# Commutative rings such that all non-units are zero divisors

The construction of the total ring of fractions shows that any commutative ring (with 1) can be put inside a ring whose non-units are zero divisors.

Examples of these rings are the products of fields.

Are there other examples? Do these rings have a standard name?

• Any finite commutative (unital) ring. Jun 21, 2021 at 13:37
• Similarly, any algebra of finite dimension over some field. Jun 21, 2021 at 13:39
• Aha, I forgot they can be non-reduced... Jun 21, 2021 at 13:39
• Related but a different question. Jun 21, 2021 at 14:16
• Also related Jun 21, 2021 at 19:26

The broadest, easily described subclass, I think, is that of strongly $$\pi$$-regular rings, which are rings that satisfy the DCC on all chains of the form $$xR\supseteq x^2R\supseteq x^3R\supseteq\cdots$$ for every $$x\in R$$.
Suppose, though, that you fabricate strongly $$\pi$$-regular rings $$R_i$$ such that there exists $$x_i\in R_i$$ satisfying $$x_i^iR\neq x_i^{i+1}R$$. Then in $$R=\prod_{i=1}^\infty R_i$$, it is easy to see that nonunits are zero divisors, but $$R$$ is not strongly $$\pi$$-regular, because the element $$x=(x_1, x_2,\ldots)$$ spoils things. So at least we can say rings with your condition are a strictly bigger class than the strongly $$\pi$$-regular rings.