#groups-rings-fields | Mathematics | Page 522

next obsidian Oct 31, 2020, 3:02 AM

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At least I didn’t see them when I checked

sturdy marsh Oct 31, 2020, 3:02 AM

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huh

next obsidian Oct 31, 2020, 3:02 AM

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And stacks treatment of divisors was waaay different

sturdy marsh Oct 31, 2020, 3:02 AM

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I haven't read through all of chapter 2 so I might have missed some stuff

next obsidian Oct 31, 2020, 3:02 AM

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It’s making me want to die but

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It’s been like that since the start of doing it so nothings changed haha

sturdy marsh Oct 31, 2020, 3:03 AM

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are you self-studying?

next obsidian Oct 31, 2020, 3:03 AM

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Die as a figurative die of course

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Yeah

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I took a course last year but like

sturdy marsh Oct 31, 2020, 3:03 AM

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are you doing it with someone else

next obsidian Oct 31, 2020, 3:03 AM

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It was very... self-driven and I got caught up in other classes

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So I have he notes to help

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But it’s primarily myself

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And yeah just me

sturdy marsh Oct 31, 2020, 3:03 AM

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oh ok

next obsidian Oct 31, 2020, 3:03 AM

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The issue is like I’m an UG

sturdy marsh Oct 31, 2020, 3:04 AM

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which year?

next obsidian Oct 31, 2020, 3:04 AM

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3rd year. So a lot of the grad students I took the class with are like further along

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Say in chapter 3

sturdy marsh Oct 31, 2020, 3:04 AM

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oh lol same

next obsidian Oct 31, 2020, 3:04 AM

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And my UG friends are like all just starting

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Or not yet started

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So I’m like in this dead zone

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It kind of ducking sucks haha

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I did like all the problems, with a few exceptions in II.1-4 all by myself

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And I wanted to die

sturdy marsh Oct 31, 2020, 3:05 AM

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yeah it was pretty much the same in my class too I think, people had very different backgrounds

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I did like all the problems, with a few exceptions in II.1-4 all by myself
@next obsidian noice

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I need to do that

next obsidian Oct 31, 2020, 3:05 AM

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It’s just lonely haha

sturdy marsh Oct 31, 2020, 3:05 AM

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probably over a summer or something

next obsidian Oct 31, 2020, 3:05 AM

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Yeah I did it over a summer

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I wish I had an advisor

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Someone I could like just meet with once a week and get clarifying stuff or like

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Some of the missing stuff that’s just you’re expected to develop idfk how

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Example: a lot of properties of scheme theoretic image for Chow’s Lemma

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I looked at EGA and stuff and there’s a loooot of like lemmas you basically need to prove

sturdy marsh Oct 31, 2020, 3:07 AM

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hartshorne/AG is super rough if youre self studying

next obsidian Oct 31, 2020, 3:07 AM

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That’s the type of stuff I feel like you learn from@others

sturdy marsh Oct 31, 2020, 3:07 AM

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even with a group

next obsidian Oct 31, 2020, 3:07 AM

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Yeah

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Idk how the first wave of grad student Hartshorne warriors did it

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Before the advent of MSE and ahit

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I can feel myself connecting to my AG ancestors at least, undertaking the “do all the problems” pilgrimage

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Lmfao

sturdy marsh Oct 31, 2020, 3:08 AM

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yeah but it might be more efficient to try other books and get back to hartshorne

next obsidian Oct 31, 2020, 3:08 AM

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Rip

sturdy marsh Oct 31, 2020, 3:08 AM

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I'm reading mumford right nor and it's pretty nice

next obsidian Oct 31, 2020, 3:08 AM

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Hmmm

sturdy marsh Oct 31, 2020, 3:08 AM

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lectures on curves on an algebraic surface

next obsidian Oct 31, 2020, 3:08 AM

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Oh not red book?

sturdy marsh Oct 31, 2020, 3:09 AM

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yeah not that

next obsidian Oct 31, 2020, 3:09 AM

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Tbh I heard Hart’s treatment of cohomology is good

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And I can maybe catch up to a few ppl at my school

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Since I’m close to being done with chapter II

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I think I’ll skip differentials and the formal scheme stuff

sturdy marsh Oct 31, 2020, 3:09 AM

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did they not do cohomology in your AG class?

next obsidian Oct 31, 2020, 3:09 AM

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We did

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But like

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Also I was lost by then

sturdy marsh Oct 31, 2020, 3:10 AM

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ah ok, understandable

next obsidian Oct 31, 2020, 3:10 AM

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I have notes so I can refer to those which is nice

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And my prof had some like

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Chad methods to do some stuff

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The like “proper” way

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Where Hartshorne has some more ad-hoc stuff

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So it’s nice to have the notes on top of Hartshorne

sturdy marsh Oct 31, 2020, 3:10 AM

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Hartshorne 3 is actually kinda easier than 2

next obsidian Oct 31, 2020, 3:11 AM

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Thank

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God

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You made my night

sturdy marsh Oct 31, 2020, 3:11 AM

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I got lost in the details in chapter 2 for a bit

next obsidian Oct 31, 2020, 3:11 AM

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I was hoping it was but didn’t ask anyone

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Since I didn’t want to be disappointed

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Haha

sturdy marsh Oct 31, 2020, 3:11 AM

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yeah early chapter 3 is super nice

next obsidian Oct 31, 2020, 3:11 AM

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Thank god yessss

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I think my friends are on like 3.3

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So I can maybe catch up hype

sturdy marsh Oct 31, 2020, 3:12 AM

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I've heard good things about Liu's book too

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are you interested in the arithmetic aspect at all or purely geometric?

next obsidian Oct 31, 2020, 3:13 AM

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So uhhhh

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Neither

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The algebra

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opencry

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Okay so realistically I was like NT bad

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But have sort of opened up to the idea recently, also uhhh

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Lots of depts are mostly arithmetic so it means I have more places to apply to

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And as for geometric, bruh this shit don’t seem like geometry to me bwahahaha

sturdy marsh Oct 31, 2020, 3:14 AM

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are we the same person lmao

next obsidian Oct 31, 2020, 3:14 AM

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I’m in a special topics course rn tho on like resolution of singularities

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And that’s helping me a bit

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Honestly, maybe Brofibration

sturdy marsh Oct 31, 2020, 3:14 AM

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I thought ANT was going to be super dry

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did a reading course on Neukirch

next obsidian Oct 31, 2020, 3:15 AM

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Legit I was like I really like algebra, I want to do Algebraic-something something

sturdy marsh Oct 31, 2020, 3:15 AM

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it was fantastic

next obsidian Oct 31, 2020, 3:15 AM

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AG was the first that satisfies that which I tried

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Hated varieties

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Then schemes had me pogging a bit

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I liked how like you do algebra locally

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And use the sheaf and topology to glue the stuff together

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And now I’m just doing my thing

sturdy marsh Oct 31, 2020, 3:16 AM

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is AG what you want to get into?

next obsidian Oct 31, 2020, 3:16 AM

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I think so

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I kind of decided it would be my thing, and my friend keeps reminding me it’s not set in stone

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But for the time being I’ve kind of hard committed to it and the schools I want to go to are mainly because of AG depts

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And by mainly I mean solely lmao

sturdy marsh Oct 31, 2020, 3:17 AM

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I'm not sure yet on what I want to do, but I'm also in the algebraic-something boat

next obsidian Oct 31, 2020, 3:18 AM

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Yeah, I haven’t tried AT yet, just too busy

sturdy marsh Oct 31, 2020, 3:18 AM

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AT is nice

next obsidian Oct 31, 2020, 3:18 AM

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And I think I’d want to try some ANT at some point

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But like also I really fuckinf love finite groups and like AAAA

sturdy marsh Oct 31, 2020, 3:18 AM

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finite groups are used in AT

next obsidian Oct 31, 2020, 3:18 AM

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I could see myself doing a lot of things in the general algebra-sphere as long as it stays away from analysis

sturdy marsh Oct 31, 2020, 3:18 AM

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equivariant homotopy

next obsidian Oct 31, 2020, 3:19 AM

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But for now I’m enjoying AG and my school is strong with it and there’s a lot of AG-related opportunities present so I’m running with that

sturdy marsh Oct 31, 2020, 3:19 AM

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AG does seem like a good option

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I also like the idea of geometric rep theory

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havent done any tho

next obsidian Oct 31, 2020, 3:20 AM

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Is that like GGT stuff?

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I know there’s ppl at my school doing geometric group theory which idk much about

sturdy marsh Oct 31, 2020, 3:20 AM

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nah not geometric group theory

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the only geometric rep theory ive seen is where people construct reps using vector bundles and stuff

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so im talking algebraic/lie groups

next obsidian Oct 31, 2020, 3:22 AM

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Ohhh

sturdy marsh Oct 31, 2020, 3:22 AM

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borel-weil-bott etc.

bleak abyss Oct 31, 2020, 3:22 AM

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Did someone say GGT?

next obsidian Oct 31, 2020, 3:22 AM

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Nope

sturdy marsh Oct 31, 2020, 3:22 AM

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nope

bleak abyss Oct 31, 2020, 3:22 AM

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Understandable have a nice day

next obsidian Oct 31, 2020, 3:22 AM

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😅

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😰

sturdy marsh Oct 31, 2020, 3:22 AM

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but anyway, that stuff looks dope

bleak abyss Oct 31, 2020, 3:23 AM

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Geometric rep theory is mad fun tbh

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I say knowing neither rep theory nor AG

sturdy marsh Oct 31, 2020, 3:23 AM

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i wanna read Ginzburg and Chriss

bleak abyss Oct 31, 2020, 3:23 AM

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Oh you remind me I need to catch up on a class Ginzburg's teaching that I'm auditing

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On Lie theory

next obsidian Oct 31, 2020, 3:24 AM

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Dami

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Just do AG instead

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Did you ditch your AG class?

bleak abyss Oct 31, 2020, 3:24 AM

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Mostly yeah, though the last few lectures I wanna try on

next obsidian Oct 31, 2020, 3:24 AM

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You said you were covering flat descent

bleak abyss Oct 31, 2020, 3:24 AM

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I'm more or less not gonna bother with descent

next obsidian Oct 31, 2020, 3:24 AM

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And didn’t know how the fibered product works

