Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:20:32Z http://mathoverflow.net/feeds/question/31460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31460/can-an-infinite-commutative-ring-have-a-finite-but-nonzero-number-of-non-nilpot Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors? Oliver 2010-07-11T20:48:12Z 2010-07-25T05:31:03Z <p>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.)</p> <p>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.</p> <p>However, I am unaware of any ring with an infinite number of zero-divisors, of which $0 &lt; n &lt; \infty$ are non-nilpotent.</p> <p>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.</p> http://mathoverflow.net/questions/31460/can-an-infinite-commutative-ring-have-a-finite-but-nonzero-number-of-non-nilpot/31471#31471 Answer by Jack Huizenga for Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors? Jack Huizenga 2010-07-11T21:46:18Z 2010-07-12T01:36:11Z <p>No, there is no such example.</p> <p>Recall that the nilradical $N$ of $R$ is the ideal of nilpotent elements. It equals the intersection of all prime ideals of $R$.</p> <p>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):</p> <p>$$D = \bigcup_{x\neq 0} \sqrt{(0:x)}$$</p> <p>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 $I$ <em>properly</em> 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.</p> <p>EDIT: Here's an easier proof in a different spirit, motivated by the preceding argument.</p> <p>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.</p> <p>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.</p>