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Let $L/K$ be a finite separable extension of a field and let $M$ be its Galois closure (i.e. the minimal degree extension of $L$ for which $M/K$ is Galois). Show that the set of embeddings (injections) $hom_K(L, M)$ of $L$ in $M$ which fix $K$ is in a natural bijection with the set of right cosets of $Γ(M : L)$ in $Γ(M : K)$.

I tried some "natural" bijections (e.g. associate $\varphi$ to $\sigma_{\varphi}$ which restricted on $Im \varphi$ behaves as taking the preimage for $\varphi$), but they can't work as I don't use the minimality of $M$. Also, when I say that $Γ(M : L)\sigma = Γ(M : L)\tau$ implies $\sigma = g\tau$ for some $g\in Γ(M : L)$, does this mean that $\sigma(m) = g(\tau(m))$ for all $m\in M$ or $\sigma(m) = \tau(g(m))$ in $M$?

Any help appreciated!

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    $\begingroup$ This is close to being a duplicate of this older question. The same mechanisms are at work, but there the target is an algebraic closure $\overline{K}$ of $M$ instead of $M$ itself. That actually makes no difference due to normality of $M/K$. $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 7:06
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    $\begingroup$ Anyway, the key ingredients are: 1) Any $K$-embedding $\phi:L\to M$ comes from a restriction of some (actually several) $K$-automorphism $\sigma\in\Gamma(M:K)$. 2) Two automorphisms of $M$ have the same restriction to $L$ if and only if they belong to the same right coset of $\Gamma(M:L)$ $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 7:11
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    $\begingroup$ So to answer your question. The minimality of $M$ does not really play a role at all. We can use any finite normal extension of $K$ containing $L$. Because $M/L$ is finite and separable, it is simple, say $M=L(\alpha)$. Let $\phi:L\to M$ is a $K$.embedding. Then normality implies that the minimal polynomial of $\alpha$ (first over $K$) has all its zeros in $M$. It follows that the homomorphic image of the minimal polynomial of $\alpha$ over $L$ must also have a zero in $M$. Therefore we can extend $\phi$ to a $K$-automorphism of $M$ et cetera. $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 7:19
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    $\begingroup$ A further point: the index of $\Gamma(M:L)$ in $\Gamma(M:K)$ is independent of the choice of $M$, because Galois correspondence implies that the said index is $[L:K]$. We can use any larger normal extension in place of $M$. $\endgroup$ – Jyrki Lahtonen Oct 14 '18 at 7:20
  • $\begingroup$ Thank you very much! $\endgroup$ – DesmondMiles Oct 15 '18 at 9:29

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