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And I was like 🤨

bleak abyss Oct 31, 2020, 3:25 AM

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And I'm gonna just deal with sheaf cohomology

next obsidian Oct 31, 2020, 3:25 AM

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Lmao

bleak abyss Oct 31, 2020, 3:25 AM

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I mean he didn't spend much time on descent

next obsidian Oct 31, 2020, 3:25 AM

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I mean that’s how you get all the real results

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I guess but fibered products are so important

bleak abyss Oct 31, 2020, 3:25 AM

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And it's possible he talked about a bunch of properties of base change during that time

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He has video lectures so I can go back and rewatch

next obsidian Oct 31, 2020, 3:26 AM

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Yeah but until you get your hands dirty it’s kinda hard to get what’s going on

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And show directly that ___ is stable under base change

sturdy marsh Oct 31, 2020, 3:26 AM

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I ate too many peanut butter cups

next obsidian Oct 31, 2020, 3:26 AM

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Rip

bleak abyss Oct 31, 2020, 3:26 AM

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There's no such thing as too many

sturdy marsh Oct 31, 2020, 3:26 AM

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I feel sick

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I will never eat a peanut butter cup again

next obsidian Oct 31, 2020, 3:27 AM

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man who will eat more peanut butter cups

bleak abyss Oct 31, 2020, 3:27 AM

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Anyway yeah Chmonkey I imagine Dima's treatment of stuff was reasonably self-contained? It's possible I just missed some of his base change material, it's possible he's not going deep into descent, or it's possible that he talked a bunch about base change at the same time

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He has since moved on to cohomology of sheaves and now this "Fourier-Mukai transform"

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So I might try to get caught up again

next obsidian Oct 31, 2020, 3:28 AM

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Bruh moment

bleak abyss Oct 31, 2020, 3:29 AM

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I also like the book zetamath mentioned once, "Lectures on Algebraic Geometry" by Gunter Harder

next obsidian Oct 31, 2020, 3:29 AM

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Hmmm

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I have never heard that one

bleak abyss Oct 31, 2020, 3:29 AM

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First volume is mostly material that builds up to AG rather than AG itself lmao

next obsidian Oct 31, 2020, 3:29 AM

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Ffs

sturdy marsh Oct 31, 2020, 3:29 AM

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just read every page of the stacks project

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EZ

next obsidian Oct 31, 2020, 3:30 AM

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I just will learn so much technical stuff

bleak abyss Oct 31, 2020, 3:30 AM

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Like it's a chapter on categories, homological algebra, sheaves, sheaf cohomology, and compact Riemann surfaces/abelian varieties

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Then volume 2 is like

next obsidian Oct 31, 2020, 3:30 AM

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I can get to the point I can abstract things so much it’s impossible to see any geometry

bleak abyss Oct 31, 2020, 3:30 AM

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Schemes, commalg, etc

next obsidian Oct 31, 2020, 3:30 AM

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Then my lack of geometric knowledge won’t matter

bleak abyss Oct 31, 2020, 3:30 AM

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And volume 1 does some amount of algebraic topology along with sheaf cohomology

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So yeah I might just say lol whatever to the class and take my own route through Harder's book

next obsidian Oct 31, 2020, 3:31 AM

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Why do you even need AG actually tho

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Are you doing ANT shit?

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I literally don’t know what you actually do besides the words modular, forms, and Automorphic

bleak abyss Oct 31, 2020, 3:32 AM

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I probably mostly don't need it? I mean there's a non-zero chance I work with Jordan Ellenberg

sturdy marsh Oct 31, 2020, 3:32 AM

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Why do you even need AG actually tho
@next obsidian to understand math memes

bleak abyss Oct 31, 2020, 3:32 AM

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In which case AG is good to know

next obsidian Oct 31, 2020, 3:32 AM

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Just learn 300 adjectives

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Then you can understand AG memes

bleak abyss Oct 31, 2020, 3:32 AM

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And like algebraic groups stuff is relevant to me I feel? Also my advisor once talked about a GIT quotient

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Idk AG creeps up in weird ways here

sturdy marsh Oct 31, 2020, 3:33 AM

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GIT also seems dope

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Mukai's book is nice

next obsidian Oct 31, 2020, 3:33 AM

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Tf is GIT

bleak abyss Oct 31, 2020, 3:33 AM

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Geometric invariant theory

next obsidian Oct 31, 2020, 3:33 AM

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Oh nvm

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Yeah

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I thought about if for a bit

bleak abyss Oct 31, 2020, 3:33 AM

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Seems like good stuff

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But yeah the thing about AG for me is that the actual "content", even abstract stuff, seems cool

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Like Hodge theory, intersection theory, algebraic groups, GIT, moduli spaces, etc

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The problems it solves are excellent

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But the foundations are just a slog and a half

next obsidian Oct 31, 2020, 3:35 AM

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Yes

sturdy marsh Oct 31, 2020, 3:35 AM

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yup, I'm doing a class which is on spectral AG. But we've just been doing stuff on quasicategories up till now

next obsidian Oct 31, 2020, 3:35 AM

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Spectral AG?!??

sturdy marsh Oct 31, 2020, 3:35 AM

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the build up takes super long

next obsidian Oct 31, 2020, 3:35 AM

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Like the fuckinf Lurie book?

sturdy marsh Oct 31, 2020, 3:35 AM

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ye

next obsidian Oct 31, 2020, 3:35 AM

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Fuck that

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I mean

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I’m doing a class that covers DM-stacks so

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I’m already getting sort of higher pilled but

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Maaaaaaaan F that

bleak abyss Oct 31, 2020, 3:36 AM

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Actually you know what fuck it I'm gonna start working through Harder's book rn

next obsidian Oct 31, 2020, 3:36 AM

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Is it comprehensible? I feel like my foundations would be too weak

sturdy marsh Oct 31, 2020, 3:36 AM

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but it's motivated by fairly concrete stuff

next obsidian Oct 31, 2020, 3:37 AM

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I guess it’s more like for how abstract that is

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I don’t see the limitations of scheme theory

sturdy marsh Oct 31, 2020, 3:37 AM

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moduli of elliptic curves

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"they have automorphisms"

next obsidian Oct 31, 2020, 3:37 AM

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To appreciate why this abstract setting is better at which point it would be like “hurr durr abstract good”

sturdy marsh Oct 31, 2020, 3:37 AM

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nah it's super well motivated

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we want moduli space + universal bundle

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use automorphism of elliptic curve to construct a twisted bundle

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so we can't have representing objects within the category of schemes

next obsidian Oct 31, 2020, 3:38 AM

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Is the class covering more of the like basics leading into it at least. Like Lurie’s book like starts out just incomprehensible unless you know a lot of this general higher category nonsense

sturdy marsh Oct 31, 2020, 3:38 AM

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you need something that remembers some more data

bleak abyss Oct 31, 2020, 3:38 AM

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Yeah my impression about the importance is just that sometimes you can't have a "moduli scheme of blah" but you have a "moduli stack of blah"

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And then if you can still do geometry on those stacks

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Then that's something that makes sense to think about

next obsidian Oct 31, 2020, 3:39 AM

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I thought Spectral AG is like even on higher shit than just stacks tho

sturdy marsh Oct 31, 2020, 3:39 AM

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Is the class covering more of the like basics leading into it at least. Like Lurie’s book like starts out just incomprehensible unless you know a lot of this general higher category nonsense
The first class did some motivating stuff, but after that we have just been doing quasicategories

bleak abyss Oct 31, 2020, 3:39 AM

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Oh I think we were both talking about your class on DM stacks

next obsidian Oct 31, 2020, 3:39 AM

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Hmmm?

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We = who?

bleak abyss Oct 31, 2020, 3:39 AM

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Brofibration and I

next obsidian Oct 31, 2020, 3:40 AM

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Wait what

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That’s in winter tho

bleak abyss Oct 31, 2020, 3:40 AM

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"nah it's super well motivated" -Brofibration

it = DM stacks

next obsidian Oct 31, 2020, 3:40 AM

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Oh

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Ohhhh lol

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I was talking about spectral Algebraic geometry

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This entire time

sturdy marsh Oct 31, 2020, 3:40 AM

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I was talking about both

bleak abyss Oct 31, 2020, 3:41 AM

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Oh I've got no idea what motivates spectral stuff

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Is it spectral in the sense of AT spectra?

sturdy marsh Oct 31, 2020, 3:41 AM

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ye

bleak abyss Oct 31, 2020, 3:41 AM

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Of cohomology theories and shit

sturdy marsh Oct 31, 2020, 3:41 AM

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yup

bleak abyss Oct 31, 2020, 3:41 AM

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I see. Yeah I've got nothing remotely close to a perspective there

sturdy marsh Oct 31, 2020, 3:42 AM

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there's multiple things that lead up to SAG, one way is moduli stuff, another way is some weird link between the bordism ring and formal group laws

next obsidian Oct 31, 2020, 3:42 AM

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Tfw

bleak abyss Oct 31, 2020, 3:42 AM

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:0

next obsidian Oct 31, 2020, 3:42 AM

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Idk enough of this stuff to have an actual discussion on it haha

bleak abyss Oct 31, 2020, 3:42 AM

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Did someone say bordism ring

sturdy marsh Oct 31, 2020, 3:43 AM

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and there's also derived AG which is closely related and motivated by intersection problems

bleak abyss Oct 31, 2020, 3:43 AM

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Yeah I've heard derived AG is actually geometry for sure

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Or I mean only inasmuch as AG in general is geometry lol

sturdy marsh Oct 31, 2020, 3:43 AM

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idk what geometry is

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right now geometry = sheaves

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or just some structure with descent data

next obsidian Oct 31, 2020, 3:44 AM

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I’m not Chad enough to have this woke biew of geometry

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Geometry to me is still triangles :(

sturdy marsh Oct 31, 2020, 3:44 AM

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Oh no I believe geometry is triangles too

next obsidian Oct 31, 2020, 3:45 AM

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But I have to pretend geometry = the types of functions on your space you consider

sturdy marsh Oct 31, 2020, 3:45 AM

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But all the courses with the word in geometry in them seem to take this viewpoint

next obsidian Oct 31, 2020, 3:45 AM

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Aka the sheaf

sturdy marsh Oct 31, 2020, 3:45 AM

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yup

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you cant have moduli spaces of triangles in the cat of schemes too

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if you want a universal triangle bundle that is

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same idea

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fml all the sugar has given me a headache

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and the rest of the peanut butter cups are staring at me smugly

bleak abyss Oct 31, 2020, 3:48 AM

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Lmao much as I'm pro peanut butter cups better to hide them away lol

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And drink some water

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Also I'm officially hijacking this channel to be AG commentary

next obsidian Oct 31, 2020, 3:50 AM

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Go to geometry

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that's the AG channel

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Algebra is for comm-alg that is secretly AG

bleak abyss Oct 31, 2020, 3:52 AM

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Well yeah the first bit of this book is category theory and homological algebra straight up lol

next obsidian Oct 31, 2020, 3:52 AM

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lol

sturdy marsh Oct 31, 2020, 3:52 AM

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just like all books should be

next obsidian Oct 31, 2020, 3:52 AM

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what kind of homological algebra tho?

