Examples of finite commutative rings that are not fields

Question

Would someone be able to provide examples of finite commutative rings that are not fields? I attempted to create commutative rings using matrices (ie permutation matrices) and integers (ie $\mathbb{Z}_p$), however, either they were not rings or they were fields.

Answer 1

$\mathbb Z_4$, why not? Beginning students may think that the field of $4$ element is this, but it is not, so this is a good example to talk about.

A finite commutative ring with no zero divisors is a field, so we have to look for zero divisors to get an example that you ask for.

Answer 2

Take $A \times B$, where $A$ and $B$ are finite commutative rings, even fields.

Answer 3

Here is a rough classification. Any finite commutative ring is in particular Artinian, and commutative Artinian rings $R$ are finite products $\prod_m R_m$ of Artinian local rings; more specifically they have finitely many maximal ideals $m$ and are the products of the localizations at each maximal ideal.

So we can reduce to the local case. If $R$ is a finite commutative local ring with maximal ideal $m$ then the quotient $R/m$ is a finite field $\mathbb{F}_q$, and $m$ is nilpotent. So, roughly speaking, finite commutative local rings look like finite fields $\mathbb{F}_q$ with some extra "nilpotent fuzz" attached. The two easiest classes of examples to describe here are $\mathbb{Z}/p^k\mathbb{Z}$ and $\mathbb{F}_p[t]/t^k$, both of which are $\mathbb{F}_p$ with some extra "nilpotent fuzz" attached. One can contemplate more complicated quotients of this sort, e.g. $\mathcal{O}_K/P^k$ where $\mathcal{O}_K$ is the ring of integers of a number field and $P$ is a prime ideal in it, which is $\mathcal{O}_K/P$ with some extra "nilpotent fuzz" attached.

Among other things, it follows that the smallest examples of finite commutative rings that are not fields have order $4$, and include $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{F}_2[t]/t^2$, as well as the simpler $\mathbb{F}_2^2$. I believe these are the only examples of order $4$.

Answer 4

What about the $\mathbb{Z}_2\times\mathbb{Z}_2$?

Answer 5

Let $X$ be a nonempty finite set and $R$ be a finite ring. Take the set $R^X$ of all mappings $f:X\rightarrow R$. This set becomes a ring with $(f+g)(x) = f(x)+g(x)$ and $(f\cdot g)(x) = f(x)\cdot g(x)$ for all $x\in X$ and $f,g\in R^X$.

Answer 6

Take any finite field $F$ and a finite dimentional $F$-vector space $V$; then the set $F\times V$ with addition $$ (a,x)+(b,y)=(a+b,x+y) $$ and multiplication $$ (a,x)(b,y)=(ab,ay+bx) $$ is easily seen to be a commutative finite ring. Taking as $F$ the field with $2048$ elements and $\dim V=1003$, I guess this is far enough from examples of type $\mathbb{Z}_m$ or variants thereof.

Answer 7

$\mathbb{Z}_n$ is a commutative ring with unity for all n. See this. But it is a field if and only if n is prime. See this different question. Therefore, $\mathbb{Z}_n$ for n composite is a commutative ring with unity but is not a field. The simplest example is $\mathbb{Z}_4$, as per GEdgar.

My abstract algebra book, Pinter page 172, says "If A is a commutative ring with unity in which every nonzero element is invertible, A is called a field." In $\mathbb{Z}_4$, 2 is not invertible.