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

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?

• What about $\Bbb R[x,y]/(xy)$? Commented Aug 5 at 17:14
• What about $\mathbb Z/6\mathbb Z$? Commented Aug 5 at 17:33
• @J.W.Tanner Yes, a simpler example Commented Aug 5 at 17:37
• To give a wide class of examples, any product of (at least two) reduced rings will satisfy your criteria. J.W. Tanner's nice example is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ via the Chinese remainder theorem. Commented Aug 5 at 17:44
• DaRT query Commented Aug 5 at 20:25

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 $$n$$.

• Generally only $\,0\,$ is nilpotent in $\,\Bbb Z/q,\,$ i.e. $\,[\,q\mid n^k\Rightarrow q\mid n\,]\iff q\,$ is squarefree. See here for this and many other characterization of "squarefree". $\ \$ Commented Aug 5 at 19:11

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 $$P^n\notin(xy)$$.

• Is it not easier to witness $(x)$ and $(y)$ are distinct primes? Hence, their intersection = their product ideal $(xy)$ is radical? Commented Aug 5 at 18:15

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 $$fg=0$$.

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) $$a = b = 0$$.

• +1, a nice class of examples indeed. (But I did in fact actually mention this in the comments. :) ) Commented Aug 7 at 3:43

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 $$\mathbb R$$.

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