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Is it like here's Tor and Ext in a like ad-hoc way

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or is it like here's chain maps, chain homotopies...

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here's derived functors

bleak abyss Oct 31, 2020, 3:54 AM

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📎 harder.png

next obsidian Oct 31, 2020, 3:54 AM

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Gamma-modules

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bwahahahaha

bleak abyss Oct 31, 2020, 3:54 AM

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This guy's notation and typesetting are... interesting

sturdy marsh Oct 31, 2020, 3:54 AM

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oh yeah there's another motivating thing for Lurie style stuff, simplicial stuff let's you do stuff even when the category is not abelian

next obsidian Oct 31, 2020, 3:55 AM

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Yeah but I like R-mod...

sturdy marsh Oct 31, 2020, 3:55 AM

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so you replace homological algebra with homotopical algebra

next obsidian Oct 31, 2020, 3:55 AM

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blech

sturdy marsh Oct 31, 2020, 3:55 AM

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it's actually not bad at all

next obsidian Oct 31, 2020, 3:55 AM

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bruh

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you don't even have a ring anymore

sturdy marsh Oct 31, 2020, 3:56 AM

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you have simplicial rings

next obsidian Oct 31, 2020, 3:56 AM

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😔

sturdy marsh Oct 31, 2020, 3:56 AM

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the thing is Ring is not an abelian category

next obsidian Oct 31, 2020, 3:56 AM

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yeah

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But R-mod is and R-mod is wonderful

sturdy marsh Oct 31, 2020, 3:56 AM

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simplicial stuff is supposed to somehow let you get over that

next obsidian Oct 31, 2020, 3:56 AM

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euch

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maybe in a years time

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I will be based (cringe) enough to adopt this view

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but for now I like me rings

sturdy marsh Oct 31, 2020, 3:57 AM

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but ringed spectra are also rings

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R-Points of Spec of the bordism spectrum correspond to 1-dim formal group laws over R

next obsidian Oct 31, 2020, 3:59 AM

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Chmonkey has left the room

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Too much for me

sturdy marsh Oct 31, 2020, 3:59 AM

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but anyway, back to finite group theory

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isaacs looks pretty good

next obsidian Oct 31, 2020, 3:59 AM

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It's cool

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I like it, apparently his writing style is nice

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and I'm not super deep into it to judge it yet, but I'm inclined to agree

sturdy marsh Oct 31, 2020, 4:00 AM

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I think I read a bit of his algebra book

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the noncommutative algebra part

next obsidian Oct 31, 2020, 4:00 AM

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He uses right-actions though

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which I think is more common among group theorists

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and even like writing f(x) as xf is not uncommon among group theorists

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So for me it's just a minor mindset shift

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but it ends up being practically inconsequential

snow cliff Oct 31, 2020, 4:01 AM

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what's the point of having two different notations then? Or how'd they come about?

next obsidian Oct 31, 2020, 4:01 AM

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The main thing is mainly for functions stuff

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it makes composition "look right"

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since f\circ g(x) is actually g(f(x))

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where as if you do the other one (x)f\circ g

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is just ((x)f)g

bleak abyss Oct 31, 2020, 4:02 AM

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Tbh I think it's morally correct

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But it's non-standard so

next obsidian Oct 31, 2020, 4:02 AM

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And it's also nice for group actions

snow cliff Oct 31, 2020, 4:02 AM

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morally correct haha

bleak abyss Oct 31, 2020, 4:02 AM

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Kinda not good to learn it since you have to unlearn it and that's annoying

next obsidian Oct 31, 2020, 4:02 AM

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for example if you denote conjugation by g^x = x^-1gx

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this is a right action

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but then (g^x)^y = g^{xy}

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and things like that

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I think in an ideal world this is actually the standard we all were brought up with

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but it isn't, so most people even if they think the other is better, just use what's most common

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but for group theory specifically I think right group actions are much better, and since they're so common more people adopt that notation

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FWIW Isaacs does use functions like most people

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it's just actions and cosets where he does right ones

snow cliff Oct 31, 2020, 4:04 AM

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gotcha

sturdy marsh Oct 31, 2020, 4:06 AM

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the font of the book is very calming lol

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or maybe group theory is just soothing

next obsidian Oct 31, 2020, 4:06 AM

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Does it do normal rep theory too?

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I learned rep thoeyr

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but can't remember any of it and want to give it another go sometime

sturdy marsh Oct 31, 2020, 4:08 AM

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there's a book called rep theory - a homological algebra point of view

next obsidian Oct 31, 2020, 4:10 AM

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I guess like morphisms from R[G] or whatever

fierce perch Oct 31, 2020, 4:10 AM

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Group theory is so difficult

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Its like solving math olympiad stuff to me

next obsidian Oct 31, 2020, 4:12 AM

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I like group theory, I certainly find it easier than AG haha

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But also my experience with group theory is that high-powered theorems make previously hard problems near trivial

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to a degree I don't feel in other subjects

sturdy marsh Oct 31, 2020, 4:13 AM

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example?

next obsidian Oct 31, 2020, 4:14 AM

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I mean Sylow makes a lot of problems about solvability or simplicity of low order groups really trivial

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Classifying groups of certain orders with Sylow and semi direct products becomes a lot easier

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You can use a lot of stuff about the structure of p-groups, etc.

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Idk, it just feels like when my friends ask me a group theory question, I'm like "oh apply __ theorem" and it's trivially done

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moreso than like problems in ring theory they would ask me or something

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Having developed a similar amount of theory, and at this point probably even more ring theory through comm. alg

sturdy marsh Oct 31, 2020, 4:17 AM

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fair enough

paper flint Oct 31, 2020, 8:43 AM

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https://cdn.discordapp.com/attachments/737043741168238673/772017689220743168/697848465819959388.png

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I have never dealt with a group of polynomials with its coefficients coming from Z_10 before and hence would like some help to understand what this group G is actually about.

carmine fossil Oct 31, 2020, 8:46 AM

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Since we are talking only wrt addition,think of 7x^2+5x+4 as (7,5,4)

paper flint Oct 31, 2020, 8:47 AM

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Addition modulo 10?

carmine fossil Oct 31, 2020, 8:47 AM

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Yes

paper flint Oct 31, 2020, 8:47 AM

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Ah okay, so identity would be having all coefficients as 0?

carmine fossil Oct 31, 2020, 8:47 AM

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Yes

paper flint Oct 31, 2020, 8:47 AM

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Okay, that makes sense. Thanks!

paper flint Oct 31, 2020, 9:16 AM

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https://cdn.discordapp.com/attachments/737043741168238673/772025618023186432/697848465819959388.png

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My gut feeling about this one is that H need not be a subgroup(since I cannot deduce any condition for closure), but I'm not completely sure.

carmine fossil Oct 31, 2020, 9:20 AM

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If G is finite,yes

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Otherwise,no

paper flint Oct 31, 2020, 9:21 AM

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How is it a subgroup when G is finite?

carmine fossil Oct 31, 2020, 9:21 AM

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Let a be in H,then a^-n=1 for some n implying 1 is in H(Identity)

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Which implies a^-1 in H(Inverse)

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Which would imply ab in H,for all a,b in H(Closure)

paper flint Oct 31, 2020, 9:23 AM

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Need some time to process this.

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Okay, so if G were to be finite, every element in G will have finite order, and the inverse of every element will have the same finite order as well. How does this add up to your first point?

carmine fossil Oct 31, 2020, 9:26 AM

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if a in H,By that condition you get a^-2 is in H,which implies a^4 is in H

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Which means all elements of form $a^{4^k}$ will be in H

cloud walrusBOT Oct 31, 2020, 9:28 AM

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DrunkenDrake:

📎 734020733595942952.png

carmine fossil Oct 31, 2020, 9:29 AM

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But,There are only finitely many elements in H

paper flint Oct 31, 2020, 9:29 AM

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Okay, makes sense. How about odd powers of a?

carmine fossil Oct 31, 2020, 9:29 AM

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Well,you get a^n=1 for some n

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Don't think you can get anything else

nova plank Oct 31, 2020, 9:30 AM

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This seems off. If a has order 3, how is a^(4^k) going to be 1 for any k?

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But the question does not specify G being finite or not, so you can give an infinite counterexample

paper flint Oct 31, 2020, 9:32 AM

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Ah, counterexample...

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Didn't even think of that.

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Although I didn't exactly process this as a finite/infinite group discrepancy so I can be excused. :p

carmine fossil Oct 31, 2020, 9:33 AM

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I don't mean to say a^4^k is 1

paper flint Oct 31, 2020, 9:33 AM

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I'll still need some time to actually understand this problem. Thanks for the help guys.

carmine fossil Oct 31, 2020, 9:33 AM

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I mean to say a^(4^k)=a^(4^l) for some k and l,k Not l

paper flint Oct 31, 2020, 9:34 AM

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@carmine fossil Yeah, but this way odd powers of a won't necessarily be in H, right? That blows up closure?

carmine fossil Oct 31, 2020, 9:34 AM

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Well,you get a^n=1 for some n
This is the only thing we are getting from this

paper flint Oct 31, 2020, 9:35 AM

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So basically it need not even be a subgroup even when G is finite?

nova plank Oct 31, 2020, 9:36 AM

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I still don't see how you get that drake. If a has order 3, no power of 4 as the exponent of a can give you 1.

carmine fossil Oct 31, 2020, 9:36 AM

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you get a^(4^1)=a

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Which means a^(4^1-1)=1

nova plank Oct 31, 2020, 9:36 AM

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Why does it mean that?

