part of gr An. Hence if f ∈ ker(ϕ) then we may assume f to be homogeneous of

degree m say:

f =

α,β∈Nn

|α|+|β|=m

λαβZ α11 · · · Z αnn Z β1n+1 · · · Z βn2n .

Now ϕ(f ) = 0 implies that

α,β∈Nn

|α|+|β|=m

λαβxαyβ ∈ (An)m−1.

Since the {xαyβ} are linearly independent over k by Proposition 8.3, λαβ = 0 for

all relevant α, β and hence f = 0. Hence ϕ is an isomorphism.

8.4. Zero divisors.

Proposition. Suppose R is a filtered ring with a separated filtration (Ri)i∈Z, such

that gr R is a domain. Then R is a domain also.

Proof. Suppose for a contradiction that r, s ∈ R are nonzero but rs = 0. Since the

filtration is separated, there exist i, j∈

Z such that r∈

Ri

\Ri−1 and s

∈Rj

\Rj−1.

But now σ(r) = r + Ri−1 and σ(s) = s + Rj−1 are nonzero in gr R and

σ(r).σ(s) = rs + Ri+j−1 = 0,

contradicting the assumption that gr R is a domain.

Corollary. (1) An is a domain.

(2) If G is a uniform pro- p group, then the Iwasawa algebra ΛG is a domain.

8.5. PBW Theorem. Let g be a finite dimensional k−Lie algebra with basis

{x1, . . . , xn}. The enveloping algebra U (g) contains the standard monomials

xα = xα1

1 xα2

2

· · ·xαn

n

for all α ∈ Nn.

Theorem (Poincare-Birkhoff-Witt). {xα : α ∈ Nn} form a basis for U (g).

Proof. Omitted.

We can deduce exactly as in Corollary 8.3 that gr U (g) is a polynomial algebra in

n variables, and it follows from Proposition 8.4 that U (g) is a domain. This works

also when char(k) > 0.



49

8.6. Proposition.

(1) Z (An) = k.

(2) An is a simple ring.

Proof. First, note that if a,b,c are elements of some ring R then

[ab,c] = a[b, c] + [a, c]b and [a,bc] = b[a, c] + [a, b]c.

Next, we prove that a, b satisfy [a, b] = 1 then [am, b] = mam−1. This is true when

m = 1 so assume inductively that m > 1 and [am−1, b] = (m − 1)am−2. Then

[am−1, b] = a[am−1, b] + [a, b]am−1 = (m − 1)aam−2 + am−1 = mam−1

as claimed. In particular, since [yi, xi] = 1, [yβii , xi] = β iyβi−1

i for all β i ∈ N.

Now let ei ∈ Nn be the element with 1 in i−th position and 0’s elsewhere.

Because xi commutes with every generator apart from yi, the above shows thatinside An,

[xαyβ, xi] = β ixαyβ−ei .

Similarly, since [xi, −yi] = 1 we see that [xαii , −yi] = αixαi−1

i so

[xαyβ, yi] = −αixα−eiyβ .

Now if r =

α,β∈Nn λαβxαyβ ∈ Z (An) then r must commute with each generator.

Hence

[r, xi] =

α,β∈Nn

β iλαβxαyβ−ei = 0

so β iλαβ = 0 for every possible i,α,β by Proposition 8.3. By considering [r, yi] weobtain αiλαβ = 0 for all i,α,β . As char(k) = 0, λαβ = 0 ⇒ αi = β i = 0 for all i, so

the only possible nonzero coefficient in r is λ0,0. Hence r ∈ k and part (1) follows.

Next, suppose that I is a nonzero two-sided ideal of An. Pick a nonzero element

r ∈ I of least degree m, say. The above equation for [r, xi] shows that

deg[r, xi] ≤ deg r − 1 < m

and similarly deg[r, yi] ≤ deg r − 1 for all i. Since I is a two-sided ideal [r, xi] and

[r, yi] must lie in I for all i. By minimality of m, these must all be zero and hence

r ∈ Z (An) = k. Since r = 0 and k is a field, 1 ∈ I and hence I = An.

Warning: this result is false in positive characteristic!8.7. Rings of differential operators. Let R be a commutative k−algebra. The

ring of differential operators of R is a subring of Endk(R). We construct it induc-

tively. Note that we have an injection a → (a : r → ar) of R into Endk(R). Thus

we can identify R with a subring of Endk(R). Define

D0(R) = {D ∈ Endk(R) : [D, a] = 0 for all a ∈ R}D1(R) = {D ∈ Endk(R) : [D, a] ∈ D0(R) for all a ∈ R}



50

D2(R) = {D ∈ Endk(R) : [D, a] ∈ D1(R) for all a ∈ R}and so on. These are k

−vector spaces D0(R)

⊆D1(R)

⊆D2(R)

⊆. . .

⊆Endk(R).

Lemma. Let P ∈ Dn(R) and Q ∈ Dm(R). Then P Q ∈ Dm+n(R).

Proof. Induct on n + m. If n + m = 0, then n = m = 0 so P, Q commute with R

and hence P Q also commutes with R, meaning that P Q ∈ D0(R).

