Wikipedia reminds me that the version of his death (staying up all night before the duel to frantically write down his mathematical ideas at the last possible moment) that I read in Men of Mathematics may have been overdramatized, apparently like a number of the stories in that book.
What definitely did happen is that Galois was worried that he might not win the duel and that he might never get the chance to redeem himself in front of his mathematical peers. We know this because of the letter he wrote before he died; we actually have it, and WP has a scan of a page: https://en.wikipedia.org/wiki/File:E._Galois_Letter.jpg
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.
Tony Rothman wrote an article in the American Mathematical Monthly which helped to separate a lot of the fact from the fiction in accounts of Galois's life. Highly recommended:
This document was also mentioned yesterday on the 'Why is the quintic unsolvable?'[1] post. The post also has some additional interesting references in the comments. Worth checking out!
Your sigma is a bijection on rational functions of a, b, not on the evaluations of those functions at particular values of a and b. In lambda notation:*
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.
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.
> Your sigma is a bijection on rational functions of a, b, not on the evaluations of those functions at particular values of a and b.
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.
> 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.
Some authors would write the following: σ is a permutation of the numbers 1, ..., n and
(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,
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.
Oh, right, subfield of the complex numbers. 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...
> 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.
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.
OK, thanks. 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. 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.
> 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.
You are right but I don't think the audience here (people trying to learn Galois theory for the first time) will understand your comment either. The original paper didn't even make clear that a_i are the coefficients and x_i are the roots … that's the level at which we need to be clarifying.
Galois had an insight that I always seemed particularly deep to me: that problems should be classified not by topic area (analysis, theory of equations, geometry) but by their underlying form.
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.
Why? Whitout a clear explanation Galois functors you miss everything... And without the fundamental theorem of algebra, it's going to be very hard to have any real life example.
I have never studied group theory, so this is way beyond what I'm ready for. But I scanned over a few sections and read to the point that I got lost.
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?
https://www.youtube.com/watch?v=x1v2tX4_dkQ
PS: He has an entire series of lectures on his channel. Highly recommend.