# Zero divisors in polynomial rings

The following is an exercise in Hungerford (Ch. III, ex. 5.6).

Let $R$ be a commutative ring with identity. If $f=a_nx^n+\dots+a_0$ is a zero divisor in $R[x]$, then there exists a nonzero $b$ in $R$ such that $ba_n=ba_{n-1}=\dots=ba_0=0$.

I can see for example that $\{g\in R[x]\mid fg=0\}$ is a nonzero ideal, so it contains a nonzero element of smallest degree. But how to show that such an element is actually a constant?

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Thanks rschwieb. The question in the other post is not an exact duplicate, but the answer from Gone proves the general case. – PatrickR Dec 21 '12 at 21:24
If you want you can delete it, but in this case you don't have to do anything. If someone really doesn't like it, they'll have to vote to delete. Please don't edit the other question's contents. – rschwieb Dec 21 '12 at 21:24
I've made a meta question on the issue of near duplicates: meta.math.stackexchange.com/questions/6941/… – Hurkyl Dec 21 '12 at 21:34
I gave here a complete solution to this question. – user26857 Dec 21 '12 at 21:58
@user26857: I cannot close it either. This question is cursed. – Jyrki Lahtonen Dec 25 '15 at 20:29