Suppose now that this is true whenever m+n < l and consider the case m+n = l.

Fix a ∈ R. By definition, [Q, a] ∈ Dm−1(R) and [P, a] ∈ Dn−1(R), so by inductive

hypothesis P [Q, a] ∈ Dm+n−1(R) and [P, a]Q ∈ Dm+n−1(R). Hence

[P Q, a] = P [Q, a] + [P, a]Q ∈ Dm+n−1(R)

and hence P Q

∈Dm+n(R) as required.

Definition. The ring of differential operators D(R) of R is the subring ∪∞n=0Dn(R)

of Endk(R).

Note that by Lemma 8.7, D(R) is a subring, and moreover, (Di(R))i∈Z is a

positive filtration on D(R), provided we assume Dk(R) = 0 for negative k.

8.8. Derivations.

Definition. A derivation of a commutative k−algebra R is a k−linear map D :

R → R satisfying the Leibnitz rule

D(ab) = aD(b) + D(a)b for all a, b∈

R.

The k−vector space of all derivations of R is denoted by Der(R).

Note also that if x ∈ R then xD : R → R is another derivation, since

(xD)(ab) = x(aD(b) + D(a)b) = a(xD)(b) + (xD)(a)b

Hence Der(R) is a left R−module.

For small values of n, we have alternative descriptions of Dn(R).

Proposition. Let R be a commutative k−algebra. Then

(1) D0(R) = {a : a ∈ R}, and

(2) D1

(R) = D0

(R) + Der(R).

Proof. (1) Since R is commutative, a ∈ D0(R) for all a ∈ R. On the other hand,

if D ∈ D0(R) then D(a) = D(a.1) = (D.a)(1) = (a.D)(1) = a(D(1)) = D(1)(a).

Hence D = D(1) as required.

(2) If D ∈ Der(R) and a ∈ R, then

[D, a](b) = (Da)(b) − (aD)(b) = D(ab) − aD(b)

= D(a)b = D(a)(b).



51

Thus [a, D] = D(a) ∈ D0(R) by (1) for any a ∈ R, showing that Der(R) ⊆ D1(R).

Hence D0(R) + Der(R)

⊆D1(R).

On the other hand, let Q ∈ D1(R) and set P = Q − Q(1) ∈ D1(R), so that

P (1) = 0. Now P ∈ D1(R) so [P, a] ∈ D0(R) for any a ∈ R and hence [[P, a], b] = 0

for any a, b ∈ R. Applying this to 1 ∈ R we get

0 = [P, a]b(1) − b[P, a](1) = (P a − aP )(b) − b(P a − aP )(1) =

= P (ab) − aP (b) − bP (a) + baP (1)

so P (ab) = aP (b)+bP (a) and P ∈ Der(R). Hence Q = P + Q(1) ∈ Der(R)+D0(R)

as required.

8.9. The case R = k[X 1, . . . , X n].

Proposition. Let R = k[X 1, . . . , X n] and let ∂ i = ∂ ∂Xi . Then

Der(R) =n

i=1

R∂ i.

Proof. It’s clear that every map D : R → R of the form D =n

i=1 f i∂ i is a

derivation. Conversely, let D ∈ Der(R). Then D(X mi ) = mX m−1i D(X i) by an easy

induction. Hence

(D −n

i=1

D(X i)∂ i)(X α1

1 · · · X αnn ) = 0

so D =

ni=1 D(X i)∂ i ∈

ni=1 R∂ i, since the monomials form a k−basis for R.

By this result and Proposition 8.8(2), we see that

D1(R) = R + R∂ 1 + . . . + R∂ n.

This is a “part” of the Weyl algebra. Indeed, we have

Theorem. D(R) =

α∈Nn R∂ α ∼= An.

Proof. Omitted.

Warning: this is false in characteristic p > 0!

8.10. An

−modules.

Proposition. (1) The natural An−module A = k[X 1, . . . , X n] is simple.

(2) A ∼= An/n

i=1 Anyi as left An−modules.

(3) No nonzero left An−module M is finite dimensional over k.

Proof. (1) Let 0 = f ∈ A. Let X α = X α11 · · · X αnn be a monomial of maximal

degree in f with nonzero coefficient λ, say. Then

yα.f = λyα.X α = α!λ = 0



52

since char(k) = 0 so 1 ∈ Anf . Hence any nonzero element of A generates A, so A

is simple.

(2) We have A = An.1 is a cyclic module, so A ∼= An/I where I = ann(1) r An.

Since yi.1 = 0 for all i = 1, . . . , n, I containsn

i=1 Anyi. Suppose r ∈ I . By

Proposition 8.3, we can write r = a + b where a = a(x1, . . . , xn) ∈ k[x1, . . . , xn] and

b ∈ni=1 Anyi. Now r.1 = a.1 = a(X 1, . . . , X n) = 0 forces a = 0 so r ∈n

i=1 Anyi

as required.

(3) Suppose M is a finite dimensional An−module of dimension t. The action

of An gives a ring homomorphism

ϕ : An → Endk(M ) ∼= M t(k).

