# Zerodivisors in polynomial rings over a non-commutative ring

Prove that if $f \in R[x]$ is a zero divisor then $\exists r(\neq 0) \in R$ s.t $rf=0$, where $R$ is a ring.

I know that for $(a_0+a_1x+ \cdots +a_nx^n)(b_0+b_1x+\cdots+b_mx^m)=(c_0+c_1x+\cdots+c_{m+n}x^{m+n})=0$ where $c_k=\sum_{i=0}^k a_ib_{k-i}$ Then $a_0$ & $a_n$ is a zero divisor. But how to prove from here now?

I have found 1 proof in Zero divisor in $R[x]$ but this is for commutative ring, what about non-commutative rings? Is the result still true or not? Give me proof or counterexamples.

• I don't think this holds in noncommutative rings. See for example Section 3 of this paper. – rogerl Oct 13 '14 at 21:25

McCoy's theorem (the theorem described in the link you used) does not hold in general for noncommutative rings.

Maybe the first example given was here:

L. G. Jones and L. Weiner, Advanced Problems and Solutions, Solutions 4419, Amer. Math. Monthly 59 (1952), no. 5, 336–337.

This lead to the following two papers defining a right McCoy ring to be a ring $R$ for which for $f,g\in R[x]$, $fg=0$ implies $fr=0$ for some $r\in R$.

P. P. Nielsen, Semi-commutativity and the McCoy condition, J. Algebra 298 (2006), no. 1, 134–141.

M. B. Rege and S. Chhawchharia, Armendariz rings, Proc. Japan Acad. Ser. A Math. Sci. 73 (1997), no. 1, 14–17.