$\text{Hey I have a general question about linear transformations as long as I confirm that } L(v_1+v_2) = L(v_1)+L(v_2), \\ \text{ and } L(\alpha v) = \alpha L(v) \\ \text{do I need to show}$ $L(\alpha v_1 + \beta v_1) = L(\alpha v_1) + L(\beta v_2) \text{Or is this already inferred}?$

@Antonios nice! What does he work in?

@Balarka good luck and yeah try to keep it off your mind in the meantime

@Jon you don't need to put the whole message in mathjax

@Jon just use your two rules a few times

remembering that the variables in the rules are arbitrary

@Daminark he works pretty broadly in (complex) algebraic geometry

Ah, should be a fun time

but also in number theory and rep theory and a bunch of other random stuff

Woot

I think I fixed it

Lol I'm taking Galois theory at the moment and while as a class it's perhaps somewhat less fun than combinatorics, I feel somehow like more than anything I've done before, this resonates with me

@LeakyNun So just use the II, and III. I is already confirmed by II and III yes?

@Daminark I think galois theory is the course that makes you realize that "pure" math works

Lol true

when does the UChicago sem end

quarter*

We're in, 8th week at the moment, so we have 9th week, 10th week, then finals week

what else are you in

We have to leave the dorms June 8th

Right now I'm taking combinatorics, complex analysis, Galois theory, and Topics in Geometric Measure Theory

nice.

hows GMT

It's very interesting for sure. I didn't quite absorb stuff as well in this class since I kinda jumped into a class where I didn't know much GMT, Fourier, or PDE

And there were a few cases where the stuff just happened. Less PDE, really just one person's talk involved it, but at the beginning, Fourier and not knowing the basics of GMT made wrestling with the class trickier

Since it didn't teach the subject from the ground up, really people just started presenting papers published in the last few years

oof haha

I'm at least semi-decent at sets with the big pieces of Lipschitz graphs property

this stuff is all foreign to me lolz

i'd like to learn it some day though

But yeah I mean, this class has been good fun but it's giving me an idea that GMT is probably not what I'm likely gonna aim for

what stuff are you more into now

There's a certain way of processing things in this subject that makes me believe I wouldn't be able to solve any problems :P

@Daminark yeah trying to do that

getting some math done in the process

How do you prove that all the roots of (zeta(z))^(1/log(z)), are contained in a finite region?

Hmm, I feel it's a bit early to say with much detail which direction I'm gonna be going but of the things I've seen so far, I feel like Galois theory and combinatorics are the two subjects where I enjoy solving problems the most

And if I had to guess, Galois theory is where I'm gonna have more luck. I've had some solutions in combo that made me proud of myself but I dunno if I would be able to have the consistent creativity of the type you need to really do that subject as my main thing

the good news about combo is that the arguments kinda come up everywhere

I don't really know of anyone working in combo on its own right now

(but htat could just be my lack of knowledge)

Hmm, so my combinatorics professor doesn't do only only combo

But like, he's in the general area of combinatorics, complexity theory, computational group theory

Theoretical compsci basically

yeah that ive seen

alg combo is a thing too

Also I know there's another theoretical compsci professor who is very combo-ish, though it seems that guy's work feels closer to math that's kinda sophisticated as opposed to clever counting arguments (mathematical logic, Fourier analysis, etc)

Yeah true algebraic combo sounds fun. It's all stuff I'd like to eventually explore

Fantastic, @Antonios. Does this mean for Ph.D.?

potentially, for master's certainly @TedShifrin

Hey Ted!

That's great, @Antonios.

Hi Demonark.

Demonark: Did you see my two book recommendations for you last night?

Oh so funny thing, I just realized that there have been so many incidents where I would find the inverse of an element in some field extension in a way more complicated than usual manner. These things often came up as, find the inverse of the root of (something), and that's so easy to do

@Ted yeah I did see the recommendations, thank you very much!

My favorite way for the inverse is Euclidean algorithm :)

Euclidean algorithm is an A+ way, I've usually done it with matrices because I just kinda process things in terms of matrices more often

matrices are gonna generally be lots more work.

But if you already know an annihilating polynomial for some $\alpha$ which has non-zero constant term, say its minimal polynomial, it's kinda slick

You'd rather invert a nontrivially large matrix than do Euclidean algorithm? With computation time, you lose big.

You don't need the inverse. I retract that.

$p(x) = a_nx^n + \ldots + a_0$, and since $p(\alpha) = 0$ you can just write $1 = \alpha(-\frac{1}{a_0}(a_n\alpha^{n-1} + \ldots + a_1)$

Yeah you only need the first column of the inverse, which in small cases that I've usually had to deal with by hand (e.g. 3x3) has often been reasonably quick

Oh, $1/\alpha$ is trivial. I'm giving you $1/p(\alpha)$ to find.

Ah, yeah that of course gets much tougher

This was a standard exam question (for some not-ridiculous cases) in the first semester of my algebra course.

@Daminark how?

so you could find the inverse of say $2+2\sqrt[3]2+\sqrt[3]4$ fairly quickly?

I could, yes.

how?

$a(x)p(x)+b(x)f(x)=1$. Reduce mod $f(x)$ and you have that $[a(x)]$ is the inverse of $[p(x)]\in F[x]/\langle f(x)\rangle$.

do we always have bezout's lemma even if we're not in a euclidean domain?

oh wait, Q[X] is a euclidean domain

what am I talking about

Um, but we are in a Euclidean domain.

