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The Binomial Theorem without middle terms:Putting prime numbers to work in algebra

Tom Marley

University of Nebraska-Lincoln

April 8, 2016

Tom Marley University of Nebraska-Lincoln

The Binomial Theorem without middle terms: Putting prime numbers to work in algebra

Rings!

RING

THEORY

Modular Arithmetic

Clock arithmetic:

8 (o’clock) + 5 (hours) = 1 (o’clock)

8 (hours) + 5 (o’clock) = 1 (o’clock)

So in clock arithmetic, we can simply write:

8 + 5 = 1.

We can also subtract: 4− 6 = 10.

As well as multiply: 4× 7 = 28 = 4.

General rule:

Divide by 12 and take remainder.

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

We can also subtract: 4− 6 =

10.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

As well as multiply: 4× 7

= 28 = 4.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

As well as multiply: 4× 7 = 28

= 4.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

Modular Arithmetic

Clock arithmetic:

8 + 5 = 1.

General rule:

The integers modulo n

We denote this number system by Z12, “the integers modulo 12”.

But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

We can do arithmetic in Z16 in the same way:

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Similarly for Zn for any integer n ≥ 1.

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.

For example, a day on Neptune lasts about 16 hours.

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

We denote this number system by Z12, “the integers modulo 12”.But there is nothing special about a clock with 12 hours.For example, a day on Neptune lasts about 16 hours.

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

10 + 10 = 20

= 4

7− 13 = −6 = 10

5× 7 = 35 = 3

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

10 + 10 = 20 = 4

7− 13 = −6

= 10

5× 7 = 35 = 3

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35

= 3

10 + 10 = 20 = 4

7− 13 = −6 = 10

5× 7 = 35 = 3

Ring axioms

In number systems like Zn, many of the familiar axioms fromarithmetic hold: For example:

a + b = b + a (commutativity of addition)

ab = ba (commutative of multiplication)

a(b + c) = ab + ac (distributive property)

Number systems that satisfy these axioms (and a couple more) arecalled Rings.

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Ring axioms

One can also subtract in rings, but not necessarily divide.

For example, in Z12 we have: 6× 3 = 6× 9.However: 3 6= 9 in Z12.

Ring axioms

One can also subtract in rings, but not necessarily divide.For example, in Z12 we have: 6× 3 = 6× 9.

However: 3 6= 9 in Z12.

Ring axioms

Polynomial and power series rings

There are many examples of rings out there in addition to Zn:

Z (the integers)

R (the real numbers)

C (the complex numbers)

It’s easy to create new rings from old. One way is to consider allpolynomials in a variable x with coefficients from a given ring R:

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

We can also consider all power series in x with coefficients from R:

R[[x ]] := {∞∑i=0

aixi | ai ∈ R ∀ i}.

Z (the integers)

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

R[[x ]] := {∞∑i=0

Z (the integers)

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

R[[x ]] := {∞∑i=0

Z (the integers)

R[x ] := {a0 + a1x + · · ·+ anxn | ai ∈ R ∀ i , n ≥ 0}.

R[[x ]] := {∞∑i=0

Polynomials and power series (cont.)

We can iterate these processes:

R[x , y ] := (R[x ])[y ]

R[x , y , z ] := (R[x , y ])[z ]

R[[x , y ]] := (R[[x ]])[[y ]]

For example, in Z[[x , y ]], we have:

(1− xy) · (1 + xy + x2y2 + · · ·+ x iy i + · · · ) = 1.

Polynomials and power series (cont.)

We can iterate these processes:

R[x , y ] := (R[x ])[y ]

R[x , y , z ] := (R[x , y ])[z ]

R[[x , y ]] := (R[[x ]])[[y ]]

For example, in Z[[x , y ]], we have:

(1− xy) · (1 + xy + x2y2 + · · ·+ x iy i + · · · ) = 1.

Quotient spaces

Quotient rings

Given a ring R and elements f , g ∈ R, we can form a quotient ringfrom R by identifying f and g ; or equivalently, by identifyingh = f − g and 0. We write this quotient ring as R/(h).

More generally, given f1, . . . , fn ∈ R, the quotient ring obtained byidentifying f1 = 0, . . . , fn = 0 is written:

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼= C.

To invert an element a ∈ R, just consider the ring:

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

For example: Z/(n) ∼=

Zn and R[x ]/(x2 + 1) ∼= C.

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn

and R[x ]/(x2 + 1) ∼= C.

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

For example: Z/(n) ∼= Zn and R[x ]/(x2 + 1) ∼=

C.

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

R[x ]/(ax − 1).

Quotient rings

R/(f1, . . . , fn).

R[x ]/(ax − 1).

What are rings good for?

They’re precious!

Applications of ring theory:

Cryptography

Error-correcting codes

3D animation

Communication networks

Number theory, algebraicgeometry, invariant theory

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

They’re precious!

Cryptography

3D animation

The Binomial Theorem

Let p be a (positive) prime integer. For the remainder of this talk,we’ll restrict our attention to rings of the form

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

Consider the Binomial Theorem in such rings:

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

R = Zp[[x1, . . . , xn]]/(f1, . . . , fr ).

