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.

  • $\begingroup$ "Factors through" just says that $f\mapsto 0$, which is trivial since $f(\eta)=0$ (you didn't say this explicitly, but you map $x\mapsto \eta$). $\endgroup$ – user138530 Oct 18 '15 at 7:13
  • $\begingroup$ @MorganRodgers yes, thats truer to what I mean than what I have written $\endgroup$ – frogeyedpeas Oct 18 '15 at 7:20
  • $\begingroup$ @MorganRodgers in your example there still is one automorphism per root, dictated by its action on the root $\alpha$ Also, I realized my question is only well posed if we consider normal extensions. Since a simple $Q[\sqrt[3]{2})]$ has no non trivial automorphisms, unless we first extend by the roots of unity $\endgroup$ – frogeyedpeas Oct 18 '15 at 7:30
  • $\begingroup$ Made changes, hopefully that clears up my intent $\endgroup$ – frogeyedpeas Oct 18 '15 at 8:00
  • $\begingroup$ Yes, much more clear now. $\endgroup$ – Morgan Rodgers Oct 18 '15 at 9:28

Conversely if $\gamma \in F[\eta]$ is a root of $f$ then the map $F[x]\to F[\eta]$, $g(x)\mapsto g(\gamma)$ factors through $F[x]/f(x)$.

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:

When composed with the inverse isomomorphism $x+f(x)\mapsto \eta$, $F[x]/f(x)\to F[\eta]$ this becomes a homomorphism $F[\eta]\to F[\eta]$ sending $\eta$ to $\gamma$.

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.

  • $\begingroup$ Thank you so much for explaining everything. Just to tidy up it's occurred to me that it's not clear that $\pi^{-1}$ even exists. Is there any reason I should believe that? I do agree that once composed we get a homomorphism that permutes the roots, and showing that $\psi$ is an isomorphism yields that the entire thing is an automorphism. So really it seems the question I should analyze is why is $\pi$ an isomorphism. I believe that reduces to showing $F[x]/f(x) = F[\eta]$ $\endgroup$ – frogeyedpeas Oct 19 '15 at 0:33
  • $\begingroup$ Since $\pi$ is an isomorphism it must be bijective. The fact it is an isomorphism is a pretty standard result in the construction of a quotient field, that if $f(x)$ is irreducible then $F[x]/f(x)$ is a field and is isomorphic to $F[\eta]$. $\endgroup$ – Morgan Rodgers Oct 19 '15 at 5:29

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