In Exercise 14.2.17, Dummit & Foote:
Let $K/F$ be any finite separable extension, and let $\alpha\in K$. Let $L$ be a Galois extension of $F$ containing $K$ and let $H\leq \text{Gal}(L/F)$ be the subgroup corresponding to $K$. Define the norm of $\alpha$ from $K$ to $F$ to be
$$ N_{K/F}(\alpha)=\prod_\sigma\sigma(\alpha), $$
where the product is taken over all the embeddings of $K$ into an algebraic closure of $F$, (so over a set of coset representatives for $H$ in $\text{Gal}(L/F)$ by the Fundamental Theorem of Galois Theory).
Why is it true that set of all embeddings of $K$ into an algebraic of closure of $F$ the same as a set of coset representatives for $H$ in $\text{Gal}(L/F)$? And should it be left coset representatives or right coset representatives?
This answer says that $\text{Hom}_F(K, \overline{F}) \cong \text{Hom}_F(K, L) \cong \text{Gal}(L/F) / \text{Gal}(L/K)$ but I do not see why the second isomorphism holds.
(Notation changed to match this context.)
Best Answer
Let $r: \text{Hom}_F(L, L) \to \text{Hom}_F(K, L)$ be the natural restriction map.
I should probably explain what I mean by this. Take an element $\sigma \in \text{Hom}_F(L, L)$. So $\sigma$ is function $L \to L$. Now let $\sigma|_K : K \to L$ be the function you get by restricting the domain of $\sigma$ to $K$. Then $\sigma|_K$ is an element of $\text{Hom}_F(K, L)$. We define $r(\sigma)$ to be $\sigma|_K$.
Lemma 1: $r$ is surjective.
Proof: Let $\tau$ be an element of $\text{Hom}_F(K, L)$. Then $\tau$ is an isomorphism from $K$ to $\tau(K)$ that fixes elements in $F$.
$L$ is a Galois extension of $F$, so $L$ is the splitting field for some polynomial $f(x) \in F[x]$ over $F$. $L$ is therefore also the splitting field for $f(x)$ over $K$.
The isomorphism $\tau: K \to \tau(K)$ induces an isomorphism $\hat \tau : K[x] \to \tau(K)[x]$, which maps a polynomial $\sum_{i=1}^N a_i x^i \in K[x]$ to the polynomial $\sum_{i=1}^N \tau(a_i) x^i \in \tau(K)[x]$.
Since all the coefficients of $f$ are in $F$, and $\tau$ fixes elements of $f$, we have that $\hat{\tau}(f) = f$.
To summarise:
Therefore, by the uniqueness theorem for splitting fields (Dummit and Foote, Chapter 13.4, Theorem 27), there exists an isomorphism $\sigma : L \to L$ that extends $\tau$.
When we say that $\sigma$ extends $\tau$, what we mean is that $\sigma|_K = \tau$, i.e. $r(\sigma) = \tau$. Thus $\tau$ is in the image of $r$. This proves that $r$ is surjective.
Lemma 2: For any $\sigma_1, \sigma_2 \in \text{Hom}_F(L, L)$, $r(\sigma_1) = r(\sigma_2)$ holds if and only there exists a $\lambda \in \text{Hom}_K(L, L)$ such that $\sigma_1 = \sigma_2 \circ \lambda$. In other words, $r(\sigma_1) = r(\sigma_2)$ holds if and only if $\sigma_1$ and $\sigma_2$ are in the same left coset of $\text{Hom}_K(L, L)$.
Proof of $\implies$. Suppose that $r(\sigma_1) = r(\sigma_2)$, i.e. $(\sigma_1)|_K = (\sigma_2)|_K$. Consider $\lambda := \sigma_2^{-1} \circ \sigma_1$. If $x \in K$, then $\sigma_1(x) = \sigma_2(x)$, so $\lambda(x) = \sigma_2^{-1}(\sigma_1(x)) = x$. Thus $\lambda$ fixes elements of $K$, i.e. $\lambda$ is in $\text{Hom}_K(L, L)$.
Proof of $\impliedby$. Suppose that $\sigma_1 = \sigma_2 \circ \lambda$ for some $\lambda \in \text{Hom}_K(L, L)$. Then for any $x \in K$, we have $\lambda(x) = x$, so $ \sigma_1(x) = \sigma_2(\lambda(x)) = \sigma_2(x) $. Hence $(\sigma_1)|_K = (\sigma_2)|_K$, i.e. $r(\sigma_1) = r(\sigma_2)$.
Proposition: There is a bijection $\vartheta : \text{Hom}_F(L, L)/ \text{Hom}_K(L, L) \to \text{Hom}_F(K, L)$, defined by $\vartheta\left(\sigma . \text{Hom}_K(L, L)\right) = r(\sigma)$.
Here, $\text{Hom}_F(L, L)/ \text{Hom}_K(L, L)$ denotes the set of left cosets of $\text{Hom}_K(L, L)$ in $\text{Hom}_F(L, L)$.
Proof: