Showing that a finite commutative ring with more than one element and no zero divisors has an identity. [duplicate]
Yes, you can prove it has an identity. Take any element $a\ne 0$, and define $f:R\to R$ by $f(x)=ax$. Since $a$ is not a zero divisor, $f$ is injective. Since $R$ is finite, we conclude $f$ is surjective as well. In other words, $f$ is a permutation on $R$. The group $S_R$ of permutations on $R$ is finite, and so the element $f$ has finite order. Thus, there is some $n$ such that the map $x\to a^nx$ (which is $n$th power of $f$) is the identity map. So $a^n$ is an identity element of $R$.
Note that all you need is one element $a$ which is not a zero divisor.