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

is some polynomial and is a map, I prefer to denote by the polynomial , rather than . Although the latter notation does have the advantage that it gives us an explicit name for the “induced map” , I can deal with referring to that map simply as .

One question I had was the following: let be an extension and be such that each is -algebraic. Is it true that ? If , this follows from one of the first theorems we covered about algebraic elements. The result follows by induction if is finite. Is the general case true, though?

Now consider the converse: if then each is -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 , then is algebraic, which implies only has algebraic elements. Why is this true? Also, if only has algebraic elements, does it follow that is algebraic? I want to know exactly which implications regarding this stuff break down when is infinite.

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

but how does this show that ? Why is ? I agree that , though — because is Galois — so .

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

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## About mlbaker

just another guy trying to make the diagrams commute.

by Artin’s theorem.

So is a subgroup of , another group with the same, finite, cardinality.

So

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 is any (possibly infinite) Galois extension, then the Galois group has a natural topology that makes it a compact topological group, and the operation

is just the closure operator . So in general, the correspondence only bijective between ClOSED subgroups of the Galois group.

In the case of finite extensions, is finite, and the topology is just the discrete topology, so every subgroup is closed. But this fact is specific to finite extensions.

@First question: It suffices to show that is a field, since it contains and is the smallest subextension of which contains . But any specific element of is just a polynomial in finitely many terms , 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 is a field when are algebraic, in the form of an algorithm which computes inverses for you.

The answers to your follow up questions about the case when 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 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 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 .

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

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

Suppose .

Let be a transcendence basis for over . The existence of a

transcendence basis is easy to prove from Zorn’s Lemma.

Then where is purely

transcendental and is algebraic.

The statement that is algebraic is equivalent to the statement

that is empty. We will show by contradiction that is empty.

Let . Then , so is algebraic over .

Write . Clearing denominators, we obtain an

algebraic dependency in , contradicting the fact that is a

transcendence basis.

Small typo: the equation in the last paragraph should be . The proof strategy should be clear in any case.