I'm learning Galois theory, and trying to prove the following theorem:
Let $K$ be a field and $L/K$ be a finite extension with $x_1,\dots,x_n\in L$ such that $L=K(x_1,\dots,x_n)$. let $M$ be a field and $\sigma:K\to M$ be an embedding. Then the number of extensions $\overline{\sigma}:L\to M$ of $\sigma$ is at most $[L\,:\,K]$ with equality if and only if, for each $i\in\{1,\dots,n\}$, $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$.
Here's a partial proof:
For each $i$, any $\sigma$-embedding $\overline{\sigma}:L\to M$ restricts to a $\sigma$-embedding $\sigma_i:K(x_i)\to M$. $0=\sigma_i(0)=\sigma_i(m_{K,x_i}(x_i))=\sigma m_{K,x_i}(\sigma_i(x_i))$, so $\sigma_i$ sends $x_i$ to a root of $\sigma m_{K,x_i}$, and, since the domain of $\sigma_i$ is $K(x_i)$, $\sigma_i$ is uniquely determined by which root of $\sigma m_{K,x_i}$ it sends $x_i$ to. Therefore, the number of $\sigma$-embeddings $\sigma_i:K(x_i)\to M$ is at most $[K(x_i)\,:\,K]$, with equality if and only if $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$.
Now one can try to induct on $n$. Here goes: we just proved the base case. Suppose that the result is true when adjoining $n$ elements to $K$, and let $L=K(x_1,\dots,x_{n+1})$ be a finite extension of $K$. By the induction hypothesis, the number of $\sigma$-embeddings $\overline{\sigma}:K(x_1,\dots,x_n)\to M$ is at most $[K(x_1,\dots,x_n)\,:\,K]$ with equality if and only if, for all $i\in\{1,\dots,n\}$, $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$. Moreover, for each $\overline{\sigma}$, the number of $\overline{\sigma}$-embeddings $\overline{\overline{\sigma}}:L\to M$ is at most $[L\,:\,K(x_1,\dots,x_n)]$ with equality if and only if $\overline{\sigma}m_{K(x_1,\dots,x_n),x_{n+1}}$ splits into distinct linear factors over $M$. Hence, by the tower law, the number of extensions $\overline{\overline{\sigma}}:L\to M$ of $\sigma$ is at most $[L\,:\,K]$ with equality if and only if for all $i\in\{1,\dots,n\}$ $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$ and $\overline{\sigma}m_{K(x_1,\dots,x_n),x_{n+1}}$ splits into distinct linear factors over $M$, where $\overline{\sigma}$ is any $\sigma$-embedding from $K(x_1,\dots,x_n)$ to $M$.
Now we just need to prove that this equality condition is equivalent to the condition that, for all $i\in\{1,\dots,n+1\}$, $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$. One direction is easy: Suppose that, for all $i\in\{1,\dots,n+1\}$, $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$. $m_{K,x_{n+1}}\in K(x_1,\dots,x_n)[X]$ and $m_{K,x_{n+1}}(x_{n+1})=0$, so $m_{K(x_1,\dots,x_n),x_{n+1}}\,|\,m_{K,x_{n+1}}$. Say $q\in K(x_1,\dots,x_n)[X]$ and $m_{K,x_{n+1}}=qm_{K(x_1,\dots,x_n),x_{n+1}}$. Then $\sigma m_{K,x_{n+1}}=\overline{\sigma}m_{K,x_{n+1}}=\overline{\sigma}(q)\overline{\sigma}(m_{K(x_1,\dots,x_n),x_{n+1}})$, so $\overline{\sigma}m_{K(x_1,\dots,x_n),x_{n+1}}\,|\,\sigma m_{K,x_{n+1}}$ in $M[X]$, so $\overline{\sigma}m_{K(x_1,\dots,x_n),x_{n+1}}$ splits into distinct linear factors over $M$.
I am stuck on the other direction.
I've finally figured it out, and the answer is extremely simple. I'm ashamed that I didn't realise this earlier. It's just symmetry. We can reorder the $x_1,\dots,x_{n+1}$ to, say, $x_2,\dots,x_{n+1},x_1$ and apply the argument in the induction step, and we get that the number of $\sigma$-embeddings $\overline{\overline{\sigma}}:L\to M$ is at most $[L\,:\,K]$ with equality if and only if $\sigma m_{K,x_2},\dots,\sigma m_{K,x_{n+1}}$ split into distinct linear factors over $M$ and $\overline{\sigma} m_{K(x_2,\dots,x_{n+1}),x_1}$ splits into distinct linear factors over $M$, where $\overline{\sigma}$ is any $\sigma$-embedding from $K(x_2,\dots,x_{n+1})$ to $M$. Combine that with the non-reordered result to conclude that equality holds if and only if for all $i\in\{1,\dots,n+1\}$ $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$, and $\sigma_1 m_{K(x_2,\dots,x_{n+1}),x_1}$ splits into distinct linear factors over $M$, and $\sigma_{n+1} m_{K(x_1,\dots,x_n),x_{n+1}}$ splits into distinct linear factors over $M$, where $\sigma_1$ is any $\sigma$-embedding from $K(x_2,\dots,x_{n+1})$ to $M$, and $\sigma_{n+1}$ is any $\sigma$-embedding from $K(x_1,\dots,x_n)$ to $M$. But this condition is equivalent to the condition that, for all $i\in\{1,\dots,n+1\}$, $\sigma m_{K,x_i}$ splits into distinct linear factors over $M$ by the argument given in my partial proof for one direction of the equality case.