#groups-rings-fields | Mathematics | Page 445

latent anvil Sep 30, 2019, 3:04 AM

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Fam find a harder problem so we can help

bleak abyss Sep 30, 2019, 3:04 AM

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If it's too easily we'll overthink it

latent anvil Sep 30, 2019, 3:05 AM

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I have the best easy group theory problem

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Prove that if G has a unique maximal subgroup then G is cyclic

bleak abyss Sep 30, 2019, 3:07 AM

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Hmm

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Hmm

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Oh I guess

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Take a maximal subgroup containing a given element

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And take a maximal subgroup not containing it?

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That's my instinct

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Actually that sounds wrong

chilly ocean Sep 30, 2019, 3:13 AM

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Proves that the center is a subgroup

Proves that and also proves that the elements all commute as well

Ah, so that's what a normal subgroup is.

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We haven't even gotten to that chapter yet

latent anvil Sep 30, 2019, 3:16 AM

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what do you think a normal subgroup is?

chilly ocean Sep 30, 2019, 3:18 AM

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I thought a normal subgroup was a subgroup where communitivity (I am a bad speller) holds

latent anvil Sep 30, 2019, 3:18 AM

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Right, so that's almost correct

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But you need to be careful about what commutativity means

chilly ocean Sep 30, 2019, 3:18 AM

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commutativity

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Okay, saving this

latent anvil Sep 30, 2019, 3:18 AM

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It does not mean that H is normal in G iff gh = hg for all h in H, g in G

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It means that for all g in G and h in H, there is some possibly different h' in H such that gh = h'G

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We write this gH = Hg

chilly ocean Sep 30, 2019, 3:20 AM

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Ah, I see what you're saying

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My book makes a remark about that very early in the chapter

latent anvil Sep 30, 2019, 3:21 AM

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Normal subgroups are very difficult for students

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You need to be careful

chilly ocean Sep 30, 2019, 3:21 AM

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That's like 3.5 chapters ahead of me

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I'm glad to see an example of it now.

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Thanks

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I'll try to not screw it up, but given that it's me, I probably will XD

latent anvil Sep 30, 2019, 3:22 AM

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Yeah, the center is a very special normal subgroup

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Subgroups H that satisfy gh = hg for all h in H are called...

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subgroups of the center!

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Or sometimes central, but it's important to realize that this only happens if H <= Z(G)

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Also screwing up is how you learn math

chilly ocean Sep 30, 2019, 3:27 AM

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Oh okay, that's how this last question is different. Got it. Little g vs big G

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Alright, last problem

latent anvil Sep 30, 2019, 3:27 AM

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What's the last problrm?

chilly ocean Sep 30, 2019, 3:40 AM

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Let G be a group, and let g \in G. Define the centralizer, Z(g), of g in G to be the subset Z(g) = {x \in G, xg = gx}. Prove that Z(g) is a subgroup of G.

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$\in$

cloud walrusBOT Sep 30, 2019, 3:40 AM

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Psychopotence:

📎 282952129797292033.png

bleak abyss Sep 30, 2019, 3:45 AM

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So if $xg = gx$ and $xh = hx$, then try to play with the expression $x(gh)$

cloud walrusBOT Sep 30, 2019, 3:45 AM

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Dami:

📎 301641201382064128.png

chilly ocean Sep 30, 2019, 3:46 AM

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Yeah, going on Wikipedia to make sure that I don't confuse center with centralizer

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Thanks @bleak abyss

bleak abyss Sep 30, 2019, 3:50 AM

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👍

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Let me know if you still need more hints

chilly ocean Sep 30, 2019, 3:55 AM

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I got it

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I just needed to look up the difference to make sure that I didn't accidentally do a repeat proof

bleak abyss Sep 30, 2019, 3:56 AM

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The art of not proving things twice is a subtle one for sure 😛

chilly ocean Sep 30, 2019, 4:02 AM

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Done with all HW due Tuesday. About to do this lab that's due on Wednesday, so that way my next due assignment will be Thursday or later

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Not looking forward to taking a midterm this Saturday

bleak abyss Sep 30, 2019, 4:03 AM

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Damn you're ahead of me

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I'm supposed to be done with chapter 2 of Silverman by Tuesday, also should hand back quizzes

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I am still not done (though almost) with chapter 1, and haven't graded quizzes

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And I have a pset due Thursday I haven't touched

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😢

chilly ocean Sep 30, 2019, 4:06 AM

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We started in the middle of August

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I better be ahead of you lol

bleak abyss Sep 30, 2019, 4:07 AM

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sanic

chilly ocean Sep 30, 2019, 4:08 AM

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I mean, this uni here starts earlier and finishes earlier than all the other unis around here

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Most start early September

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Or late August

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Because of how my schedule is set up, I don't have classes past late November hype

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Just finals

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I hope to God that final for Psych Stats is not 7:30-9:30am

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FTS, I will go to the Office of Disability to take it 9:30-11:30 or 9:00-11:00

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Going to a calc 3 final to arrive for 8am was brutal

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I can't do that

bleak abyss Sep 30, 2019, 4:12 AM

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Ouch

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Yeah I had an 8AM or 8:30AM final once and it was disgusting

chilly ocean Sep 30, 2019, 4:13 AM

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Maybe I should try for academia...

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If only for the flexibility in hours

old lava Sep 30, 2019, 4:34 AM

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what's a really good textbook for abstract algebra?

latent anvil Sep 30, 2019, 4:35 AM

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what's your background?

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I really really like aluffi but it's tough if you've never seen a proof before

old lava Sep 30, 2019, 4:37 AM

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I'm in high school trynna learn some higher level math. I've done a lot of math competitions

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I've learned thru calc 3, lin alg and diff eq

stone fulcrum Sep 30, 2019, 4:38 AM

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I read Fraleigh, and it's easy but is a bit lacking from what I've heard

latent anvil Sep 30, 2019, 4:38 AM

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I was in pretty much the same spot when j first read aluffi (was a high school student)

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Probably worse at math, no linear algebra or math competitions

old lava Sep 30, 2019, 4:38 AM

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I've learned some group theory, like rings, fields, groups, and stuff

latent anvil Sep 30, 2019, 4:38 AM

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And I still got a lot out of it

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I guess that's my recommendation

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It's pretty abstract though

stone fulcrum Sep 30, 2019, 4:39 AM

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If you are the genius, you can try Dummit and Foote. You won't miss a single thing, but it's a trek

old lava Sep 30, 2019, 4:39 AM

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Alright I'll check those out, thanks guys!

latent anvil Sep 30, 2019, 4:39 AM

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I like Dummit and foote :(

stone fulcrum Sep 30, 2019, 4:39 AM

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I like it too but it's not a first time read for most people

old lava Sep 30, 2019, 4:39 AM

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I'm trynna take Galois Theory soon but my Abstract Algebra is a bit lacking atm

latent anvil Sep 30, 2019, 4:40 AM

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You should focus on other things first

old lava Sep 30, 2019, 4:40 AM

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do you guys have pdfs of those books?

latent anvil Sep 30, 2019, 4:40 AM

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Yee

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Idk if it's okay to post here

stone fulcrum Sep 30, 2019, 4:40 AM

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You seem to know some theory already so you might actually be comfy with dummit and foote

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Go ahead

old lava Sep 30, 2019, 4:40 AM

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ight

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yea I know like definitions, but I'm still kinda loosey goosey with using them

latent anvil Sep 30, 2019, 4:41 AM

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Google algebra Chapter 0 pdf

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You should do problems

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I have some algebra problem sets if you want something more structured than random problem sets

old lava Sep 30, 2019, 4:41 AM

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yes that'll help a lot, thank you!

latent anvil Sep 30, 2019, 4:41 AM

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https://math.berkeley.edu/~tb65536/algebra_materials/index.html

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Let me know if you have any comments or there are typos or anything

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Also rings are currently a wip, but we have two weeks or so that aren't up yet, so let me know if you get to that (groups should take a while though)

bleak abyss Sep 30, 2019, 5:23 AM

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Oh wait is it time for commentary on algebra books?

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Why did none of you fucks ping me

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Aluffi is fine for learning some category theory with the algebra, but that's prob not as high of a selling point as one might expect, and apparently the exercises are just bad

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Dummit and Foote is fine, has everything, in a way effortless to read since he spells stuff out a lot but that makes him gigaboring

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So I generally don't like it, but there is a sort of appeal so I might say it's fine but not my taste rather than just bad

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Jacobson "Basic Algebra I" is what I like

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Also Lang's not too bad but he takes more effort to read

latent anvil Sep 30, 2019, 5:32 AM

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Aluffi's problems aren't great

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But I really like the exposition

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I think the actual category theory is a distraction, it's more of a philosophy about using general definitions/caring about maps/universal properties

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Which is helpful for me personally

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Then again, it's the only algebra book I've used consistently (I read d&f for ring /field/galois theory but it's not the same jumping in later on)

bleak abyss Sep 30, 2019, 5:37 AM

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Yeah D&F may very well be more tolerable for that stuff tbh

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For group theory though

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He's so

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fucking

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slow

latent anvil Sep 30, 2019, 5:45 AM

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I liked aluffi more, it just went into more depth than my class had time for

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I was reading it on the side

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He does everything with modules pretty early on, and introduces like noetherian rings and uses them, which takes extra time but pays off imo

fickle brook Sep 30, 2019, 6:19 AM

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uhhhhhhhhhhhh my brain is hecked atm

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i'm stuck writing up another solution for my pset and this one seems like it's either easy and i'm missing it or wrong and reliant on an unstated assumption

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let G be a group, A ⊆ G, denote by N_G(A) the normalizer of A in G (ie the set of all x s.t. a ∈ A implies xax^-1 ∈ A)

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i'm trying to show N_G(A) is closed under inversion but it's not coming together for some reason

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thonk

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maybe i hecked the definition of the normalizer

latent anvil Sep 30, 2019, 6:31 AM

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Take a in A. Then a in xAx^(-1), so a = x b x^(-1) for some b in A. Then x^(-1) a x = b in A