Now ϕ(y1)ϕ(x1) − ϕ(x1)ϕ(y1) = I t×t. Since tr(U V ) = tr(V U ) for any matrices

U, V , taking traces we conclude that t = 0, so M = 0 as required.

Part (3) is a special case of Bernstein’s Inequality . (1) and (3) are false in positive

characteristic, and only (2) remains valid.

8.11. Solutions of PDEs. Suppose we have a finite set of elements P 1, . . . , P m ∈An. Then we can consider the set of simultaneous partial differential equations

(‡) P 1.f = P 2.f = . . . = P m.f = 0

where f ∈ A = k[X 1, . . . , X n]. The An−module associated with this system of

equations is M P 1,...,P m = An/m

i=1 AnP i.

Proposition. The k

−vector space of polynomial solutions of (

‡) is isomorphic to

HomAn(M P 1,...,P m , A).

Proof. We can define a left An−module map θf : An → A by θf (1) = f . If f ∈ A

is a solution, then P i ∈ ker(θf ) for all i, so we obtain a map θf : M P 1,...,P m → A.

Conversely, if α : M P 1,...,P m → A is a map of left An−modules, then α(1) ∈ A is

a solution of the system (‡). This gives a bijective correspondence.

One can also look for solutions in other spaces B, which are An−modules - the

solution space will be HomAn(M P 1,...,P m , B).

For example, when k = R, we can take B = C∞(U ), the set of all infinitely

differentiable functions f : U

→R, for some open set U

⊆Rn. This is a left

An−module, with the action given by

(xi.f )(a1, . . . , an) = aif (a1, . . . , an) and

yi.f =∂f

∂ai.



53

9. Dimensions

Throughout, k denotes a field. Again, modules will be on the left.

9.1. Hilbert polynomials. We start by considering finitely generated graded

modules

N =∞i=0

N i

for the commutative polynomial algebra S = k[X 1, . . . , X n] = ⊕∞i=0S i. Here S i is

the space of all homogeneous polynomials of degree i.

Pick a homogeneous generating set {u1, . . . , um} for N , where deg uj = kj, say.

Then

N =

m

j=1

ujS =

∞

i=0

m

j=1

ujS i−kj so N i =

m

j=1

ujS i−kj

for all i ≥ 0. Since each graded component S i of S is finite dimensional over k, we

see that each N i must itself be finite dimensional over k.

Definition. The Poincare series for N is the power series

P (N, t) =

∞i=0

(dim N i) ti ∈ Z[[t]].

Theorem (Hilbert,Serre). Let S and N be as above. Then

(1) There exists f (t)∈

Z[t] such that

P (N, t) =f (t)

(1 − t)n.

(2) Let d ∈ N be the order of the pole at t = 1 of P (N, t). Then there exists

ϕ(t) ∈ Q[t] of degree d − 1, such that for large enough i, dim N i = ϕ(i).

(3) There exists χ(t) ∈ Q[t] of degree d, such that for large enough i,

ij=0

dim N j = χ(i).

Proof. (1) Proceed by induction on n. When n = 0, N is a finitely generated

k−module, and hence dimk N < ∞. Hence N i = 0 for large enough i, so P (N, t) isclearly a polynomial.

Now, assume n > 1 and consider the map θ : N → N given by θ(v) = X nv.

Since R is commutative, this is a map of R−modules. Since X n has degree 1, θ

maps N i to N i+1. Let K i = ker(θ|N i) and let Li+1 = coker(θ|N i) = N i+1/X nN i.

Let K = ⊕∞i=0K i and L = ⊕∞

i=0Li where we set L0 = N 0. Then K = ker(θ)

and L = coker(θ). Note that both K and L are finitely generated S −modules since

S is Noetherian and N is finitely generated. Also, both are killed by X n, and are



54

hence finitely generated graded k[X 1, . . . , X n−1]−modules. By induction, P (K, t)

and P (L, t) have the required form:

P (K, t) =a(t)

(1 − t)n−1and P (L, t) =

b(t)

(1 − t)n−1

for some a(t), b(t) ∈ Z[t]. Since N i/K i ∼= X nN i as k−vector spaces, we see that

dim N i − dim K i = dim N i+1 − dim Li+1.

Multiply this equation by ti+1 and sum over i to get

tP (N, t) − tP (K, t) = P (N, t) − P (L, t),

since P (N, 0) = P (L, 0). Hence

P (N, t) =P (L, t) − tP (K, t)

1−

t=

b(t) − ta(t)

(1−

t)n

as required.

(2) By cancelling powers of (1 − t) if necessary, we may assume that

f (t) = a0 + a1t + . . . + asts ∈ Z[t]

satisfies f (1) = 0, so that P (N, t) = f (t)(1 − t)−d with d the order of the pole of

P (N, t) at t = 1.

Now, (1 − t)−1 = 1 + t + t2 + . . .. Repeated differentiation gives

(1 − t)−d =∞

m=0

d + m − 1

d − 1

tm, so

P (N, t) =s

j=0

ajtj∞

m=0

d + m − 1d − 1

tm =∞i=0

min(s,i)j=0

ajd + i − j − 1

d − 1 ti.

