Decomposition of finite algebras over finite fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:26:19Z http://mathoverflow.net/feeds/question/61533 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61533/decomposition-of-finite-algebras-over-finite-fields Decomposition of finite algebras over finite fields Franz Lemmermeyer 2011-04-13T10:51:56Z 2011-04-20T17:19:53Z <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#61534 Answer by Charles Matthews for Decomposition of finite algebras over finite fields Charles Matthews 2011-04-13T11:04:52Z 2011-04-13T11:04:52Z <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#62437 Answer by David Tweedle for Decomposition of finite algebras over finite fields David Tweedle 2011-04-20T17:19:53Z 2011-04-20T17:19:53Z <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>