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 arxr and bsxs 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 arbs 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.
Read more about this topic: Gauss's Lemma (polynomial)
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