Going well, thanks! Quarter is beginning to finish up so exams may not be a fun time but soon after that it'll be nice

which exams will you have?

Combinatorics this coming Tuesday, and then the following week complex analysis on Monday + Galois on Tuesday

good luck!

Thanks!

I hope your Galois exam will not be like that exercise :P

Lol, I think our professor isn't quite as ridiculous, if the midterm was any indication

Ugh D&F hint made me take way longer on this problem than I would've otherwise

It was a good idea but we already solved another problem that expedited this so much

The first part was showing that if $D$ is a square-free integer which can be written as a sum of squares, then $\mathbb{Q}(\sqrt{D})$ is contained in some $L$ whose Galois group is $\mathbb{Z}/4$

I once had a problem in Atiyah-Macdonald were the hint was a lot less elegant and more complicated than my solution

I always get scared I did something wrong when that happens

Second part was showing that the converse was true. Now, the path that Dummit and Foote takes is basically saying aight, you can write $L = K(\sqrt{a+b\sqrt{D}})$ (where $K = \mathbb{Q}(\sqrt{D})$, so show that $\mathbb{Q}(\sqrt{a^2 - b^2D})$ is another intermediate extension so it must be equal to $\mathbb{Q}(\sqrt{D})$.

Now we had an earlier exercise that section saying that if $F$ is a field containing $n^{th}$ roots of unity (characteristic not dividing $n$), then $F(\sqrt[n]{a}) = F(\sqrt[n]{b}) \iff a = b^ic^n$ for some $c\in F$

And they were like, okay use that to conclude that $D$ can be written as a sum of squares

In the process of doing that I computed that $\alpha = \sqrt{a+b\sqrt{D}}$ had minimal polynomial $x^4 - 2ax^2 + a^2 - b^2D$

This was in order to say that $\sqrt{a-b\sqrt{D}}$ was another element in our field, and so the Galois automorphism taking $\alpha$ to this (order 4 since we're $\mathbb{Z}/4$ and there's only one element of order 2 already given by $\alpha\mapsto -\alpha$) cannot fix $\sqrt{a^2 - b^2D}$, so that can't be in $\mathbb{Q}$

@Mathein: That reminds me of a (geometric) linear algebra problem I had in the first homework set in my Honors multivariable math course. One of my brilliant first-year students turned in a far superior solution, ignoring the hint, afraid that her solution was wrong. She finished a Ph.D. at Chicago a year ago. :)

But we had a previous problem already establishing that if you had an irreducible polynomial of the form $x^4 + ax^2 + b$, then the Galois group is $\mathbb{Z}/4$ iff $b(a^2 - 4b)$ is a square. That just immediately takes care of this

@TedShifrin I had a few other cases where I ignored hints, but my solution was equally complex to the proposed solution

I hate it when problems force you do something a particular way and you can't get creative

Sometimes authors miss better solutions than students find (unswayed by a crummy hint).

It's a good argument that one shouldn't overdo giving hints. That's one of my main complaints about Guillemin's style, particularly in Guillemin & Pollack. They give way too many hints and ruin problems.

for the record, the problem was this: Let $A$ be a commutative ring. Show that a polynomial $f \in A[x]$ is a unit iff the constant term is a unit in $A$ and all the coefficients of the higher terms are nilpotent

Rip. Well, in this particular case I can sorta buy why they made that suggestion, since they don't want to assume everyone did that classification problem from before (tbh this wasn't slick so much as, I already did some of this work in another problem)

Oh, that's a standard problem. It's certainly been discussed here not long ago (perhaps by you, perhaps by Leaky).

the difficult part is proving the part where we assume $f$ is a unit

Demonark: I don't know why they can't refer to the previous problem. I like doing that. Unless they're trying to suggest an alternative approach.

the solution proposed was some weird induction over the coefficients

They didn't really seem to ask you to use this solution, just that they said it's one possibility, and their hint also referred to a problem but from the same section instead of a previous one

Oh, I can see you'd end up following that path if you just follow your nose.

Their hint could say, "See Exercise blah.blah" (from before) "or use Exercise bloh.bloh" (from this section). ... Or maybe they forgot.

my solution is just this: if $A$ is an integral domain, then this is obvious because then degrees add so all units in the polynomial ring over an integral domain are constant. If $A$ is a general commutative ring, then for any prime ideal $\mathfrak{p}$ in $A$, $A/\mathfrak{p}$ is an integral domain, so if we reduce a unit in $A[x]$ to $A/\mathfrak{p}[x]$, we get a constant polynomial, so all higher coefficients are in $\mathfrak{p}$.

Since $\mathfrak{p}$ was arbitrary, all higher coefficients are in every prime ideal, so they are nilpotent

You mean $f$?

That's also possible. Dummit and Foote has a lot of exercises so it's possible they're not keeping track

Wait. I don't follow. What's your first sentence?

lol

strange mistake

I meant $A$ an integral domain

Demonark: It was one reason I realized I should write the solutions manual to my last two books before publishing the books. (And I rather lavishly edited my diff geo notes after writing a solutions manual for myself, as that made me realize things could be improved/reorganized.)