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You don't have that a^(-1) is in H

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Oh, obviously it does mean that

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I meant why is it in H

carmine fossil Oct 31, 2020, 9:41 AM

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mb

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It doesn't

paper flint Oct 31, 2020, 9:43 AM

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So what exactly should my argument look like? Should I produce a counterexample using a finite group G?

nova plank Oct 31, 2020, 9:44 AM

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It can be an infinite group

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And maybe look at our discussion here, and choose an H that contains a, a^(-2), a^4, a^(-8), etc. But not a^(-1).

paper flint Oct 31, 2020, 9:45 AM

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Oh, okay. Thanks again.

paper flint Oct 31, 2020, 10:34 AM

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https://cdn.discordapp.com/attachments/764073465241534484/772045458809421824/697848465819959388.png

nova plank Oct 31, 2020, 10:49 AM

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@paper flint It says more than 3 elements of order 2, but then you only choose 3: a, b, and c

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You can start with a, b, c, and d

paper flint Oct 31, 2020, 11:04 AM

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Oh, when I was writing it in my notebook I did try the 4 element case but I had ab,bc,cd and da. Obviously ab≠bc, ab≠da for the same reasons, but again I couldn't figure out the uniqueness of ab and cd, or that of ac and bd.

chilly ocean Oct 31, 2020, 11:08 AM

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just to be more clear, it is not the case that an abelian group with 3 elements has at least 7 elements of order 2, ~~consider (Z2)^3~~

paper flint Oct 31, 2020, 11:10 AM

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Yeah, that seems reasonable. I just got a bit mixed up because of 'more than three'.

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,preamble

pallid ember Oct 31, 2020, 11:11 AM

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just take a,b,c d and work with ab ac and ad

chilly ocean Oct 31, 2020, 11:11 AM

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wait i'm a dumbass lol, counterexample should be (Z2)^2

paper flint Oct 31, 2020, 11:12 AM

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AH CRAP. Sorry for opening this preamble up here, the channel switched too quickly.

pallid ember Oct 31, 2020, 11:12 AM

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i got u dw

paper flint Oct 31, 2020, 11:12 AM

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Thanks!

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Aah yes, that totally makes sense!!! Tf I should've written da as ad right from the get go to see it more clearly haha.

pallid ember Oct 31, 2020, 11:13 AM

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yea

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careful tho theres a sneaky case in there

paper flint Oct 31, 2020, 11:14 AM

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There is? I assume a,b,c,d to be distinct non-identity elements, and G is given to be Abelian. Anything I'm missing?

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But a is its own inverse, and inverse of an element is unique.

pallid ember Oct 31, 2020, 11:15 AM

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yea thats not true

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i meant u could have ab=c

paper flint Oct 31, 2020, 11:15 AM

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Yeah, this was the roadblock I hit in the 3 element scenario.

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How do I fix this?

pallid ember Oct 31, 2020, 11:16 AM

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split it into cases

chilly ocean Oct 31, 2020, 11:16 AM

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i recall doing problems like this in algebra class, have you considered the fact that multiplication is a bijection? EDIT: but i could be totally off, just a thought

pallid ember Oct 31, 2020, 11:16 AM

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ab=c and ab doesnt equal c

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if ab does equal c then u have ab=c, ac=b and bc=a

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so u can just use ad bd and cd

paper flint Oct 31, 2020, 11:17 AM

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@chilly ocean I haven't studied isomorphisms/homomorphisms yet 😅

chilly ocean Oct 31, 2020, 11:17 AM

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sorry, i can getting my terms mixed up, i meant bijection (well, i guess this is equivalent to "cancellation works")

paper flint Oct 31, 2020, 11:17 AM

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Thanks, @pallid ember , I'll give that a try now.

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Hmmmm I'm not sure I'm following @chilly ocean

chilly ocean Oct 31, 2020, 11:18 AM

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yeah sorry, i am not really making sense

paper flint Oct 31, 2020, 11:18 AM

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Ah nw. Thanks for the example.

steep hull Oct 31, 2020, 12:51 PM

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||Start with distinct a,b of order 2. We know that each non-identity element in <a,b> has order 2. If there are more than 3 elements of order 2, there exists a c of order 2 not in <a,b>. Since c is not in <a,b>, c<a,b> and <a,b> are disjoint. This gives four more elements of order 2.||

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||In general, the 2-torsion of an abelian group is the maximal boolean subgroup of the group||

paper flint Oct 31, 2020, 1:52 PM

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Thank you! I really don't understand your second statement yet, but the first one makes sense to me.

next obsidian Oct 31, 2020, 9:00 PM

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Zheng lmfao what a statement

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I wonder if you could prove this via some convoluted proof that says if you have more than 3 distinct order 2-subgroups you have to have at least 7

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Which doesn't involve you directly constructing more order 2 elements

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Oh wait, you can directly show you have (Z/2Z)^3 as a subgropu

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use the fact that if H,N are normal subgroups with trivial intersection then HN is iso to H x N

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take two order 2 elements a,b consider <a> iso to <a> x

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then that has 3 order 2 elements, so take c outside of it

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then <a><c> iso to (<a> x )<c> iso to <a> x x <c>

steep hull Nov 1, 2020, 1:03 AM

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||The overkill proof is just the Fundamental Theorem of Abelian Groups||

sturdy marsh Nov 1, 2020, 5:21 AM

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📎 unknown.png

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heh

smoky cypress Nov 1, 2020, 5:22 AM

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thonk

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Now the law requires me to ask for the definition of WeakAss(M)

sturdy marsh Nov 1, 2020, 5:23 AM

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Let M be an R-mod

chilly ocean Nov 1, 2020, 5:23 AM

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weak ass GWcorbinHolyFuck

sturdy marsh Nov 1, 2020, 5:23 AM

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WeakAss(M) is the set of prime ideals minimal over the annihilators of elements of M

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weak ass :GWcorbinHolyFuck:
@chilly ocean it's gotta be intentional

paper flint Nov 1, 2020, 6:26 AM

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https://cdn.discordapp.com/attachments/737043741168238673/772342013324165130/697848465819959388.png

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https://cdn.discordapp.com/attachments/737043741168238673/772345134120960030/697848465819959388.png

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(this isn't particularly well-written, sorry about that)

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I need help solving this problem.

chilly ocean Nov 1, 2020, 6:29 AM

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I'm not sure I follow the last two sentences

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But, are you familiar with the statement that a subgroup of a cyclic group is cyclic?

paper flint Nov 1, 2020, 6:30 AM

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Haven't covered the chapter on cyclic groups yet, but that sounds reasonable.

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I was a bit incoherent at the end, my bad.

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The basic idea was just to show that the subgroup of Z_n generated by a non-identity even element from it has all its members even.

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Other than that I just covered the trivial subgroup and the group itself. I don't know what to do if an odd element is included in the subgroup.

golden pasture Nov 1, 2020, 6:42 AM

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whotf names things weakass hank hank hank

carmine fossil Nov 1, 2020, 6:42 AM

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the same guy who made brainfuck

golden pasture Nov 1, 2020, 6:42 AM

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hank

carmine fossil Nov 1, 2020, 6:43 AM

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And that" fucking concrete introduction to algebra"

pallid ember Nov 1, 2020, 6:45 AM

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maybe try finding a injection from H to itself that maps evens to odds and vice versa @paper flint

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*bijection

paper flint Nov 1, 2020, 6:45 AM

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PandaCry How

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Can I not go by cases?

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(I'm a newb to the subject so I always like breaking down proofs to simplest possible cases)

carmine fossil Nov 1, 2020, 6:47 AM

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Or show that a subgroup is generated by the smallest non zero element in it

next obsidian Nov 1, 2020, 6:47 AM

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Cases are trash

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As in, you should try to figure out ways to always reduce the number of cases you check IMO

carmine fossil Nov 1, 2020, 6:48 AM

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And then show the order is even,(if the element is odd) and you are done

next obsidian Nov 1, 2020, 6:48 AM

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Like if there’s 4 cases but you can WLOG to only show 2 that’s chad

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And also a good habit to get into, since this is moreso the habit of reducing things to easier stuff

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Which is a good habit to get into

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Especially in algebra IMO where if you quotient by the right thing you can get a lot of mileage by “assuming WLOG that...”

paper flint Nov 1, 2020, 6:49 AM

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I do use WLOG arguments for reducing redundant cases.

next obsidian Nov 1, 2020, 6:49 AM

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I mean idk, sometimes the two cases are mega trivial to check so like whatever

paper flint Nov 1, 2020, 6:49 AM

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In this particular problem I can't figure out what cases to work with.

next obsidian Nov 1, 2020, 6:50 AM

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But if you can avoid a cases argument I think it’s better to

pallid ember Nov 1, 2020, 6:50 AM

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Can I not go by cases?
@paper flint yea u consider 2 cases

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no odd numbers and at least 1 odd number

next obsidian Nov 1, 2020, 6:50 AM

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Also I haven’t put thought into what drunken drake suggested, but if it what he says does work then it’s probably useful to figure out why it works and how the argument runs

paper flint Nov 1, 2020, 6:50 AM

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I did the no odd numbers bit. Now I need to get down the at least 1 odd number bit I suppose.

next obsidian Nov 1, 2020, 6:51 AM

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Oh right, I remember the problem I think

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Right, I think cases is fine here I guess

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I didn’t really even consider this cases lol

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If it has no odds you’re just legit done

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“It has only evens”

pallid ember Nov 1, 2020, 6:52 AM

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yep

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nothing to prove

paper flint Nov 1, 2020, 6:52 AM

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Kekw I wrote a paragraph for that.

next obsidian Nov 1, 2020, 6:52 AM

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Right, then yeah this is fine

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Whaaa?

