Points of confusion in Galois theory

As I type up the third PMATH 442 assignment, I’m also slowly making my way through the notes, making sure all the arguments are correct, and attempting to understand them. I thought I would make a post outlining some of my thoughts on the material, including points of confusion. I invite anyone to leave a comment. Once I can understand everything enough to speak coherently about it (and this week is over), I promise I will give another (hopefully fruitful) Scattered Comments post for PMATH 442 (I know another one for PMATH 753 is long overdue, but things just got a bit busy, which is to be expected since it is currently the middle of the term).

I’ve fixed a few (minor) typos so far, and revised some notation, for example if

$p(x) = a_0 + \ldots + a_n x^n \in F_1[x]$

is some polynomial and $\varphi : F_1 \to F_2$ is a map, I prefer to denote by $p^{\varphi}(x)$ the polynomial $\varphi(a_0) + \ldots + \varphi(a_n)x^n \in F_2[x]$, rather than $\varphi'(p)$. Although the latter notation does have the advantage that it gives us an explicit name for the “induced map” $F_1[x] \to F_2[x]$, I can deal with referring to that map simply as $q \mapsto q^{\varphi}$.

One question I had was the following: let $K/F$ be an extension and $S \subseteq K$ be such that each $s \in S$ is $F$-algebraic. Is it true that $F[S] = F(S)$? If $S = \{ \alpha \}$, this follows from one of the first theorems we covered about algebraic elements. The result follows by induction if $S$ is finite. Is the general case true, though?

Now consider the converse: if $F(a_1,\ldots,a_\ell)=F[a_1,\ldots,a_\ell]$ then each $a_i$ is $F$-algebraic (equivalently, the extension is algebraic). This is apparently called Zariski’s Lemma, and was used near the beginning of PMATH 764 to prove Hilbert’s Nullstellensatz. We did not prove it in PMATH 442. The converse of the more general statement appears in my notes: if $F(S) = F[S]$, then $F(S)/F$ is algebraic, which implies $S$ only has algebraic elements. Why is this true? Also, if $S$ only has algebraic elements, does it follow that $F(S)/F$ is algebraic? I want to know exactly which implications regarding this stuff break down when $S$ is infinite.

I’m also not that satisfied with the proof of the Fundamental Theorem of Galois Theory presented today. We had

$H \overset{\psi^{-1}}{\longmapsto} \mathrm{Fix} \; H \overset{\psi}{\mapsto} \mathrm{Gal}(K/\mathrm{Fix} \; H) \geq H$

but how does this show that $\psi \circ \psi^{-1} = \mathrm{id}_{\mathcal{G}}$? Why is $\mathrm{Gal}(K/\mathrm{Fix} \; H) = H$? I agree that $L = \mathrm{Fix}(\mathrm{Gal}(K/L))$, though — because $K/L$ is Galois — so $\psi^{-1} \circ \psi = \mathrm{id}_{\mathcal{F}}$.

I will probably post more later. Also, my camera is back from repair 😀

just another guy trying to make the diagrams commute.
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7 Responses to Points of confusion in Galois theory

1. David McLaughlin says:

$|\mathrm{Gal}(K/\mathrm{Fix} \;H)| = [K : \mathrm{Fix} \; H] = | H |$ by Artin’s theorem.
So $H$ is a subgroup of $\mathrm{Gal}(K/\mathrm{Fix} \; H)$, another group with the same, finite, cardinality.
So $H = \mathrm{Gal} (K/\mathrm{Fix} \; H) = \psi \circ \psi^{-1} (H)$

2. Erik Crevier says:

Re: the last question. You might want to look at the Galois correspondence for infinite extensions to see what breaks down in the general case. To summarize: In general, if $K/F$ is any (possibly infinite) Galois extension, then the Galois group $Gal(K/F)$ has a natural topology that makes it a compact topological group, and the operation

$H\mapsto Gal(K/Fix H)$

is just the closure operator $H\mapsto \overline{H}$. So in general, the correspondence only bijective between ClOSED subgroups of the Galois group.

In the case of finite extensions, $Gal(K/F)$ is finite, and the topology is just the discrete topology, so every subgroup is closed. But this fact is specific to finite extensions.

3. Erik Crevier says:

@First question: It suffices to show that $F[S]$ is a field, since it contains $S$ and $F(S)$ is the smallest subextension of $K/F$ which contains $S$. But any specific element of $F[S]$ is just a polynomial in finitely many terms $s_1,...,s_n\in S$, which puts you back in the finite case.

Also, I think it is enlightening to actually write out in some gory computational detail the proof that $F[s_1,...,s_n]$ is a field when $s_1,...,s_n\in K$ are algebraic, in the form of an algorithm which computes inverses for you.

4. Erik Crevier says:

The answers to your follow up questions about the case when $S$ is infinite follow similarly from their finite counterparts.

This is something that happens rather often. It is even lurking just beneath the surface in the infinite Galois correspondence: The topology on $Gal(K/F)$ is basically just described by gluing together the discrete topologies on all its finite quotients (search keyword: profinite topology). This is exactly dual to the way that an infinite algebraic extension can be understood by gluing together what you know about its finite sub-extensions.

As an exercise, try to come up with a construction of the algebraic closure of a field that “respects” this way of viewing things as much as possible. This is a rather vague statement, and open to some interpretation, but here’s a concrete way of interpreting: If your base field $F$ is one that you can actually compute things in algorithmically (eg the rationals, or finite fields), then your construction should cleanly translate into code which does the same for $\overline{F}$.

(spoiler alert: I’m secretly warming you up for the compactness theorem in model theory.)

5. mlbaker says:

Thank you both very much; that clarified things a lot.

6. David Jao says:

Suppose $F(S) = F[S]$.

Let $T$ be a transcendence basis for $F(S)$ over $F$. The existence of a
transcendence basis is easy to prove from Zorn’s Lemma.

Then $F \subset F(T) \subset F(S) = F[S]$ where $F(T)/F$ is purely
transcendental and $F[S]/F(T)$ is algebraic.

The statement that $F(S)/F$ is algebraic is equivalent to the statement
that $T$ is empty. We will show by contradiction that $T$ is empty.

Let $t \in T$. Then $1/t \in F[S]$, so $1/t$ is algebraic over $F[T]$.
Write $1/t = f(t_1, \ldots, t_k)$. Clearing denominators, we obtain an
algebraic dependency in $T$, contradicting the fact that $T$ is a
transcendence basis.

• David Jao says:

Small typo: the equation in the last paragraph should be $0 = f(1/t, t_1, \ldots, t_k)$. The proof strategy should be clear in any case.