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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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The

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 is a of $R$ can be expressed as the union of some prime ideals the radicals of the annihilators of individual nonzero elements of $R$ (see below for a precise statement)Atiyah-MacDonald Prop. Now either one of these primes $P$ of 1.15):

$R$ is not equal to $D = \bigcup_{x\neq 0} \sqrt{(0:x)}$$Here$N$, or there (0:x)$ is only one such prime an ideal, and its radical is the intersection of all the primes containing itequals . Thus $N$. In the former caseD$is a union of ideals, each of which contains the nilradical$N$. If any of these ideals$I$properly contains$N$, then if$N$is infinite , then$N \subset P$is a proper containment and thus we conclude$P\setminus I\setminus N$is also infinite . In the latter case, every zero divisor is nilpotent. So (since it isn't possible to have contains a nonzero finite number of non-nilpotent zero-divisors but infinitely many nilpotents. The precise statement referred to above is the following: let$X$be the collection whole coset of prime ideals$P$with the following property: there exists$f\in R$such that$P$is minimal in the set of prime ideals containing$(0:f)$. Then the set of zero divisors of N$), and hence $R$ D\setminus N$is the union of the primes in$X$. (Atiyah-MacDonald, Exinfinite.4.9) 4 added 359 characters in body The nilradical$N$of$R$is the intersection of all prime ideals of$R$. On the other hand, the set of zero-divisors contains the is a union of the minimal primes some prime ideals of$R$. R$ (see below for a precise statement). Now either some minimal prime one of these primes $P$ of $R$ is not equal to $N$, or there is only one minimal such prime and it equals $N$. In the former case, if $N$ is infinite, then $N \subset P$ is a proper containment and thus $P\setminus N$ is also infinite. In the latter case, every zero divisor is nilpotent. So it isn't possible to have a nonzero finite number of non-nilpotent zero-divisors but infinitely many nilpotents.

The precise statement referred to above is the following: let $X$ be the collection of prime ideals $P$ with the following property: there exists $f\in R$ such that $P$ is minimal in the set of prime ideals containing $(0:f)$. Then the set of zero divisors of $R$ is the union of the primes in $X$. (Atiyah-MacDonald, Ex. 4.9)

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