Bourbaki's proof of normal basis theorem Part 2 Let $K/k$ be a finite Galois extension of a field $k$, $G$ its Galois group.
The normal basis theorem states as follows.

There exists an element $\alpha$ of $K$ such that $\{\sigma(\alpha)\ |\ \sigma \in G\}$ is a $k$-linear basis of $K$.

This is equivalent to the assertion that the $k$-algebra $K$ is isomorphic to the group algebra $k[G]$ as $k[G]$-modules.
Bourbaki says that $K\otimes_k K$ is isomorphic to $K\otimes_k k[G]$ as $K\otimes k[G]$-modules(https://math.stackexchange.com/questions/1213617/bourbakis-proof-of-normal-basis-theorem-part-1).
To prove the normal basis theorem, they refer to the following proposition in Chapter 8 of the other volume which I don't have.

Let $A$ be an algebra over a field $k$. Let $M_1$ and $M_2$ be $A$-modules both of which are finite dimensional over $k$. Let $K/k$ be an extension of $k$. Suppose $K\otimes M_1$ and $K\otimes M_2$ are isomorphic as $K\otimes A$-modules. Then $M_1$ and $M_2$ are isomorphic as $A$-modules.

How do you prove this?
 A: You note that the Bourbaki volume you're missing is about semi-simple algebras over a field, so let's assume $A$ is semisimple.  Then $A$ is a product of matrix algebras over division rings and the simple left $A$-modules are exactly the simple left ideals one gets by taking all elements whose entries are nonzero in one particular column of one particular matrix factor.  The algebra $K \otimes A$ is the same product of matrix algebras but with the division rings tensored up.  The left ideals have the same description.
The upshot of all of this is that $K \otimes -$ then induces a bijection between the isomorphism classes of simple left $A$-modules and the isomorphism classes of simple left $K \otimes A$-modules.
So if $K \otimes M_1 \simeq K \otimes M_2$ then the decompositions of these modules into direct sums of simple modules agree and these simple modules are just the simple modules from $M_1$ and $M_2$ tensored up, so the decompositions of $M_1$ and $M_2$ agree.  Hence $M_1 \simeq M_2$.
Problem: The obvious problem here is that if $k$ has characteristic $p$ and $K/k$ has degree $p$ then $k[G]$ is not semisimple.  There are other proofs of the normal basis theorem that don't go the representation theory route, so the result certainly still holds in these situations.  Maybe Bourbaki wasn't considering this case (do they assume characteristic $0$?) or maybe the result about $A$ is true without assuming that $A$ is semisimple.  I'm not sure.
