# Reduced is equivalent to non-singular for commutative rings

Intro I've been (not successfully) trying to educate myself a bit about singular/nonsingular modules and I came by a proposition on Wikipedia that I find really interesting.

For commutative rings, being nonsingular is equivalent to being a reduced ring.

Question Can anyone give me some hints on how to prove this?

My attempts I didn't really get far. I understand that the domain is non-singular. Because it has no non-zero zero divisors.

If I denote $$Z$$ the set of zero-divisors then clearly annihilators are subsets of $$Z$$. I even read a stronger statement in some article saying that that Annihilators in reduced rings are intersections of minimal primes (minimal primes are contained in $$Z$$ for reduced rings)

Also, it seems like a good direction to investigate when are ideals contained in $$Z$$ essential (which is a question I stumped upon also in different settings in past, so seems like an interesting problem on its own)

Let $$R$$ be a commutative ring.
Lemma $$\operatorname{Nil}(R) \subseteq \mathcal{Z}(R)$$
Proof. Let $$r \in R$$ be nilpotent. Let $$x \in R \setminus \{0\}$$ be arbitrary. Let $$n \geq 0$$ be maximal such that $$r^n x \neq 0$$ (such an $$n$$ exists because $$r$$ is nilpotent and $$r^0 x = x \neq 0$$). Then $$r^{n+1} x = 0$$, so $$r^n x \in \operatorname{ann}(r) \cap Rx$$ tells us that $$\operatorname{ann}(r) \cap Rx \neq 0$$. We conclude that $$\operatorname{ann}(r)$$ is essential, so $$r \in \mathcal{Z}(R)$$. $$\square$$
This gives one direction of the proof – if $$R$$ is nonsingular, then $$R$$ is reduced. Here's the other direction (by contrapositive):
Suppose $$R$$ is not nonsingular, and let $$z \in \mathcal{Z}(R) \setminus \{0\}$$. Then $$\operatorname{ann}(z) \cap Rz \neq 0$$, so there is some $$r \in R$$ such that $$rz \neq 0$$ and $$rz^2 = 0$$. But then $$(rz)^2 = r(rz^2) = 0$$, so $$rz$$ is a nonzero nilpotent element, and thus $$R$$ is not reduced. $$\square$$