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Does that make sense?

warped flicker Sep 30, 2019, 8:17 AM

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Ahoy friends, does anyone happen to have a pdf for Algebra 2e by Artin? I can only find one for the 1st edition and it would be much appreciated to have or know I can't find

tribal pasture Sep 30, 2019, 9:24 AM

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Can somebody explain the difference between centralizer and conjugacy class?

mild laurel Sep 30, 2019, 9:24 AM

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What are you confused about?

tribal pasture Sep 30, 2019, 9:24 AM

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The definition seem particularly the same

wind steeple Sep 30, 2019, 9:24 AM

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One depends of all g in G, the other only for one g in G

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Btw I think you're thinking about the stabilizer of the conjugacy action and not the orbit, isn't it ?

tribal pasture Sep 30, 2019, 9:26 AM

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So Cl(a) = {gag^{-1}, for all g in G}

wind steeple Sep 30, 2019, 9:26 AM

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Yes

tribal pasture Sep 30, 2019, 9:27 AM

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whereas C_G(a) = {g in G | ga = ag}

wind steeple Sep 30, 2019, 9:27 AM

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Oh mb

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Ok

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One is in X wheras the other is in G

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X = G here

tribal pasture Sep 30, 2019, 9:28 AM

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But I can rewrite C_G(a) = {g in G | ga = ag} = {g in G | a^{-1}ga = g} = {a^{-1}ga for all g in G}

wind steeple Sep 30, 2019, 9:28 AM

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Non

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The last equality is false

tribal pasture Sep 30, 2019, 9:28 AM

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Oh yep nvm

wind steeple Sep 30, 2019, 9:29 AM

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You can have a^-1ga = h for any h in G

tribal pasture Sep 30, 2019, 9:29 AM

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One more thing

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Can you explain intuititively what the class terms are supposed to represent in the Class equation

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Like my professor just used an intuitive argument as for why |G| = |Z(G)| + sum ....

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But I cannot recall what was the explanation for the summed part

wind steeple Sep 30, 2019, 9:48 AM

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Intuitively, the quotients measure how an element in G doesn't commute with other elements

fickle brook Sep 30, 2019, 9:53 AM

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man i really am checking all the "lazy textbook writer" boxes aren't i

📎 unknown.png

hot lake Sep 30, 2019, 9:55 AM

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wait until you do "Proposition 17. Proof : See exercise 13.b"

fickle brook Sep 30, 2019, 9:55 AM

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this is a pset

hot lake Sep 30, 2019, 9:55 AM

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"exercsie 13.b : Prove proposition 17"

fickle brook Sep 30, 2019, 9:55 AM

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sdghjsdfkghsldfg

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HDGLSDFJK

wind steeple Sep 30, 2019, 9:55 AM

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What's [G,G] ?

fickle brook Sep 30, 2019, 9:55 AM

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commutator subgroup of G

wind steeple Sep 30, 2019, 9:55 AM

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The derivate subgroup ?

fickle brook Sep 30, 2019, 9:56 AM

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le sous-groupe généré par tous les commutateurs dans G

wind steeple Sep 30, 2019, 9:56 AM

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Y

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Ok

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That's a direct property

somber bramble Sep 30, 2019, 11:15 AM

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@fickle brook I did that on an exam once

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“this is not possible as otherwise it would violate the result from exercise (another one on the same exam)”

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with a small justification

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both were kinda similar, about the existence or nonexistence of biholomorphic maps between certain sets

chilly ocean Sep 30, 2019, 1:46 PM

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I quoted my homework once on a test.

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I got full credit

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(I did the HW problem right of course.)

solemn quartz Sep 30, 2019, 5:21 PM

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When I'm trying to factor x3-4x2+x+6 by using the value of a root that makes the entire equation equal to 0, how would it actually work?

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for example, the equation above is factored by (x-2) because the equation equals 0 when x=2. but why is this method working?

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I understand that x=2 is one of the root values which makes the equation equal 0. but WHAT does that have to do with the fact that it is factored by (x-2)?

fickle brook Sep 30, 2019, 5:29 PM

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do you mean x**^3 - 4x^**2 + x + 6?

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also, pretty sure this doesn't belong here

wind steeple Sep 30, 2019, 5:32 PM

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(-1) is an obvious root

brisk granite Sep 30, 2019, 5:32 PM

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If $H \trianglelefteq G$ and $G \trianglerighteq A \trianglerighteq H$, then do I know $A/H \trianglelefteq G/H$?

cloud walrusBOT Sep 30, 2019, 5:32 PM

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Cabbage:

📎 375866732226543616.png

latent anvil Sep 30, 2019, 5:37 PM

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Yes, this is the lattice isomorphism theorem

solemn quartz Sep 30, 2019, 5:41 PM

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@fickle brook oh yeah I meant that. not sure why my superscript didn't work on this chat(it worked on Google Doc). sorry about posting it to the wrong channel. did'nt know what channel it belonged to

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maybe it belongs to #prealg-and-algebra

fickle brook Sep 30, 2019, 5:46 PM

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yes

brisk granite Sep 30, 2019, 5:58 PM

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(g⁢N)⁢(A/N)⁢(g⁢N)-1=(g⁢A⁢g-1)/N=A/N

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Why is this true?

mild laurel Sep 30, 2019, 5:59 PM

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Which step are you unsure about

brisk granite Sep 30, 2019, 6:00 PM

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(g⁢N)⁢(A/N)⁢(g⁢N)-1=(g⁢A⁢g-1)/N

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That one

mild laurel Sep 30, 2019, 6:01 PM

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Well first think about what (gN)^{-1} is

brisk granite Sep 30, 2019, 6:01 PM

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(g^-1)N

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um, could you elaborate on what I should think about please?

whole bridge Sep 30, 2019, 6:02 PM

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isnt A normal in G?? so that should be it..

brisk granite Sep 30, 2019, 6:03 PM

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why?

mild laurel Sep 30, 2019, 6:04 PM

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I think he meant to ask if N was normal in A

brisk granite Sep 30, 2019, 6:08 PM

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Ok, I think I see it now

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Thank you

sullen island Oct 1, 2019, 1:24 PM

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https://math.stackexchange.com/questions/181538/a-bit-of-direct-sum-confusion

Mathematics Stack Exchange

A bit of direct sum confusion

So I'm reading through Serre's "Linear Representations of Finite Groups," and I'm a bit confused by what's probably a fairly minor point. However, subsequent proofs are hinging on it, so I figure ...

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i'm having understanding the accepted answer, how is there a bijection between the vectors in $\mathrm{Ker}(p)$ and the projections of V onto W?

cloud walrusBOT Oct 1, 2019, 1:25 PM

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xy:

📎 300157326696906752.png

sullen island Oct 1, 2019, 1:27 PM

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if you have 2 vectors, $v_1 = w_1 + u$ and $v_2 = w_2 + u$, then $p(v_1) = w_1$ and $p(v_2) = w_2$, but both correspond to vector $u$ in $\mathrm{Ker} (p)$, so how is that bijective?

cloud walrusBOT Oct 1, 2019, 1:27 PM

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xy:

📎 300157326696906752.png

sullen island Oct 1, 2019, 1:28 PM

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i came across this while learning representation theory and thought this would be the most appropriate channel to post in (since theres no rep theory channel)

fading wagon Oct 1, 2019, 1:30 PM

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hmm

sullen island Oct 1, 2019, 1:33 PM

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:(

fading wagon Oct 1, 2019, 1:33 PM

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I'm still trying to read it

sullen island Oct 1, 2019, 1:34 PM

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yeah no worries take your time

fading wagon Oct 1, 2019, 1:34 PM

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I worry about myself not sleeping

sullen island Oct 1, 2019, 1:34 PM

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i can screenshot the relevant pages from serre as well if you want me to

fading wagon Oct 1, 2019, 1:35 PM

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might be helpful

sullen island Oct 1, 2019, 1:36 PM

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📎 Screenshot_20191001-213555_Adobe_Acrobat.jpg

fading wagon Oct 1, 2019, 1:48 PM

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I'm just lost on what a complement is

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hold on

sullen island Oct 1, 2019, 1:48 PM

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okay

fading wagon Oct 1, 2019, 1:49 PM

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hmm, It seems that a complement is just any subspace W' that has the direct sum W+W'=V

sullen island Oct 1, 2019, 1:49 PM

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mm hmm

fading wagon Oct 1, 2019, 1:50 PM

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So, for every projection of V onto W, consider the kernel, that would be W', so that is a map from projections to complements

sullen island Oct 1, 2019, 1:51 PM

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yeah

fading wagon Oct 1, 2019, 1:51 PM

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Obviously, for different projections, we have different complements

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so it is injective

sullen island Oct 1, 2019, 1:51 PM

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?!

fading wagon Oct 1, 2019, 1:51 PM

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okay, maybe we should go a bit slower

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Let's say we have different projections

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then they cannot have the same kernel, because, that is sufficient to determine the projection

sullen island Oct 1, 2019, 1:52 PM

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different projections as in p maps V to different W's?

fading wagon Oct 1, 2019, 1:52 PM

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Different projections from V to W

sullen island Oct 1, 2019, 1:53 PM

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i thought there was only one unique projection from V to W

fading wagon Oct 1, 2019, 1:54 PM

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there are multiple

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consider f(x, y)=(x, 0) and f(x, y)=(x+y, 0)

sullen island Oct 1, 2019, 1:55 PM

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oh okay

fading wagon Oct 1, 2019, 1:55 PM

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the kernel for the first is (0, y), the kernel for the second is (-y, y)

sullen island Oct 1, 2019, 1:56 PM

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mm hmm

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so they have different kernels

fading wagon Oct 1, 2019, 1:56 PM

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yeah

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that's the main idea

sullen island Oct 1, 2019, 1:57 PM

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but then claiming that the kernel is isomorphic to the image of projection seems a little wrong

fading wagon Oct 1, 2019, 1:58 PM

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yeah, that's quite wrong

sullen island Oct 1, 2019, 1:58 PM

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so wheres the bijective correspondence ? 😅

fading wagon Oct 1, 2019, 1:59 PM

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from kernels of projections, to complements