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You can assume H is a subgroup

pallid ember Nov 1, 2020, 6:52 AM

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lol wut

next obsidian Nov 1, 2020, 6:52 AM

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If it has no odds, it is only evens, so it fits

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Okay now assume it has at least 1 odd, show it’s half even half odd

pallid ember Nov 1, 2020, 6:52 AM

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this man does not believe in the excluded middle

paper flint Nov 1, 2020, 6:53 AM

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I'm just being pedantic I guess, stating things like 'an element of a finite group has a finite order' which now I realise isn't even relevant.

next obsidian Nov 1, 2020, 6:53 AM

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Oh lmao

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Yeah it isn’t about the order even right?

paper flint Nov 1, 2020, 6:53 AM

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Nope

next obsidian Nov 1, 2020, 6:53 AM

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It’s about it’s value where you like take the most natural set of represnetatives

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Aka 1 through n I guess

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(Altho I’m partial to 0 through n -1)

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Oh they did do that one

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😎

paper flint Nov 1, 2020, 6:54 AM

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Ah, I see. Also, if it has one odd and one even member, must it be equal to the group itself(just a guess)?

carmine fossil Nov 1, 2020, 6:54 AM

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No

next obsidian Nov 1, 2020, 6:54 AM

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No

pallid ember Nov 1, 2020, 6:54 AM

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nope

next obsidian Nov 1, 2020, 6:54 AM

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The other case is it’s half even half odd

carmine fossil Nov 1, 2020, 6:54 AM

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Take Z_30,and subgroup generated by 5

paper flint Nov 1, 2020, 6:54 AM

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Ah, okay.

pallid ember Nov 1, 2020, 6:55 AM

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u mean Z10?

paper flint Nov 1, 2020, 6:57 AM

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Z_30 works too, no?

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Anyway thanks for the help everyone!

pallid ember Nov 1, 2020, 7:05 AM

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f(x)= x+O mod n works @paper flint

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O is an odd number

paper flint Nov 1, 2020, 7:06 AM

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Ah niceeeeeee

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Thanks again!!!

latent anvil Nov 1, 2020, 7:11 AM

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Does anyone want a group theory problem?

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Too bad I'm telling it anyway

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Determine the number of semidirect products of A6 and C2 up to isomorphism

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(I'm not asking for help, I just figured this out with a friend and it was a lot of fun)

chilly ocean Nov 1, 2020, 7:17 AM

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What is C2? (Well, not that I want to do this problem anyway, but)

latent anvil Nov 1, 2020, 7:17 AM

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Cyclic group of order 2, sorry

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C2 = Z/2Z

sturdy marsh Nov 1, 2020, 7:45 AM

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The problem is pretty much just counting order 2 automorphisms of A6 (mod ones that induce the same semidirect product)

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Conjugation by a 2-cycle is an order two aut, and all of them induce isomorphic semidirect products

latent anvil Nov 1, 2020, 7:46 AM

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~~most of the difficulty is in the parenthetical there~~

sturdy marsh Nov 1, 2020, 7:46 AM

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as all 2 cycles are conjugate

latent anvil Nov 1, 2020, 7:49 AM

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I agree that this is a family of isomorphic semidirect products

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In fact they're all S6

sturdy marsh Nov 1, 2020, 7:51 AM

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yeah but there's more right

latent anvil Nov 1, 2020, 7:51 AM

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Yup

sturdy marsh Nov 1, 2020, 7:51 AM

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there are ones that dont come from conjugation from S6

latent anvil Nov 1, 2020, 7:51 AM

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One other trivial example I'll give you, A6 × C2 :P

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Note that you didn't consider conjugation by all elements of order 2 in S6

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Just the 2 cycles

sturdy marsh Nov 1, 2020, 7:54 AM

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yeah there are other ones too

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but im trying to figure out the outer one

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outer to S6 that is

latent anvil Nov 1, 2020, 7:54 AM

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Right

sturdy marsh Nov 1, 2020, 7:54 AM

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cant remember how it was done

latent anvil Nov 1, 2020, 7:54 AM

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This is why it's interesting for A6

sturdy marsh Nov 1, 2020, 7:55 AM

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was there just one?

latent anvil Nov 1, 2020, 7:55 AM

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Out(S6) = C2

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So there's one up to translation by Inn(S6)

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Here's an example of how to construct an outer aut: S5 has six 5-sylow subgroups. Acting on them gives an embedding of S5 into S6 as a transitive subgroup. This has index 6!/5! = 6, so acting on the cosets gives a map S6 -> S6, which turns out to be a non inner automorphism

sturdy marsh Nov 1, 2020, 7:58 AM

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noice

latent anvil Nov 1, 2020, 7:58 AM

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Also I did use gap for this

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Relevant: https://mail.gap-system.org/pipermail/forum/2011/003357.html

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There's a couple explicit facts about the automorphism and the conjugacy structure of S6/Aut(S6) which I didn't want to do by hand lol

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Like for instance you need to know there's a non inner automorphism of order 2

sturdy marsh Nov 1, 2020, 8:04 AM

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looks fun, will give it a shot in the morning

latent anvil Nov 1, 2020, 8:04 AM

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Yeah I would recommend. Took me a while though

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Have fun :)

sturdy marsh Nov 1, 2020, 8:05 AM

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oh shoot I get to sleep an extra hour tonight

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they should do this whole time switch thing for entire days

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where we just pretend a day didnt happen

latent anvil Nov 1, 2020, 8:07 AM

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I think that's called a blackout

quaint ivy Nov 1, 2020, 8:36 PM

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someday we'll just pretend 2020 didn't happen

tawny gust Nov 2, 2020, 12:33 AM

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https://www.youtube.com/playlist?list=PLi01XoE8jYoi3SgnnGorR_XOW3IcK-TP6

YouTube

Abstract Algebra

Abstract Algebra deals with groups, rings, fields, and modules. These are abstract structures which appear in many different branches of mathematics, includi...

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what do you guys think of this playlist

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does that cover the broad topic of abstract algebra or is that only an introduction

scarlet estuary Nov 2, 2020, 12:37 AM

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it basically just gives definitions and simple examples

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it is not a substitute for reading a textbook and/or taking a course.

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and certainly not a substitute for doing exercises

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(most of the learning in mathematics, especially intro courses, happens in exercises)

tawny gust Nov 2, 2020, 12:39 AM

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true

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im gonna need a good abstract algebra textbook then lol

scarlet estuary Nov 2, 2020, 12:42 AM

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https://discordapp.com/channels/268882317391429632/716264872018706443/718931426346795099

thorn delta Nov 2, 2020, 2:49 AM

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Suppose that $G$ is finite nilpotent. Let $g$ be a nontrivial element of the center of $G$ of order $p$ for some prime dividing the order of $G$. Why is it true that a sylow $p$-subgroup of $G/\langle g \rangle $ must be normal?

cloud walrusBOT Nov 2, 2020, 2:50 AM

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kxrider:

📎 195237017049759744.png

thorn delta Nov 2, 2020, 2:58 AM

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nvm, this is part of a proof by induction, and this was an application of the induction hypothesis lol

wispy glen Nov 2, 2020, 5:05 AM

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$F$ is a field. $$\frac{F[x]}{(g(x)h(x))} \cong \frac{F[x]}{(g(x))} \times \frac{F[x]}{(h(x))}$$ is this true or something similar?

cloud walrusBOT Nov 2, 2020, 5:05 AM

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gelearndeutsch:

📎 751854211486187621.png

carmine fossil Nov 2, 2020, 5:14 AM

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Just construct a surjective homomorphism from $
{F[x]} \text{ to } \frac{F[x]}{(g(x))} \times \frac{F[x]}{(h(x))}$

chilly ocean Nov 2, 2020, 5:14 AM

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chinese remainder theorem?

carmine fossil Nov 2, 2020, 5:14 AM

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And find the kernel of that

cloud walrusBOT Nov 2, 2020, 5:15 AM

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DrunkenDrake:

📎 734020733595942952.png

carmine fossil Nov 2, 2020, 5:30 AM

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Similar to crt

latent anvil Nov 2, 2020, 5:53 AM

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I don't think this is true in general?

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take g = h squarefree

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Then the lhs has a nilpotent and the right doesn't

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CRT says this is true when g and h are coprime but it can fail in general

carmine fossil Nov 2, 2020, 5:54 AM

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Is g and h being coprime a sufficient condition?

latent anvil Nov 2, 2020, 5:55 AM

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Yes, because that implies (g) + (h) = 1

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That's no longer true in F[x, y] or with more variables though. F[x, y]/(xy) and F[y] × F[x] ≈ F[x,y]/(x) × F[x, y]/(y) are not isomorphic but x and y are coprime

wispy glen Nov 2, 2020, 5:58 AM

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@chilly ocean crt has solved the problem. thanks

paper flint Nov 2, 2020, 6:25 AM

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https://cdn.discordapp.com/attachments/764073465241534484/772707203597467658/697848465819959388.png

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Is this proof okay?

thorn delta Nov 2, 2020, 6:26 AM

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yes

paper flint Nov 2, 2020, 6:26 AM

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(one thing I'd like to ask here is about the uniqueness of inverse across various subgroups: suppose I have some a\in G which is present in its subgroups H and K as well. Is it guaranteed that a^(-1)\in G is the same as a^(-1)\in H and a^(-1)\in K?)

#

yes
I see. Thanks.

thorn delta Nov 2, 2020, 6:35 AM

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to address ur parenthetical remark, when a is in G, a^{-1} is not just the inverse of a in H or K, but rather the inverse of a in G.

paper flint Nov 2, 2020, 6:39 AM

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Yeah, but that is sufficient to guarantee it's also the inverse of a in H and K, right?

#

To make my question precise: is it possible for b≠c such that ab=e in H and ac=e in K?

chilly ocean Nov 2, 2020, 6:42 AM

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Perhaps you can try proving it 🙂

paper flint Nov 2, 2020, 6:42 AM

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https://cdn.discordapp.com/attachments/764073465241534484/772710989095108628/697848465819959388.png

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Is this proof okay?

chilly ocean Nov 2, 2020, 6:43 AM

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Yeah

paper flint Nov 2, 2020, 6:43 AM

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Perhaps you can try proving it 🙂
I couldn't write anything down. Intuitively, I feel the inverse should be uniquely defined since it's a property of the operation itself and has nothing to do with the elements present in the set/group.

chilly ocean Nov 2, 2020, 6:44 AM

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You are familiar that, if you are just working in one group, the inverse is unique right?

thorn delta Nov 2, 2020, 6:45 AM

#

sure, that looks good.

is sufficient to guarantee it's also the inverse of a in H and K, right?
by definition, the inverse of a in H is the inverse of a in G. if an element a of G is in H then a^{-1}, a's inverse in G is also in H. Same with K

paper flint Nov 2, 2020, 6:45 AM

#

You are familiar that, if you are just working in one group, the inverse is unique right?
@chilly ocean Yes.

