Krdytkyu

Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $Votimes_k Kcong Wotimes_k K$ as $K$-representations, do we have that $Vcong W$?

Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?

In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?

Any reference for these questions would be much appreciated.

In the case of finite group with semisimple group algebra (i.e. $mathrm{char}, k = 0$ or $mathrm{char}, k$ is coprime to $|G|$), the claim should be true, i.e. $Votimes_k K simeq Wotimes_k K$ implies $V simeq W$.

In the case $mathrm{char}, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $langle chi_i, chi rangle$ where $chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $chi_i$ runs over irreducible chracters, as in the case $k=mathbb{C}$.

Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $mathrm{char}, k$.

However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $rho(g)$ when $rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)

In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then $Mcong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).

For finite field extensions the proof is very short. If $Motimes_kKcong Notimes_kK$ as $Aotimes_kK$-modules, then, restricting to $A$, $Motimes_kKcong Notimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $Motimes_kKcong M^n$, the direct sum of $n$ copies of $M$, and $Notimes_kKcong N^n$. So $M^ncong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$Mcong N$.