We deduce that for all i = s, s + 1, . . .,

(†) dim N i = a0

d + i − 1

d − 1

+ a1

d + i − 2

d − 1

+ . . . + as

d + i − s − 1

d − 1

.

The right hand side of this expression can be rearranged as ϕ(i) for some polynomial

ϕ(t) ∈ Q[t]. Note also that

ϕ(t) =f (1)

(d − 1)!td−1 + lower degree terms.

Since f (1)= 0, we see that ϕ(t) has degree d

−1.

(3) Since ab = a−1b−1+ a−1b

, we see that

ij=0

d + j − 1

d − 1

=

d + i

d

.

Replacing i by j in (†) and summing from j = 0 to i, we obtain

ij=0

dim N j = a0

d + i

d

+ a1

d + i − 1

d

+ . . . + as

d + i − s

d

= χ(i)



55

for some polynomial χ(t) ∈ Q[t] of degree d. This is valid whenever i ≥ s.

9.2. Definition. Let S = k[X 1, . . . , X n] and let N = ⊕∞

i=0N i be a finitely gener-ated graded S −module. The objects appearing in Theorem 9.1 have names:

• The Hilbert polynomial of N is ϕ(t) ∈ Q[t].

• The Samuel polynomial of N is χ(t) ∈ Q[t].

• The dimension of N is the integer d = d(N ).

• The multiplicity of N is the integer m(N ) = d!a where a is the leading

coefficient of χ(t).

Our next goal will be to define dimension and multiplicity for finitely generated

modules over almost commutative algebras. For this, we first need to make a

digression to study Rees rings and good filtrations.

9.3. Rees rings and modules. Let R be a filtered ring with filtration (Ri)i∈Z,and let M be a filtered left R−module with filtration (M i)i∈Z.

Definition. The Rees ring R of R is a subring of the ring of Laurent polynomials

R[t, t−1]: R =i∈Z

Riti ⊆i∈Z

Rti = R[t, t−1].

The Rees module M of M is the abelian group

M =i∈Z

M iti

where the action of R is given by on homogeneous components by Riti × M jtj → M i+jti+j

riti , mjtj → mirjti+j .

Note that t ∈ R is a central regular element, since 1 ∈ R1 always. There is a

certain amount of interplay between the Rees ring of R and the associated graded

ring gr R.

Lemma. Let R and M be as above. Then

(1) R/t R ∼= gr R as rings,

(2)

M/t

M ∼= gr M as left gr R−modules,

(3) R/(t − 1) R ∼= R as rings,(4) M/(t − 1)M ∼= M as left R−modules.

Proof. We will only prove the result for the rings, leaving the modules as an exercise.

(1). We have an isomorphism of abelian groups

R/t R =

i∈Z Riti

i∈Z Ri−1ti∼=i∈Z

Ri/Ri−1∼= gr R.

It can be checked that this is also a ring isomorphism.



56

(2). Define a ring homomorphism π : R → R by π(

riti) =

ri. Since

π(Riti) = Ri we see that π is onto. Clearly t

−1

∈ker(π). Check that in fact

ker(π) = (t − 1) R. The result follows.

So R is a ring which has both R and gr R as epimorphic images. Therefore,

sometimes gr R is thought of as a ”deformation” of R. Also note that if R is right

(or left) Noetherian, then so are both R and gr R.

Thanks are due to Lorenzo Di-Biagio for pointing out that an earlier version of

the following Lemma was incorrect.

9.4. Lemma. Suppose R is an almost commutative algebra with finite generating

set {x1, . . . , xn}. Equip R with the positive filtration as in Example 7.1(3). Then

R = R0 ⊕ R1t ⊕ R2t2 ⊕ · · ·

is right and left Noetherian.

Proof. We observe first that k[t] is a subring of R and that R is generated as

a k[t]−algebra by {tx1, tx2, . . . , t xn}. Note that we can widen the definition of

”almost commutative k−algebra” in 7.4 to the case when k is any commutative

ring, and that Proposition 7.4 will still be valid.

Since we can write [txi, txj] as a k[t]−linear combination of {tx1, . . . , t xn}, we

see that R is an almost commutative k[t]−algebra. Hence gr R is a quotient of

k[t][X 1, . . . , X n] by Proposition 7.4, and therefore is Noetherian by Theorem 1.14.

Hence

R is right and left Noetherian, by Corollary 7.8.

This result will be useful to us in what follows.

9.5. Good filtrations.

Definition. Let R be a filtered ring and let M be a left R−module.

• A filtration (M i) on M is said to be good iff the Rees module M is finitely

generated over R.

• Two filtrations (M i) and (M i) on M are algebraically equivalent (or just

equivalent) if there exist c, d ∈ Z such that

M i ⊆ M i+c and M j ⊆ M j+d for all i, j ∈ Z.

Note that if (M i) is a good filtration, then gr M ∼= M /tM is finitely generated

over gr R and M ∼= M/(t − 1)M is finitely generated over R, by Proposition 9.3.

Proposition. Let R be a filtered ring and let M be a left R−module.

