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When studying behaviour of linear representations of finite groups under extension of fields, I came up across two natural questions, which I couldn't solve (Reference: Representations of finite groups: Algebra and Arithmatic by S. Weintraub).

Let $R$ be a ring which is $F$-algebra ($F$ is a field), $E$ be an extension of $F$, set R'=E\otimes_F R. If $M$ is an $R$-module, them M'=E\otimes_F M is an R'-module.

Definition: If M' is an R' module that is isomorphic to $E\otimes_F M$ for some $R$-module $M$, then M' is said to be defined over $F$.

(If I have understood this definion correctly, then in terms of matrix representations of groups, it says: if $\rho\colon G\rightarrow GL(n,E)$ is a matrix representation of a finite group $G$, we say that it is defined over $F$ if there is $A\in GL(n,E)$ such that $A\rho(g)A^{-1}\in GL(n,F), \forall g\in G$.)

With this set up, an irreducible representation of $G$ over $F$ may decompose over $E$. Assume $G$ is finite group, and $char(F)>0$, $(char(F),|G|)=1$, $E$ is an extension of $F$.

Question 1: If $\rho_1, \rho_2, \cdots, \rho_r$ is the set of all inequivalent irreducible representations of $G$ over $F$, and if $\rho_i$ decomposes over $E$ as $\rho_{i1}\oplus \rho_{i2} \oplus\cdots \oplus \rho_{it_{i}}$ ($1\leq i\leq r$), is the set $\{ \rho_{ij} \colon 1\leq i\leq r, 1\leq j\leq t_i\}$ is complete set of irreducible representations of $G$ over $E$?

Question 2: If M' is an irreducible R' module, that is not defined over $R$, does some multiple of M', say dM' ($d\in \mathbb{N}$) is defined over $R$?


The second question, I tried, seems to have negative answer; I am not sure:

for example, consider a complex $1$-dimensional representation of $C_4=\langle x\colon x^4\rangle$ given by $x\mapsto [\sqrt -1]$. This representation is clearly not defined over $\mathbb{R}$. Some multiple of this representation (say $d$) will be the map $x\mapsto \sqrt-1 I_d$; still this representation (of dimension $d$) is not defined over $\mathbb{R}$, since this matrix $\sqrt-1 I_d$ is central, its conjugate in $GL(d,\mathbb{C})$ can not be in $GL(d,\mathbb{R})$.

In the paragraph before Prop: 4.17 in the book mentioned above (p.33), author may want to say positive answer to question 2, but I couldn't understood.

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    To add to Jyrki's comment: the only obstruction to qn 2 having a positive answer are the character values. The field generated by the character of $M'$ is the same as that generated by the character of $dM'$ for any nonzero $d$. If that field is not contained in $F$, then you have no chance. But if it is, then there is indeed such a $d$, and this is the theory of Schur indices. I echo Jyrki's recommendation of Curtis and Reiner, especially the 2nd vol. of their two volume work, the chapter on "Rationality Questions". The situation $({\rm char}F,|G|)=1$ is no different from ${\rm char}F=0$.2018-06-20

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