Suppose you have a field $F$, and a value $\eta$ such that $f$ is a polynomial of minimal degree, with coefficients in $F$ such that $f(\eta)=0$. Then it is claimed that for every unique root $q$ of $f$ there exist a unique automorphism $\phi F(\eta) \rightarrow F(\eta)$ fixing $F$ (assuming implicitly $F(\eta)$ is a normal extension I believe) such that $\phi(\eta) = q$
In simple terms, if we have an extension field, then the set of all base-field fixing automorphism corresponds to the set of roots of $f$.
This problem was given in a textbook, but their proof just doesn't make sense:
Let $f(x) = \sum a_i x^i $ And consider an $F$ fixing automorphism $\phi F[\eta] \rightarrow F[\eta]$. On applying $\phi$ to the equation $f(\eta) = 0$ we end up with $\sum a_i \phi(\eta)^i = 0 $ Thus, $\phi(\eta)$ is always a root. (This makes sense now comes the part that escapes me)
Conversely if $\gamma \in F[\eta]$ is a root of $f$ then the map $F[x] \rightarrow F[\eta]$, $g(x)\rightarrow g(\gamma)$ factors through $F[x]/f(x)$. When composed with the inverse isomomorphism $x + f(x) \rightarrow \eta, F[x]/f(x) \rightarrow F[\eta] $ this becomes a homomorphism $F[\eta] \rightarrow F[\eta]$ sending $\eta$ to $\gamma$
So in analyzing that sentence I realized that I don't know what "then the map $F[x] \rightarrow F[\eta]$, $g(x)\rightarrow g(\gamma)$ factors through $F[x]/f(x)$" means. The stacking of the two arrows means nothing to me, and I don't understand what it means for a map to factor through the "division" of a field by a polynomial. Also, where did $g$ come from? It's never referenced earlier and I don't understand what purporse it serves.
And after that my next follow up is what is the isomorphism given in: " When composed with the inverse isomomorphism $x + f(x) \rightarrow \eta, F[x]/f(x) \rightarrow F[\eta] $" and I have a feeling the rest follows easily.
Alternative idea:
So given the automorphisms $\phi$ are invertible, and $F$ fixing, then if $\phi(q_1 )= \phi(q_2)$ for any roots $q_1, q_2$ of $f(x)$ then $q_1 = q_2$. Thus we conclude that since $\phi(q)$ is a root for each root q, if $\phi(q) \ne q$. Then it follows that $I, \phi, \phi^2 ... \phi^{\text{# of distinct roots}-1}$ are each different automorphisms. Thus we can say that for each root there exists at least one corresponding automorphism. Dictated by which root that automorphism maps $\eta$ to. The only problem here is to prove that there exists at least ONE non trivial automorphism.
This just says we can define the map from the polynomial ring $F[x]$ to $F[\eta]$ since $\gamma \in F[\eta]$. The stacking of the two arrows: the first arrows gives the structures the map is between, and the second gives the rule used to calculate the image of an arbitrary element $g$.
"Factors through" is a weird term; I think the meaning is that, we have this map $F[x] \to F[\eta]$, then we are going to factor this map to consider it as a map $F[x]/f(x) \to F[\eta]$. I will call this map $\varphi:\ F[x]/f(x) \to F[\eta]$, $x+f(x) \mapsto \gamma$. Now the next statement:
This is strange wording; but I think it means, we have the isomorphism $\pi: \ x+f(x)\mapsto \eta$, $F[x]/f(x)\to F[\eta]$. Now if we compose $\varphi \circ \pi^{-1}$, this maps $\eta \mapsto x + f(x) \mapsto \gamma$.
I don't see uniqueness addressed in this, nor do I see verification that $\varphi$ is an isomorphism, but this should be relatively easy to complete.