(1) A filtration (M i) on M is good if and only if there exist k1, k2, . . . , ks ∈ Zand m1 ∈ M k1 , m2 ∈ M k2 , . . . , ms ∈ M ks such that

M i = Ri−k1m1 + Ri−k2m2 + · · · + Ri−ksms for all i ∈ Z.

(2) All good filtrations on M are equivalent.



57

Proof. (1) If the graded module M is finitely generated, it has a finite homogeneous

generating set

{tk1m1, . . . , tksms

}say, with mj

∈M kj . Then the i

−th homogeneous

component of M is

tiM i = Ri−k1ti−k1(tk1m1) + · · · + Ri−ksti−ks(tksms), so

M i = Ri−k1m1 + Ri−k2m2 + · · · + Ri−ksms.

Conversely, any filtration of this form is good, since {tk1m1, . . . , tksms} is then

a generating set for M .

(2) Take two good filtrations (M i) and (M i). Then we have

M i = Ri−k1m1 + · · · + Ri−ksms for all i

M j = Rj−l1m1 +

· · ·+ Rj−lrm

r for all j.

We can find c ∈ Z such that mt ∈ M lt+c for all t = 1, . . . , r. Then M i ⊆ M i+c for

all i, and similarly there exists d ∈ Z such that M i ⊆ M i+d for all i.

One consequence of this result is that any finitely generated module has some

good filtration: just take a generating set X = {x1, . . . , xs} and set the integers ki

to be 0, so that M i = Rix1 + . . . + Rixs = RiX (the standard filtration on M ) is

good.

9.6. Dimension theory for almost commutative algebras. Let R be an al-

most commutative algebra with generating set {x1, . . . , xn}. Let (Ri) be the usual

filtration on R, so that gr R is a quotient of the polynomial algebra k[X 1, . . . , X n]

by Proposition 7.4. Let M be a finitely generated left R−module and let (M i) besome good filtration on M .

Note that each M i is finite dimensional over k by Proposition 9.5(1), since each

Ri is. Also, gr M is a finitely generated graded gr R−module by Lemma 9.3 and

hence a finitely generated graded k[X 1, . . . , X n]−module. As such, it has a Samuel

polynomial χ(t) defined in (9.2). Note that

χ(i) =i

j=0

dim(M j/M j−1) = dim M i − dim M −1

for sufficiently large i.

Definition. The dimension d(M ) of M is the degree d of χ(t). The multiplicitym(M ) is d!a where a is the leading coefficient of χ(t). Thus d(M ) = d(gr M ) and

m(M ) = m(gr M ).

At first sight, this depends on the choice of good filtration (M i) on M . However,

let (M i) be another good filtration on M and let χ(t) be the corresponding Samuel

polynomial. By Proposition 9.5(2), there exists c ∈ N such that

M j−c ⊆ M j ⊆ M j+c for all j ∈ Z,



58

so dim M j−c ≤ dim M j ≤ M j+c. Hence, for large enough j,

χ

( j − c) ≤ χ( j) ≤ χ

( j + c).Since the behaviour of a polynomial for large enough j is determined by its leading

term, we see that dimension and multiplicity is well defined.

9.7. Examples.

(1) Let R = k[X 1, . . . , X n] and let (Ri) be the usual positive filtration on R,

given by Ri = k.{X α : |α| ≤ i}. Then dim Ri is the number of monomials

in n variables of length at most i which is well known to be

dim Ri =

n + i

n

=

(i + n)(i + n − 1) · · · (i + 1)

n!=

in

n!+ o(in−1).

So if we view R as an R−module by multiplication, we can read off d(R) = nand m(N ) = 1.

(2) Let R = An with the usual filtration. Viewing R as a filtered module

over itself with the same filtration, we see that gr R = k[X 1, . . . , X 2n] by

Corollary 8.3, so d(R) = 2n and m(R) = 1.

(3) Let R = An and let M = k[X 1, . . . , X n] be the natural An−module with

the usual filtration. Then d(M ) = n and m(M ) = 1.

(4) Let R be any almost commutative k−algebra and M a finitely generated

R−module. Then d(M ) = 0 if and only if dim M < ∞.

9.8. Theorem. Let R be an almost commutative k−algebra and let M be a finitely

generated R−module. Suppose N is a submodule of M . Then(1) d(M ) = max{d(N ), d(M/N )}.

(2) If d(N ) = d(M/N ) then m(M ) = m(N ) + m(M/N ).

Proof. Let (M i) be a good filtration on M . Give N the subspace filtration (N i)

and M/N the quotient filtration (M/N )i. First, we prove that these filtrations are

good.

Note that N i ≤ M i and M i/N i = M i/(N ∩ M i) ∼= (M i + N )/N . A computation

similar to the proof of Proposition 7.7 shows that

N

≤ M and

M

N ∼=M /

N

as left R−modules.

Now R is left Noetherian by Lemma 9.4, so both N and M/N are finitely gener-

ated R−modules, as (M i) is good. Hence the filtrations on N and M/N are good

and we can use them to compute dimensions and multiplicities.

