Decomposition of finite algebras over finite fields - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T22:26:19Zhttp://mathoverflow.net/feeds/question/61533http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/61533/decomposition-of-finite-algebras-over-finite-fieldsDecomposition of finite algebras over finite fieldsFranz Lemmermeyer2011-04-13T10:51:56Z2011-04-20T17:19:53Z
<p>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. </p>
<p>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:</p>
<p><b>Is there a simple proof that $A_p$ is a direct sum of finite fields and some easily described nil rings?</b></p>
<p>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.</p>
http://mathoverflow.net/questions/61533/decomposition-of-finite-algebras-over-finite-fields/61534#61534Answer by Charles Matthews for Decomposition of finite algebras over finite fieldsCharles Matthews2011-04-13T11:04:52Z2011-04-13T11:04:52Z<p>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.</p>
http://mathoverflow.net/questions/61533/decomposition-of-finite-algebras-over-finite-fields/62437#62437Answer by David Tweedle for Decomposition of finite algebras over finite fieldsDavid Tweedle2011-04-20T17:19:53Z2011-04-20T17:19:53Z<p>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.</p>
<p>A more general situation is:</p>
<p>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.</p>