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It's injective as we shown

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and from complements to projection, it's also injective

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so it's bijective

sullen island Oct 1, 2019, 2:00 PM

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how do you get that from complements to projections it is injective?

sullen island Oct 1, 2019, 2:33 PM

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i think i understand:
Let $p_i$ be i distinct projections. Define an equivalence relation $\sim_i := x \sim_i y \iff p_i(x) = p_i(y)$
Denote $$K:= {[x]{\sim_i} \ \vert \ p_i(x) = 0}$$ i.e. let K be the set of all kernels of the i different projections
Denote $$R := { [x]{\sim_i} \ \vert \ p_i(x) \neq 0 }$$
Thus R is the set of all images of projection.
Thus $R \cong K$

cloud walrusBOT Oct 1, 2019, 2:34 PM

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xy:

📎 300157326696906752.png

sullen island Oct 1, 2019, 2:35 PM

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correct me if im wrong

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this seems like the most obvious and intuitive way to interpret what the author is saying

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@fading wagon

mild laurel Oct 1, 2019, 2:41 PM

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Are you defining a different K and R for each p_i?

sullen island Oct 1, 2019, 2:41 PM

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no

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K consists of all the kernels

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so it has i distinct equivalence classes (kernels of each p_i)

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similarly for R

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and since both of them have cardinality i they are isomorphic i.e. the set of all projections can be bijected to the set of complements

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right?

mild laurel Oct 1, 2019, 2:43 PM

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Well R isn't really the set of images of the projection

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Also, in most cases there will be infinitely many projections

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Definitely in this case since our vector spaces are over C

sullen island Oct 1, 2019, 2:44 PM

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why isnt R the set of images of different projections?

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and yeah how can i compensate for the infinite case ?

mild laurel Oct 1, 2019, 2:46 PM

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I mean maybe you meant to write p([x]) but

sullen island Oct 1, 2019, 2:46 PM

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wait, oops, R should be p(x)

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yeah

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but it still doesnt compensate for infiniteness 🤔

mild laurel Oct 1, 2019, 2:47 PM

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But this still isn't really what Serre is talking about

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The images for every single p_i should be the exact same

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The image should always be W in this case

sullen island Oct 1, 2019, 2:48 PM

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hmmm

mild laurel Oct 1, 2019, 2:48 PM

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Serre is talking about a bijection between the actual projections p_i and complements of W

sullen island Oct 1, 2019, 2:48 PM

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how ?

mild laurel Oct 1, 2019, 2:49 PM

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In exactly the way he desribes

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If you start with a projection p onto W

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Then the kernel of p will give you a complement of W

sullen island Oct 1, 2019, 2:49 PM

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uh huh

mild laurel Oct 1, 2019, 2:49 PM

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Similarly, if you start with a complement of W

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You can get a projection onto W by the way he describes

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Write every element of V as w + w' and have the function be p(w + w') = w

sullen island Oct 1, 2019, 2:51 PM

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lets say v1 = w1 + w' and v2 = w2 + w'

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then p(v1) = w1 and p(v2) = w2

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and these correspond to the same w'

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so its not a one to one correspondence 🤔

mild laurel Oct 1, 2019, 2:52 PM

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So?

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How is that a function from the complements of W and the projections onto W?

sullen island Oct 1, 2019, 2:52 PM

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wait, oh

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i thought he was talking about literally the projected vectors

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okay i think i understand it

chilly ocean Oct 1, 2019, 8:51 PM

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Quote from my abstract algebra teacher today: "The identity is the identity."

stone fulcrum Oct 1, 2019, 8:52 PM

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And God forgive us if it isn't

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The identity is unique, that's a good quote

potent lynx Oct 1, 2019, 8:56 PM

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your still at the beginning right?

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or is that just a random quote

chilly ocean Oct 1, 2019, 10:46 PM

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He was saying: "The identity is the identity (map)"

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But it was understood clearly that he implied maps because we're talking about functions.

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But it was still hilarious to hear this

stone fulcrum Oct 1, 2019, 10:55 PM

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No that's something you need to say. You've got to point out what your identity is.

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Even if the identity is always called the identity lol

fierce flower Oct 1, 2019, 11:10 PM

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am I dumb for not understand in the slightest what this is about

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What level is it at?

stone fulcrum Oct 1, 2019, 11:18 PM

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Abstract algebra 1 and 2 are typical pure math uni courses

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Download a pdf for it! Take a look.

latent anvil Oct 1, 2019, 11:47 PM

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@fierce flower high school algebra is at #prealg-and-algebra

rain burrow Oct 2, 2019, 2:25 PM

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hi

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if I have a matrix A of size m x n

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and I vector x of size n x 1

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and the equation Ax = 0

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is there a theory to know the number of basis for the nullspace A?

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also the size of basis?

bleak abyss Oct 2, 2019, 2:28 PM

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This belongs more in #linear-algebra

rain burrow Oct 2, 2019, 2:29 PM

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Okay, thanks.

barren delta Oct 2, 2019, 6:55 PM

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I'm struggling to find a counterexample to the statement "If a group G is such that every proper subgroup is cyclic, then G is cyclic"

cloud walrusBOT Oct 2, 2019, 6:55 PM

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kickpuncher:

📎 150339192923226112.png

barren delta Oct 2, 2019, 6:55 PM

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as a counterexample

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but I can't find any others

mild laurel Oct 2, 2019, 6:57 PM

#

try similar things

barren delta Oct 2, 2019, 6:57 PM

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well, yeah but I was hoping to find some different ones 😄

mild laurel Oct 2, 2019, 6:58 PM

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Hm, I'm not so sure there are any other nice examples

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Okay I lied, I think the symmetric group on 3 elements is an example

barren delta Oct 2, 2019, 7:00 PM

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ohhhh wait

#

Klein-4 works as a counter example, yes?

stone fulcrum Oct 2, 2019, 7:00 PM

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Z2×Z2 is a beautiful example

mild laurel Oct 2, 2019, 7:00 PM

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klein 4 is Z_2 x Z_2

barren delta Oct 2, 2019, 7:01 PM

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u wot

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LOL

mild laurel Oct 2, 2019, 7:01 PM

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https://en.wikipedia.org/wiki/Klein_four-group

Klein four-group

In mathematics, the Klein four-group is a group with four elements, in which each element is self-inverse (composing it with itself produces the identity)
and in which composing any two of the three non-identity elements produces the third one.
It can be described as the symm...

barren delta Oct 2, 2019, 7:01 PM

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all I have in my notes is {e a b c} with those properties

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but I guess.....

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that means it's isomorphic to z2 x z2

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right?

stone fulcrum Oct 2, 2019, 7:02 PM

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It's pretty easy to show by Cayley tables that there are only two groups of order 4. They are Z4 and Z2×Z2

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Sometimes we call that second one V4 instead

barren delta Oct 2, 2019, 7:05 PM

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so wait, to clarify, @mild laurel (also lol at your handle) you're saying Klein-4 IS Z2 x Z2 in the sense that it's isomorphic to Z2 x Z2 or that it literally is that group?

woven delta Oct 2, 2019, 7:05 PM

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I think Z_2x Z_p may work generally

stone fulcrum Oct 2, 2019, 7:05 PM

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We say two groups are "the same" when they're isomorphic

barren delta Oct 2, 2019, 7:06 PM

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ok thanks for clarifying

woven delta Oct 2, 2019, 7:06 PM

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Oh lmao

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No

stone fulcrum Oct 2, 2019, 7:06 PM

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Any difference between two iso groups is not something that algebra cares about

woven delta Oct 2, 2019, 7:06 PM

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I'm silly

mild laurel Oct 2, 2019, 7:06 PM

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Yeah

barren delta Oct 2, 2019, 7:06 PM

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which... I mean, I guess yeah now that I think about it

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right Kay, yes

mild laurel Oct 2, 2019, 7:06 PM

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We only really ever care about groups up to isomorphism

woven delta Oct 2, 2019, 7:06 PM

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Those are cyclic

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https://danaernst.com/an-infinite-non-cyclic-group-whose-proper-subgroups-are-cyclic/

Dana C. Ernst

An infinite non-cyclic group whose proper subgroups are cyclic

The Fall semester of 2013 just ended and one of the classes I taught was abstract algebra. The course is intended to be an introduction to groups and rings, although, I spent a lot more time discussing group theory than the latter. A few weeks into the semester, the students ...

woven delta Oct 2, 2019, 7:32 PM

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This is a good example

thorn delta Oct 2, 2019, 7:51 PM

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there is the circle group as well

woven delta Oct 2, 2019, 7:54 PM

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No, that's not an example

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@thorn delta

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You have the group of all nth roots of unity inside it

thorn delta Oct 2, 2019, 7:55 PM

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wait what do you mean by that? Its subgroups are the nth roots of unity, which are cyclic.....

woven delta Oct 2, 2019, 7:55 PM

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No

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You have the prufer p groups for example

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Lots of other stuff

thorn delta Oct 2, 2019, 7:56 PM

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oh wow never knew. huh

uncut girder Oct 2, 2019, 8:02 PM

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Given an elementary abelian group of order p^(n-1)

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How many ways can you choose a basis

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Apparently the answer is

#

📎 Screenshot_20191002-155549_Chrome.jpg

woven delta Oct 2, 2019, 8:02 PM

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What do you mean elementary?

mild laurel Oct 2, 2019, 8:02 PM

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what do you mean basis lmao

uncut girder Oct 2, 2019, 8:02 PM

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Product of Z/p s

woven delta Oct 2, 2019, 8:02 PM

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Oh

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I mean, isn't this just asking the size of the automorphism group?

