Galois Theory for Beginners (2010) [pdf]
math.jhu.edu
math.jhu.edu
https://www.youtube.com/watch?v=x1v2tX4_dkQ
PS: He has an entire series of lectures on his channel. Highly recommend.
It is hard to overdramatize the events of Galois' life. He lived during a revolution, he went out into the streets to protest and fight, he was arrested multiple times for his outrageous political speech, and he laid down his life for his political beliefs. A friend of mine who survived a chronic and often-fatal disease in his teens used to quip, "Well, if I were Galois, I'd have already made my best contributions and died by now." Galois indeed burned fast and bright.
https://www.maa.org/sites/default/files/pdf/upload_library/2...
It says any permutation sigma of x_1, ..., x_n can be extended to a bijection of Q(x_1, ..., x_n) defined by
sigma f(x_1, ..., x_n) = f(sigma x_1, ..., sigma x_n).
But I don't see how this definition can be consistent. For example, let f(a, b) = a - b
g(a, b) = a/a + b/b = 2
x_1 = 5
x_2 = 3
sigma x_1 = x_2
sigma x_2 = x_1
Then according to the formula: sigma f(5, 3) = sigma (5 - 3) = sigma 2 = f(3, 5) = -2
But sigma g(5, 3) = sigma 2 = g(3, 5) = 2
Contradiction?sigma f = \a b -> f (sigma a) (sigma b) = f b a
That means sigma f = \a b -> b - a. On the other hand, g := \a b -> 2. So sigma g = g.
Looking again at your equality:
> sigma f(5, 3) = sigma (5 - 3) = sigma 2 = f(3, 5) = -2
We should read sigma f(5, 3) as (sigma f)(5,3) = (3 - 5) = -2. Note: sigma f is not equal to sigma (5 - 3), because f is not the same rational function as "5 - 3"!
* This is an abuse of notation, because a and b are bound variables inside the lambda. In this case, being more precise would probably be less clear.
I see, this is the key point I was missing. But now I am confused as to whether Q(x_1, x_2) is supposed to be a subfield of the real numbers or a field of rational functions of x_1, x_2.
The latter. However, wherever you get `x_1` and `x_2`, if `\{x_1, x_2\}` is algebraically independent over `\mathbb Q`, then `\mathbb Q(x_1, x_2)` is isomorphic to a field of rational functions. This allows you to realise the same ground field inside many different larger fields.
σf(x_1,...,x_n) = f(σx_1,...,σx_n)
is certainly consistent and I'm not sure what you were hoping to show with your example functions.I'm really not sure what σ(2) is supposed to represent? σ acts on rational functions like f and g, so
σf(5, 3) = f(3, 5) = -2
and σg(5,3) = g(3, 5) = 2
Of course σf and f might not be equal, but I don't see how that is a contradiction? Happy to try and clear things up.This is what I was missing. But the paper also says σ is an automorphism of Q(x_1, ... x_n). Which is weird, since I thought Q(x_1, ... x_n) was a subfield of the reals (not a field of rational functions). So I still don't get what's going on.
Sigh, I think I've forgotten more since school than I thought.
No field of rational functions here, that's waaaay off given what Stillwell intends to do.
Also, subfields of reals are a bit restrictive, don't you think?
Right-o!
> Now sigma is supposed to be an automorphism on that subfield. Which means sigma does take scalar values as arguments, contrary to what stablemap said...
There's that ol' abuse of notation going on here. The first sigma is just a permutation on the roots: {x_i} -> {x_i}. In particular, this skinny sigma is not defined on scalars.
It embiggens into a second sigma that's your automorphism: Q({x_i}) -> Q({x_i}). This fat sigma now maps scalars.
Now identify the first and second sigmas, and the abuse is complete.
