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I was reading the fundamental theorem of Galois theory. Here's an excerpt.

Theorem. Let $E/F$ be a finite Galois extension, then $$ \varphi: K \mapsto Aut(E/K) $$ and $$ \psi: H \mapsto E^H$$ are bijections between the set of the intermediate fields and subgroups of $Gal(E/F)$.

Partial proof. We show $\psi \varphi$ is the identity map.

Let K be an intermediate field, then $E/K$ is a Galois extension, thus $ |Aut(E/K)| = [E:K] $. Let $B = E^{Aut(E/K)}$, then $B \supset K$. $E/B$ is also a Galois extension, and $$ [E:B] = |Aut(E/K)| = [E:K] $$ So $B = K$, that is $\psi \varphi $ is the identity map.

I'm not sure with the last argument. We know that $[B:K] = [E:K] / [E:B] = 1$. But does this always imply $B=K$?

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  • $\begingroup$ Yes (equal, or isomorphic, depends on the context). $\endgroup$ – Bernard Apr 16 '16 at 1:12
  • $\begingroup$ @Bernard isomorphic, of course. What about equality? $\endgroup$ – Qian Apr 16 '16 at 1:15
  • $\begingroup$ Wait a sec. Why are they isomorphic? I only know they're isomorphic as vector spaces. $\endgroup$ – Qian Apr 16 '16 at 1:17
  • $\begingroup$ $\psi\varphi$ is a field homomorphism because $\operatorname{Aut}(E/K)$ denotes the $K$-algebra automorphisms of $E$. $\endgroup$ – Bernard Apr 16 '16 at 1:29
  • $\begingroup$ @Bernard Can you elaborate more? I don't get it. $\endgroup$ – Qian Apr 16 '16 at 1:42
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Since $B$ contains $K$, it has the structure of a vector space over $K$. We know $K \subseteq B$, and we want to show that $B \subseteq K$.

The dimension of $B$ over $K$ is $1$, so there exists a basis of $B$ over $K$ consisting of a single element. In other words, there exists a $v \in B$ with the property that every element of $B$ can be written as $kv$ for some $k \in K$. In particular, $1 \in B$, so $1 = k_0v$ for some $k_0 \in K$. Hence $v$ is the inverse of $k_0 \in K$. Hence $v \in K$.

Now every element of $B$ is a product of two elements of $K$, hence every element of $B$ is in $K$.

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  • $\begingroup$ I tacitly used the fact that for fields $K \subseteq B$, $1_K = 1_B$ (and I just refer to that thing as $1$). But this is easy to see. $\endgroup$ – D_S Apr 16 '16 at 1:22
  • $\begingroup$ Thanks! Can I have an example where $v \neq 1$ can be a basis? $\endgroup$ – Qian Apr 16 '16 at 1:36
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    $\begingroup$ If $K$ is a field, then any nonzero element of $K$ is a basis for $K$ as a vector space over itself. $\endgroup$ – D_S Apr 16 '16 at 3:05

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