Principal Ideal Domains and Unique Factorization Domains

March 31, 2009

A Principal Ideal Domain (PID) is an integral domain A in which every ideal is principal. Ifa ∈ A is a generator of a nonzero ideal I in A the other generators of I are A∗ · a (the elementsof A that are associate to a, i.e., multiples of a by a unit in A. The nonzero prime ideals of A aregiven by irreducible elements of A, i.e., elements a ∈ A such that for every factorization a = b · ceither b or c is a unit. Often one says that a prime element of a ring is an element that generatesa prime ideal, so one might capture the above discussion by saying that in a PID prime elementsand irreducible elements are the same. If k is a field the ring of polynomials over k in one variable,A = k[t], is a PID.

Lemma 1 In a PID if p is a prime element, and p | a · b then either p | a or p | b.

A Unique Factorization Domain (UFD) is an integral domain A in which every nonzero elementa admits a factorization

a = up1 · p2 · . . . · ps

where u ∈ A∗, s ≥ 0, and the pi are irreducible elements; moreover, this factorization is “unique”in the evident sense of uniqueness: you can change the order of the irreducible elements, you canchange each of the irreducible elements by replacing them by associate irreducible elements andyou then can (i.e., must) change the unit u to suit.

Theorem 1 A PID is a UFD

Let A be a PID. Step one: Every nonzero element a ∈ A Has a decomposition into a productof irreducible elements as above. Consider S the set of principal ideals in A that have generatorsnot admitting prime factorizations. Find a maximal ordered set of ideals in S and consider theirunion—since A is a PID it is generated by a single element a which must also not have a primefactorization. But considering any factorization a = b · c you argue to reach a contradiction.

Step two: Uniqueness. Assume two factorizations of teh same nonzero element and lop off primefactors by Lemma ??.

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Definition 1 If A is a UFD, K its field of fractions, and p an irreducible element of A (now takenup to multiplication by a unit) define the function ordp by setting ordp(0) = ∞ and—for a ∈ K∗

ordp(a) =: the exponent of p in the unique factorization of a.

That is,ordp : K → Z ∪ {∞}

is defined by (setting ordp(0) = ∞ and) so that that

a = u ·∏p

pordp(a)

for u an appropriate unit, for all a ∈ K∗.

Definition 2 If A is a UFD, K its field of fractions, extend the above definition of ordp to poly-nomials f(X) = anXn + . . . + a0 by setting ordp(f) = the minimum of the ordp of the coefficientsai. If f is not identically zero, define the content of f(X) (an element of K∗moduloA∗) to be

cont(f) :=∏p

pordp(f).

To say that cont(f) = 1 is to say that f ∈ A[X] and the gcd pof its coefficients is 1. We may writefor any f not identically zero,

f = cont(f) · f1

where f1 has content one. Also cont(a · f) = a · cont(f) for any a ∈ K∗ and f ∈ K[X].

Lemma 2 (Gauss):cont(f · g) = cont(f) · cont(g)

Proof: It suffices to show that cont(f1 · g1) = 1 or ordp((f1 · g1) = 0 for all p. This we do.

Lemma 3 Let A be a UFD. The irreducible elements of A[X] are given (up to multiplication byunits in A) by irreducible polynomials of K[X] of content 1, and irreducible elements of A (viewedas polynomials of degree zero in A[X]).

Proposition 1 Let A be a UFD. Then A[X] is a UFD.

Corollary 2 Let A be a UFD. Then A[X1, X2, . . . , Xn] is a UFD.

Lemma 4 (Eisenstein): Let A be a UFD and p an irreducible element of A. If

f(X) = anXn + · · ·+ aiXi + · · ·+ a0 ∈ A[X] ⊂ K[X]

has the property that

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• ordp(an) = 0,

• ordp(ai) > 0 for i < n,

• ordp(a0) = 1,

then f(X) is irreducible in K[X].

Corollary 3Gal(Q(µp)/Q) ' (Z/pZ)∗.

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