Is there a finite non-regular ring which is a direct product of local rings with atleast one not a field?

Question

Let $R$ be a commutative ring with unity. $R$ is regular if for each $a\in R$, there exists $b\in R$ such that $a=aba$. It is local it has a unique maximal ideal.

(James L. Fisher , Finite Principal Ideal Ring, Canad. Math. Bull. Vol. 19 (3), 1976) proves the following:

THEOREM 2. Let $R$ be a finite principal ideal ring. Then $R$ is an ideal direct sum $R_1\oplus\cdots \oplus R_k\oplus N$ where $R_i, i = l,\ldots, k$ are primary principal ideal rings and $N$ is a nilpotent principal ideal ring.

I am trying to check for examples of commutative rings satisfying this results but most of them are regular ring.

My question: is there an example of a finite non-regular ring which is direct product of local rings one of which is a field and the other is not a field?

Answer 1

Yes, such rings exist, and I will give you a quick recipe and idea for how to build these gadgets. It is known that commutative unital ring $A$ is von Neumann regular if and only if $A$ is both Artinian (zero-dimensional) and reduced (so in particular every prime ideal is maximal and $A$ admits no non-zero nilpotents). Because finite rings are automatically Artinian (you cannot have infinitely many non-equal ideals of a finite ring --- use the power set $\mathcal{P}(A)$ to give an upper bound on how many such ideals you can have) you can generate finite non-regular commutative unital rings $A$ by violating the reduced requirement that von Neumann regularity forces.

Here is an explicit example. Fix an integer prime $p$, an integer $n$ with $n \geq 2$, and consider the ring $$ A := \mathbb{F}_{p} \times \frac{\mathbb{Z}}{p^n\mathbb{Z}}. $$ Both $\mathbb{F}_p$ and $\mathbb{Z}/p^n\mathbb{Z}$ are local rings (as $\mathbb{F}_p$ is a field and $\mathbb{Z}/p^n\mathbb{Z}$ has unique maximal ideal $(p) = \mathfrak{m}$) but $\mathbb{Z}/p^n\mathbb{Z}$ is not a field. In particular, $p$ is nilpotent of nilpotency index $n$ in $\mathbb{Z}/p^n\mathbb{Z}$. Thus, because $A$ is non-reduced, $A$ is a product of local rings (one a field and one not a field) that is finite and not von Neumann regular.

Answer 2

Let's put aside finiteness for a second.

You have two things going for you:

1. Every single local ring that isn't a division ring is not regular
2. A product is regular iff the rings involved are regular

So any product of local rings is an example (sometimes infinite) as long as you pick one of them to not be a division ring.

Convenient examples of nonfield local rings are, for example:

1. Localizations of a nonfield PID at any maximal ideal (e.g. $\mathbb Z$ localized at the complement of $(2)$)
2. Quotients of a PID by a power (greater than 1) of a maximal ideal (e.g. $\mathbb Z/(4)$)

To make it finite, you just choose finite rings. (Like something in the second point above.)

Just so my answer suggests a finite local ring which hasn't already been suggested: $F_2[x,y]/(x,y)^2$ is a finite local ring, where $F_2$ is the field of two elements.