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This question contains a characterization of zero divisors in $A[x]$ for a commutative ring $A$ with identity. But obviously the trick used in the proof does not work anymore for $A[x_1,...,x_n]$ mainly because whatever ordering we use for $\mathbb{Z}^n$ for $n\geq 2$ will not give us a clean expression like $a_nb_m=0$ when we worked in $A[x]$. Thus the inductive steps do not work anymore. I would like to know what kind of characterization can we obtain for zero divisors in $A[x_1,...,x_n]$.

William Sun
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    I just found out that doing induction on the number of indeterminants might work very well in this case. If you have an alternate answer please also post it here, thank you. – William Sun Sep 15 '19 at 19:06

2 Answers2

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McCoy's theorem extends as corollary to multivariate polynomials with very little work.

In $R[x, y] = R[x][y]$ a zero divisor $f$ is annihilated by an element $g\in R[x]$, using McCoy's theorem.

Writing $f = \sum f_i y^i$ with $f_i \in R[x]$, we see by comparing coefficients that $f_i g = 0$.

Then let $f' = \sum f_i x^{n_i} \in R[x]$ be a polynomial where the $n_i \in \mathbb{N}$ are chosen large enough so that the $f_i$ do not 'interact.'

Clearly $gf' = 0$, and McCoy's theorem now shows that $r f' = 0$ for some $r \in R$. We deduce that $r f_i = 0$ for all $i$, so also $rf = 0$.

luxerhia
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Badam Baplan
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Well, use the lexicographic ordering on the polynomial ring $R=K[x_1,\ldots,x_n]$.

Given two polynomials $f,g\ne 0$ in $R$. Suppose $x^\alpha$ and $x^\beta$ are the largest monomials involved in $f$ and $g$ w.r.t. the lex ordering, respectively.

Then the largest monomial involved in $fg$ w.r.t. the lex ordering is $x^{\alpha+\beta}$, which shows that $fg\ne 0$.

This is a direct extension of the case of one variable.

Wuestenfux
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