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And isn't this actually a linear algebra question

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#linear-algebra

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It's asking the cardinality of GL_{n-1}(Z/(p))

mild laurel Oct 2, 2019, 8:06 PM

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That's an interesting question tho

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think about choosing each column one at a time

woven delta Oct 2, 2019, 8:19 PM

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Oh, so at first there are p^n -1 choices of basis elements

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Then you choose the first one, and there are p^n - p

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And so on

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That's actually pretty easy

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is that what you meant @mild laurel ?

mild laurel Oct 2, 2019, 8:22 PM

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Yea

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Still pretty cool I think

woven delta Oct 2, 2019, 8:22 PM

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Does this actually count the number of elements in GL_n though?

mild laurel Oct 2, 2019, 8:22 PM

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I think so?

woven delta Oct 2, 2019, 8:22 PM

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Or do you have to mod out by something

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Hmm

mild laurel Oct 2, 2019, 8:23 PM

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All your elements are already in Z/pZ, not sure what else you have to mod out by

woven delta Oct 2, 2019, 8:23 PM

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Maybe order

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Are we considering basis or ordered basis?

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I guess this takes order into account actually

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Yeah it should be fine

whole bridge Oct 2, 2019, 8:26 PM

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yes size of GL_n(Z/pZ) is product of p^n - p^i

upbeat burrow Oct 2, 2019, 9:12 PM

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With modules, what is the difference between a generating set and a basis?

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I'm not seeing how these two things are different.

mild laurel Oct 2, 2019, 9:12 PM

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Look at the two definitions

upbeat burrow Oct 2, 2019, 9:13 PM

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lol

chilly ocean Oct 2, 2019, 9:13 PM

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hey idiots

upbeat burrow Oct 2, 2019, 9:13 PM

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Generating set: every element of a module M can be written as a linear combination of the generating set.

mild laurel Oct 2, 2019, 9:14 PM

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Okay but actually, it's really the same as in linear algebra

chilly ocean Oct 2, 2019, 9:14 PM

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is $K[x]/(x-a)\cong K$ froall a\in K?

cloud walrusBOT Oct 2, 2019, 9:14 PM

#

how do you pronounce Žižek?:
Compile Error! Click the errors reaction for details. (You may edit your message)

📎 603330685184245762.png

mild laurel Oct 2, 2019, 9:14 PM

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Yes

chilly ocean Oct 2, 2019, 9:14 PM

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me?

mild laurel Oct 2, 2019, 9:14 PM

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yes

upbeat burrow Oct 2, 2019, 9:14 PM

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How is that different than a basis?

chilly ocean Oct 2, 2019, 9:14 PM

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ok

mild laurel Oct 2, 2019, 9:14 PM

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What's the definition of a basis

chilly ocean Oct 2, 2019, 9:14 PM

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MAXIMAL LINEARLY INDEPENDENT SET

upbeat burrow Oct 2, 2019, 9:14 PM

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Isn't it the same?

mild laurel Oct 2, 2019, 9:14 PM

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That's the linear algebra definition, doesn't work too well for modules

chilly ocean Oct 2, 2019, 9:15 PM

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oh there's no difference basis and genset

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for rings and modules

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atleast most of the time

mild laurel Oct 2, 2019, 9:15 PM

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This is a case where there's a difference

chilly ocean Oct 2, 2019, 9:15 PM

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sorry i wasn't paying attention ill leave u 2 alone

mild laurel Oct 2, 2019, 9:15 PM

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Just like in linear algebra

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A basis has to be linearly independent

upbeat burrow Oct 2, 2019, 9:16 PM

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Oh right

woven delta Oct 2, 2019, 9:16 PM

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@chilly ocean you're quotienting by a maximal ideal

upbeat burrow Oct 2, 2019, 9:16 PM

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But a generating set need not be linearly independent?

mild laurel Oct 2, 2019, 9:16 PM

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Just like in linear algebra

chilly ocean Oct 2, 2019, 9:16 PM

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ah yes good point @smoky cypress

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i meant to @ horse

mild laurel Oct 2, 2019, 9:16 PM

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Just add 0 to your set of generators or something, now it's not linearly independent, but it's still a generating set

woven delta Oct 2, 2019, 9:16 PM

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But you can make an explicit isomorphism

chilly ocean Oct 2, 2019, 9:16 PM

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yeah but that's hard

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er

woven delta Oct 2, 2019, 9:17 PM

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Nah

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It's pretty easy actually

chilly ocean Oct 2, 2019, 9:17 PM

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i mean elements of a field are clearly not killed yeah

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and everything of deg \geq 1 is killed

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so whatever

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it's clearly K

woven delta Oct 2, 2019, 9:18 PM

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All the elements look like b + (x-a)

chilly ocean Oct 2, 2019, 9:18 PM

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mhm

woven delta Oct 2, 2019, 9:18 PM

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So the isomorphism is mapping b to b + (x-a)

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Yeah

chilly ocean Oct 2, 2019, 9:19 PM

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i need to know $F[x]/(a_1)\oplus F[x]/(a_2) \oplus F[x]/(a_3)\cong F^3$ if they are all degree 1

cloud walrusBOT Oct 2, 2019, 9:19 PM

#

how do you pronounce Žižek?:

📎 603330685184245762.png

woven delta Oct 2, 2019, 9:19 PM

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Yes

chilly ocean Oct 2, 2019, 9:19 PM

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whihc is obvious

woven delta Oct 2, 2019, 9:19 PM

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This is the case

upbeat burrow Oct 2, 2019, 9:19 PM

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@mild laurel Okay, I think I understand now. Thanks for the clarification!

woven delta Oct 2, 2019, 9:19 PM

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This is just linear algebra lol

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#linear-algebra

chilly ocean Oct 2, 2019, 9:20 PM

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what if its like $F[x]/(a_1)$ and its degree 3

cloud walrusBOT Oct 2, 2019, 9:20 PM

#

how do you pronounce Žižek?:

📎 603330685184245762.png

woven delta Oct 2, 2019, 9:20 PM

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Think about it

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You get basis elements as 1, x, x^2

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So F^3

mild laurel Oct 2, 2019, 9:21 PM

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both of these questions were just #linear-algebra lmao

chilly ocean Oct 2, 2019, 9:21 PM

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that's what all algebra is

woven delta Oct 2, 2019, 9:21 PM

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Haha

chilly ocean Oct 2, 2019, 9:21 PM

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i want to say that all these in the last thing i posted are the same for all a_1 of degree 3

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which I believe is true

upbeat burrow Oct 2, 2019, 9:21 PM

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I never covered modules in linear. Sorry for asking in the wrong channel 😅

chilly ocean Oct 2, 2019, 9:21 PM

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since they all are F^3

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and finally all things like $F[x]/(a_1)\oplus F[x]/(a_2)$ are iso if a_1 degree 1 and a_2 degree 2

cloud walrusBOT Oct 2, 2019, 9:22 PM

#

how do you pronounce Žižek?:
Compile Error! Click the errors reaction for details. (You may edit your message)

📎 603330685184245762.png

mild laurel Oct 2, 2019, 9:22 PM

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Nah I mean you definitely don't learn modules in linear, but what you asked, you can just think of vector spaces and the difference becomes clear

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Which happens quite a bit in module theory

woven delta Oct 2, 2019, 9:22 PM

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Can't you factor F[x]/(p) in terms of the irreducible parts of p?

chilly ocean Oct 2, 2019, 9:22 PM

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uhm

woven delta Oct 2, 2019, 9:23 PM

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So over an Algebraically closed field there's a nice answer

chilly ocean Oct 2, 2019, 9:23 PM

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so you mean over alg closed they are all like F^n

woven delta Oct 2, 2019, 9:23 PM

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Otherwise there's still a nice answer, but it doesn't factor as cleanly

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No, it's always F^n

upbeat burrow Oct 2, 2019, 9:23 PM

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So a basis is a generating set, but not necessarily vice-versa?

chilly ocean Oct 2, 2019, 9:23 PM

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yes new nivk

woven delta Oct 2, 2019, 9:23 PM

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But it splits cleaner over Algebraically closed

chilly ocean Oct 2, 2019, 9:23 PM

#

with rings a lot of people don't distinguish

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basis vs genset

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ya it's not alg closed tho

woven delta Oct 2, 2019, 9:24 PM

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But yeah, generally if you quotient by a degree k you get a k dimensional vector space

upbeat burrow Oct 2, 2019, 9:24 PM

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Why don't they distinguish between them? My prof said the difference was important.

chilly ocean Oct 2, 2019, 9:24 PM

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it is for vector spaces

woven delta Oct 2, 2019, 9:25 PM

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For modules basis are pretty rare

chilly ocean Oct 2, 2019, 9:25 PM

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yeah basis of a module means its a free module

upbeat burrow Oct 2, 2019, 9:25 PM

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Right

woven delta Oct 2, 2019, 9:25 PM

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But finitely generated modules are very important

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So we care more about generating sets

upbeat burrow Oct 2, 2019, 9:26 PM

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I think I'm going to have to find some examples.

chilly ocean Oct 2, 2019, 9:26 PM

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degree k you get a k dimensional vector space

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these are quotients as rings tho

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i need to know what the rings are

woven delta Oct 2, 2019, 9:27 PM

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👀

chilly ocean Oct 2, 2019, 9:27 PM

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are minimal polynomials of matrices always irreducible

woven delta Oct 2, 2019, 9:27 PM

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You know what they are

chilly ocean Oct 2, 2019, 9:27 PM

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this makes it work

woven delta Oct 2, 2019, 9:27 PM

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As rings

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So that's why the story for Algebraically closed is nice

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Cause it splits as rings

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So it's literally F^k as a ring

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But you can think of polynomial multiplication

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And that gives you the ring structure

chilly ocean Oct 2, 2019, 9:29 PM

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but the poly isn't irreducible and F isn't closed

woven delta Oct 2, 2019, 9:29 PM

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It's not too hard to figure out

