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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. We say $f = \sum\limits_{i=0}^n r_ix^i \in R[x]$ is

primitiveif $\langle r_0,r_1,…,r_n\rangle = R$, i.e., the ideal generated by the coefficients of $f$ is $R$.Prove that $fg$ is primitive iff $f$ and $g$ are both primitive.

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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.

Thanks very much in advance,

And have a good day.

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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:

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

This works for that situation:

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

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