Next, we have

dimN j

N j−1+ dim

(M/N )j(M/N )j−1

= dimM j

M j−1



59

so summing over j, we see that for large enough i,

(*) χN (i) + χM/N (i) = χM (i).

Hence χN + χM/N = χM as polynomials, since (∗) holds for infinitely many i.

Since the leading coefficients of the Samuel polynomials are positive, we cannot

have cancellation, so (1) follows. Moreover, if d(N ) = d(M/N ) then the leading

coefficients add, and we obtain (2).

9.9. Proposition. Let R be an almost commutative algebra with d(R) = n.

(1) If M is a finitely generated left R−module, then d(M ) ≤ n.

(2) If R is a domain and 0 = I r R then d(R/I ) < n.

Proof. From Theorem 9.8(1), we see that d(Rm

) = d(R) = n for any m ≥ 1.Since M is finitely generated, it’s a left R−module quotient of Rm for some m, so

d(M ) ≤ d(Rm) = n, proving (1).

Next, pick 0 = x ∈ I . Since R/I R/Rx, we see that d(R/I ) ≤ d(R/Rx) by

Theorem 9.8(1). Suppose for a contradiction that d(R/Rx) = d(R) = n. Since

R is a domain, Rx ∼= R as left R−modules, so d(Rx) = d(R/Rx). But now by

Theorem 9.8(2), m(Rx) + m(R/Rx) = m(R) so m(R/Rx) = 0, a contradiction.

Hence d(R/Rx) < n as required.

9.10. Bernstein’s Inequality.

Theorem (Bernstein, 1972). Suppose char(k) = 0 and let M be a nonzero finitely generated left An(k)−module. Then d(M ) ≥ n.

Proof. (Joseph)

Pick a finite generating set X and let (M i) be the standard filtration on M

given by M i = (An)iX . This filtration is good and positive. Let χ(t) be the

Samuel polynomial of gr M with respect to this filtration. Then for large enough i,

χ(i) = dim M i by the remarks in (9.6).

Claim: dim Ri ≤ dim Homk(M i, M 2i) = (dim M i)(dim M 2i).

Assuming the claim is true, for large enough i, dim Ri ≤ χ(i)χ(2i). Since

dim Ri =

i+2n2n

is a polynomial in i of degree 2n, χ(t)χ(2t) must be a polyno-

mial of degree ≥ 2n. But it has degree 2d(M ), so d(M ) ≥ n as required.To prove the claim, consider the k−linear map

ϕi : Ri → Homk(M i, M 2i)

r → (m → rm)

It’s sufficient to show that ϕi is injective. Equivalently, we must show that if

0 = r ∈ Ri then rM i = 0. Proceed by induction on i. When i = 0, r ∈ k\0 is a

unit so rM 0 = M 0 = kX = 0 since M = 0.



60

Now assume i > 0 and suppose for a contradiction that 0 = r ∈ Ri is such

that rM i = 0. Then r /

∈k = Z (An) by Proposition 8.6(1), so [r, u]

= 0 for some

u ∈ {x1, . . . , xn, y1, . . . , yn}. We’ve seen in the proof of Proposition 8.6 that

[r, u] ∈ Ri−1,

so [r, u]M i−1 = 0 by induction. But uM i−1 ⊆ M i and rM i = 0 so

[r, u]M i−1 = (ru − ur)M i−1 = 0,

a contradiction. The result follows.

Note that when n = 1, this says that any nonzero finitely generated An−module

is infinite dimensional over k, by Example 9.7(4). This is also the content of Propo-

sition 8.10(3).



Noetherian AlgebrasMichaelmas 2004

Example Sheet 1

Throughout this sheet, k denotes an arbitrary field.

1. Let R1, . . . , Rn be rings. The direct product R = R1 × · · · × Rn is a ring with componentwise

multiplication and addition. The identity element of R is given by (1, 1, . . . ,1). We can view each

Ri as an additive subgroup of R which is closed under multiplication. However, Ri is not a subring

of R (unless n = 1) since the identity elements are different.

Show that R is left Noetherian if and only if each Ri is left Noetherian. Show that the same result

holds with Noetherian replaced by Artinian.

2. Show that the ring

Z Q

0 Q

is right, but not left Noetherian. Find an example of a ring which is

left Noetherian but not right Noetherian.

3. Let R be a ring and let M be a Noetherian left R-module. Suppose ϕ : M → M is an R-modulehomomorphism. Show that if ϕ is surjective, then it is also injective. Deduce that if R is a left

Noetherian ring with ab = 1 for some a, b ∈ R then ba = 1.

4. By considering trailing coefficient ideals show that a commutative ring R is Noetherian if and only

if the power series ring R[[x]] is Noetherian.

5. Let R be a ring and suppose a ∈ R is nilpotent. Show that 1 + a is a unit. If R is commutative,

show that the sum of a unit and a nilpotent element is a unit. Is this true if R is noncommutative?

6. Let R be a commutative ring and let a = a0 + a1x + . . . + anxn ∈ R[x].

(a) Show that a is nilpotent if and only if each ai is nilpotent.

(b) Show that a is a unit if and only if a0 is a unit in R and ai is nilpotent for each i ≥ 1.

