$B/K$ is a field extension of degree 1 implies $B = K$?

Question

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 $B=K$?

Answer 1

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 $K$.