(a + b)p =

p∑i=0

(p

i

)ap−ibi

= ap + pap−1b +p(p − 1)

2ap−2b2 +

p(p − 1)((p − 2)

6ap−3b3 + · · ·

= ap + bp

Fermat’s (Little) Theorem

Fermat’s Theorem

Let p be a prime. Then ap = a for any a ∈ Zp.

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp. Then

(a + 1)p = ap + 1p

= a + 1.

Fermat’s Theorem

Proof

The theorem clearly holds when a = 0 and a = 1. Suppose ap = afor some a ∈ Zp.

Then

(a + 1)p = ap + 1p

= a + 1.

Fermat’s Theorem

Proof

(a + 1)p = ap + 1p

= a + 1.

Fermat’s Theorem

Proof

(a + 1)p = ap + 1p

= a + 1.

A special subring

Key Observation

The setRp := {ap | a ∈ R}

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

The relationship between the ring R and the subring Rp providesimportant clues about the structure of R.

A special subring

Key Observation

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp.

Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp. Finally, −(ap) = (−a)p ∈ Rp.

A special subring

Key Observation

is a subring of R.

Proof

First observe 0 = 0p and 1 = 1p, so 0, 1 ∈ Rp. Now letap, bp ∈ Rp. Then ap + bp = (a + b)p ∈ Rp andapbp = (ab)p ∈ Rp.

Finally, −(ap) = (−a)p ∈ Rp.

A special subring

Key Observation

is a subring of R.

Proof

A special subring

Key Observation

is a subring of R.

Proof

Curve singularities

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]

∼= R[[x ]].

For the cusp: R[[x , y ]]/(y2 − x3) ∼= R[[t2, t3]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the node:

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the node:

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the node:

R[[x , y ]]/(y2 − x3 − x2) ∼= R[[x , y ]]/(y2 − x2(x + 1))

∼= R[[x , y ]]/((y − cx)(y + cx)).

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Coordinate rings

For the parabola:

R[[x , y ]]/(y − x2) ∼= R[[x , x2]]∼= R[[x ]].

For the node:

where c =√x + 1 = 1 + 1

2x −18x

2 + · · · .

Smoothness

Definition

Let f (x , y) = 0 define a curve C . We say C is smooth ornonsingular at the origin if its coordinate ring k[[x , y ]]/(f ) isisomorphic to a polynomial ring in one variable. Otherwise, C issingular at the origin.

Remark

If k = R then C is smooth at (0, 0) if and only if ∂f∂x and ∂f

∂y don’tboth vanish at the origin. That is, C is smooth at (0, 0) if and onlyif there is a unique well-defined tangent line to C at the origin.

Smoothness

Definition

Let f (x , y) = 0 define a curve C . We say C is smooth ornonsingular at the origin if its coordinate ring k[[x , y ]]/(f ) isisomorphic to a polynomial ring in one variable. Otherwise, C issingular at the origin.

Remark

If k = R then C is smooth at (0, 0) if and only if ∂f∂x and ∂f

∂y don’tboth vanish at the origin. That is, C is smooth at (0, 0) if and onlyif there is a unique well-defined tangent line to C at the origin.

An illustrative example

Question

Given a ring R = Zp[[x1, . . . , xd ]]/(f1, . . . , fr ), how can we decide ifR is smooth at the origin? That is, how can we tell ifR ∼= Zp[[y1, . . . , ys ]]?

Consider R = Zp[[x ]]. Then Rp = Zp[[xp]].

Note that every element in f ∈ R can be written uniquely in theform:

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Question

Consider R = Zp[[x ]].

Then Rp = Zp[[xp]].

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Question

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Question

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.

For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7. Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Question

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

for some c0, . . . , cp−1 ∈ Rp.For example, let p = 3 and f = 2 + x + 2x3 + x4 + 2x5 + x7.

Then

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

Question

f = c0 + c1x + c2x2 + · · ·+ cp−1x

p−1

f = (2 + 2x3) · 1 + (1 + x3 + x6)x + (2x3)x2.

An illustrative example, cont.

Thus R = Zp[[x ]] has a basis over Rp = Zp[[xp]]. Namely,

{1, x , x2, . . . , xp−1}.

The same holds for R = Zp[[x1, . . . , xd ]].

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].Then

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2? No!

Suppose {1, t2} is part of the basis. Then

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

Now consider the coordinate ring of the cusp over Z2:R = Z2[[x , y ]]/(y2 − x3) ∼= Z2[[t2, t3]].

ThenR2 ∼= Z2[[t4, t6]].

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

Does R have a basis over R2?

No!

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

Suppose {1, t2} is part of the basis.

Then

t6 · 1 + t4 · t2 = 0.

{1, x , x2, . . . , xp−1}.

R2 ∼= Z2[[t4, t6]].

t6 · 1 + t4 · t2 = 0.

Kunz’s Theorem

Theorem (Ernst Kunz, 1969)

R has a basis over Rp if and only if R ∼= Zp[[x1, . . . , xr ]].

That is, a curve, surface, solid, etc. over Zp is smooth at theorigin if and only if its coordinate ring R has a basis over Rp.

Kunz’s Theorem is in some sense the very beginning of the story ofthe study of singularities over Zp. This continues to be a veryactive area of research today.

Theorem (Avramov-Hochster-Iyengar-Yao, 2012)

If there exists any nonzero R-module which has a finite basis overRp then R ∼= Zp[[x1, . . . , xr ]].