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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.

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up vote 4 down vote accepted

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.

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Thanks. I think your proof is correct. Please wait for a few days before I accept it. Regards, –  Makoto Kato Aug 20 '12 at 3:59

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