Is there a ring with zero divisors but no nilpotents? [duplicate]

Question

A ring $R$ (henceforth assumed to be commutative and with unity $1$) has a zero divisor $a$ iff $a$ is nonzero and there exists a nonzero element $b$ such that $a*b=0$. Also, by definition, a nilpotent element $n$ is a nonzero element of $R$ such that some positive integer power of $n$ is $0$. My question is, is there a ring $R$ such that $R$ has zero divisors but no nilpotent elements?

Answer 1

A simple example is the ring $R=\mathbb Z/6\mathbb Z$.

This ring has elements represented by $0$, as well as $1$ and $5$; the latter two are self-inverses.

The rest of the elements are zero divisors, because $2\times3\equiv3\times4\equiv0\pmod6$

but they are not nilpotent because $2^n\equiv\pm2, 3^n\equiv3$, and $4^n\equiv4\pmod6$,

for $n>0$, where the sign in front of $2$ depends on the parity of $n$.

Answer 2

Let be $A=\Bbb R[x,y]/(xy)$.

Crearly, $A$ has zero divisors because $xy=0$ in $A$. Let's see that it has no nilpotent elements.

Assume that $P\in\Bbb R[x,y]$ and $P^n\in (xy)$ for some $n\ge 1$. Let

$$P(x,y)=\sum_{j,k} a_{j,k}x^jy^k$$ $$P(x,y)^n=\sum_{j,k}b_{j,k}x^jy^k$$ but $b_{j,k}$ is $0$ whenever $j=0$ or $k=0$.

If $a_{j,k}\neq 0$ and $j=0$, then $b_{0,nk}=a_{0,k}^n\neq 0$, so $P^n\notin(xy)$. Also, if $a_{j,k}\neq 0$ and $k=0$, $P^n\notin(xy)$.
Thus, if $P\notin(xy)$, then $P^n\notin(xy)$.

Answer 3

An example coming from analysis would be $R=C(X)$, where $X$ is a compact Hausdorff space with more than one connected component.

For instance $R=C([0,1]\cup[2,3])$. Any function supported on one of the components is a divisor of zero. Explicitly, $f=1_{[0,1]}$ and $g=1_{[2,3]}$ are nonzero elements of $R$ with $fg=0$.

Answer 4

Many good examples already, but one very simple example not yet mentioned: the direct product ring $\mathbb{Z} \times \mathbb{Z}$, or more generally $R \times S$ for any nontrivial nilpotent-free rings $R$, $S$. There are non-trivial zero divisors in the product ring: for instance, $(1,0)\cdot(0,1) = (0,0)$. But there are no non-trivial nilpotents: if $(a,b)^n = 0$, that means $(a^n,b^n) = (0,0)$, so (since $\mathbb{Z}$ is nilpotent-free) $a = b = 0$.

Answer 5

More generally, the ring of sequences such as $\mathbb R^{\mathbb N}$ has zero divisors but no nilpotents. If one "quotients out by all the zero divisors" (in a suitable sense), one can get a proper ordered field extension of $\mathbb R$.

Answer 6

As an example, split-complex numbers. In this ring there are no nilpotents, but there are zero divisors, for instance, $1+j$ and $1-j$: $$(1+j)(1-j)=1^2-j^2=0$$