(c) Deduce that J (R[x]) = N (R[x]).

7. Let R be a commutative Noetherian ring and let f ∈ R[[x]]. Prove that f is nilpotent if and only if

all its coefficients are nilpotent.

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8. Let R = k[x1, x2, . . .] be the polynomial ring with countably many indeterminates over and let I

be the ideal generated by all the elements xii, i ≥ 1. Show that R/I is not Noetherian and that its

prime radical is not nilpotent.

9. Let R be a ring.

(a) Show that R has a prime ideal.

(b) Show that R has a minimal prime, with respect to inclusion.

10. Let R be a ring with a prime ideal P . Let P [x] = {a0 + a1x + . . . + anxn ∈ R[x] : ai ∈ P for all i}.

Show that P [x] is a prime ideal in R[x]. If P is maximal, is P [x] maximal?

11. Show that the free algebra kx, y is not left Noetherian.

12. Let G be a group. If H 1 H 2 . . . is a strictly ascending chain of subgroups of G, show that the

left ideals I j of the group algebra kG generated by {h− 1 : h ∈ H j} form a strictly ascending chain

I 1 I 2 . . . .

13. Show that any subgroup or quotient of a poly-{cyclic or finite} group is poly-{cyclic or finite}.

14. (a) Let g be a finite dimensional Lie algebra over k. An ideal of g is a k-subspace h such that

[g, h] ≤ h. Suppose g has an ideal h of codimension 1 and that U (h) is left Noetherian. Use the

noncommutative version of Hilbert’s Basis Theorem to show that U (g) is also left Noetherian.

(b) The Lie algebra g is solvable if it has a chain of subspaces

0 = g0 ≤ g1 ≤ . . . ≤ gn = g

such that [gi, gi] ≤ gi−1 for each i = 1, . . . , n. Suppose g is a finite dimensional solvable Lie

algebra. Show that U (g) is left and right Noetherian.

15. Let A be a commutative ring. Suppose An injects into Am as A-modules. Must n ≤ m ?

Comments to [email protected].

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Example Sheet 2

1. Let D be a division ring and let R = M n(D). Express R as a direct sum of minimal nonzero right

ideals and also as a direct sum of minimal nonzero left ideals. Show that R is a simple ring, that

is, the only (two-sided) ideals are 0 and R.

2. Let R be a right Artinian ring and let x ∈ R be a right regular element, that is,

rann(x) := {r ∈ R : xr = 0}

is zero. Show that x must be a unit. Deduce that a right Artinian ring with no zero divisors must

be a division ring.

3. Find an example of a right Artinian ring which is not left Artinian.

4. Let R be a finite dimensional k−algebra, where k is an algebraically closed field and let M be a

simple R−module. Show that EndR(M ) ∼= k.

5. (Maschke) Let G be a finite group and let k be a field, so kG is left and right Artinian. Show that

kG is semisimple if and only if char k |G|.

6. Let G be a finite group and let k be a field with char k |G|. By considering the centre of kG,

show that the number of isomorphism classes of simple right kG−modules is at most the number

of conjugacy classes of elements of G.

7. Let G be a finite p-group and let k be a field of characteristic p. Let J be the Jacobson radical of

kG. Show that J = ker(ϕ) where ϕ : kG → k is the trivial representation.

8. Let R be a ring. An element e ∈ R is an idempotent if e2 = e. Suppose e ∈ R is a central idempotent.

(a) Show that 1 − e is also a central idempotent.

(b) Show that eR is a ring with the same multiplication as in R, but with identity element e.

(c) Show that R ∼= eR × (1 − e)R.

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9. A commutative ring is local if it has a unique maximal ideal. Show that a commutative ring is

Artinian if and only if it is the direct product of finitely many Artinian local rings.

10. Let R, S be commutative rings and let ϕ : R → S be a ring homomorphism. Let P S be a prime

ideal. Show that ϕ−1(P ) R is also a prime ideal. Is this true if R and S are not assumed to be

commutative?

11. Let k be a field and let I = (xy,y2) k[x, y]. Find two distinct primary decompositions of I .

12. Let R be a commutative ring and let M be a Noetherian R−module. Show that R/ Ann(M ) is also

Noetherian.

13. Let R be a commutative Noetherian ring and let M be a finitely generated R−module. Show that

Ann(M ) =

P ∈Ass(M )

P.

14. Let D be the set of zero divisors in a commutative Noetherian ring R. Show that

D =

P ∈Ass(R)

P.

A ring R is called von Neumann regular if for any a ∈ R there exists x ∈ R such that axa = a.

15. Let R be a ring. Prove that:

(a) If M is a semisimple R−module, then EndR(M ) is von Neumann regular.(b) R is von Neumann regular if and only if every principal right ideal generated by an idempotent.

(c) A right Noetherian ring is von Neumann regular if and only if it is semisimple Artinian.

(d) A commutative ring is von Neumann regular if and only if it is semiprime and every prime

ideal is maximal.

16. Let R be a von Neumann regular ring. Show that every finitely generated right ideal in R is

generated by an idempotent.


