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 $\rho_6=\rho_1\otimes\mathbb{F}_{p^6}$.

is this statment for any coprime number m,n replace of 2,3 , and mn replace of 6 true? Thank you!

  • 1
    $\begingroup$ The isomorphism $\rho_6=\rho_2 \otimes \mathbb F_{p^6}$ gives an action of $Frob_{p^2}$ on the representation space of $\rho_6$ that commutes with the $G$ action, say $F_2$. Similarly, the isomorphism $\rho_6=\rho_3 \otimes \mathbb F_{p^6}$ gives an action of $Frob_{p^3}$ on the representation space, say $F_3$. If $F_2$ and $F_3$ commute, we are done. We simply take $F_3 F_2^{-1}$, which is an action of $Frob_p$ that commutes with the $G$ action. The action of $G$ on the fixed subspace is the desired $\rho_1$, by Hilbert $90$. But there's no reason they should commute in general. $\endgroup$
    – Will Sawin
    Oct 12, 2012 at 17:08

1 Answer 1


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 any field.

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.

On the other hand, the hypothesis absolutely irreducible is necessary for the method. I don't know if it is necessary for your question...

  • $\begingroup$ Do you mean that if $\rho_6$ is not absolutely irreducible then the stament may be false? $\endgroup$
    – TOM
    Oct 12, 2012 at 16:57
  • $\begingroup$ I mean that I don't know it it is true. Perhaps someone will come up with a better proof that mine which will cover all cases. $\endgroup$
    – Joël
    Oct 12, 2012 at 17:13

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