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sure, that looks good.
by definition, the inverse of a in H is the inverse of a in G. if an element a of G is in H then a^{-1}, a's inverse in G is also in H. Same with K
@thorn delta Aah okay, thanks!

thorn delta Nov 2, 2020, 6:46 AM

#

np.

paper flint Nov 2, 2020, 6:47 AM

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The centralizer of an element of a group need not be Abelian, right?

thorn delta Nov 2, 2020, 6:50 AM

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according to what u just proved, the center is contained in all of the centralizers. Whenever the center is a proper subgroup of one of the centralizers, that centralizer is non abelian.

paper flint Nov 2, 2020, 6:50 AM

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That makes sense!

#

Another question(I seem to have too many, but I'll stop after this one): If G is the symmetry group of a circle, then it has elements of every finite order as well as infinite order. Proof: For a finite order n, we mark n equally spaced points on the circle. Then under rotation, one of the points will have order n. This takes care of all finite orders. I'm not sure what to do in the infinite order case.

chilly ocean Nov 2, 2020, 6:55 AM

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Hint: characterize all angles for which rotation by that angle has finite order (maybe this is not that helpful)

thorn delta Nov 2, 2020, 6:56 AM

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right, so for finite order, you have the nth roots of unity. for infinite order, you need to find a rotation s.t. repeated applications of that rotation never returns to 1 (on the complex plane)

paper flint Nov 2, 2020, 6:57 AM

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Aah makes sense. Thanks!!!

thorn delta Nov 2, 2020, 6:57 AM

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i was about to say xd

thorn delta Nov 2, 2020, 8:13 AM

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anyone have a hint on this? Let $G$ be a group and $H$ be a subgroup of $G$. Suppose that $[H, [G,G]] = {e}$. Then show that $[[H,H],G] = {e}$.

cloud walrusBOT Nov 2, 2020, 8:13 AM

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kxrider:

📎 195237017049759744.png

latent anvil Nov 2, 2020, 8:23 AM

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weird

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Makes me think of the hall witt identity

chilly ocean Nov 2, 2020, 8:27 AM

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i would probably just try spamming random substitutions tbh

latent anvil Nov 2, 2020, 8:27 AM

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Oh actually [H, G'] = 1 is very concrete

chilly ocean Nov 2, 2020, 8:27 AM

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yeah, it means they commute

latent anvil Nov 2, 2020, 8:28 AM

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Right

#

So H <= C_G(G') which means H' <= C_G(G')', and we want to show H' <= Z(G)

#

So it suffices to show C_G(G')' <= Z(G)

#

And in fact this is equivalent by taking H = C_G(G') in the statement

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So if x and y both commute with all elements of the form g h g^-1 h^-1 then x y x^-1 y^-1 commutes with all of G

thorn delta Nov 2, 2020, 8:32 AM

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what is C_G?

latent anvil Nov 2, 2020, 8:33 AM

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The centralizer. For a subgroup K of G, C_G(K) = { g in G : gk = kg for all k in K }

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If y is in C_G(G') and x is arbitrary then x y x^-1 = x y x^-1 y^-1 y = y x y x^-1 y^-1 = (yx) y (yx) ^-1, so x^-1 y x commutes with y

#

not sure if this helps

#

then if x, y in C_G(G') we get [x, y] = x y x^-1 y^-1 = y^-1 x y x^-1 = x^-1 y^-1 x y = y x^-1 y^-1 x

#

I think

#

Wait I think this might actually follow from the hall witt identity

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So if x and y both commute with all elements of the form g h g^-1 h^-1 then x y x^-1 y^-1 commutes with all of G
This follows I mean

#

Hall Witt says
${\displaystyle [x,y^{-1},z]^{y}\cdot [y,z^{-1},x]^{z}\cdot [z,x^{-1},y]^{x}=1.}$

cloud walrusBOT Nov 2, 2020, 8:41 AM

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shamrock:

📎 228562950703546368.png

latent anvil Nov 2, 2020, 8:42 AM

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Where $[a, b, c] = [[a, b], c]$

cloud walrusBOT Nov 2, 2020, 8:42 AM

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shamrock:

📎 228562950703546368.png

latent anvil Nov 2, 2020, 8:43 AM

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If x and y are in the centralizer of G' then $[y, z^{-1}, x]$ and $[z, x^{-1}, y]$ are both trivial, since they're the commutator of something in G' and something which commutes with all of G'

cloud walrusBOT Nov 2, 2020, 8:43 AM

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shamrock:

📎 228562950703546368.png

latent anvil Nov 2, 2020, 8:45 AM

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This means $[x, y^{-1}, z]^y = 1$, so $[[x,y^{-1}],z] = [x, y^{-1}, z] = 1$. Since $y$ was arbitrary we get that $[x, y]$ commutes with $z$ for any $z$ and any $x, y\in C_G(G')$

cloud walrusBOT Nov 2, 2020, 8:45 AM

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shamrock:

📎 228562950703546368.png

latent anvil Nov 2, 2020, 8:45 AM

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@thorn delta here's a proof depending on a really weird identity that I happened to know

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the superscripts denote conjugation, idk if you're familiar with that notation. The identity is very easy to prove, just expand and cancel terms

thorn delta Nov 2, 2020, 8:49 AM

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i see, thanks. I was actually trying to apply this identity from an earlier exercise to see if i could get anywhere:
[ab,c] = [b,c]^a[a,c]
but idk idk

chilly ocean Nov 2, 2020, 8:49 AM

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I have never really liked these group theory problems. They just seem like random symbol manipulations to me

latent anvil Nov 2, 2020, 8:50 AM

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I mean this one is very much symbol manipulation lol

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The hall witt identity is, I mean

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But I don't think all group theory problems are like that

paper flint Nov 2, 2020, 9:36 AM

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I need to prove that every finite group with more than one element has an element of prime order. I think I can easily prove this for finite cyclic groups, but how to go about the general case?(or should I approach this by contradiction instead)?

carmine fossil Nov 2, 2020, 9:38 AM

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Did gallian not give you hints?

paper flint Nov 2, 2020, 9:38 AM

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None.

carmine fossil Nov 2, 2020, 9:38 AM

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Ok, That's a pretty significant theorem

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It's called cauchy's theorem

paper flint Nov 2, 2020, 9:39 AM

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Hmmmm, I'll think harder about it.

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Can you drop the hints/solution with a spoiler?

latent anvil Nov 2, 2020, 9:39 AM

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This isn't cauchy's theorem

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It is much easier

chilly ocean Nov 2, 2020, 9:39 AM

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Do you mean an element of each order p for each prime p dividing the order of G?

#

That would be cauchy's theorem

latent anvil Nov 2, 2020, 9:40 AM

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Grim, think about reducing from the general case to the case of a cyclic group

carmine fossil Nov 2, 2020, 9:40 AM

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Ok,nvm mb

#

Just take a random element

#

A power of that random element will have prime order

paper flint Nov 2, 2020, 9:42 AM

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catThink

carmine fossil Nov 2, 2020, 9:42 AM

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(Say a^n=1 and p be a prime factor a^(n/p) has prime order)

woven delta Nov 2, 2020, 9:42 AM

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Arbitrary element

paper flint Nov 2, 2020, 9:42 AM

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How do I reduce the general case to cyclic alone? It need not be the case that every finite group has a cyclic subgroup I guess? @latent anvil

latent anvil Nov 2, 2020, 9:42 AM

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no liquid

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It is the case grim

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Random element

woven delta Nov 2, 2020, 9:43 AM

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Arbitrary nonidentity element

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Why random?

latent anvil Nov 2, 2020, 9:43 AM

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r a n d o m

paper flint Nov 2, 2020, 9:43 AM

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Hmmm, I haven't covered the chapter on cyclic groups yet.

woven delta Nov 2, 2020, 9:43 AM

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Gross

latent anvil Nov 2, 2020, 9:43 AM

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well think about it

paper flint Nov 2, 2020, 9:43 AM

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Okay, then this becomes straightforward.

carmine fossil Nov 2, 2020, 9:43 AM

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You are arbitrary

latent anvil Nov 2, 2020, 9:44 AM

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you know G has some nontrivial element a

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Can you use a to construct a nontrivial cyclic subgroup of G?

paper flint Nov 2, 2020, 9:45 AM

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Hmmm, makes sense. It would have the order 2 at the very least, and since it is finite, at most as much as the order of the group itself.

#

So the order of this arbitrary element could be a prime itself, or...

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I don't know. PandaCry

carmine fossil Nov 2, 2020, 9:47 AM

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Just write a few groups down

paper flint Nov 2, 2020, 9:47 AM

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Okay.

pallid ember Nov 2, 2020, 10:01 AM

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mniip

paper flint Nov 2, 2020, 10:27 AM

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I can't figure it out.

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pandaOhNo

chilly ocean Nov 2, 2020, 10:28 AM

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which groups are you familiar with?

paper flint Nov 2, 2020, 10:30 AM

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For finite groups, I'm mostly familiar with U(n)(multiplication mod n with its elements as positive integers less than n and coprime to n), Z_n(addition mod n with elements as 0,..,n-1), dihedral groups primarily.

chilly ocean Nov 2, 2020, 10:31 AM

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actually nvm, you said you can prove it for finite cyclic groups, what's your proof?

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also, do you see why showing that every group has a cyclic subgroup is sufficient to prove the claim (after you prove the claim for finint cyclic groups)?

paper flint Nov 2, 2020, 10:36 AM

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Uhhhh I initially had the contorted idea that a cyclic group of order n generated by an element a will have powers of a through 1 to n. For even n, a^(n/2) has a prime order(2)...

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My brain's jammed. Need to think about this later. sad

pallid ember Nov 2, 2020, 10:38 AM

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it's easier than you think

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dont overthink it

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I thought you didnt cover cyclic groups yet?

paper flint Nov 2, 2020, 10:40 AM

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No, I haven't.

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The definition was dropped in the current chapter on subgroups, but the next chapter is dedicated to cyclic groups exclusively.

pallid ember Nov 2, 2020, 10:43 AM

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yea you dont need cyclics for this anyway

paper flint Nov 2, 2020, 10:44 AM

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Can you give the answer(with a spoiler)? 😓

chilly ocean Nov 2, 2020, 10:45 AM

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You barely need the notion of cyclic groups

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And anyway, you mentioned you already know them (Z_n)

pallid ember Nov 2, 2020, 10:58 AM

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|| let |G|=m and |a|=n, a in G, if n is prime then youre done. Otherwsie n has a prime factor p, so n=pq, and thus a^q must have order p.||

upper inlet Nov 2, 2020, 4:18 PM

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lemon doing group theory??

