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By a theorem of Ganesan, if a commutative ring not a domain has only finitely many zero-divisors, then the ring must be finite. (There are analogous results for non-commutative rings.)

There are plenty of examples of infinite rings with a finite number of nonzero nilpotents. There are also plenty of examples of infinite rings with an infinite number of zero-divisors, all of which are nilpotent.

However, I am unaware of any ring with an infinite number of zero-divisors, of which $0 < n < \infty$ are non-nilpotent.

Can anyone give an example or explain why this can't happen. I am mostly interested in the commutative case, but non-commutative examples would be interesting too.

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Do you want finitely many non-nilpotent zero divisors, or finitely many nilpotent zero divisors? (Just making sure the question statement is right) –  Jack Huizenga Jul 11 '10 at 21:38
    
Finitely many NON-nilpotent zero-divisor. I apologize for all the "nots" in this question. Even I am getting confused reading through it. –  Oliver Jul 11 '10 at 21:44

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up vote 10 down vote accepted

No, there is no such example.

Recall that the nilradical $N$ of $R$ is the ideal of nilpotent elements. It equals the intersection of all prime ideals of $R$.

On the other hand, the set $D$ of zero-divisors of $R$ can be expressed as the union of the radicals of the annihilators of individual nonzero elements of $R$ (Atiyah-MacDonald Prop. 1.15):

$$D = \bigcup_{x\neq 0} \sqrt{(0:x)}$$

Here $(0:x)$ is an ideal, and its radical is the intersection of all the primes containing it. Thus $D$ is a union of ideals, each of which contains the nilradical $N$. If any of these ideals $I$ properly contains $N$, then if $N$ is infinite we conclude $I\setminus N$ is also infinite (since it contains a whole coset of $N$), and hence $D\setminus N$ is infinite.

EDIT: Here's an easier proof in a different spirit, motivated by the preceding argument.

Suppose $x,y\in R$, such that $x$ is nilpotent and $y$ is a zerodivisor. I claim $x+y$ is a zerodivisor. Let $z\neq 0$ be such that $yz=0$. If $xz=0$, we are done. Otherwise, let $n$ be the smallest number such that $x^nz=0$ (which happens for some $n$ since $x$ is nilpotent). Then $x^{n-1}z\neq 0$ but $x(x^{n-1}z)=0$, so $(x+y)x^{n-1}z=0$. Thus $x+y$ is a zerodivisor.

Now if $y$ is not nilpotent, $x+y$ is not nilpotent since the nilradical $N$ is an ideal. It follows that the coset $N+y$ consists entirely of nonnilpotent zerodivisors, so if $N$ is infinite then there are infinitely many nonnilpotent zerodivisors.

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Are you sure this works in the absence of any Noetherianness hypothesis? (As I recall, A-M do most of this chapter assuming that primary decompositions exist in the rings under consideration, since they have not introduced the Noetherian condition yet.) –  Charles Staats Jul 11 '10 at 23:58
    
I haven't done the exercise in A-M, but they they are not assuming the zero ideal is decomposable (which is the assumption you need in order to say that the zero divisors are the union of the associated primes). In fact, it's a second part of the exercise to show that "If the zero ideal has a primary decomposition, show that $X$ (defined above) is the set of associated primes of 0." I'll think about how to do the exercise, but I'm pretty sure its OK. –  Jack Huizenga Jul 12 '10 at 0:14
    
I've rewritten the argument to no longer rely on the previously mentioned exercise; it now relies on a much more basic result from A-M which clearly does not require Noetherianness. –  Jack Huizenga Jul 12 '10 at 1:07
    
And I'm now seeing that the fact I previously used--the set of zero-divisors is a union of prime ideals--is also stated in a more crude form in Ex. 1.14 of A-M. The suggested proof requires Zorn's Lemma in the non-Noetherian case. –  Jack Huizenga Jul 12 '10 at 1:15
    
Thank you! Both of your proofs are quite enlightening. –  Oliver Jul 12 '10 at 17:28

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