Gauss’s Lemma

February 3, 2010

(Read 11.3 of [A]).

1 Gauss’s Lemma for Q:

Define the content of a polynomial in Q[t]. Define primitive polynomial (∈ Z[t]). Any (nonzero) polynomial in Q[t] may be written uniquely (up to multiplication by ±1) as a product of its content (∈ Q∗) times a primitive polynomial (∈ Z[t]). The product of two primitive polynomials (f times g) is again primitive. (Proof: If not, then there is a p such that f · g mod p is zero even though f mod p and g mod p are nonzero. But Z/pZ[t] is an .) A consequence of this is that the mapping Q[t] − {0} −→ Q∗ is a “homomorphism,” (i.e., multiplicative). There is a one:one correspondence between the (equiv- alence classes of) irreducible elements of the Q[t] and the (equivalence classes of) irreducible elements of the ring Z[t]. To go from the former to the latter, divide by the content. Discuss Unique Factorization.

Corollary 1 (An Irreducibility Criterion) Let f(t) ∈ Z[t]. Then a sufficient condition for f to be irreducible in Q[t] is for there to exist a prime number p such that the reduction f¯(t) ∈ Fp[t] be irreducible.

2 More detail and more generality

A Domain (PID) is an integral domain A in which every ideal is principal. If a ∈ A is a generator of a nonzero ideal I in A the other generators of I are A∗ · a (the elements of A that are associate to a, i.e., multiples of a by a in A. The nonzero prime ideals of A are given by irreducible elements of A, i.e., elements a ∈ A such that for every factorization a = b · c either b or c is a unit. Often one says that a prime element of a ring is an element that generates a , so one might capture the above discussion by saying that in a PID prime elements

1 and 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 element a 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 can change each of the irreducible elements by replacing them by associate irreducible elements and you then can (i.e., must) change the unit u to suit.

Theorem 2 A PID is a UFD

Let A be a PID. Step one: Every nonzero element a ∈ A Has a decomposition into a product of irreducible elements as above. Consider S the set of principal ideals in A that have generators not admitting prime factorizations. Find a maximal ordered set of ideals in S and consider their union—since A is a PID it is generated by a single element a which must also not have a prime factorization. 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 prime factors by Lemma ??.

Definition 1 If A is a UFD, K its field of fractions, and p an of A (now taken ∗ up 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 Y a = u · pordp(a) p 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- n nomials f(X) = anX + ... + a0 by setting ordp(f) = the minimum of the ordp of the coefficients ∗ ∗ ai. If f is not identically zero, define the content of f(X) (an element of K moduloA ) to be Y cont(f) := pordp(f). p

2 To say that cont(f) = 1 is to say that f ∈ A[X] and the gcd pof its coefficients is 1. We may write for 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 by units in A) by irreducible polynomials of K[X] of content 1, and irreducible elements of A (viewed as polynomials of degree zero in A[X]).

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

Corollary 3 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

n i f(X) = anX + ··· + aiX + ··· + a0 ∈ A[X] ⊂ K[X] has the property that

• ordp(an) = 0,

• ordp(ai) > 0 for i < n,

• ordp(a0) = 1, then f(X) is irreducible in K[X].

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