We know that, if $p$ is a prime number and $k$ is a natural number, then there is just one finite field $F$, such that $|F|=p^{k}$.


How about finite commutative local ring?


Let $R$ and $S$ be local rings with maximal ideal $m\_{1}$ and $m\_{2}$, respectively, such that $|R|=|S|$ and $|m\_{1}|=|m\_{2}|$. Is it true that $R\simeq S$?


In the other words, let $p$ be a prime number and $k,n$ ($k\geq n$) be natural numbers. Can we conclude that there exists just one finite commutative local ring of order $p^{k}$, with maximal ideal of order $p^{n}$?



Finite commutative local ring
No, the cardinalities alone are not enough to identify a local commutative ring. For example, you can take $R = \mathbb F\_p[x]/\langle x^3 \rangle$ and $S = \mathbb F\_p[x,y] / \langle x^2, xy, y^2\rangle$, which both have cardinality $p^3$ with maximal ideal of cardinality $p^2$.


A more refined invariant you could ask for is the cardinality of all powers of the maximal ideal instead of just the $0$th and $1$st powers, which is the same information as the Hilbert function. This will distinguish the above two rings because in the first $m\_R^2$ has cardinality $p$ but in the second $m\_S^2 = 0$ has cardinality $1$.


However, again, the Hilbert series alone is not enough to distinguish isomorphism classes of finite local rings. For example, $\mathbb F\_p[x,y]/ \langle x^2, y^2\rangle$ and $\mathbb F\_p[x,y] / \langle x^3, xy, y^2\rangle$ both have the same Hilbert function, but they are not isomorphic. In general, the problem of classifying finite local rings is impractical for large prime powers.


