# The ring of finite algebras over Z/p^n

Here is a idea concerning the classification of finite rings (commutative, unital). Related question: Classification of finite commutative rings.

Every finite ring is a direct product of finite algebras over $\mathbb{Z}/p^n$ for some prime power $p^n$ (Chinese Remainder Theorem). Now fix such a prime power. If $R,S$ are finite algebras over $\mathbb{Z}/p^n$, then $R \otimes_{\mathbb{Z}/p^n} S$ and $R \times S$ are also finite algebras over $\mathbb{Z}/p^n$. If $\mathcal{A}$ is the set of isomorphism classes of finite algebras over $\mathbb{Z}/p^n$, then $\mathcal{A}$ becomes a commutative semiring with addition $\times$, zero element $0$, multiplication $\otimes$ and identity $\mathbb{Z}/p^n$. What is known about the associated ring $\overline{\mathcal{A}}$? Note that the map $\mathcal{A} \to \overline{\mathcal{A}}$ is not injective (I don't think that $\mathcal{A}$ is cancellative), but perhaps this loss of information makes it possible to give a desciription of this ring.

What about varying $n$? For every $k \leq n$ we may regard $\mathcal{A}_k$ as an ideal of $\mathcal{A}_n$, and $R \mapsto R/p^k$ is a projection $\mathcal{A}_n \to \mathcal{A}_k$. Does this help to describe $\mathcal{A}_n$ or $\overline{\mathcal{A}_n}$ recursively?

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What if p^n=2? Then this should be an odd ring structure generated by the finite Boolean algebras, about which more is known. – Greg Muller Sep 21 '10 at 13:12
But Boolean algebra is not the same thing as algebra over Z/2. – Tom Goodwillie Sep 21 '10 at 19:08

Not an answer, just a longish comment, I hope correct:

Let's start with the case $p^n=p$, so this is about finite algebras over a field. Let's replace $\mathbb Z/p$ by any field $k$ and ask the same question. The additive monoid of isomorphism classes of such algebras (under cartesian product) does have the cancellation property, because of the way that these rings decompose as products of local rings; it's the free abelian monoid on the set of iso classes of local finite algebras. Let's specialize to the case when $k$ is algebraically closed. Then the tensor product of local algebras is local, so that the semiring is determined by a tensor-product monoid of local algebras, and so is the associated ring.

If you go on to simplify the ring more by inverting these local algebras, you get the integral group ring of some abelian group: the group obtained by adjoining inverses to that monoid of local algebras. I wonder what this group is like. It maps onto the group of positive rational numbers by sending a local algebra to its vector space dimension.

Added: Come to think of it, there's a much bigger invariant than this: To a local algebra A with max ideal m, associate the polynomial $p_A(T)=1 + dim(m/m^2)T+dim(m^2/m^3)T^2+\dots$. This gives a homomorphism onto a much bigger group.

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Back to the original question with some thoughts: As a unital ring $\cal A_n$ maps onto $\cal A_{n-1}$ by the functor $\otimes_{\mathbb Z/p^n}\mathbb Z/p^{n-1}$. This map has as a left inverse a nonunital ring map, given by the forgetful functor. In fact, this is a unital splitting of $\cal A_n$ as product of $\cal A_{n-1}$ and another ring, call it $\cal \tilde A_n$, because the image of that map from $\cal A_{n-1}$ is the ideal generated by an idempotent in $\cal A_n$, the element $e_n$ determined by the algebra $\mathbb Z/p^{n-1}$. The ring $\cal \tilde A_n$ is the quotient $\cal A_n/e_n\cal A_n$. So what can you say about $\cal \tilde A_n$?

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There is a preprint by Manjul Bhargava with the title Mass formulae for algebras over $\mathbf{Z}_p$ and $\mathbf{F}_p$; I haven't seen it but I suspect that it might have a bearing on your question.

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