For a finite local ring $R$, why $|R|$ is a prime power.

Question

When I read about finite local rings, every one firstly suppose that $|R|=p^t$ where $R$ is a finite local ring and $p$ is a prime. I don't understand why this can be supposed ?

Answer 1

There is a unique maximal ideal $M$ in $R$. Then $k=R/M$ is a field and so has prime power order. The ideal $M$ is nilpotent. Each $M^i/M^{i+1}$ is a finite-dimensional vector space; its order is a power of $|k|$. As $M^n=0$ for some $n$, this proves that the order of $R$ is a power of $|k|$.

Answer 2

Let $R$ be a local ring with maximal ideal $m$.

$R/m$ is a field $K$ of cardinality $p^k$ for some prime $p$ and integer $k$.

Thus $p^k \cdot 1 \in m$. Moreover, since $m$ is also the only prime ideal of $R$ (for finite rings, maximal ideals and prime ideals coincide - see my commet below), it must be the nilradical, and so $p^k\cdot 1$ must be nilpotent, so $(p^k\cdot 1)^n= 0$ for some $n$.

But $(p^k\cdot 1)^n= p^{kn}\cdot 1$. Therefore, the exponent of the group $(R,+)$ is a prime power (it divides $p^{kn}$).

It's well known that the order of an abelian group divides a power of its exponent, and therefore the order of $R$ must be a prime power as well.

Answer 3

Here's a different argument, with some similarities to Max's, which may appeal to those who are unfamiliar with Artinian rings.

The characteristic of $R$ is a positive integer $m$. If $m$ is a prime power, then $R$ is an $m$-torsion abelian group whose order divides some power of $m$: $|R|$ is a prime power.

Otherwise $m=st$ with $s$, $t>1$ and coprime. Then $R=sR+tR$. The ideals $sR$ and $tR$ of $R$ are proper and add to $R$. They cannot both be contained in the maximal ideal of $R$: contradiction.