Let $K_1$ and $K_2$ be finite Galois extensions of $k$, set $G_k = \text{Gal}(K_1 K_2/k)$, and $H= \text{Gal}(K_1 K_2/K_1 \cap K_2)$. I want to prove that as $k$ algebras, $$K_1 \otimes_k K_2 \cong \prod_{G_k/H} K_1 K_2.$$ I particular, I am trying to show that if we choose one $\sigma_i$ from each of the cosets in $G_k/H$, and define $\tilde{\sigma_i}: K_1 \otimes K_2 \to K_1 K_2$ by linearly extending $$a \otimes b \mapsto a \sigma_i(b)$$ to all of $K_1 \otimes K_2$, then $$\Sigma = \prod_{\sigma_i \in G_k/H} \tilde{\sigma}_i: K_1 \otimes_k K_2 \to \prod_{G_k/H} K_1 K_2$$ is an isomorphism.
I know that as vector spaces over $k$, both sides have the same dimension. Furthermore, I know that $\ker \tilde{\sigma}_i$ is maximal in $K_1 \otimes_k K_2$ because its image is a field. I also know that $K_1 \otimes_k K_2$ has a Jacobson radical of zero, so if $\mathfrak{n}_i$ are all the maximal ideals in $K_1 \otimes_k K_2$, then $$K_1 \otimes_k K_2 \cong \prod \frac{K_1 \otimes K_2}{\mathfrak{n}_i}.$$ Therefore, if I could show that $\ker \tilde{\sigma}_i$ are all distinct, the isomorphism would follow. The only issue is that when I suppose that $\ker \tilde{\sigma}_i = \ker \tilde{\sigma}_j$, I don't know how to find a $h \in H$ such that $\sigma_i = h \sigma_j$. Does anyone have any hints as to how I could find this or ideas for how else to prove that this map is an isomorphism?
Suppose that $\ker \tilde{\sigma}_i = \ker \tilde{\sigma}_j = \mathfrak{m}$. Then $$K_1 K_2 \cong \text{im}(\tilde{\sigma}_i) \cong \frac{K_1 \otimes K_2}{\mathfrak{m}} \cong \text{im}(\tilde{\sigma}_j) \cong K_1 K_2,$$ so there exists an isomorphism $\tau \in \text{Gal}(K_1 K_2/k)$ such that $\tilde{\sigma}_i = \tau \circ \tilde{\sigma}_j$. Therefore, for any $a \in K_1$, $\tilde{\sigma}_i(a \otimes 1) = (\tau \circ \tilde{\sigma}_j)(a \otimes 1)$, so $a = \tau(a)$. In other words, $\tau \in \text{Gal}(K_1 K_2/K_1)$. Furthermore, for any $b \in K_2$, $\tilde{\sigma}_i(1 \otimes b) = (\tau \circ \tilde{\sigma}_j)(1 \otimes b)$. Therefore, $\sigma_i(b) = \tau(\sigma_j(b))$ for all $b \in K_2$, so $\mu = \sigma_i \sigma_j^{-1} \tau^{-1} \in \text{Gal}(K_1 K_2/K_2)$. Hence $\sigma_i = \mu \circ \tau \circ \sigma_j$, and $\mu \circ \tau \in \text{Gal}(K_1 K_2 /K_1 \cap K_2) = H$. Therefore, all the $\ker \tilde{\sigma}_i$ are all distinct, so $\Sigma$ is an isomorphism.