2




Example Sheet 3

1. Suppose R is a commutative ring, S is a multiplicatively closed subset of R and M is a finitely

generated R−module. Show that

(a) M S = 0 if and only if there exists s ∈ S with M s = 0.

(b) Ann(M )S = Ann(M S ).

2. Let R be a commutative ring and let S be a multiplicatively closed subset of R. Show that the

prime ideals of RS are in 1-1 correspondence with the prime ideals of R which are contained in R\S .

3. Let I be an ideal in a commutative ring R. Show that S = 1 + I is a multiplicatively closed subset

of R and that I S is contained in the Jacobson radical of RS .

4. Let M be an module for a commutative ring R. Show that the following are equivalent:

(a) M = 0,

(b) M P = 0 for all prime ideals P ,

(c) M Q = 0 for all maximal ideals Q.

5. Let ϕ : M → N be an R−module map for a commutative ring R. Show that the following are

equivalent:

(a) ϕ is injective,

(b) ϕP : M P → N P is injective for every prime ideal P ,

(c) ϕQ : M Q → N Q is injective for every maximal ideal Q.

Is the same true if ”injective” is replaced by ”surjective”?

6. Let R be a commutative ring. The support of an R−module M is defined to be the set Supp(M )of prime ideals P of R such that M P = 0. Show that

(a) If 0 → A → B → C → 0 is an exact sequence of R−modules, then

Supp(B) = Supp(A) ∪ Supp(C ).

(b) If M is finitely generated, then Supp(M ) consists of all the prime ideals containing Ann(M ).

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7. Let R be a commutative ring. Suppose that for every prime ideal P , the local ring RP is semiprime.

Show that R is semiprime. If RP is an integral domain for every prime P , must R be an integral

domain, too?

8. Let R be a ring and let M be a right R−module. Let X and Y be submodules of M and let S bea right divisor subset of R. Show that

(a) (X + Y )S = X S + Y S , and

(b) (X ∩ Y )S = X S ∩ Y S . viewed as submodules of M S .

9. Find a ring R with elements a, b ∈ R such that ab = 1 but ba = 1. Show that S = {1, a , a2, . . .}

satisfies the right Ore condition and that ass(S ) = 0, but that no element of S \{1} is right regular.

10. Let R =

Z Z

0 Z

and let P =

Z Z

0 0

. Show that P is a prime ideal in R. Also, show that

S = R\P is multiplicatively closed but is not a right Ore set. Prove that S is a left divisor set andthat S R ∼= Q.

11. Let R be a ring and let M be a right R−module. Show that

(a) If X e Y and Y e M then X e M .

(b) If X ≤ M then there exists Y ≤ M such that X ∩ Y = 0 and X ⊕ Y e M . (Hint: Zorn.)

(c) M is semisimple if and only if any proper submodule of M is not essential.

(Hint for (⇐): do this first when M is cyclic.)

12. Let S be a right Ore subset of a ring R consisting of regular elements. Let A r R and let B r RS .

Show that A e R if and only if AS e RS , and that B e RS if and only if B ∩R e R. Use this togive another proof of the second part of Goldie’s Theorem which does not use the notion of closed

right ideals. (Show that if R satisfies conditions (1),(2),(3) of Theorem 6.6 and if S is the set of all

regular elements of R, then RS is semisimple Artinian.)

13. Let R be a prime ring (so 0 is a prime ideal of R). Show that any nonzero two-sided ideal I of R is

essential as a right ideal (and as a left ideal).

14. Let R be a prime right Noetherian ring. Show that R has a classical right ring of quotients isomorphic

to M n(D) for some division ring D and some integer n ≥ 1.

15. Let R be a ring and S a right Ore subset of R consisting of regular elements. Suppose that RS isright Noetherian. Show that if I is a two-sided ideal of R then I S is a two-sided ideal of RS .

16. Let R be a simple right Noetherian ring. Show that the classical right ring of quotients Q exists.

Show also that the centre of R is equal to the centre of Q and is a commutative field.


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Example Sheet 4

A topological group is a topological space G which is also a group, such that the maps

m : G × G → G and

x , y → xy

c : G → G

x → x−1

are continuous, when we equip G×G with the product topology.

1. Let R be a negatively filtered ring and let M be a negatively filtered R−module.

(a) Show that M is a topological group under addition with respect to the filtration topology.

(b) Show that M is Hausdorff if and only if the filtration on M is separated.

2. Show that the ring of p−adic integers Z p is compact.

3. Let R be a commutative Noetherian ring and let I be an ideal of R.

(a) Show that the Rees ring of R with respect to the I −adic filtration is also Noetherian.

(b) (Artin-Rees Property) Let A be another ideal of R. Show that there exists n ∈ N such that

A ∩ I n ⊆ AI.

(c) Suppose R is local with maximal ideal I . Using Nakayama’s Lemma, or otherwise, show that

∞n=0

I n = 0.

4. Suppose R is a commutative Noetherian integral domain and I is a proper ideal of R. Show that

∞n=0

I n = 0.

(This is Krull’s Intersection Theorem .)

5. Let R be a commutative ring and let D1, D2 ∈ Der(R). Show that [D1, D2] ∈ Der(R).

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