The character of a representation of a finite group when changing the underlying field Recently, I am reading the text "Linear Representations of Finite Groups" by J. P. Serre.
In the Lemma $12$ of Chapter $12$:
Let $K$ be a field of characteristic zero, and $G$ be a finite group. Let $L$ be a finite extension of $K$ containing all the desired eigenvalues occurring in any representation of $G$ over $K$ (Indeed, it can be chosen to be some cyclotomic extension of $K$). Suppose that $(V,\rho)$ is a (finite dimensional) representation of $G$ over $L$ with character $\chi$. Then via restricting scalars one can obtain a representation of $G$ over $K$, say $(V_K,\rho_K)$.
Then the author claims that $\text{Tr}_{L/K}(\chi)$ is the character of $(V_K,\rho_K)$, where $\text{Tr}_{L/K}$ denotes the trace associated with the field extension $L/K$.
Suppose the extension degree of $L/K$ is $d$, and the dimension of $V$ over $L$ is $\text{ }n$. I try to write down $\{u_i e_j\}_{1\leq i\leq d, 1\leq j\leq n}$ as a $K$ basis of $V$, where $\{u_i\}_{1\leq i\leq d}$ is a $K$ basis of $L$, and $\{e_j\}_{1\leq j\leq n}$
a $L$ basis of $V$. But I fail to obtain the result in the claim above.
Maybe I misunderstand the construction of $(V_K,\rho_K)$, so I also wonder the concrete construction of $(V_K,\rho_K)$.
I'd appreciate all your help!
 A: Let $\{u_s\}_{1\leq s\leq d}$ be a $K$ basis of $L$, and $\{e_t\}_{1\leq t\leq 
 n}$ be a $L$ basis of $V$. Suppose that
$\rho(g)(e_j)=\sum_i\rho_{i,j}(g)e_i$,
where the coefficients $\rho_{i,j}(g)$ are in $L$. Then using the basis $\{u_i\}_{1\leq i\leq d}$ of $L$, we write
$\rho_{i,j}(g)=\sum_{k}a_{ij}^k(g)u_k$,
where the coefficients $a_{ij}^k(g)$ are in $K$. Now we have
$\rho(g)(u_se_j)=u_s\rho(g)(e_j)=\sum_i\rho_{i,j}(g)u_se_i=\sum_i(\sum_{k}a_{ij}^k(g)u_ku_s)e_i$,
Let $\varphi$ be the character of $\rho_K$. Then $\varphi(g)$ must be the sum of the coefficients of $u_s$ in the expansion of $\sum_{k}a_{jj}^k(g)u_ku_s$, but by the definition of $\text{Tr}_{L/K}$,
$\text{Tr}_{L/K}(a_{jj}^k(g)u_k)$ is exactly the sum of the coefficient $b_{jk}^s(g)$ of $u_s$  in the product of $u_sa_{jj}^k(g)u_k$, for all $1\leq s\leq d$. Consequently, we obtain
$\varphi(g)=\sum_{s,j,k}b_{jk}^s(g)=\sum_{j,k}\sum_s b_{jk}^s(g)=\sum_{j,k}\text{Tr}_{L/K}(a_{jj}^k(g)u_k)=\sum_{j}\text{Tr}_{L/K}(\sum_ka_{jj}^k(g)u_k)=\sum_j\text{Tr}_{L/K}(\rho(g)_{jj})=\text{Tr}_{L/K}(\chi(g))$.
