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.
(https://en.wikipedia.org/wiki/Singular_submodule#Examples)
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)
Best Answer
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. $\square$