$k$-embeddings sends splitting fields to themselves I have been reading from splitting fields from Lang's Algebra textbook. This is a statement which the author does not justify:
Let $k$ be a field, $\overline{k}$ be its algebraic closure and $\alpha_1, \dots, \alpha_n \in \overline{k}$ . If $\sigma : E:=k(\alpha_1 ,\alpha_2 , \ldots, \alpha_n) \to \overline{k}$ is a $k$-embedding and $\sigma (\alpha_i) \in E$ for each $i=1,\ldots, n$ then $\sigma (E) \subseteq E$.
Here's what I have done in order justify this:
We proceed by induction on $n$.
If $n=1$ we suppose that $E=k(\alpha_1)$. If $[k(\alpha_1) : k] = n$ then $E = \text{span} \{ 1, \alpha_1, \ldots , \alpha_1 ^{n-1}\}$. Then $\sigma (E)= \text{span} \{ 1, \sigma (\alpha_1) , \ldots, (\sigma (\alpha_1))^{n-1}\}$. Clearly, $\sigma (E) \subseteq E$.
Suppose that my claim is true for all $n=m$.
Let $\alpha_1, \ldots , \alpha_m, \alpha_{m+1}\in \overline{k}$ and suppose $E=k(\alpha_1, \alpha_2, \ldots, \alpha_m, \alpha_{m+1})$. Define $F=k(\alpha_1, \ldots , \alpha_m)$. Then $E=F(\alpha_{m+1})$. However, $\sigma (\alpha_i)$ may not be in $F$ and I cannot use induction anymore.
I am also looking for ways to prove it with/without induction. Any help will be appreciated!
 A: Think about what the elements of $E = k(\alpha_1, ... , \alpha_n)$ look like.  They are sums of the form
$$c \alpha_1^{e_1} \cdots \alpha_n^{e_n}$$
for $e_i \geq 0$ and $c \in k$.  So it is obvious that if $\sigma$ sends each $\alpha_i$ into $E$, then it sends every element of $E$ back into $E$.
A: The main thing is, you are not using the fact that it is a $k-$ embedding. That is it fixes $k$.
We have that $E=k(\alpha_{1},...,\alpha_{n})$.
Then you take $\alpha_{1}$ and look at the minimal polynomial. It is of degree say $n_{1}$. Then $\{1,\alpha_{1},...,\alpha_{1}^{n_1-1}\}$ is a $k-$basis for $k(\alpha_{1})$ over $k$
Now you take the minimal polynomial for $\alpha_{2}$ over $k(\alpha_{1})[X]$. Then it is of degree say $n_{2}$. Then again $\{1,\alpha_{2},...,\alpha_{2}^{n_2-1}\}$ is a $k-$basis for $k(\alpha_{1},\alpha_{2})$ over $k(\alpha_{1})$.
Similarly you proceed to conclude that any element $\beta$ in $E$ can be expressed as a finite linear combination of monomials in $\alpha_{1},...,\alpha_{n}$
i.e. $$\beta=\sum_{I}c_{I}\alpha^{I}$$
Where the sum is taken over multi-indices $I=(i_{1},...,i_{n})$ . and $\alpha^I=\alpha_{1}^{i_{1}}\alpha_{2}^{i_{2}}...\alpha_{n}^{i_{n}}$.
For an arbitrary subset $\mathcal{S}\subset \bar{k}$ , if $\sigma(S)\subseteq S$ then $\sigma(k(S))\subseteq k(S)$.
When $\mathcal{S}$ is algebraic over $k$, In that case $k[\mathcal S]=k(\mathcal{S})$. And since you are taking only finite sums of monomials whose multi-index are finitely supported, it boils down to the previous case of finite set. ($k[S]$ denotes the polynomial ring over $k$ with variables in $S$.)
That is any element of $k(\mathcal{S})$ say $\beta$ can be written as a finite sum
$\beta= \sum_{I}c_{I}S^{I}$ . Where $I=(i_{s})_{s\in\mathcal{S}\text{ such that all but finitely many}\,i_{s}=0\}}$ . That is you are taking multi-indices with finite support and if say $i_{s_{1}},...,i_{s_{m}}$ be the non-zero indices in $I$ , then $S^{I}=s_{1}^{i_{s_{1}}}s_{2}^{i_{s_{2}}}...s_{m}^{i_{s_{m}}}$.
Then as $\sigma(S)\subset S$, you have $\sigma(k(S))\subset k(S)$ as all the elements in the sum are in $k(S)$.
