Gauss's Lemma (polynomial) - Proofs of The Primitivity Statement

Proofs of The Primitivity Statement

An elementary proof of the statement that the product of primitive polynomials over Z is again primitive can be given as follows.

Proof: Suppose the product of two primitive polynomials f(x) and g(x) is not primitive, so there exists a prime number p that is a common divisor of all the coefficients of the product. But since f(x) and g(x) are primitive, p cannot divide either all the coefficients of f(x) or all those of g(x). Let a_rxr and b_sxs be the first (i.e., highest degree) terms respectively of f(x) and of g(x) with a coefficient not divisible by p. Now consider the coefficient of xr+s in the product. Its value is given by

This sum contains a term a_rb_s which is not divisible by p (because p is prime, by Euclid's lemma), yet all the remaining ones are (because either i > r or j > s), so the entire sum is not divisible by p. But by assumption all coefficients in the product are divisible by p, leading to a contradiction. Therefore, the coefficients of the product can have no common divisor and are thus primitive. This completes the proof.

A cleaner version of this proof can be given using the statement from abstract algebra that a polynomial ring over an integral domain is again an integral domain. We formulate this proof directly for the case of polynomials over a UFD R, which is hardly different from its special case for R = Z.

Proof: Let S,T be primitive polynomials in R, and assume that their product ST is not primitive, so that some noninvertible element d of R divides all coefficients of ST. There is some irreducible element p of R that divides d, and it is also a prime element in R (since R is a UFD). Then the principal ideal pR generated by p is a prime ideal, so R/pR is an integral domain, and (R/pR) is therefore an integral domain as well. By hypothesis the projection R→(R/pR) sends ST to 0, and also at least one of S,T individually, which means that p divides all of its coefficients, contradicting primitivity.

The somewhat tedious bookkeeping in the first proof is simplified by the fact that the reduction modulo p kills the uninteresting terms; what is left is a proof that polynomials over an integral domain cannot be zero divisors by consideration of the leading coefficient of their product.