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So you split the poly into irreducible components

chilly ocean Oct 2, 2019, 9:30 PM

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ya

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so they arent all iso

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if u change a_1 in F[x]/a_1

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like u can have differnet deg 3 polys and the quotients are not iso

woven delta Oct 2, 2019, 9:30 PM

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Yeah

chilly ocean Oct 2, 2019, 9:31 PM

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dang

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this isn't what i want

woven delta Oct 2, 2019, 9:31 PM

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Depending on shit with reducibility

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You basically get products of integral domains

chilly ocean Oct 2, 2019, 9:32 PM

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i think you would just get F^n \otimes F^m etc

woven delta Oct 2, 2019, 9:32 PM

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And the number of integral domains is determined by number of irreducible components

chilly ocean Oct 2, 2019, 9:32 PM

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yeah

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they are all free modules ?

woven delta Oct 2, 2019, 9:32 PM

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Yeah

chilly ocean Oct 2, 2019, 9:32 PM

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so even better than int domain

woven delta Oct 2, 2019, 9:32 PM

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But as rings you just have to consider F[x]/(p) for p irreducible

chilly ocean Oct 2, 2019, 9:32 PM

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or something idk

woven delta Oct 2, 2019, 9:33 PM

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And you understand the whole story

chilly ocean Oct 2, 2019, 9:33 PM

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what i want to show is that matrices with the same characteristic and minimal polys are similar

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when they are 3x3

woven delta Oct 2, 2019, 9:33 PM

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Oh lol

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You should have come out and said that

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Instead of have me go on a shitty inane rant

chilly ocean Oct 2, 2019, 9:34 PM

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lol

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it is obvious if degree of minimal is 1

woven delta Oct 2, 2019, 9:35 PM

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This is shit @bleak abyss likes lmao

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So maybe he'll want to explain

bleak abyss Oct 2, 2019, 9:35 PM

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Did someone say "explain"?

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hyper

woven delta Oct 2, 2019, 9:35 PM

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👀

chilly ocean Oct 2, 2019, 9:38 PM

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im sure u can just write out some normal form

woven delta Oct 2, 2019, 9:50 PM

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Yeah, Jordan normal form

mild laurel Oct 2, 2019, 9:51 PM

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~~smith normal form~~

woven delta Oct 2, 2019, 9:52 PM

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We're doing square

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So Jordan is enough

chilly ocean Oct 2, 2019, 10:05 PM

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ok well no

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it's rational canonical form since you don't know it's alg closed

woven delta Oct 2, 2019, 10:06 PM

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Oh you're right

chilly ocean Oct 2, 2019, 10:07 PM

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idk how to show this if the deg of min poly is 2

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it is probably obvious

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gonna make some coffee i guess

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if deg is 2 u have 2 parts to the rcf, you only know what one is

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i'm sure there's some divisibility conditions with the char poly to make the other polynomial in the module structure work

left scroll Oct 3, 2019, 2:57 AM

#

This may be a bit of a silly question, but what does it mean to "take the character of the representation you are being offered and contract it against the character of the representation r"? This pops up in the book Group Theory in a Nutshell & it looks like that means summation over equivalence classes of the quantity: (number of members in an equivalence class c)(conjugate of character of equivalence class c in representation r)(character of the representation you're being offered). Only thing I can think of is Tensor Contraction, but that has nothing to do with this, it seems.

#

Oh, hmm, it could be Tensor Contraction, actually. (conjugate of character of equivalence class c in representation r) would have a downstairs index and (character of the representation you're being offered) would have an upstairs index, allowing for a contraction of those two that's then summed over equivalence classes c with factor (number of members in an equivalence class c). That's still super unclear in context because they use the upstairs & downstairs index notation very inconsistently & don't use it in this particular case where it'd make sense, haha.

#

They also never introduce tensor contraction before that point in the book, lol, so there's a tendency to pull stuff out of a hat there. Sorry for the wall of text.

hot lake Oct 3, 2019, 7:57 AM

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could they possibly be talking about the inner product of two representations ?

upbeat burrow Oct 4, 2019, 4:10 AM

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R^n isomorphic R^m if and only if n = m

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R^n is an R-module

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Can anyone assist with this?

latent anvil Oct 4, 2019, 4:16 AM

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What are the assumptions on R?

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I know a proof if it's an integral domain

upbeat burrow Oct 4, 2019, 4:16 AM

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R is commutative with identity

latent anvil Oct 4, 2019, 4:16 AM

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ooh

upbeat burrow Oct 4, 2019, 4:17 AM

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Going the backwards direction seems obvious

latent anvil Oct 4, 2019, 4:17 AM

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I'm thinking you could take a maximal ideal of R and tensor with R/m? Maybe that's stupid though

upbeat burrow Oct 4, 2019, 4:17 AM

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I was thinking of the maximal ideal thing

latent anvil Oct 4, 2019, 4:17 AM

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That should reduce to vector spaces

upbeat burrow Oct 4, 2019, 4:17 AM

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Is there a way without tensors?

latent anvil Oct 4, 2019, 4:19 AM

#

prove directly that if R^n ≈ R^m as R modules then (R/I)^n ≈ (R/I)^m as R/I-modules for any ideal I of R?

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I'm just tensoring to get that

upbeat burrow Oct 4, 2019, 4:20 AM

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I'm not really familiar with tensors

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I don't understand them that well.

latent anvil Oct 4, 2019, 4:21 AM

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Yeah you don't need to know them

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I'm just using them to prove that if R^n ≈ R^m as R modules then (R/I)^n ≈ (R/I)^m as R/I-modules for any ideal I of R

upbeat burrow Oct 4, 2019, 4:25 AM

#

I'm not sure I see how to prove that though.

latent anvil Oct 4, 2019, 5:39 AM

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You have projection maps R^n -> R^n/(I×…×I) which reduce mod I on each factor, and similarly for m

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And R^n/(I×...×I) ≈ (R/I)^n

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So you get maps R^n -> (R/I)^n and R^m -> (R/I)^m

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I think it can put these together to get what you want

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/shrug

wind cypress Oct 4, 2019, 4:01 PM

#

So I had an English exam today, no need to know about math, but it had this article which I found fascinating, posted it earlier to general chat: http://users.jyu.fi/~tojusaar/What Does Less Than or Equal Really Mean.pdf

Don't know if it's really algebra, but fascinating how such simple thing as < or <= can have had mathematics scratching their heads for over 50 years.... I learned a little from it, perhaps someone else will too?

uncut girder Oct 4, 2019, 8:28 PM

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Your English exam had a math paper?

#

That's cool

wind cypress Oct 4, 2019, 8:52 PM

#

The English exam was meant for math students, the idea was to be able to reference mathematical paper, and understand stuff in it. Not all students of course have same paper to analyze.

chilly ocean Oct 4, 2019, 11:02 PM

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What English class is this?

bleak finch Oct 4, 2019, 11:05 PM

#

You know this book?

#

http://abstract.ups.edu/aata/index.html

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Is there something like it for modules?

mild laurel Oct 4, 2019, 11:07 PM

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I'm not sure what exactly you mean by "like"

bleak finch Oct 4, 2019, 11:11 PM

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The book is very approachable and builds up group, ring, and field theory step-by-step

#

Is there something like that for modules?

mild laurel Oct 4, 2019, 11:36 PM

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I mean every book aims to do that

woven delta Oct 5, 2019, 12:56 AM

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Except for bourbaki

potent lynx Oct 5, 2019, 1:11 AM

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except for rudin

onyx lance Oct 5, 2019, 2:24 AM

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Are there any interesting examples of canonical/basis-independent operations on tensors besides the trace & determinant of a matrix?

golden pasture Oct 5, 2019, 3:22 AM

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#linear-algebra

fading wagon Oct 5, 2019, 3:23 AM

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@onyx lance Quite crowded there, maybe go #❓how-to-get-help

woven delta Oct 5, 2019, 4:36 AM

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I think abstract Algebra is fine for this

fading wagon Oct 5, 2019, 4:40 AM

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any question?

#

@woven delta

woven delta Oct 5, 2019, 4:44 AM

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Nah, I'm saying I think #groups-rings-fields is fine for marivas question

meager flint Oct 5, 2019, 7:15 PM

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Let X and Y be subgroups of a group G. Let f be a function from X x Y to G such that f((x,y))=xy. Is there a bijection between any two fibers of the map??

woven delta Oct 5, 2019, 7:51 PM

#

This sounds like the Zappa szep product lmao

#

The image that is

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Also saying let it be a function with that property is weird

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There is exactly one function with that property

#

The answer is trivially no because the map is not surjective

#

Hmm

#

If your group is commutative this is true

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But that's not saying much

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(well assuming you disregard the trivial counterexample of empty fiber)

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Cause if it's commutative the map on XxY is a homomorphism

#

Hey @mild laurel

#

Any thoughts?

latent anvil Oct 5, 2019, 8:06 PM

#

What if we consider the action of X×Y on im f = XY?

