# 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 $p^{n}$?

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No. Consider $\mathbb{F}_p[\epsilon]/\langle\epsilon^4\rangle$ and $\mathbb{F}_p[\epsilon_1,\epsilon_2]/\langle \epsilon_1^2, \epsilon_2^2\rangle$. –  Kevin Ventullo Oct 19 '12 at 16:52
See mathoverflow.net/questions/98883/finite-local-rings for more info around this. Vote to close as almost duplicate and/or too localized. –  quid Oct 19 '12 at 16:54
A still 'smaller' example could be Z/(p^2) and F_p [t]/(t^2). –  quid Oct 19 '12 at 16:57

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.
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.