Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 leads me to believe I've made a mistake somewhere. If so, I apologize.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.