# Theorem regarding $K$-embeddings of finite extension $L/K$ into normal extension $M/L$

The following theorem is stated in my notes:

Theorem

Let $L/K$ be a finite extension, and let $\theta : L \hookrightarrow M$ with $M/L$ normal. Let $L' = \theta(L) \subseteq M$. Then

1. The number of distinct $K$-embeddings $L \hookrightarrow M$ is at most $[L:K]$, with equality iff $L/K$ is separable.

2. $L/K$ is normal iff every $K$-embedding $\phi: L \hookrightarrow M$ has image $L'$ iff every $K$-embedding is of the form $\phi = \theta \circ \alpha$ for some $\alpha \in \mbox{Aut}(L/K)$

My issue is with the second part of the theorem. Isn't it necessary for $\theta$ to be a $K$-embedding, rather than just an embedding, for this to work? The proof starts of by observing facts that include:

"Any $K$-embedding $\phi : L \hookrightarrow M$ with image $L'$ gives rise to an automorphism $\alpha$ of $L/K$ such that $\phi = \theta \circ \alpha$. Conversely, any $\phi$ of this form is a $K$-embedding with image $L'$."

I don't see how this can be true unless $\theta$ is a $K$-embedding. By definition $\phi$ fixes $K$, and as a $K$-automorphism $\alpha$ fixes $K$.

Am I misunderstanding something, or should my notes specify that $\theta$ is a $K$-embedding?

Thanks.

EDIT: Follow up question

I have the following definition:

"Let $L/K$ and $L'/K$ be field extensions. A $K$-embedding of $L$ into $L'$ is an embedding which fixes $K$".

I'm not exactly sure what "fixes $K$" means here. I first interpreted it as "if $\psi$ is a $K$-embedding of $L$ into $L'$, then $\psi(k) = k$ for all $k$ in $K$." But this only makes sense if $K$ is 'actually' a subfield of both $L$ and $L'$, i.e. if the elements of $K$ as elements of $L$ and $L'$ have the same "labels". What I really want is that $\psi$ sends each element of $K$ in $L$ to the element of $K$ in $L'$ that corresponds to it under isomorphism.

Thinking of these as abstract field extensions, let $\phi$ give the extension of $K$ to $L$ and $\phi'$ give the extension of $K$ to $L'$. If $\psi$ is a $K$-embedding from $L$ to $L'$, does this mean that $\psi(k) = \phi' (\phi^{-1}(k))$ for all $k$ in $\phi(K) \cong K$?

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It seems like you're right. –  Dylan Moreland Dec 5 '11 at 20:20
Great, thank you. –  Anon Dec 5 '11 at 20:39
Please note that I've updated the original post with a follow up question. –  Anon Dec 5 '11 at 20:56
But when you write $L/K$, you are saying that $K \subset L$. If I have an embedding $f\colon K \to M$ of $K$ in some other field, then I do think it's correct to say "$g\colon L \to M$ is an embedding over (extending) $f$". But that's different. –  Dylan Moreland Dec 5 '11 at 20:58
As for the follow up question: it means $\psi\circ \phi=\phi'$. –  user18119 Dec 5 '11 at 21:00