Base change by a finite extension Let $L/K$ be a finite extension of fields
Now let $T$ and $X$ be $L$-varieties (we can take $T$ affine if you want) I would like to know if it is true that
$$
Hom_L(T\times_K L,X) = Hom_L(T,X)^n
$$
It is quite obvious if $T$ and $X$ are affine indeed in this case is $T = Spec(A)$ and $X = Spec(L[X_1,\ldots,X_d]/(f_1,\ldots,f_r))$ then, since $L \cong K^n$ we have $T = Spec(A \otimes_K L) \cong Spec(A^n)$ and so :
$$\begin{align}
Hom_L(T\times_K L,X) & = Hom_L(L[X_1,\ldots,X_d]/(f_1,\ldots,f_r),A^n) \\
                     & = \{ (x_1,\ldots,x_d) \in (A^n)^d \;|\; f_i(x_1,\ldots,x_d) = 0 \forall i \in \{ 1,\ldots,r \} \} \\
                     & = \{ (x_1,\ldots,x_d) \in A^d \;|\; f_i(x_1,\ldots,x_d) = 0 \forall i \in \{ 1,\ldots,r \} \}^n \\
                     & = Hom_L(L[X_1,\ldots,X_d]/(f_1,\ldots,f_r),A)^n \\
                     & = Hom_L(T,X)^n
\end{align}
$$
I'm wondering (assuming that what I just wrote is correct) how can one (if one can) generalize this for arbitrary $X$ ?
**Edit : ** beware there seems to be something wrong in the affine argument, i'm still wondering if the result is true though.
 A: This is not an answer, but I am not eligible to comment yet.
It seems that there is problem in the proposed argument in the affine case.
Let {$e_1,...,e_n$} be a basis of $L$ over $K$.
Then, an element in $Hom_L(L[X_1,...,X_d]/(f_1,...,f_r),A\otimes _K L)$ should be given by elements
$$ \Sigma_{i} (x_{i1}e_i),...,\Sigma_{i} (x_{id}e_d)$$
where $x_{ij}\in A$, such that $f_k(\Sigma_{i} (x_{i1}e_i),...,\Sigma_{i} (x_{id}e_d))=0$
But note that the $L$-algebra structure on $A\otimes_K L$ acts on the right (i.e., on the elements $e_i$). Thus it seems that we cannot single out each tuple  $(x_{11},...,x_{1d}),...,(x_{n1},...,x_{nd})$ and apply $f_k$ to them them separately.
A: If $L/K$ is a Galois extension, then $L\otimes_K L =L[G]$ where $G$ is the Galois group of $L/K$. As $L[G]\simeq L^n$ as $L$-algebra, we have 
$T\times_K L=T\times_L (L\otimes_K L) \simeq T\times_L (L^n)$ is $n$ disjoint copies of $T$. So $\mathrm{Hom}_L(T\times_K L, X)$ is the product of $n$ copies of $\mathrm{Hom}_L(T, X)$. 
In general, the set of morphisms $\mathrm{Hom}_L(T\times_K L, X)$ is canonically identified with $\mathrm{Hom}_K(T, \mathrm{Res}_{L/K}(X))$, where $\mathrm{Res}_{L/K}(X)$ is the Weil restriction of $X$ from $L$ to $K$ (it exists if $X$ is quasi-projective over $L$ for example). If your "equality" holds canonically, then as 
$$\mathrm{Hom}_L(T, \mathrm{Res}_{L/K}(X)\times_K L)=\mathrm{Hom}_K(T, \mathrm{Res}_{L/K}(X)),$$
this implies that $\mathrm{Res}_{L/K}(X)\times_K L$ is isomorphic to $X^n$ over $L$. 
But this is not truet in general (unless $X$ is an affine space). You can take $X=\mathbb G_{m, L}$ and consider a purely inseparable (or separable but non Galois) extension $L/K$. In the purely inseparable case, $\mathrm{Res}_{L/K}(X)$ has a non-trivial unipotent subgroup, so it can not become multiplicative after extension. In the separable case, $\mathrm{Res}_{L/K}(X)$ is a torus over $K$, it becomes split only over a Galois extension containing $L$, while $X^n$ is already split over $L$. 
