# Some subtleties on zerodivisors in polynomial rings

This question is based on two answers I saw. I want to know if my reasoning is correct in each case. For me, $A$ is always a commutative ring with $1 \ne 0$.

In the first answer this proposition was given:

Let $f_1,\dots,f_t\in A[X]$ be nonzero. If there is $0 \ne g\in A[X]$ such that $gf_1=\cdots=gf_t=0$, then there is $0 \ne a\in A$ such that $af_1=\cdots=af_t=0$.

Assuming the case $t = 1$ which is McCoy's theorem, I think I can prove this as follows:

Proof: Let $\displaystyle F := \sum_{i=1}^t f_i X^{\sum_{j < i} (1 + \deg f_j)}$. Then $gf_1 = \ldots = gf_t = 0 \implies gF = 0 \implies aF = 0$ for some $0 \ne a \in A \implies af_1 = \ldots = af_t = 0$. The last implication is because every coefficient of each $f_i$ appears exactly once as a coefficient of $F$.$\newcommand{\N}{\mathbb{N}}$

Question 1: Is the proof above correct?

Moving on to the next question, in the other answer this proposition was given:

(*) Let $A$ be a commutative $\N$-graded ring. Let $f \in A$ be a zerodivisor. Then there is some $0 \ne a \in A$ homogeneous such that $af = 0$.

I think the proof given in that answer has a fatal flaw, which was pointed out as a comment but with no response. Specifically $\deg f_ig < \deg g$ may not be true. For polynomial rings in $1$ variable it's ok because $f_i = a_ix^i$ where $a_i$ has degree $0$ and $x^i$ is a nonzerodivisor, so $\deg a_ig < \deg g$ and $a_ig \ne 0$. So my second question is:

Question 2: Is (*) true? If not, what is a counterexample?

One note is that (*) is true if $A$ is Noetherian (since then every zerodivisor is in some associated prime, which is the annihilator of a homogeneous element). So a counterexample would have to be non-Noetherian.

• 1) looks right; I don't know about 2) – 57Jimmy Jul 31 '16 at 20:53

Question 2: The general case follows from the Noetherian case. Indeed, suppose $f\in A$ is a zerodivisor, with $gf=0$ for some nonzero $g$. Let $B$ be the subring of $A$ generated by the homogeneous parts of $f$ and $g$. Then $B$ is a finitely generated ring, and hence is Noetherian. Since $B$ is generated by homogeneous elements, it is a graded subring of $A$. By the Noetherian case, there exists some homogeneous $0\neq a\in B\subseteq A$ such that $af=0$.
• Thanks for the answer! I think the proof for 2) should be fine. The only things left to check are that $B$ is $\mathbb{N}$-graded, and being homogeneous in $B$ implies being homogeneous in $A$. Could you say a bit on why these are true? – Jay Jul 31 '16 at 21:09
• $B$ is just the subring of $A$ generated by a bunch of homogeneous elements, so it can inherit $A$'s grading. Explicitly, the $n$th graded piece of $B$ is just the abelian group generated by those products of the generators of $B$ which have degree $n$. – Eric Wofsey Jul 31 '16 at 21:10