$\textbf{No}$ full answers please!
I've been working on a class problem for several days now, and I'm completely stuck, mainly because I feel like there's too few assumptions given in the problem statement. Here's the problem for context.
Let $K_{/F}$ be a finite extension of subfields of $\mathbb{C}$ and let $G = \text{Aut}_{F}K$. Let $\sigma_1,\sigma_2,\dots,\sigma_n$ be all of the distinct embedding of $K \hookrightarrow \mathbb{C}$ that fix elements of $F$. For $\alpha \in K$, define $\text{Tr}_{K/F}(\alpha) = \sum_{i=1}^n \sigma_i(\alpha)$.
a) For $\alpha \in K$, let $f = X^m + f_{m-1}X^{m-1}+\dots+f_0 \in F[X]$ be the monic irreducible polynomial of $\alpha$ over $F$. Prove that $\text{Tr}_{F(\alpha)/F} (\alpha) = -f_{m-1}$.
b) Show that $\text{Tr}_{K/F}(\alpha) \in F$.
c) Express $\text{Tr}_{K/F}(\alpha)$ in terms of the coefficients of $f$.
So the first thing that strikes me as weird is that this definition of the trace doesn't seem to align with any definition in other textbooks. For an arbitrary algebraic extension $K_{/F}$, they define $\text{Tr}_{K/F}(\alpha) = [K:F(\alpha)] \sum_{i=1}^n \sigma_i(\alpha)$, the only exception being that $K_{/F}$ is Galois, in which case each $\sigma_i \in \text{Gal}(K_{/F})$, and $\text{Tr}_{K/F}(\alpha) = \sum_{\sigma \in \text{Gal(K/F})} \sigma_i(\alpha)$.
The problem is that $K/F$ need not be Galois. If $K/F$ was normal, then this would actually work, as since the prime subfield of $\mathbb{C}$ is $\mathbb{Q}$, we conclude that $\text{Char}(F)=0$, and so $K/F$ would be Galois.
The second issue is that since $K/F$ isn't necessarily Galois, I don't know how to prove b), which would otherwise be a 3 line proof.