Finite commutative local ring 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 $p^{n}$?
 A: 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.
A: Another type of example is provided by group algebras of finite Abelian $p$-groups over fields of characteristic $p.$ If $G$ is a finite Abelian $p$-group and $F$ is a field of characteristic $p$, then $FG$ is a commutative local ring. Now, for example, let $G$ be an elementary Abelian $p$-group of order $p^{2}$ and $H$ be a cyclic group of order $p^{2}.$
Take $F$ to be the field of $p$ elements. Then $FG$ and $FH$ are both commutative local rings with $p^{p^{2}}$ elements. They are not isomorphic, because $FH$ contains a unit of multiplicative order $p^{2},$ but $FG$ does not.
