most recent 30 from http://mathoverflow.net 2013-05-21T13:30:55Z http://mathoverflow.net/feeds/question/109463 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109463/representation-over-finite-field-and-field-extension TOM 2012-10-12T15:28:52Z 2012-10-12T16:00:43Z

<p>Let G be a group, p a prime number , $\mathbb{F}_p$ finite field. Suppose we have a representation $\rho_6$ of G into $GL_n(\mathbb{F}_{p^6})$ , a representation $\rho_2$ of G into $GL_n(\mathbb{F}_{p^2})$, a representation $\rho_3$ of G into $GL_n(\mathbb{F}_{p^3})$, Do we have the following statement ? : If $\rho_6=\rho_2\otimes \mathbb{F}_6$, $\rho_6=\rho_3\otimes \mathbb{F}_6$ , then there exist a representaion $\rho_1$ of G into $GL_n(\mathbb{F}_p)$, such that $\rho_6=\rho_1\otimes\mathbb{F}_{p^6}$.</p> <p>is this statment for any coprime number m,n replace of 2,3 , and mn replace of 6 true? Thank you!</p>

http://mathoverflow.net/questions/109463/representation-over-finite-field-and-field-extension/109466#109466 Joël 2012-10-12T16:00:43Z 2012-10-12T16:00:43Z

<p>The answer is yes if $\rho_6$ is absolutely irreducible. Here is a proof when $p>n$ using pseudo-characters. The trace of $\rho_j$, for $j=2,3,6$, is a pseudo-character $T_j$ of $G$ to $\mathbb F_{p^j}$ of dimension $n$. For $g$ in $G$, we have $T_6(g)=T_2(g)=T_3(g) \in \mathbb F_{p^2} \cap \mathbb F_{p^3} = \mathbb F_p$. Hence $T_6$ is a pseudo-character absolutely irreducible of $G$ to $\mathbb F_p$ and such a pseudo-character over a finite field always comse from a representation $\rho_1$, which clearly satisfies your condition since two absolutely irreducible representations with the same trace are isomorphic over <em>any</em> field.</p> <p>Of course the same argument work when 2,3 are replaced by any relatively prime integers. Also I have supposed $p>n$ just for the convenience of using the classical theory of pseudo-characters, which works well under this condition, but in the general case Chenevier has a general theory (that he calls "determinant" -- see on his webpage at polytechnique) which works without this condition, and with which the proof above can be adapted without difficulties.</p> <p>On the other hand, the hypothesis absolutely irreducible is necessary for the method. I don't know if it is necessary for your question...</p>