1
$\begingroup$

This question already has an answer here:

The question is the following:

Let $f$ be a polynomial of the ring $R[x_1, \ldots, x_n]$, with $R$ any ring, and let $\mathrm{cont}(f)$ be the ideal generated by the coefficients of $f$. Why if $\mathrm{cont}(f)$ contains a non-zerodivisor of $R$, then $f$ is a non-zerodivisor of $R[x_1, \ldots, x_n]$?

$\endgroup$

marked as duplicate by user26857 commutative-algebra May 22 at 21:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Hint: When you write that non zero divisor $a$ as a linear comnbination, this gives you a recipe for a polynomial $g$ such that one coefficient of $f\cdot g$ is $a$. Now assume $h\cdot f=0$. Then also $hfg=0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.