Question

Is there a classification of finite commutative rings available? If not, what are the best structure theorem that are known at present? All I know is a result that every finite commutative ring is a direct product of local commutative rings (this is correct, right?) in some paper which computes the size of the general linear group over that ring.

Answer 1

Yes, a finite ring $R$ is a finite direct sum of local finite rings. As a first step, for each prime $p$ there is a subring $R_p$ of $R$ corresponding to the elements annihilated by the powers of $p$. $R_p$ is then an algebra over $\mathbb{Z}/p$. $R_p$ then resembles an algebra over $\mathbb{Z}/p$ and it could be one, but it can also have a more complicated structure as an abelian $p$-group (see below). This step generalizes to maximal ideals: For each maximal ideal $m$, $R_m$ is the subring of elements annihilated by $m^n$ for some $n$, and $R$ is the direct sum of these subrings, which are local rings.

It is not difficult to write down a rough partial classification of of local finite rings. If $R$ is local with maximal ideal $m$, it is resembles an algebra over the finite field $F = R/m$; the associated graded ring is such an algebra. If you choose a basis $x_1,\ldots,x_n$ for $m/m^2$, then $R$ or its associated graded is a quotient of the polynomial ring $F[\vec{x}]$ in which only finitely many monomials are non-zero. You can make a diagram of these non-zero monomials; they can be any order ideal in the $n$-dimensional orthant. Or, in basis-independent form, $R$ has a length, which is the largest nonvanishing power of $m$, and each $m^k/m^{k+1}$ is some quotient of the $k$th symmetric power of the generating vector space $V = m/m^2$.

After that, the non-zero monomials may be linearly dependent (and never mind that $R$ might be more complicated than its associated graded). Informally, there will be an endless stream of partial results and there will never be a complete classification when the length of the local ring is 3 or more. To see this, suppose that $m^4 = 0$, and suppose that $m^3$ is only one dimension shy of $S^3(V)$. Then the ring is defined by an arbitrary symmetric trilinear form in $V$. These make a "wild" sequence of algebraic varieties, in the same sense that people say that the representation theories of certain rings are wild. For instance, I think (not quite sure) that it is NP-hard to determine when two such trilinear forms are equivalent. NP-hardness is not by itself rigorously equivalent to no classification, but informally the classification is an intractable mess.

If the nonvanishing monomials in $R$ are linearly independent, then it is a toric local ring. Toric local rings are certainly a tractable class of finite rings.

The situation is similar to non-commutative $p$-groups, which are also wild and will never be classified. In both cases, certain classes have a nice structure. It is also interesting to make estimates for how many there are.

Note: Corrected per comment.

Answer 2

As always one should check out the OEIS for questions of this type. In this case see http://oeis.org/A027623

Answer 3

This is a very interesting question related to the Hilbert scheme $Hilb^n(\mathbb A^d)$ classifying $n$ points in affine space $\mathbb A^d$. I don't think there is a classification but there is an estimate for the number of commutative rings of order $\leq N$. It is

$$exp[\frac{2}{27} \frac{log(N)^3}{(log 2)^2} \; +O(log(N)^{\frac {8}{3}})] \quad for N\to \infty $$

The proof of this result due to Bjorn Poonen and of many related interesting theorems is in his article

You will also find astonishing conjectures in the article like:

The fraction of local rings of order $\leq N$ among all commutative rings $A$ of order $\leq N$ tends to 1. Same limit 1 for the fraction of rings "of characteristic 8" in the sense that $8 . 1_A =0$ but $4 .1 _A \neq 0$.

Answer 4

The characterization of Artinian rings is relevant of course. See also the book "Finite commutative rings and their applications" and this web page.

Answer 5

Assuming that by finite you mean "the underlying set is finite", then the very first thought that comes to mind is that the underlying additive group of your ring has to be finite. I would hope that you are then able to look up or work out what one might be able to say about finite abelian groups, and presumably that might speed you on your way.

Maybe I've misunderstood your terminology?