Zero divisors in polynomial rings with several indeterminants.

Question

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 $A[x_1,...,x_n]$.

Answer 1

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 $rf = 0$.

Answer 2

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.