# Examples of finite commutative rings that are not fields

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.

• – lhf Oct 9 '18 at 13:08
• Here is the DaRT search result set . At present, it's just several quotients of $\mathbb Z$ and then a finite quotient of $F_2[x,y]$. – rschwieb Oct 9 '18 at 13:10
• $\mathbf Z/p^k\mathbf Z$, $p$ prime, $k>1$. – Bernard Oct 9 '18 at 13:21
• And of course, any number of combinations of finite products of rings of the types anyone mentions will be another example. – rschwieb Oct 9 '18 at 15:28

## 5 Answers

$$\mathbb Z_4$$, 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.

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

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

• Would that be a commutative ring? Multiplication is not commutative on matrices. – Madhav Nakar Oct 9 '18 at 13:06
• You are right. I've edited my answer. – José Carlos Santos Oct 9 '18 at 13:08

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 $$f,g\in R^X$$.

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.