I'm trying to understand / solve the following problem:
Let $ L \subset \mathbb{C} $ be a field and $ L \subset L_1 $ its finite extension ($ [L_1 : L] = m $). Prove that there are exactly $ m $ disintct embeddings
$$ \sigma_1, \dots, \sigma_m: L_1 \hookrightarrow \mathbb{C} $$
which are identity on $ L $.
I can use Abel's theorem to reduce the problem to the case where $ L_1 = L(a) $. And now I've been given a hint that if that's the case, then the number of such embeddings is equal to the number of roots of $f(x) \in L[x]$ minimal for $ a $.
I can't really think of an answer why that is. I would appreciate some hints
Let $f(x)$ be the minimal polynomial of $a$ over $L$, it then has to be of degree $m$.
This means, we can express $a^m$ by $L$-linear combination of lower powers $a^k$ ($k<m$), and by definition, $a$ doesn't satisfy any other algebraic equations that doesn't follow from $f(a)=0$.
Now, if we replace the root $a$ by an indeterminant $x$, we see that $L[x]/(f)\cong L(a)=L_1$. (More directly, you can prove that the kernel of the evaluation $L[x]\to L(a),\ x\mapsto a$ is just the ideal $(f)$.)
This could be read as if the root $a$ could equally well be just a 'formally adjoint element' $x$ requested to satisfy $f(x)=0$.
Since $f$ is irreducible in $L[x]$, and has no multiple roots (because $\gcd(f,f')=1$), it has $m$ roots in $\Bbb C$. Say these are $z_1,\dots,z_m$.
Now, if $\sigma:L_1\to\Bbb C$ is an embedding, then $f(a)=0$ implies $f(\sigma(a))=0$ as $f$ is a homomorphism, i.e. $\sigma(a)=z_k$ for some $k$.
And conversely, given any root $z_k$ of $f$ in $\Bbb C$, we can define $\sigma_k$ to take $a\mapsto z_k$, and as it wants to be a homomorphism, it needs to satisfy $$\sigma_k(\sum_{j=0}^{m-1}\lambda_ja^j)=\sum_{j=0}^{m-1}\lambda_j{z_k}^j$$ with $\lambda_j\in L$, which already defines $\sigma_k$.