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The non-zero elements of the minimal prime ideals of a noetherian commutative ring are zero-divisors.

The proof of this fact I know of uses primary decomposition. Are there alternative (e.g. more direct) proofs ?

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Let $p\subset A$ be a minimal prime and let $s\in p$. Then $pA_p$ is the only prime ideal in $A_p$ and so $s/1$ is nilpotent here. Hence there is a $t\in A-p$ such that $ts^n=0$, and hence $s$ is a zero divisor. –  J.C. Ottem Jun 9 '11 at 21:48
If $x$ is in a minimal prime, multiply it by one element from each of the other minimal primes. The result is in the intersection of the minimal primes, which is the nilradical, so some power is zero. –  Graham Leuschke Jun 9 '11 at 21:50
Nice arguments. Thanks to both for your answers. –  Ralph Jun 10 '11 at 5:54
Dear Graham, I have a problem with your argument. Suppose there is just one other minimal prime and you take in it a non-zero element $y$. You might then have $(xy)^2=x^2y^2=0$ but this doesn't prove that $x$ is a zero-divisor because $y^2$ might be zero. Am I missing something? –  Georges Elencwajg Jun 10 '11 at 7:09
By the way, J.C.'s argument shows that the noetherian hypothesis is superfluous. –  Georges Elencwajg Jun 10 '11 at 12:36

1 Answer 1

up vote 4 down vote accepted

Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$.

If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors).

If $X$ has more than one irreducible component, then do the following: Choose $f_1,\dots,f_m$ such that each $f_i$ is contained in exactly one minimal prime ideal and for each minimal prime there is (exactly) one $f_i$ contained in it. This choice ensures that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent. (The point is, that if their product vanished on an irreducible component, then one of them would have to, but then it would be the one that vanishes on that component). Now take an arbitrary element $g$ in any of the minimal primes and for simplicity assume it is from the one corresponding to $f_1$. (We allow $g$ to be contained in other primes as well, but that has no consequence). Then the following claim implies that $g$ is a zero-divisor.

Claim Let $g,f_2,\dots,f_t$ be such that $g\cdot f_2\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $g$ is a zero-divisor.

Proof Let $n$ be the smallest non-negative integer for which there exists a $t-1$-uple $(n_2,\dots,n_t)\in\mathbb N^{t-1}$ such that $g^n\cdot f_2^{n_2}\cdots f_t^{n_t}=0$. Observe that $n$ exists and $n\geq 1$ by the assumptions. Then $g\cdot (g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t})=0$, but $g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t}\neq 0$. $\square$

Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. Also, it clarifies the unclear step pointed out by Georges. In fact, it seems one needs to do a little yoga to get the result. In any case algebra=geometry. :)

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Yes, Sándor, algebra and geometry are equal, but geometry is more equal :-) –  Georges Elencwajg Jun 10 '11 at 9:11
Sándor, thanks for your answer. If I'm right the part "It is easy to see that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent" says that in Georges' example above, $y^2 = 0$ is not possible. I would be happy if you could give me some more hint, how to show that $f_2 \dotsm f_t$ is not nilpotent. –  Ralph Jun 10 '11 at 9:23
Ralph, you are right. I need to add more love to this. Thanks for pointing that out. –  Sándor Kovács Jun 10 '11 at 10:10
Georges, I'm with you. Geometry is a Pig. –  Sándor Kovács Jun 10 '11 at 10:11
Thanks for clarification. Your proof is also a fine example of an application of the prime avoidance principle (used to choose the $f_i$'s). –  Ralph Jun 10 '11 at 11:18

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