Galois extension and Tensor product The following theorem is proved in Bourbaki's algebra.
They use the technique of Galois descent.
I'd like to know the proof without using it if any.
Theorem
Let $K$ be a field.
Let $\Omega/K$ be an extension.
Let $N/K$ and $L/K$ be subextensions of $\Omega/K$.
Suppose $N/K$ is a (not necessarily finite) Galois extension and $L \cap N = K$.
Then the canonical homomorphism $\psi:L\otimes_K N \rightarrow LN$
is an isomorphism.
 A: First assume $N/K$ is finite Galois. Then $LN/L$ (inside $\Omega$) is finite Galois and the natural map $\mathrm{Gal}(LN/L)\rightarrow\mathrm{Gal}(N/K)$ is injective with image $\mathrm{Gal}(N/N\cap L)=\mathrm{Gal}(N/K)$, i.e., it is an isomorphism. So, $[LN:L]=[N:K]=\dim_L(L\otimes_KN)$. The map $L\otimes_KN\rightarrow LN$ is surjective in any case because $LN/L$ is algebraic, and since both sides are $L$-vector spaces of the same (finite) dimension, the map is an isomorphism. 
In general, write $N=\bigcup_iN_i$ as a directed union of finite Galois extensions $N_i/K$. Then $L\otimes_KN=\varinjlim L\otimes_KN_i$, $LN=\bigcup_iLN_i$, and $L\otimes_KN\rightarrow LN$ can be identified with the direct limit of the isomorphisms $L\otimes_KN_i\cong LN_i$, and so is itself an isomorphism. 
To see that $LN=\bigcup_iLN_i$, first observe that the RHS is clearly contained in the LHS. Conversely, if $\alpha\in LN$, then $\alpha$ is a polynomial (with coefficients in $L$) in elements $\beta_1,\ldots,\beta_r\in N$. If $i$ is such that $\beta_j\in N_i$ for all $j$, then $\alpha\in LN_i$.
It worries me a bit that the sacred text (Bourbaki) uses something like Galois descent to prove this...it leads me to believe I've made a mistake somewhere. If so, I apologize. 