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🤣 🤣

next obsidian Nov 2, 2020, 6:37 PM

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Ted you should derive a formula which tells you the order of g^a given the order of g

chilly ocean Nov 2, 2020, 6:37 PM

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Ted you should give a talk

chilly ocean Nov 2, 2020, 8:04 PM

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Hi, I have a question about group theory : can someone tell me how to have an intuition of what is a linear representation and irreductible ones ?
I find hard to internalize it

chilly ocean Nov 2, 2020, 8:36 PM

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isn't there a usual example that we can think of every time we wonder what is a representation ?

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Like S_3 acting on the vector space R^3 by permuting the basis vectors e_1, e_2, e_3
@open torrent but here for instance, where is the notion of representation ?

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me neither

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and that is the problem haha

cloud walrusBOT Nov 2, 2020, 8:44 PM

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Viburnum:

📎 731361064108818443.png

chilly ocean Nov 2, 2020, 8:45 PM

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and this matrix is the representation of (12)?

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ok I will think about that, thanks for you time

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actually I am studying it for physics

#

I'm following a group theory math lecture that focus on application in particles physic

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It is said that "an elementary particle is an irreductible representation of Poincare group" (what a sentence !), I think it is a central notion that I need to understand haha

thorn delta Nov 3, 2020, 2:37 AM

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i know its true that given a group $G$, a subgroup $H$, and a normal subgroup $K$ contained in $H$, we have that $H/K < Z(G/K)$ if and only if $[H,G] < K$. Is there any true statement like $H/K = Z(G/K)$ if and only if $[H,G] = K$?

cloud walrusBOT Nov 3, 2020, 2:37 AM

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kxrider:

📎 195237017049759744.png

thorn delta Nov 3, 2020, 2:39 AM

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Suppose that [H,G] = K. Let gK be in the center of G/K. Then for all x in G, [g, x] is in K which is contained in [H,G]. So we have [g,x] is in [H,G], but i don't think that is enough to conclude that g is in H sad

chilly ocean Nov 3, 2020, 2:51 AM

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This is even false for abelian groups, I think

thorn delta Nov 3, 2020, 2:53 AM

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rip, okay

pallid ember Nov 3, 2020, 10:05 AM

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why u lauffin @upper inlet

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🤨

flat marsh Nov 3, 2020, 5:28 PM

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I am finding the cosets of {1,-1} in {1,-1,i,-i} under multiplication and am I correct in saying there are only 2 distinct cosets since the coset formed by multiplying by i and -i are identical?

carmine fossil Nov 3, 2020, 5:28 PM

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They are the same

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Yes(Doesn't matter if you take left or right cosets in this case)

chilly ocean Nov 3, 2020, 8:29 PM

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Hi ! Someone is familiar with birdtracks here ?

next obsidian Nov 4, 2020, 3:58 AM

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If inside the tensor product $M\otimes N$, the simple tensors $m\otimes n = m'\otimes n$ for all $n \in N$, can we conclude that $m = m'$?

cloud walrusBOT Nov 4, 2020, 3:59 AM

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Chmonkey:

📎 146419489083228160.png

sturdy marsh Nov 4, 2020, 4:01 AM

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No

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tensor with 0 module

next obsidian Nov 4, 2020, 4:01 AM

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Right

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lmao

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😔

sturdy marsh Nov 4, 2020, 4:02 AM

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xD

#

do you have a geometric interpretation of what it means for a ring to be cohen-macauley?

next obsidian Nov 4, 2020, 4:03 AM

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Nope

#

I haven't really even gone into what the ring-theoretic definition is, it's about regular sequences right?

sturdy marsh Nov 4, 2020, 4:03 AM

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yup

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regular sequences sorta correspond to complete intersections right?

#

nvm

#

in the cohen macauley case they would I think

astral maple Nov 4, 2020, 4:11 AM

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Hi, this might be a simple question but I'm having a hard time with the topic at the moment

#

Hi lol

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I should show that the ring Z[1/2, i*sqrt(5)] consists of elements of the form a + bisqrt(5) with a,b \in Z[1/2]

#

Yeah I don't know how to alternatively write the left one

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I understand the definition but idk how to use it in a proof

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Z[1/2, i*sqrt(5)]

#

So we're looking at the smallest subset of C that contains Z and {1/2, isqrt(5)} right?

#

Yeah but how can I use that definition for an inclusion proof

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Well subring instead of subset

#

More concrete

#

Oh right perfect thanks

#

Is it ok if I ping you if I run into problems with either inclusion?

#

I think I have a good starting point

astral maple Nov 4, 2020, 5:03 AM

#

Ok so i was able to show that a + bisqrt(5) with a,b \in Z[1/2] is a subset of Z[1/2, isqrt(5)]

#

But I cannot find a general element of Z[1/2, isqrt(5)] to show that it can be described using a + bisqrt(5)

#

Any ideas?

next obsidian Nov 4, 2020, 5:07 AM

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How are you defining this ring?

#

You should be able to say this is the same as (Z[1/2])[isqrt(5)] but proving that depends on the definitions you're using

#

oh, I see you've done it via smallest subring

#

Ok so i was able to show that a + bisqrt(5) with a,b \in Z[1/2] is a subset of Z[1/2, isqrt(5)]
@astral maple Once you do this, it suffices to show that the set of a + bisqrt(5) where a,b in Z[1/2] form a subring

#

You have already shown this is a subset of Z[1/2,isqrt(5)], but if you show it's also a subring then Z[1/2,isqrt(5)] is a subset of it by virtue of being thew smallest subring containing 1/2 and isqrt(5)

#

Note that 1/2 and isqrt(5) are also in the set of a + bisqrt(5) where a,b in Z[1/2] if you want to be very thorough

#

Really, in general you should show that Z[x,y] = Z[x][y]

#

The definition of Z[x,y] in words is "smallest subring containing Z,x, and y"

#

while the definition of Z[x][y] in words is "smallest subring containing Z[x], and y" while Z[x] is "smallest subring containing Z, and x"

#

you can show the two are subsets of each other really really easily using this "smallest subring containing..." nonsense

astral maple Nov 4, 2020, 5:15 AM

#

Oh wow I find that approach very nice

#

With showing that it’s a subring

#

I’m trying to find the definition we used in the lecture since I always work with the book and it’s really inadequate in this case

next obsidian Nov 4, 2020, 5:16 AM

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Hiding here is the fact that you are embedding Z as a subring of C

#

Which is why the first thing I asked is how you even defined Z[1/2,isqrt(5)]

#

you can make sense of this in a way that doesn't include an embedding into C by means of a quotient of Z[x] which also let's you do the same sort of thing and show that Z[a,b] = Z[a][b]

astral maple Nov 4, 2020, 5:18 AM

#

Yes I think we also used the definition with the general form of the elements

#

I think

next obsidian Nov 4, 2020, 5:18 AM

#

But it's important to understand what's really going on here

astral maple Nov 4, 2020, 5:18 AM

#

Which would make the Z[a,b] = Z[a]Z[b] rather easy

#

But much less elegant

next obsidian Nov 4, 2020, 5:18 AM

#

This for example is important if you want to make sense of F_2[i]

#

You really mean you just adjoin some arbitrary element i which satisfies i^2 = -1 = 1 in F_2

#

although actually F_2[i] might be really really bad now that I think of it because of char 2 shenanigans

astral maple Nov 4, 2020, 5:21 AM

#

It’s in German but essentially the definition with minimal subring and general elements

next obsidian Nov 4, 2020, 5:21 AM

#

Right

#

Yeah

#

so this is the minimal subring

astral maple Nov 4, 2020, 5:22 AM

#

Yeah

next obsidian Nov 4, 2020, 5:22 AM

#

This lends itself to this proof nicely

astral maple Nov 4, 2020, 5:22 AM

#

And the second equality was proven

next obsidian Nov 4, 2020, 5:22 AM

#

just show that S[a,b] = S[a][b]

#

in general

#

Given an embedding S as a subset of some ambient ring R

astral maple Nov 4, 2020, 5:22 AM

#

I mean you can also use the second one

#

Second definition

next obsidian Nov 4, 2020, 5:23 AM

#

I mean you could

#

but then you'd have to show the two are equivalent

astral maple Nov 4, 2020, 5:23 AM

#

And that would make the proof trivial

next obsidian Nov 4, 2020, 5:23 AM

#

Under some natural choice of identificato9in

astral maple Nov 4, 2020, 5:23 AM

#

Oh yeah that has been shown

next obsidian Nov 4, 2020, 5:23 AM

#

Oh

#

lol

astral maple Nov 4, 2020, 5:23 AM

#

It’s essentially trivial lol

#

You just turn two sums into one

#

I’m still surprised by the approach with the proving that it’s a subring lol

#

It’s so elegant

astral maple Nov 4, 2020, 6:30 AM

#

Ok now I have to show that Z[1/2, isqrt(5)] is a Euclidean ring with respect to the norm 2^{2n}(a^2 + 5b^2) for minimal n such that 2^na, 2^nb \in Z

#

If we write a = k/2^m, b = l/2^p with k, l in Z we can see that n = max(m, p)

#

I’m not sure that helps

next obsidian Nov 4, 2020, 6:36 AM

#

Okay so

#

this works for a lot of things

#

one of them is Z[i]

#

I'm not going to bother looking at the actual formula, I'm ALMOST certain this should work

#

but if you have p, and want to divide by q and need the remainder r

#

p/q might not exist in Z[1/2, isqrt(5)] but it does exist in C

#

if you think about it hard enough, you'll realize Z[1/2, isqrt(5)] inside of C forms a lattice, it's a grid of points

#

so p/q falls inside of one of these sort of "boxes" which Z[1/2,isqrt(5)] forms

#

so you want to estimate p/q by the closest point of Z[1/2, isqrt(5)]

#

Let this be called r' for now, the closet point

#

then p/q - r' should be small, hopefully less than 1 I think?

