# Decomposition of finite algebras over finite fields

Let $K$ be a number field, $Z_K$ its ring of integers, and $p$ a rational prime number. Then $A_p = Z_K/(p)$ is a finite ${\mathbb F}_p$-algebra. Using ideal arithmetic in $Z_K$ and the Chinese remainder theorem it is easily checked that $A_p$ is the direct sum of finite fields if $p$ is unramified, and has additonal nil-rings as components if $p$ is ramified.

This decomposition result looks simple enough to have a direct and not too complicated proof. Basing it on the close relation between $A_p$ and ${\mathbb F}_p[X]/(f)$, where $f$ is the minimal polynomial of a generator of $K$, brings in problems with primes dividing the discriminant of $f$. Thus let me state my main question explicitly:

Is there a simple proof that $A_p$ is a direct sum of finite fields and some easily described nil rings?

In addition, I'd be grateful for pointers to the relevant literature, in particular to classification theorems of which the result above is a special case.

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## 2 Answers

The general form of the answer is easy to anticipate: a finite commutative ring is artinian. It will be a product of finite local rings, of characteristic that is a prime number. So your question is really which local rings you get in this case. I believe a certain amount can be done by linear algebra.

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A good starting point, at least from a number theory perspective, is the book "Local fields" by J.P. Serre. Chapter 1, Section 5 deals with the question you ask. It says that if $$pZ_K = \prod_{\mathfrak{p}|p} \mathfrak{p}^{e_\mathfrak{p}},$$ then $$A_p \cong \prod_{\mathfrak{p}|p} Z_K/\mathfrak{p}^{e_\mathfrak{p}}.$$ The nil-rings are exactly described by the prime ideals $\mathfrak{p}$ of $Z_K$ which have $e_\mathfrak{p}>1$. I think that this book is very useful, but I'm sure that this topic is also dealt with in algebraic number theory textbooks, or standard algebra textbooks.

A more general situation is:

If we assume that $L,K$ are fields with $[L:K]=n$, $L/K$ separable, $A\subset K$, with fraction field equal to $K$, and $A$ is a Dedekind domain. Further assume that the integral closure of $A$ in $L$, say $B\subset L$, is also a Dedekind domain. Then the proof of the fact you want relies on the factorization of $pB$ as a product of prime ideals in $B$, and the Chinese remainder theorem or "Approximation Lemma" (in Serre's book). For a better explanation, read Serre's book.

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