Galois theory without functor... Without fundamental theorem of algebra. Are you kidding?
Galois Theory for Beginners (2010) [pdf]
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Re: Galois Theory for Beginners (2010) [pdf]
#22Earlier quoted context omitted.
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 tim…
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.
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.
Re: Galois Theory for Beginners (2010) [pdf]
#23Earlier quoted context omitted.
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 contra…
> 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.
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?
Re: Galois Theory for Beginners (2010) [pdf]
#24Galois theory without functor... Without fundamental theorem of algebra. Are you kidding?
It's a five page article in the Monthly, but I'd hesitate to include either topic even in a book.
Re: Galois Theory for Beginners (2010) [pdf]
#25Can 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 Th…
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 tim…
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.
Re: Galois Theory for Beginners (2010) [pdf]
#26Earlier quoted context omitted.
> 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.
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?
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...
Re: Galois Theory for Beginners (2010) [pdf]
#27Earlier quoted context omitted.
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?
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...
> 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.
Re: Galois Theory for Beginners (2010) [pdf]
#28Galois 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.
Re: Galois Theory for Beginners (2010) [pdf]
#29Earlier quoted context omitted.
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 tim…
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.
> 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.
Re: Galois Theory for Beginners (2010) [pdf]
#30Galois theory without functor... Without fundamental theorem of algebra. Are you kidding?
It's a five page article in the Monthly , but I'd hesitate to include either topic even in a book.
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.