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It's transitive

woven delta Oct 5, 2019, 8:07 PM

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I'm not so sure it is a well defined action

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Cause of the way multiplication works in XxY

latent anvil Oct 5, 2019, 8:07 PM

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It is. (x, y)z = xzy^(-1)

woven delta Oct 5, 2019, 8:07 PM

#

Oh ok

latent anvil Oct 5, 2019, 8:07 PM

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You can use this action + orbit stabilizer to prove |XY| = |X|*|Y|/|X\cap Y|

woven delta Oct 5, 2019, 8:07 PM

#

This is now looking a whole lot like Zappa szep lmao

latent anvil Oct 5, 2019, 8:08 PM

#

By which I mean I did that on my homework thus week lol

woven delta Oct 5, 2019, 8:08 PM

#

Oh ok

latent anvil Oct 5, 2019, 8:10 PM

#

If the fibers are all in bijection, they have to have the same size as the fiber of e, which is |X \cap Y|

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The fiber of z in XY is the same as the set of (x, y) such that (x, y) • 1 = z

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I think this implies it

#

Let G act transitively on S. For elements s, t of S, define G(s, t) = { g in G : g • s = t }. We show all G(s, t) are in bijection. Since the action is transitive, for each s, t we have an element τ_{s, t} of G(s, t). Note that taking inverses gives a bijection between G(s, t) and G(t, s). Thus it suffices to show G(s, t) and G(s, s) = Stab(s) are in bijection. Define f : Stab(s) -> G(s, t) by f(g) = τ_{s, t} g

#

Really what's going on is that a transitive action gives us a groupoid where all objects are isomorphic

#

Anyways this should prove that the nonempty fibers of that function are all in bijection

#

@meager flint i made a couple reductions but it should work

meager flint Oct 5, 2019, 8:28 PM

#

@latent anvil thank you, for real

latent anvil Oct 5, 2019, 8:28 PM

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Np

#

Where did this come up?

#

Oh here's a much simpler proof

#

Suppose a in X and b in Y. Then

f^(-1)(ab) =
{ (x, y) in X×Y : xy = ab } =
{ (x, y) in X×Y : a^(-1)x = by^(-1) } =
{ (x, y) in X×Y : exists s in X\cap Y. s = a^(-1)x and s = by^(-1) }
{ (as, bs^(-1)) : s in X\cap Y }

#

This last set has the same size as X\cap Y

meager flint Oct 5, 2019, 8:42 PM

#

Algebra class, studying for an exam

#

Asked to prove that |XY|=|X||Y|/X\capY|

#

I had assumed that bit and proved the claim, but I was stuck on how to do this part

latent anvil Oct 5, 2019, 8:46 PM

#

Do you know the orbit stabilizer theorem?

#

There's a much nicer proof in terms of group actions

meager flint Oct 5, 2019, 8:48 PM

#

After googling it yes lol

#

Didn't know about group actions

latent anvil Oct 5, 2019, 8:49 PM

#

Ah, they're pretty great

#

One of the most important things about groups, possibly the most important

#

You have an action of X×Y on XY by (x, y)•z = x z y^(-1). The orbit of 1 under this action is all of XY, since you can get from 1 to any z = ab by (a, b^(-1))•1 = z, and so the orbit stabilizer theorem says |XY| = |X×Y|/|Stab(1)|. But Stab(1) = { (x, y) : xy^(-1) = 1 } is the set { (a, a) : a in X\cap Y }, and so it has the same size as X\cap Y. Thus |XY| = |X×Y|/|Stab(1)| = |X||Y|/|X\cap Y|

gentle ibex Oct 5, 2019, 8:59 PM

#

I'm relatively new to the concept of abstract algebra.. and I have a question about whether or not the law of annihilation (a*0=0) holds for all rings, or only holds in rings that satisfy the multiplicative cancellation law?

#

I understand the proof for the law of annihilation using additive cancellation law, but that's not a ring law, so I don't see how it can hold for all rings...? But also don't understand why it would only hold in rings that satisfy the multiplicative cancellation law?

stone forum Oct 5, 2019, 9:00 PM

#

a= a*1=a*(1+0)=a*1 + a*0=a+ a*0

#

subtract a from both sides, you get a*0 = 0

gentle ibex Oct 5, 2019, 9:01 PM

#

Yes, but that's just using the additive cancellation law, right?

stone forum Oct 5, 2019, 9:01 PM

#

" But also don't understand why it would only hold in rings that satisfy the multiplicative cancellation law"

#

it doesn't have to satisfy multiplicative cancellation law

#

additive cancellation is one of the axioms, yes

#

for every a there is a -a such that a+ (-a)=0

gentle ibex Oct 5, 2019, 9:02 PM

#

My main question is... does the law of annihilation hold in all rings or only those that satisfy the multiplicative cancellation law?

stone forum Oct 5, 2019, 9:02 PM

#

all rings

gentle ibex Oct 5, 2019, 9:02 PM

#

Why's that? The additive cancellation law isn't a ring law, is it?

stone forum Oct 5, 2019, 9:02 PM

#

those which satisfy multiplicative cancellation law are called integral domains

#

it follows from the ring laws

gentle ibex Oct 5, 2019, 9:03 PM

#

oh ok

#

So you can prove the additive cancellation law using ring laws?

#

I thought you proved the annihilation law

stone forum Oct 5, 2019, 9:04 PM

#

a+b=a+c

#

(-a)+ a+b = (-a) + a + c

#

b=c

gentle ibex Oct 5, 2019, 9:04 PM

#

Right, that seems very obvious, thank you haha

#

I see now

#

So you can say that you prove the additive cancellation law using ring laws... and therefore you can prove the law of annihilation using ring laws (via the additive cancellation law) and thus the law of annihilation holds for all rings?

stone forum Oct 5, 2019, 9:05 PM

#

yes

gentle ibex Oct 5, 2019, 9:06 PM

#

Thank you so much! Sorry if it seemed so trivial!

#

I just have one other question... I'm aware of the proof of the statement "if ab=0 then either a=0 or b=0" using multiplicative inverses, but can you prove this just using the ring laws and the cancellation law for multiplication?

stone forum Oct 5, 2019, 9:07 PM

#

no

#

that's independent of ring axioms

gentle ibex Oct 5, 2019, 9:07 PM

#

sorry I edited my question

stone forum Oct 5, 2019, 9:07 PM

#

rings which satisfy it are called integral domains

#

rings which satisfy cancellation law for multiplication the same as those which satisfy your property

#

they're called integral domains

#

it's a good exercise to prove their equivalence

gentle ibex Oct 5, 2019, 9:09 PM

#

So you can't prove "if ab=0 then a=0 or b=0" using the ring laws and the cancellation law for multiplication?

stone forum Oct 5, 2019, 9:09 PM

#

no

#

for example Z/6Z doesn't satisfy this

#

2*3=0

#

but neither 2 nor 3 is 0

#

but it's still a ring

#

satisfies all the ring axioms

gentle ibex Oct 5, 2019, 9:10 PM

#

I see

#

I am being asked to prove it using just these laws though... which is why I am very confused.. because I understand how to do it if they asked for field laws but not ring laws..

stone forum Oct 5, 2019, 9:11 PM

#

perhaps you misunderstood the question?

#

because you can't prove it

gentle ibex Oct 5, 2019, 9:11 PM

#

Is derive different to prove?

#

In this context?

stone forum Oct 5, 2019, 9:12 PM

#

means the same thing

#

prove= derive= demonstrate=convince etc etc

gentle ibex Oct 5, 2019, 9:12 PM

#

Because it's saying to use the ring laws and the cancellation law for multiplication to derive "if a*b=0 then a=0 or b=0"

stone forum Oct 5, 2019, 9:12 PM

#

oh you can do that

#

you have an additional axiom

#

law of cancellation implies this for rings

#

in fact, as I said before, they're equivalent conditions

#

one proves another and vice versa

gentle ibex Oct 5, 2019, 9:13 PM

#

What are equivalent conditions, sorry? I didn't 100% understand before

stone forum Oct 5, 2019, 9:13 PM

#

A and B are equivalent if A implies B and B implies A

#

they're logically the same thing

gentle ibex Oct 5, 2019, 9:16 PM

#

How does the law of cancellation imply this for rings?

#

(multiplicative cancellation)

stone forum Oct 5, 2019, 9:17 PM

#

what does law of cancellation say

#

write it out explicitly

gentle ibex Oct 5, 2019, 9:17 PM

#

If ac=bc and c != 0 then a=b

stone forum Oct 5, 2019, 9:18 PM

#

ok

#

now use this to prove that if ab=0 then a or b=0

#

it's not that hard

gentle ibex Oct 5, 2019, 9:19 PM

#

I don't know how to word it

#

if a = 0 then ab=0

#

if a != 0 then b = 0

stone forum Oct 5, 2019, 9:20 PM

#

ok

gentle ibex Oct 5, 2019, 9:22 PM

#

But that's not a real derivation, is it?

stone forum Oct 5, 2019, 9:22 PM

#

if a != 0 then b = 0

#

how does that follow?

#

this is what you want to prove

#

now prove it

gentle ibex Oct 5, 2019, 9:23 PM

#

I know how to prove it using multiplicative inverses, but you can't use them, as I'm just allowed to use ring laws and multiplicative cancellation law, no?

#

If I were to prove it using multiplicative inverses, you could say a has a multiplicative inverse a^-1 where a*a^-1 = 1

#

and a^-1 * ab = a^-1 * 0

#

using associativity of multiplication you can say (a^-1*a) * b = a^-1 * 0

#

so 1 * b = a^-1 * 0

#

b = a^-1 * 0

#

b = 0

#

Using the law of annihilation at the end

stone forum Oct 5, 2019, 9:26 PM

#

you don't have multiplicative inverses

gentle ibex Oct 5, 2019, 9:26 PM

#

Exactly, which is why I don't know what to do :L

stone forum Oct 5, 2019, 9:26 PM

#

use ab=ac and a!=0 => b=c

#

you need to get it into this form

#

put in the correct letters and numbers

gentle ibex Oct 5, 2019, 9:26 PM

#

0 = ac and a!=0

#

so c must be 0?

#

therefore b=0 since b=c?

stone forum Oct 5, 2019, 9:29 PM

#

what's b

gentle ibex Oct 5, 2019, 9:29 PM

#

b=c

stone forum Oct 5, 2019, 9:29 PM

#

ok

#

i dont understand you

gentle ibex Oct 5, 2019, 9:29 PM

#

I don't understand me either

stone forum Oct 5, 2019, 9:29 PM

#

Assume ab=0 and a!=0

#

you want to prove that b=0

gentle ibex Oct 5, 2019, 9:30 PM

#

yes

stone forum Oct 5, 2019, 9:30 PM

#

hint: ab=a*0

gentle ibex Oct 5, 2019, 9:30 PM

#

and using multiplicative cancellation, b=0?

