[monfrere]

Can anyone help me understand what is happening at the bottom of page 23 (page 3 of the PDF)?

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?

[topynate]

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.

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.

[monfrere]

> 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.

[JadeNB]

> 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.

[bollu]

Q(x_1, x_2) is the smallest field containing Q, x_1, x_2. So, it's a subfield of the reals, assuming x_1, x_2 \in R

[stablemap]

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,

(σf)(x_1, x_2) = f(x_{σ(1)}, x_{σ(2)}) = f(x_2, x_1) = x_2 - x_1

[rkowalick]

I am a little confused by your example, but the definition

  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.

[monfrere]

> σ acts on rational functions like f and g

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.

[ky3]

You're doing fine. Do cheer up.

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?

[monfrere]

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...

[ky3]

> Oh, right, subfield of the complex numbers

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.

[ky3]

Great observation.

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.

[monfrere]

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.

[ky3]

> OK, thanks.

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.

[waqf]

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.

[ky3]

The HN audience isn't the intended audience for this paper by Stillwell, who wrote for other math profs like himself.

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.