Ah, actually, I like your proof, @Mathein. That's a good approach.

Yeah if I don't forget immediately this will be my canonical proof now

Agreed, @MikeM.

Sorta like why I used to keep a folder of things that I figured out ... because I realized after a year of grad school that I did forget so many clever things.

That folder got trashed when I retired, too.

I realized that but never fixed it

how are you good people

@BalarkaSen are we?

have you forgotten to sleep, again, Balarka?

Lol, sorta along those lines, recently I discovered that I should write many more notes to myself when I read, especially if it's something difficult

Dami got the obvijoke before me

sniping noises

@TedShifrin My sleep's a horrible mess and I'm kind of sick now

O great.

There was one step in the GMT paper I was reading whose correctness took me an inordinate amount of time to convince myself of

Then in my lecture I completely forgot why it was correct and when someone asked I just blanked for like, 10 minutes

For comparision, this is the hint in Atiyah-Macdonald: If $f=a_nx^n + \dots + a_0$ and $g=b_mx^m+\dots+b_0$ is the inverse of $f$, show by induction on $r$ that $a_n^{r+1}b_{m-r}=0$

This is why we write notes to self.

Can anyone tell me

5 minutes after the lecture was over I was just like oshitthat'swhy

Unlikely

Right, @Mathein. Without trying too hard I can imagine a recursive argument they have in mind.

if you raise a (complex function) to a complex variable will any new roots be present other than the ones that the (complex function) had?

I'm blessed that for all my lectures some classmate either TeXs his notes or someone writes them very neatly down and scans them and uploads them to some google drive

We're back to the issue that things aren't defined as functions, @geocalc.

@Mathein same

Well

what do you mean?

Remember, we had a long conversation about how $a^z$ has infinitely many values very often.

yeah

you have to choose a branch

Second quarter of analysis there were two people TeXing notes independently, this quarter there's someone TeXing them for complex, someone started TeXing them for Galois but it was going slow at the beginning and D&F had everything so he stopped

OK ... So can $e^z$ ever be $0$?

Last year second two quarters of analysis someone TeX'd notes

no

Also second quarter of algebra had TeX'd notes

So if $a\ne 0$, can $a^z$ ever be $0$?

no

It's a lot of work TeXing class notes, Demonark.

Does that answer your question, @geocalc?

Oh for sure, I have no idea how they do it but I'm so grateful, especially in those classes where we weren't following any source especially closely

yeah, so no new roots

Right.

I've heard of some people live-TeXing, the people I know mostly just write up quick notes on the spot and then TeX notes later, perhaps changing up the presentation to be more understandable

im just annoyed because i keep graphing stuff and it looks like there's a zero but

I wrote most of my books by just sitting and TeXing. Occasionally, I had things in front of me, or I did calculations on paper first.

i think the plotter is just not precise enough or something

I've considered doing that, apparently they say it really helps them understand things better, but also time is very scarce

for my algebra lectures, they did a cool system: the lecturer wrote on a tablet and spoke into a mike. The tablet was projected onto the wall so we could see it. Then they edited the recordings from him writing on the tablet and the recordings from the mike. And now all students of the Heidelberg university (or anyone who knows the page and makes an account) can access those recordings

Same reason I suggested to students that they recopy their class notes to learn better, Demonark.

I can better follow a lecture if I don't have to write anything down

I know someone who managed to TeX the diagram for the naturality of the Snake lemma live

I actually have realized I follow better if I take some notes.

which is pretty impressive

The only class so far where I've been taking notes is combinatorics, because it goes really fast and has a large volume of content. Also he assigns all his exercises during the lecture so at minimum you have to have all those written down

@Mathei I am still impressed by that argument lol

It's the "right" argument.

It's always good in math to reduce to a simple case where it's easier :)

And then use a fundamental result about intersection of all prime ideals.

I agree. It's always good in math to think about prime ideals :P

Well, reducing to a case where you know the answer easily is a good technique :)

true

Also I remember in analysis, we had to show that $\frac{\sin(x)}{x}$ is not in $L^1(\Bbb R)$. There was some hint on how to do estimations with the harmonic series, but my proof I ignored that and went like this: Suppose that $\frac{\sin(x)}{x}$ is in $L^1$, then its Fourier transform exists and is continuous. But $\frac{\sin(x)}{x}$ is (up to some constant) the Fourier transform of the function that is $1$ on some interval and $0$ everywhere else.

By the Fourier inversion formula, that would also be the Fourier transform of $\frac{\sin(x)}{x}$, but clearly that's not continuous.

Well, I could see assigning that question long before you had done anything about Fourier transforms.

yeah, but we even computed that Fourier transform before

oh, weird ... I would use that as a basic example early on to show that Riemann integrable is not necessarily good enough.