# Galois Closure and a natural bijection

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!

• 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 $M/K$. – Jyrki Lahtonen Oct 14 '18 at 7:06
• 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 $\Gamma(M:L)$ – Jyrki Lahtonen Oct 14 '18 at 7:11
• 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. – Jyrki Lahtonen Oct 14 '18 at 7:19
• 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 $M$. – Jyrki Lahtonen Oct 14 '18 at 7:20
• Thank you very much! – DesmondMiles Oct 15 '18 at 9:29