(σf)(x_1, ..., x_n) = f(x_{σ(1)}, ..., x_{σ(n)});
nothing truly different, but possibly more comfortable. Another thing to say is that a field automorphism can't move the base field Q: all we can do and all we have done is move around these formal variables. So with the rest of your notation,
(σf)(x_1, x_2) = f(x_{σ(1)}, x_{σ(2)}) = f(x_2, x_1) = x_2 - x_1
The American Math Monthly is a journal by college professors for college professors. The readership is expected to be familiar with the topic. Groundbreaking results are published elsewhere. Whew. Unfortunately, neither is AMM a place for professors to summarize the contents of a 14-week course for adult learners.
Let's use your observation to illuminate 2 abuses of notation that happen all the time between those in the know.
The first abuse is not explicitly calling out that the coefficients a_i are restricted. The polynomial of which the a_i are the coefficients must be irreducible. That is omitted in the paper, but is typically understood. For otherwise, the field extension doesn't work, as you've found out.
When the a_i denote an irreducible, then the roots x_i are all outside Q.
And then there is no contradiction.
Your example uses (x-5)(x-3) which has all roots in Q--the diametric opposite--which is why sigma breaks down.
Digression: If you know some Haskell, you'll notice that a permutation on the roots basically fmaps to a (field) endomorphism on Q(all x_i). But here the converse is also true (exceptional in Haskell, except for trivial cases): every such endo comes from a permutation. (end of digression)
The second abuse is in the title. This is really "(My Opinion on) How to Teach Galois Theory to Undergrads" with a subtitle of "By Jettisoning the Fundamental Theorem and Focusing Exclusively on Quintic Unsolvability." The subtitle is omitted and the title shortened and de-colloquialized to read "Galois Theory for Beginners." This is all part of the prestigious mathematical tradition because ink, paper, and papyrus once upon a time were terribly scarce. Sorry about that.
Quintic Unsolvability is like FLT. The big prize is not the Yes/No answer but the VIP theorems--the statements of which are neither as easy to explain nor understand as QU nor FLT--used to nail down some pesky boolean.
So throwing out the FT of GT shortchanges the undergrad. It especially shortchanges the math-aware software professional who would appreciate experiencing the galois correspondence which later morphs into an adjunction in category theory. Quite cool.
GT has pedagogical messiness like inseparable extensions which can be skipped on a first pass. As a royal road to FTGT, I recommend the approach of fixing all fields as subfields of the complex numbers. See Postnikov's Foundations of Galois Theory available on google books the last time I checked. Nice exercises too.
p.s. (Galois) adjunctions are like a general theory of "How to Run Anything Backwards Even When There's No Chance in Hell." That's the power of math for you.
To members of HN audience who want to learn GT, I recommend the freely available book by Postnikov.
p.s. Fwiw, the paper did make clear the distinction between the a_i, x_i, and the alpha_i. See the second page proper of TFA.
No worries.
> Your answer has allowed me to follow the paper a little further, though I think your answer may be inconsistent with the other two answers I got.
I was trying to be helpful.
That said, in lieu of the irreducibility criterion, you could stay in the original context where all the variables a_i, x_i, and alpha_i are indeterminates so that all the extensions are higher-ranked rational function fields over Q.
So Q(a1,a2) is a field in 2 indeterminates and Q(x1,x2) is an extension of that.
They are isomorphic to subfields of C, but we don't think of them as subfields of C because there don't come with canonical embeddings. You have to choose the algebraically independent transcendentals.
> Also I haven't been able to show myself that if the x_i are the roots of an irreducible polynomial, then Q(x_1,...,x_n) is symmetric with respect to x_1,...,x_n, in the sense claimed in the paper without proof.
There's no claim. The paper's just defining what it means for a field extension to be "symmetric w.r.t." to the adjoined elements.
Galois was ahead of his time.[1]
[1] You could say he was ahead by a century, to quote a famous song.
The category-theoretic accretion to Galois theory is a much later addition; Galois certainly didn't think in those terms, and I think that it is not obligatory for an expository (or even a mathematical!) treatment of the subject to do so.
The part that surprised me is that it seems to focus on rational numbers. I always assumed that Galois Groups were focused on more abstract concepts of sets. Is Galois Theory mostly about rational number (or even real numbers), or is the author just using the rationals to keep the paper focused on beginners?