#

multiply through by q

#

then you get that p - r'q should be less than q

#

Try and make this idea work I guess

astral maple Nov 4, 2020, 6:41 AM

#

I’m slowly understanding the approach

#

But I’m not sure why it works

next obsidian Nov 4, 2020, 6:42 AM

#

The idea is just what I said

#

Division with remainder is like

#

approximating the actual division

#

you go "okay so p/q might not actually exist in this ring, but like I can get close enough"

#

and you just want the remainder to be small

#

so you can just cheat and go to C where p/q DOES exist

astral maple Nov 4, 2020, 6:43 AM

#

Yeah it’s just difficult because I’ve just been introduced to the topics

next obsidian Nov 4, 2020, 6:43 AM

#

then if you can estimate p/q by an element of the ring closely

#

once you clear denominators things work

astral maple Nov 4, 2020, 6:44 AM

#

📎 image0.jpg

#

This is the definition we’re working with

#

Now the multiplicativity is clear so we just need to show that phi is a degree function

next obsidian Nov 4, 2020, 6:45 AM

#

Right, which is that the euclidean algorithm works

#

which is what I described above

#

i used different letters

astral maple Nov 4, 2020, 6:46 AM

#

p and q are g and f in the above definition right?

next obsidian Nov 4, 2020, 6:46 AM

#

right

#

so we want to divide g by f is the idea

astral maple Nov 4, 2020, 6:46 AM

#

Yup

next obsidian Nov 4, 2020, 6:46 AM

#

so under this notation

#

the element of Z[1/2,isqrt{5}] which best estimates g/f is called q

#

so d(g/f - q) < 1 is our desire

#

then when you clear denominators you should get that d(g - qf) < d(f)

astral maple Nov 4, 2020, 6:47 AM

#

Oh yes

next obsidian Nov 4, 2020, 6:47 AM

#

Oh, so the remainder is going to be g - qf

#

so you can write this as like g = qf - (g -qf)

#

so g -qf = r

astral maple Nov 4, 2020, 6:48 AM

#

Makes sense yup

next obsidian Nov 4, 2020, 6:48 AM

#

but d(r) < d(f)

astral maple Nov 4, 2020, 6:49 AM

#

Showing (i) is just a matter of using the definition right?

#

And what happens in case r = 0

#

Which would mean that our point lies on one of the edges of the boxes (from the “grid” you mentioned)

next obsidian Nov 4, 2020, 6:51 AM

#

Right

#

Exactly

#

By edge we actually mean a vertex here though

astral maple Nov 4, 2020, 6:52 AM

#

Yeah I meant edge in a visual sense hahah

#

Vertex

next obsidian Nov 4, 2020, 6:52 AM

#

Anyway, this should be sufficient?

#

The reality is I haven’t done this specific case with that specific norm

#

But I’ve done similar stuff with very similar norms

astral maple Nov 4, 2020, 6:53 AM

#

No yeah this should probably be the approach

#

Thanks a lot

next obsidian Nov 4, 2020, 6:53 AM

#

No worries

astral maple Nov 4, 2020, 6:53 AM

#

I’m new to these problems so it helps to have someone offer insights

next obsidian Nov 4, 2020, 6:53 AM

#

You should keep this “pretend fractions exist” thing

#

It can come in handy

astral maple Nov 4, 2020, 6:54 AM

#

I see

latent anvil Nov 4, 2020, 7:16 AM

#

@next obsidian F2[i] isn't reduced lol

#

It's the dual numbers over F2

next obsidian Nov 4, 2020, 7:32 AM

#

I know Sham

#

That's why I said "there might actually be issues here lol"

astral maple Nov 4, 2020, 8:54 AM

#

I can understand the proof sketch now but I’m still having difficulties formalizing the thoughts

#

I mean the entire “pretending fractions exist”

#

And the geometric intuition

wooden ember Nov 4, 2020, 3:07 PM

#

Does anyone think they could help me in #help-8 ?

woven obsidian Nov 4, 2020, 3:57 PM

#

I want to show that the function f(x,y)=x/y, f(0,0)=0, defined on the curve parametrized by (t^3,t^2) is not a polynomial function, but I'm not sure how to do that

#

It seems like it should be straightforward, but I don't really have any ideas except assuming f(x,y) is the restriction of some polynomial F and trying to approach y=0. Not sure how to do it though

hot lake Nov 4, 2020, 4:06 PM

#

are you asking if t is in k[t²,t³] ?

eager kindle Nov 4, 2020, 9:42 PM

#

https://cdn.discordapp.com/attachments/697587362019934228/773658230055370802/unknown.png

#

Okay, I got all the way to subring, but what is the maximal ideal?

#

It looks like it has a whole bunch, everything with a denominator divisible by q where q is a prime not equal to p looks like an ideal of Z_(p), right?

#

Oh wait that's not right, 1/2 + 1/2 takes you outside that

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Okay so I don't think it has ANY ideals except the trivial.

delicate bloom Nov 4, 2020, 9:46 PM

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$p \bZ_{(p)}$ is an ideal

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there are many more ideals similar to this

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dum bot 😠

eager kindle Nov 4, 2020, 9:47 PM

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Is Tekkitbot asleep?

cloud walrusBOT Nov 4, 2020, 9:47 PM

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Merosity:

📎 622029249053982720.png

delicate bloom Nov 4, 2020, 9:47 PM

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lol good morning texit

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it's been hella laggy lately I wonder why

eager kindle Nov 4, 2020, 9:48 PM

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Okay but it claims only one ideal of Z_(p) is maximal.

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Which strikes me as... well, wrong.

delicate bloom Nov 4, 2020, 9:48 PM

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yeah

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the other ideals look like p^k Z_(p)

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try to reason out why these are ideals first at least

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and then which ones are contained in which

vital quail Nov 4, 2020, 9:49 PM

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just think about what aZ_(p) becomes

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if say a = bp^k with b coprime to p

eager kindle Nov 4, 2020, 9:51 PM

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Okay so you end up with things of the form pa, if you're saying cancel the denominator

vital quail Nov 4, 2020, 9:51 PM

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so basically the b in bp^k gets ||swallowed by the denominator since it can have b no problem||

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but it can't ||swallow the p power since it can't have any ps ||

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@eager kindle no no

eager kindle Nov 4, 2020, 9:52 PM

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Oh I get what you're saying now

vital quail Nov 4, 2020, 9:52 PM

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use another variable sure I mean

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I just mean if you have nZ_(p)

eager kindle Nov 4, 2020, 9:53 PM

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But yeah, whole numbers with a factor of p is what I was getting at

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I notated weird

vital quail Nov 4, 2020, 9:53 PM

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with n = p^k * m

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with m not having any p factors

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then the m will get swallowed

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but the p^k will remain since the denominators can't have any p factors

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so it becomes p^kZ_(p)

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does that vibe with you

eager kindle Nov 4, 2020, 9:54 PM

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Okay so the maximal ideal is specifically pZ_(p)

vital quail Nov 4, 2020, 9:54 PM

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catthumbsup

eager kindle Nov 4, 2020, 9:54 PM

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Okay

vital quail Nov 4, 2020, 9:54 PM

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and indeed if you quotient you get a field

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Fp

eager kindle Nov 4, 2020, 9:54 PM

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Thank you. 🙂

vital quail Nov 4, 2020, 9:55 PM

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np

astral galleon Nov 4, 2020, 10:50 PM

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Is the nth root of unity not isomorphic to real numbers under addition?

next obsidian Nov 4, 2020, 10:53 PM

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what

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an nth root of unity is an element

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the real numbers are a field

chilly ocean Nov 4, 2020, 11:07 PM

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if by "nth root of unity" you mean "group of nth roots of unity," no

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catThink

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not even isomorphic in Set catThimc

astral galleon Nov 4, 2020, 11:56 PM

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Oof thx

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Wait @chilly ocean just to be clear the group of nth roots of unity is not isomoprhic to the real numbers under addition?

chilly ocean Nov 5, 2020, 12:03 AM

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that's what i said

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there isn't even a bijection between the two

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one is finite

astral galleon Nov 5, 2020, 12:05 AM

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Ahh ok just checking sorry discord didnt load for me

next obsidian Nov 5, 2020, 12:09 AM

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One of them is infinite the other has n things.

chilly ocean Nov 5, 2020, 2:26 AM

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Something something ker f to M to im f is an SES something something direct product

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(no idea if this is right)

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Probably not lmao (I meant the nonsense that I wrote)

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Oh

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I think this follows fairly trivially from definitions

maiden ocean Nov 5, 2020, 2:46 AM

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maybe im being an idiot but how can you use the frobenius automorphism to find the roots of the polynomial here?

📎 unknown.png

chilly ocean Nov 5, 2020, 2:52 AM

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Yeah, it's the obvious thing to do

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I guess by frobenius x^2=x, so evaluating the polynomial will give you something of the form x+a (ie there is one root)?

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But this sounds wrong... Or maybe not?

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Okay, what's the obvious thing to do with the generators of the image?

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Yes

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Now suppose f(m)=n, how to decompose m?

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Okay, n is in the image, so the coefficients of the pullbacks is obvious right?

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You don't need f injective

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Sure, they are not equal

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Just think a little harder

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Maybe I misunderstood what you meant

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Qcrually

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I know that f(m) = f(sum over pullback + sum over kernel generators)
@chilly ocean

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What are the sums here?

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(I'll get on my computer)

cloud walrusBOT Nov 5, 2020, 3:10 AM

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slimvesus:

📎 98936686641156096.png

chilly ocean Nov 5, 2020, 3:10 AM

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in fact, for any beta_j

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this sort of works still

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okay, here's a clue, m and that sum you have agree under the mapping m, so they agree up to...?

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yeah, they agree up to an element in the kernel

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$\phi(x)=\phi(y)\implies \phi(x-y)=0$

cloud walrusBOT Nov 5, 2020, 3:15 AM

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88ddda:

📎 403811418614398978.png

next obsidian Nov 5, 2020, 3:17 AM

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I was gonna say snake lemma

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Although on second thought what I had in mind woildnt quite work

chilly ocean Nov 5, 2020, 3:22 AM

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i guess i should learn more about these diagram stuff at some point

paper flint Nov 5, 2020, 5:01 AM

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https://cdn.discordapp.com/attachments/764073465241534484/773773424005742613/697848465819959388.png

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Is this proof okay?

next obsidian Nov 5, 2020, 5:03 AM

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This works, but you don't need the first step

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you don't have to show there's no elements of even order, since you never use that later

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From this proof however you should be able to see hopefully how group has even order <===> odd number of elements of order 2