Wedderburn's theorem for $\mathbb{Q}G$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:05:41Z http://mathoverflow.net/feeds/question/119658 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119658/wedderburns-theorem-for-mathbbqg Wedderburn's theorem for $\mathbb{Q}G$ Diego Sulca 2013-01-23T14:01:42Z 2013-01-24T13:37:14Z <p>Let $G$ be a finite group and let $\mathbb{Q}G=M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)$ be the decomposition of $\mathbb{Q}G$ as a product of rings of matrices over divisions rings. Let $Z_i$ be the center of $D_i$. Then each $Z_i$ is an algebraic extension of $\mathbb{Q}$. The question is: how much is it known about these algebraic number fields?. More precisely, for which family of groups $G$ it is known that each $Z_i$ is a Galois extension of $\mathbb{Q}$ and in such a case what can we say about the corresponding Galois groups?.</p> <p>Any comments and references will be strongly appreciated. </p> http://mathoverflow.net/questions/119658/wedderburns-theorem-for-mathbbqg/119670#119670 Answer by Aakumadula for Wedderburn's theorem for $\mathbb{Q}G$ Aakumadula 2013-01-23T15:54:44Z 2013-01-24T13:37:14Z <p>The short answer is yes, the centres $Z_i$ are contained in the cyclotomic extension $E_n= {\mathbb Q}(e^{(2\pi i)/n})$ where $n$ is the order of the group. As pranavk says, we need only prove that the simple factors of the centre of ${\mathbb Q}[G]$ are contained on $E_n$. But the centre is spanned by the averages $C_x =\sum gxg^{-1}$ where the sum is over all the elements of $G$. Given an absolutely irreducible representation $V$ of $G$ of dimension $r$ , each $C_x$ acts by a scalar $\lambda$ say. By taking traces, you see that $\lambda$ is a rational multiple of the trace of $x$; since $x$ has order dividing $n$, all its eigenvalues in $V$ are $n$-th roots of unity, and hence the trace of $x$ lies in $E_n$. </p> <p>Now,in your notation, each $M_{n_i}(D_i)$ after tensoring (over $Z_i$) with the algebraic closure of ${\mathbb Q}$ is of the form $End (V)$ for some absolutely irreducible representation $V$ of $G$, hence each $Z_i$ (which as pranavk observed, lies in the centre of ${\mathbb Q}[G]$) lies in $E_n$. </p> <p>I believe all this is worked out in Serre's book on finite groups (Springer notes) and is well known to experts (I am not one!). </p> http://mathoverflow.net/questions/119658/wedderburns-theorem-for-mathbbqg/119672#119672 Answer by Yazdegerd III for Wedderburn's theorem for $\mathbb{Q}G$ Yazdegerd III 2013-01-23T16:10:25Z 2013-01-23T16:18:40Z <p>The simplest case of your question is the case where $G$ is the cyclic group of order $n$, it is known that $\mathbb{Q}G\simeq\mathbb{Q}[X]/(X^n-1)$. As $X^n-1=\Pi_{d|n}\Phi_d(X)$ where $\Phi_d(X)$ is the $d$-th cyclotomic polynomial, it follows that $\mathbb{Q}G\simeq\Pi_{d|n}\mathbb{Q}(\zeta_d)$ where $\zeta_d\in\mathbb{C}$ is a primitive $d$-th root of unity.</p> <p>Edit: This is also a very particular but concrete case of the idea of @Aakumadula.</p>