# Product of two primitive polynomials

I'm having troubles with one of the problems in the book Introduction to Commutative Algebra by Atiyah and MacDonald. It's on page 11, and is the last part of the second question.

Given $$R$$ a commutative ring with unit. Let $$R[x]$$ be the ring of polynomials in an indeterminate $$x$$ with coefficients in $$R$$. We say that a polynomial $$f = \sum\limits_{i=0}^n r_ix^i \in R[x]$$ (with coefficients $$r_0, r_1, \ldots, r_n$$) is primitive if $$\langle r_0,r_1,...,r_n\rangle = R$$, i.e., the ideal generated by the coefficients of $$f$$ is $$R$$.

Let $$f$$ and $$g$$ be two polynomials in $$R[x]$$. Prove that $$fg$$ is primitive iff $$f$$ and $$g$$ are both primitive.

The $$\Rightarrow$$ part is easy. Say, $$f = \sum\limits_{i=0}^n r_ix^i$$, $$g = \sum\limits_{i=0}^m s_ix^i$$; then $$fg = \sum\limits_{i=0}^{m+n} c_ix^i$$, where $$c_k = \sum\limits_{i + j = k}r_is_j$$. Since $$fg$$ is primitive, there exists a set of $$\{\alpha_i\} \subset R$$, such that $$\sum\limits_{i=0}^{m+n} \alpha_ic_i = 1$$, to prove $$f$$ is primitive, I just need to write all $$c_i$$'s in terms of $$r_i$$'s, and $$s_j$$'s, then group all $$r_i$$ accordingly, rearranging it a little bit, and everything is perfectly done. And the proof of the primitivity of $$g$$ is basically the same.

The $$\Leftarrow$$ part is just so difficult. Say $$f = \sum\limits_{i=0}^n r_ix^i$$, $$g = \sum\limits_{i=0}^m s_ix^i$$ are both primitive, then there exists $$\{\alpha_i\}; \{\beta_i\} \subset R$$, such that $$\sum\limits_{i=0}^{n} \alpha_ir_i = 1$$, and $$\sum\limits_{i=0}^{m} \beta_is_i = 1$$. At first, I thought of multiplying the two together, but it just didn't work. So, I'm stuck since I cannot see any way other than multiplying the two sums together. I hope you guys can give me a small push on this.

And have a good day.

• William Messing and Victor Reiner, A universal coefficient theorem for Gauss's Lemma, arxiv.org/abs/1209.6307 , give several proofs of this, one of which is surely to your liking. Jun 7 '13 at 13:11
• A proof is given in Wikipedia, here Jun 7 '13 at 13:12
• Thank you guys so much, I get it now. :d Jun 7 '13 at 13:17
• Sep 3 '20 at 9:35

To summarize the WP reference I gave in a comment: supposing $fg$ is not primitive, form the quotient of $R$, and consequently $R[x]$, by any maximal ideal (any prime ideal will do too) of$~R$ containing all coefficients of$~fg$. Then $fg$ is killed but neither $f$ nor $g$ is; however this is impossible in $K[x]$ where $K$ is the quotient field (or quotient integral domain) of $R$ by the mentioned ideal, since $K[X]$ is an integral domain when $K$ is.

The $\Leftarrow$ part of this falls under "Gauss's lemma", for which Wikipedia has a nice proof here:

http://bit.ly/18TjLgb

This proof should, however, be extended to an arbitrary ring.

This works for that situation:

http://bit.ly/19PhGDs

In order to prove the other direction by contrapositive, suppose that $f=\sum_{i=0}^n r_i x^i$ is not a primitive polynomial. Then each coefficient of its product with $g=\sum_{i=1}^n s_ix^i$ is a member of the ideal generated by the coefficients of $f$. That is, we may state that if $f\, g=\sum_{i=1}^{n+m} t_ix^i$, then $\langle t_1,t_2,...t_{m+n} \rangle \subseteq \langle r_1,r_2,...r_{n} \rangle$. It follows that $f\,g$ is not a primitive polynomial.

EDIT: never mind, guess I'm late for the party

• The part "$fg$ primitive $\implies$ $f$ and $g$ primitive" is the trivial part both you and the question deal with. If $f$ or $g$ would be divisible be a non-unit in $R$, then so would $fg$, that's all. The question is about the other direction, which is Gauss's lemma. Jun 7 '13 at 13:29
• Thanks Marc, will fix my question accordingly. Jun 7 '13 at 15:45