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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.

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  • $\begingroup$ See also mathoverflow.net/questions/7133/… $\endgroup$ – lhf Oct 9 '18 at 13:08
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    $\begingroup$ Here is the DaRT search result set . At present, it's just several quotients of $\mathbb Z$ and then a finite quotient of $F_2[x,y]$. $\endgroup$ – rschwieb Oct 9 '18 at 13:10
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    $\begingroup$ $\mathbf Z/p^k\mathbf Z$, $p$ prime, $k>1$. $\endgroup$ – Bernard Oct 9 '18 at 13:21
  • $\begingroup$ And of course, any number of combinations of finite products of rings of the types anyone mentions will be another example. $\endgroup$ – rschwieb Oct 9 '18 at 15:28
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$\mathbb Z_4$, 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.

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

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

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  • $\begingroup$ Would that be a commutative ring? Multiplication is not commutative on matrices. $\endgroup$ – Madhav Nakar Oct 9 '18 at 13:06
  • $\begingroup$ You are right. I've edited my answer. $\endgroup$ – José Carlos Santos Oct 9 '18 at 13:08
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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 $f,g\in R^X$.

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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.

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