#

is that sufficient to say?

stone forum Oct 5, 2019, 9:30 PM

#

correct

#

just mention that this is because a!=0

#

one of the conditions for multiplicative cancellation

gentle ibex Oct 5, 2019, 9:31 PM

#

Right, ok.

stone forum Oct 5, 2019, 9:32 PM

#

🐵

gentle ibex Oct 5, 2019, 9:32 PM

#

So if I were to write this out as an explicit proof... "Prove that if ab=0 then a=0 or b=0 using ring laws & multiplicative cancellation law":

if a=0 then by law of annihilation we are done as a * b = 0 * b = 0
if a!=0 - this is the condition for multiplicative cancellation law - then if ab=0, ab=a*0, so b=0 (by multiplicative cancellation)

stone forum Oct 5, 2019, 9:34 PM

#

you don't need to consider the case a=0

#

because then it's already proved

#

there's nothing more to say

#

you want to prove that either a=0 or b=0

#

you see what I mean?

gentle ibex Oct 5, 2019, 9:34 PM

#

ok

#

Is it not good practice to distinguish the two though?

stone forum Oct 5, 2019, 9:34 PM

#

wizard

#

distinguish what

gentle ibex Oct 5, 2019, 9:35 PM

#

when a=0 and when a!=0

stone forum Oct 5, 2019, 9:35 PM

#

you can, but just say if a=0 the theorem is proved

gentle ibex Oct 5, 2019, 9:35 PM

#

ok

stone forum Oct 5, 2019, 9:35 PM

#

or you could leave it implicit

gentle ibex Oct 5, 2019, 9:35 PM

#

gotcha

stone forum Oct 5, 2019, 9:35 PM

#

don't mention it at all

gentle ibex Oct 5, 2019, 9:35 PM

#

Thanks so much for the help 🙂

stone forum Oct 5, 2019, 9:35 PM

#

but better mention it if you're a beginner

#

no problem

gentle ibex Oct 5, 2019, 9:35 PM

#

👍

stone forum Oct 5, 2019, 9:35 PM

#

donate to my patreon

gentle ibex Oct 5, 2019, 9:36 PM

#

Where?

stone forum Oct 5, 2019, 9:36 PM

#

that was a joke

gentle ibex Oct 5, 2019, 9:37 PM

#

😆

#

Hey, 1 last thing

#

You gave this example:

#

"for example Z/6Z doesn't satisfy this
2*3=0
but neither 2 nor 3 is 0
but it's still a ring
satisfies all the ring axioms"

#

If this is true, how can I have just proven the statement if ab=0 then a=0 or b=0?

#

Is it because I used the multiplicative cancellation law also?

stone forum Oct 5, 2019, 9:43 PM

#

yes

gentle ibex Oct 5, 2019, 9:43 PM

#

And that is not a ring law?

stone forum Oct 5, 2019, 9:43 PM

#

Z/6Z doesn't satisfy multiplicative cancellation

#

so no

#

it's not a ring low

gentle ibex Oct 5, 2019, 9:43 PM

#

👌 Thanks 🙂

stone forum Oct 5, 2019, 9:43 PM

#

multiplicative cancellation and ab=0 => a or b=0 are equivalent

gentle ibex Oct 5, 2019, 9:43 PM

#

I see

stone forum Oct 5, 2019, 9:43 PM

#

rings which satisfy the condition are called integral domains

#

Z/6Z is not an integral domain

gentle ibex Oct 5, 2019, 9:43 PM

#

Why say "use the ring laws" as well then?

stone forum Oct 5, 2019, 9:43 PM

#

Z is

#

because you're working with rings

#

you can't prove the equivalence without using ring laws

gentle ibex Oct 5, 2019, 9:44 PM

#

Which equivalence?

#

Sorry if that's stupid I know you've said it a lot

#

but I don't understand exactly what equivalence you are referencing to

stone forum Oct 5, 2019, 9:46 PM

#

(if a!=0 and ab=ac, then b=c) <=> (if ab=0, then a=0 or b=0)

#

the conditions in parentheses are equivalent

#

for any given ring

#

one implies the other and vice versa

gentle ibex Oct 5, 2019, 9:46 PM

#

ok

#

what ring laws do you use to show this equivalence?

#

Is it the annihilation law?

#

because you can say ab=a*0=0

#

Because I understand the annihilation law is essentially a derived ring law from additive inverse

stone forum Oct 5, 2019, 9:48 PM

#

yes

#

that's one direction of the equivalence

#

"Because I understand the annihilation law is essentially a derived ring law from additive inverse
"

#

no

gentle ibex Oct 5, 2019, 9:49 PM

#

Why not?

stone forum Oct 5, 2019, 9:49 PM

#

consider Z[X]

#

polynomial ring

#

it doesn't have inverses

#

yet it's a domain

gentle ibex Oct 5, 2019, 9:49 PM

#

I thought you said (-a) + a + c = (-a) + a + b so b=c

stone forum Oct 5, 2019, 9:49 PM

#

ohhh

#

sorry

#

misread you

#

yeah yeah

gentle ibex Oct 5, 2019, 9:49 PM

#

that's proving additive cancellation law

stone forum Oct 5, 2019, 9:49 PM

#

you're right

gentle ibex Oct 5, 2019, 9:49 PM

#

then use that to prove annihilation law

#

ok

stone forum Oct 5, 2019, 9:50 PM

#

disregard what i said

gentle ibex Oct 5, 2019, 9:50 PM

#

phew haha

#

I thought I had lost it o-o

stone forum Oct 5, 2019, 9:50 PM

#

thought you said multiplicative inverses

gentle ibex Oct 5, 2019, 9:50 PM

#

nono dw

stone forum Oct 5, 2019, 9:50 PM

#

it's late

#

im getting tired

gentle ibex Oct 5, 2019, 9:50 PM

#

ikr same

#

I should look at this in the morning. Thanks so much for the help again. I'll review it all in the morning when it'll hopefully make a lot more sense

#

I've only been studying abstract algebra for a couple of days :l

latent anvil Oct 6, 2019, 4:46 PM

#

Suppose N_G(P) = N_G(Q) for all sylows P, Q. Can there be more than one sylow subgroup?

woven delta Oct 6, 2019, 4:53 PM

#

No

#

@latent anvil there is exactly one p sylow subgroup iff the subgroup is normal

latent anvil Oct 6, 2019, 4:54 PM

#

Yee

#

Oh look at the sylows of the normalizer

woven delta Oct 6, 2019, 4:54 PM

#

Yeah

latent anvil Oct 6, 2019, 4:55 PM

#

Cool, ty

#

This gives a proof of "sylows normal" from "Z(G/Z(G)) = G/Z(G)" without using the big nilpotent group equivalence

#

Since G/Z(G) abelian implies Inn(G) abelian implies that for any sylows P, Q = gPg^(-1), and x in N_G(P), xQx^(-1) = x(gPg^(-1))x^(-1) = g(xPx^(-1))g^(-1) = gPg^(-1) = Q, so N_G(P) <= N_G(Q), and since this holds for all sylows P, Q there's only one normalizer

woven delta Oct 6, 2019, 5:02 PM

#

Hmm

#

I agree

#

But I think Inn(G) abelian is a pretty strong condition anyway

latent anvil Oct 6, 2019, 5:08 PM

#

It is

#

It says you're a product of p-groups P satisfying P' <= Z(P) (not for the same p)

woven delta Oct 6, 2019, 5:10 PM

#

Oh lmao

#

You're right

latent anvil Oct 6, 2019, 5:10 PM

#

My friend is working through a paper on the probability that two randomly chosen elements of a finite group commute, and this came up

woven delta Oct 6, 2019, 5:12 PM

#

That seems like a pretty hard problem

#

Is it a ggt paper?

latent anvil Oct 6, 2019, 5:13 PM

#

It is. The 5/8 bound is easy enough that it's on a homework set for the algebra class I run, but apparently you can establish a lot more

#

No I don't think so, just a lot of finite group theory

#

https://projecteuclid.org/download/pdf_1/euclid.pjm/1102785075

Rusin : What is the probability that two elements of a finite grou...

Project Euclid - mathematics and statistics online

#

"the main insight is: (probability of two elements commuting) = (number of conjugacy classes)/|G| = (number of irreducible characters)/|G|
and the sum of the squares of the degrees of the irreducible characters is |G|
and the number of degree 1 characters is the size of the abelianization G/[G,G]
if you combine that all together (page number 244), you get a relation between the probability and the size of the derived group [G,G]"

#

And there's cool stuff like this

📎 image-4.png

#

Somehow it's just impossible to get between 7/16ths and 1/2

woven delta Oct 6, 2019, 5:16 PM

#

Wow

#

That's pretty fucking weird

latent anvil Oct 6, 2019, 5:16 PM

#

Ikr

latent anvil Oct 6, 2019, 8:21 PM

#

Can I check a commutative algebra thing?

#

Suppose f is a polynomial in x1,...,xn,y over a field k. Further suppose f is monic in y, so monic if you think of it as an element of k[x1,...,xn][y]. Then for any polynomial g in k[x1,...,xn,y], there are polynomials q, r in k[x1,...,xn,y] such that g = qf + r and deg_y r < deg_y f. You can prove this by doing division with remainder in k(x1,...,xn)[y] to get q, r which are rational functions in x1,...,xn, do induction on deg_y q to prove it is a polynomial, and then write r = g - qf to conclude it is also a polynomial

#

Does that sounds plausible?

#

I wrote it out on a whiteboard but feel uncomfortable with it

#

This is part of the proof of the noether normalization lemma but my book just says "by polynomial long division"

wind steeple Oct 6, 2019, 8:35 PM

#

forget about polynomial

#

in a ring A

#

a polynomial f in A[y]

#

you can do euclidean division by f if f is monic

#

you can proove like the proof you wrote

#

and it's less weird since you consider only elements of a ring A

#

since k[x1,...,xn] is a ring, it works

latent anvil Oct 6, 2019, 9:19 PM

#

Oh cool

#

Does A need to be an integral domain (so you can take the field of fractions)?

latent anvil Oct 6, 2019, 10:20 PM

#

More commutative algebra that I'm not seeing

#

Suppose A is a finitely generated k-module and also an algebra over k. Then there are finitely many maximal ideals of A

#

Anybody have any insight here?

