Equivalence of group representations under base change Let $G$ be a group and $\rho_{1},\rho_{2}:G\rightarrow \operatorname{GL}(V)$ be two representations of $G$ over a field $k$. Let $K/k$ be a field extension. Suppose that $\rho_{1}\otimes_{k}K$ is equivalent to $\rho_{2}\otimes_{k}K$. Is it true that $\rho_{1}$ and $\rho_{2}$ are equivalent?
I know the answer is positive if $k$ is an infinite field. What if the case when $k$ is a finite field?
Here is a proof of the case when $k$ is an infinite field: Consider the matrix representations. By assumption, there is an $A\in \operatorname{GL}_{n}\left(K\right)$ such that $\rho_{1}\left(g\right)A=A\rho_{2}\left(g\right)$ for all $g\in G$. Then, write
$$
A=e_{1}A_{1}+\cdots +e_{m}A_{m}
$$
for some $A_{i}\in M_{n}\left(k\right)$ and $\left\{e_{i}\right\}$ is $k$-linear independent subset of $K$. By the linearly independence of $e_{i}$, we have $\rho_{1}\left(g\right)A_{i}=A_{i}\rho_{2}\left(g\right)$ for all $g\in G$ and $i$. Then, consider the polynomial
$$
f\left(x_{1},\ldots,x_{m}\right)=\det\left(x_{1}A_{1}+\cdots +x_{m}A_{m}\right)\in k\left[x_{1},\ldots,x_{m}\right].
$$
Since $k$ is infinite, we may choose some non-zero $\left(a_{1},\ldots,a_{m}\right)\in k^{m}$ such that $f\left(a_{1},\ldots,a_{m}\right)\neq 0$. Then, set
$$
B=a_{1}A_{1}+\cdots +a_{m}A_{m}\in\operatorname{GL}\left(k\right).
$$
Then, $\rho_{1}\left(g\right)B=B\rho_{2}\left(g\right)$ for all $g\in G$ and thus $\rho_{1}$ is equivalent to $\rho_{2}$.
Thanks in advanced.
 A: As commented by Jyrki Lahtonen, I will post the completed answer as follows.
Write $n=\dim_{k}V$. We first deal with the case when $k$ is a field with at least $n$ elements. In the following, we consider the matrix representations. By assumption, there is an $A\in \operatorname{GL}_{n}\left(K\right)$ such that $\rho_{1}\left(g\right)A=A\rho_{2}\left(g\right)$ for all $g\in G$. Then, write
$$
A=e_{1}A_{1}+\cdots +e_{m}A_{m}
$$
for some $A_{i}\in M_{n}\left(k\right)$ and $\left\{e_{i}\right\}$ is $k$-linear independent subset of $K$. By the linearly independence of $e_{i}$, we have $\rho_{1}\left(g\right)A_{i}=A_{i}\rho_{2}\left(g\right)$ for all $g\in G$ and $i$. Then, consider the polynomial
$$
f\left(x_{1},\ldots,x_{m}\right)=\det\left(x_{1}A_{1}+\cdots +x_{m}A_{m}\right)\in k\left[x_{1},\ldots,x_{m}\right].
$$
Since $k$ has at least $n$ distinct elements, we may choose some non-zero $\left(a_{1},\ldots,a_{m}\right)\in k^{m}$ such that $f\left(a_{1},\ldots,a_{m}\right)\neq 0$. Then, set
$$
B=a_{1}A_{1}+\cdots +a_{m}A_{m}\in\operatorname{GL}\left(k\right).
$$
Then, $\rho_{1}\left(g\right)B=B\rho_{2}\left(g\right)$ for all $g\in G$ and thus $\rho_{1}$ is equivalent to $\rho_{2}$.
Next, if $k$ does not have $n$ elements, then consider a finite extension $E$ of the finite field $k$ such that $E$ has at least $n$ elements. Then, the same arguments implies $\rho_{1}\otimes_{k}E$ is equivalent to $\rho_{2}\otimes_{k}E$ as a representations over the field $E$. Now, we may assume $K/k$ is a finite extension. Write $\left[K:k\right]=d$ and $\left\{e_{i}\right\}_{i=1}^{d}$ be a $k$-basis of $K$. Then, as an $k\left[G\right]$-modules, we have
$$
\bigoplus_{i=1}^{d}V_{1}\otimes e_{i}\cong V_{1}\otimes_{k}K\cong V_{2}\otimes_{k}K\cong \bigoplus_{i=1}^{d}V_{2}\otimes e_{i}.
$$
Now, we decompose $V_{1}+\cdots+V_{1}$ ($d$ many copies) and $V_{2}+\cdots+V_{2}$ into indecomposable $k\left[G\right]$-modules. By the Kurll-Schmidt theorem, counting the indecomposable summands on both sides, we conclude that $V_{1}\cong V_{2}$ as $k\left[G\right]$-modules.
