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Let $I$ be a finitely generated ideal of a commutative ring $R$. Assume every element of $I$ is a zero divisor. Does then exist a $x \neq 0$ in $R$ with $xI=0$?

This is true if $0$ is a decomposable ideal, for example if $R$ is noetherian. I wonder if we actually need this. Doesn't it sound plausible? The problem is that we cannot just multiply the elements which kill the generators of $I$, the product can vanish.

Martin Brandenburg
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1 Answers1

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No, for a counterexample see Exercises 2-2-6,7 pp. 62-63 in Kaplansky: Commutative Rings (excerpted below). See also the discussion following Theorem 82 p.56.

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Bill Dubuque
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  • Could you explain the notation? I mean what are $Z(A)$ and $<(A)$? – Martin Brandenburg Oct 21 '10 at 14:39
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    @Martin: $ Z(A) $ is the set of zero-divisors on the $R$-module $A$, i.e. the elements of $R$ that annihilate some nonzero element of $A$. The latter is presumably a misprint for the former. – Bill Dubuque Oct 21 '10 at 14:55