Any polynomials over a field is ...

right, I got confused for a second

it happens

Wait 'til you're my age.

hmm, i'm not very good at extended euclidean algorithm

In my algebra course you had to get good at it.

I see

Of course, I made up reasonable questions for exams.

But that's something any CAS can do easily. At least Mathematica sure can.

I see

Lol maybe you were thinking of $\mathbb{Z}[x]$ or something? Though usually it goes the opposite way if anything, you try to assume it's a PID and then... erm.. rip

@Daminark I was thinking of $\Bbb Z[\sqrt[3]2]$

Well, you do it in $\Bbb Q(\root3\of 2)$ and see if the answer lives with integral coefficients. :P

oh wait, what am I thinking again

$\Bbb Q(\sqrt[3]2)$ is a bloody field

Maybe a moratorium on thinking would be good.

Yup, it's $\Bbb Q[X]/\langle X^3-2\rangle$. :D

:c

just prove everything by bloody cayley-hamilton

good luck on the computational side of things

since when did we become constructivists

puts Leaky on permanent ignore

@TedShifrin :(

hey @Leaky, do you have time for one more question?

also, hi Ted

sure

alright, so: let $G=N\rtimes H$ be the semi-direct product. My book says that each $g\in G$ can be written uniquely as $(n,e)\cdot(e,h)$. However, why should this be unique? Can’t we just have $g=nh=n’h’$?

$\rtimes$ \rtimes

ah thanks

also, hi Sha.

@ShaVuklia show that if $nh=e$, then $n=h=e$

and I sort of forgot the group law for semi-direct product

@LeakyNun I don't see why this should necessarily hold

wait

It's just a twisted version of a direct product. You still get uniqueness because there's no overlap of the two subgroups.

by $nh$, do you mean the semi-direct multiplication?

right, $(n,e) \cdot (e,h)$

you need $G = NH$ with $N \cap H = \lbrace e \rbrace$

or you could use ted's idea

I'm not entirely sure how to use Ted's hint, but why would we want to look at $nh=e$, @Leaky?

Do the direct product case first, @Sha.

If you have $g=nh=n'h'$, what has to happen?

we get $(n^{-1}n',h^{-1}h')=(e,e)$?

The idea is exactly the same for forming the direct sum of vector spaces, if you've done this in linear algebra!

How did you get to that?

wait it's probably wrong, let me redo it:p

@ÍgjøgnumMeg we don't even need commutativity here :P

So we have $(n,h)=(n',h')$, and then $(n^{-1},h^{-1})(n',h')=(e,e)$

oh wait

@Leaky the principle is the same though, the lack of overlap forces uniqueness

If ordered pairs are equal, then you're done.

why can't I think

jesus

But start with $nh = n'h'$ in the big group.

extensionality of ordered pairs

I assume big group means semi-direct product?

I said to understand the direct product case first.

oh I thought it was settled

@ShaVuklia "But"

It's no different in the semi- case.

right let me see

oh I should have written $(n,e)(e,h)=(n',e)(e,h')$

You're still missing the main point in understanding it all, I think.

well, we get $n^{-1}n'=e$, so $n=n'$?

If you have abstract groups and do the direct product, then it's like products of sets. Of course there's a unique decomposition.

If you have actual subgroups, when do you have the possibility that the whole group is isomorphic to their direct product?

if it satisfies three properties, which are

the subgroups are disjoint, they are commutative in a way, and it's surjective in a way

if $N \cap H = \lbrace e \rbrace$ then $nh = n^\prime h^\prime$ gives $(n^\prime)^{-1}n = h^\prime h^{-1}$. The left-hand side is where? And the right-hand side is where?

Disjoint except for the identity. Yes.

Thanks for doing the exercise for us, @ÍgjøgnumMeg.

I've been trying to get @Sha to figure this out.

growls loudly

flips table

sorry my bad, I'm just too slow:p

dabbing gently

I'm not angry with you, @Sha.

It's the same thing on main where somebody has to interfere with my questions/hints to show off and get rep.

what is meant by "the left-hand side is where"? as in, the left hand side is $(n')^{-1}n=(n'^{-1}n,e)$, so it's like in the first component?

so the closed points of Spec(R[X]) is a mess

Again, forget ordered pairs for a while.

I was just nudging, sorry!

$n$ lives where, $h$ lives where?

$n$ lives in $N$, and $h$ in $H$. oh and they are 'disjoint'

So if $nh = n'h'$ ...

yea then $n=n'$ and $h=h'$

Whoa. How?

because $(n')^{-1}n=e$:p

throws up hands in disgust

anyhow, so the semi-direct product guarantees disjointness?

oh

what

Can we actually just get this nailed down. How do you get from my equation to the result?

Without just saying so.

Okay, so we have $nh=n'h'$. We can then write $n'^{-1}n=h'h^{-1}$, and since $N\cap H=\{e\}$, we know that $n'^{-1}n=h'h^{-1}=e$, and hence $n'=n$ and $h'=h$

OK, thank you.

(Y)

Semidirect, despite your notation, is the same idea. Except now you have $(nh)(n'h')$ isn't $(nn')(hh')$.