#

for k a field (algebraically closed if it matters)

latent anvil Oct 6, 2019, 11:29 PM

#

I think I see how to prove it

stone forum Oct 6, 2019, 11:33 PM

#

how did you prove it? @latent anvil

latent anvil Oct 6, 2019, 11:40 PM

#

Okay well keep in mind I said I think I see how to prove it

#

Not that I did prove it for sure

#

I need to write it out more clearly, every time I think about it I get twisted around. It's equivalent to say that for any ideal I of k[x1,...,xn], if the subset Z(I) of affine n-space is infinite then k[x1,...,xn]/I is infinite dimensional over k. Then(?) for some j there are infinitely many a in k such that (xj - a) intersects I nontrivially. Then(?) the map k[t] -> k[x1,...,xn]/I sending t to xj has a big image

#

Obviously this isn't rigorous

#

I might need to handle I = 0 separately

#

sorry if this is totally wrong lol

latent anvil Oct 7, 2019, 12:00 AM

#

I can prove that for any I as above, there is some j such that either Z(xj - a) \cap Z(I) is infinite for some a in k, or that Z(xj - a) \cap Z(I) is nonempty for infinitely many a in k

#

In the first case, I think I can do induction on n (the dimension of the ambient space)?

latent anvil Oct 7, 2019, 2:32 AM

#

📎 fin_dim.png

#

🎊 🎊 🎊 🎊 🎊

#

my idea was basically right

#

@stone forum sorry it took 3 hours to actually put into words

#

way simpler proof:

#

📎 unknown.png

#

oops, that should say Gamma(X) -> Gamma(X)/\bigcap_{n=0}^N I_n

wind steeple Oct 7, 2019, 5:25 AM

#

I don't see why A needs to be an integral domain here but I remember a condition like this on a @latent anvil

latent anvil Oct 7, 2019, 5:33 AM

#

I'm not sure how you could pass to the field of fractions otherwise

#

I don't understand localization in non-integral domains super well

#

Oh, I think it still works

#

since A is a f.g. k-algebra, you have A = k[x1, ..., xn]/I. Given a monic f in A[y], you can lift it to a monic f' in k[x1, ..., xn][y], and then given g in A[y] lift it similarly and do long division in k[x1,...,xn][y]

#

wait no this only works for an A which is a finitely generated algebra over a field. I forgot other kinds of rings exist lmao

#

I think you can drop the finitely generated assumption, since you can write A as a quotient of some really big polynomial ring, and when you want to divide g by f, there can only be finitely many variables mentioned in f and g, so you can do the long division in the subring generated by those variables

#

so this works for all A which contain a field as a subring

#

I think this will work for any quotient of an integral domain actually

#

is every ring a quotient of an integral domain?

brisk granite Oct 7, 2019, 6:32 AM

#

Given $G \trianglerighteq K \trianglerighteq H$, how would I show $(G/H)/(K/H) \cong G/K$

cloud walrusBOT Oct 7, 2019, 6:32 AM

#

Cabbage:

📎 375866732226543616.png

brisk granite Oct 7, 2019, 6:33 AM

#

I think this is one of the isomorphism theorems, but I don't know which one

wind steeple Oct 7, 2019, 6:34 AM

#

@latent anvil no it works for arbitrary unitary ring A

#

Just by induction

mild laurel Oct 7, 2019, 6:34 AM

#

Second isomorphism theorem I think it's usually called @brisk granite

wind steeple Oct 7, 2019, 6:35 AM

#

@brisk granite just write the diagram down and check that the only morphism you can define works

brisk granite Oct 7, 2019, 6:35 AM

#

uh, the diagram? Edit: got it, thx

latent anvil Oct 7, 2019, 6:40 AM

#

@wind steeple by induction on what?

#

Oh the degree of the polynomial, just like how you prove F[x] is a Euclidean domain

#

I was trying to reduce to the case of a field

#

I forgot you can just prove things lmao

wind steeple Oct 7, 2019, 7:25 AM

#

On the degree

#

Yes

brisk granite Oct 7, 2019, 11:46 AM

#

If Ik a group is solvable, do I know it has a composition series that is a solvable series?

hot lake Oct 7, 2019, 12:07 PM

#

only for finite groups

brisk granite Oct 7, 2019, 12:44 PM

#

oh, could you explain why?

hot lake Oct 7, 2019, 12:47 PM

#

because some infinite abelian groups don't have a finite composition series

#

and iirc composition series have to be finite

brisk granite Oct 7, 2019, 12:52 PM

#

why does it hold for the finite case tho?

hot lake Oct 7, 2019, 12:58 PM

#

because every finite abelian group has a finite composition series

fading wagon Oct 7, 2019, 1:00 PM

#

is finitely generated sufficient?

hot lake Oct 7, 2019, 1:01 PM

#

hmm probably

#

yeah it should work with finitely generated groups

chilly ocean Oct 7, 2019, 1:05 PM

#

Hello, I am trying to prove that this integral defines an inner product in the space of second degree polynomials

📎 unknown.png

#

I am having issues with one of the axioms, <x,x> >=0 for all x belonging to P2

#

I tried the following

fading wagon Oct 7, 2019, 1:06 PM

#

try substituting in f, f?

chilly ocean Oct 7, 2019, 1:06 PM

#

i did

#

📎 unknown.png

#

I must have done something wrong somewhere

#

(my prof told me that when i have integrals from -1 to 1 i can just get rid of the odd degree polynomial expressions)

#

(Im pretty sure id get the same results if i didnt do that so dw about it)

cloud walrusBOT Oct 7, 2019, 1:10 PM

#

leoli1:

📎 193026762026778624.png

#

leoli1:

📎 193026762026778624.png

chilly ocean Oct 7, 2019, 1:12 PM

#

Can I just say that's always greater or equal to zero? i'm legitimately not sure which is why i developed it

#

sorry if its obvious 😛

#

Oooh, I realize now that is true when i look it up

#

For real valued functions

#

Ok, yeah, that's fair lol

#

Wait, so if the integrand https://media.discordapp.net/attachments/496784958430380033/630754099553370122/193026762026778624.png

#

is non negative, is the defined integral from -1 to 1 also necessarily non negative?

queen vine Oct 7, 2019, 1:37 PM

#

Yes this is because of the monotonicity property of the integral, i.e. for functions $f\leq g$ you have $\int_{-1}^{1} f(x)dx\leq \int_{-1}^{1}g(x)dx$

cloud walrusBOT Oct 7, 2019, 1:37 PM

#

leoli1:

📎 193026762026778624.png

kindred compass Oct 7, 2019, 1:44 PM

#

How did they get that f(x) = 0 for all x in F if the kernel is everything?

📎 image0.png

chilly ocean Oct 7, 2019, 1:45 PM

#

Thank you @queen vine

#

I'm so rusty on calculus

queen vine Oct 7, 2019, 1:47 PM

#

🙂

cloud walrusBOT Oct 7, 2019, 1:47 PM

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leoli1:

📎 193026762026778624.png

kindred compass Oct 7, 2019, 1:48 PM

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I thought the kernel is the set of r(x) in F(x)/(f(x)) such that r(alpha) = 0..

brisk granite Oct 7, 2019, 1:54 PM

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Is this statement even true? If $|G/H|$ is of a prime power order, then don't I know that there is only one sylow p-subgroup S and, hence, that H is not a maximal normal subgroup?

📎 Screen_Shot_2019-10-07_at_7.21.06_PM.png

cloud walrusBOT Oct 7, 2019, 1:54 PM

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Cabbage:

📎 375866732226543616.png

brisk granite Oct 7, 2019, 2:03 PM

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nvm, the sylow p subgroup would be G/H. but wouldn't Z(G/H) also be normal in G/H?

faint elm Oct 7, 2019, 3:29 PM

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Yes, center is always normal subgroup

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Let $M$ be a flat module, and let $ 0 \rightarrow X \rightarrow Y \rightarrow M \rightarrow 0$ be an exact sequence. The for every module $N$ we have that $0 -> N \otimes X \rightarrow N \otimes Y \rightarrow N \otimes M \rightarrow 0$ is an exact sequence

cloud walrusBOT Oct 7, 2019, 3:32 PM

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emme:

📎 578598429471801344.png

faint elm Oct 7, 2019, 3:32 PM

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Do you have any hints?

little folio Oct 7, 2019, 3:55 PM

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imagine it's not exact. what's the first term on the left?

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@faint elm

faint elm Oct 7, 2019, 3:59 PM

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Ok so there exists a module $N$ s.t. $0 \rightarrow N \otimes X \rightarrow N \otimes Y \rightarrow N \otimes M \rightarrow 0$ is not exact, this means that $N \otimes X \rightarrow N \otimes Y$ isn't injective. Right?

cloud walrusBOT Oct 7, 2019, 3:59 PM

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emme:

📎 578598429471801344.png

little folio Oct 7, 2019, 4:01 PM

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how much homological algebra have you seen? I am assuming you are familiar with the left derived functors of the tensor product

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the point is you have a long exact sequence including the Tor groups

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and the first term on the left is gonna be Tor(M,N)

faint elm Oct 7, 2019, 4:04 PM

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No, I've seen no derived functors so far. I just know basic commutative algebra, like first 2 chapters of atiyah

little folio Oct 7, 2019, 4:04 PM

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ah

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that's the problem

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you might wanna read it from dummit & foote

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10.5 and 17.1

faint elm Oct 7, 2019, 4:08 PM

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I'll take a read thank you. Do you think it's impossible to answer without derived functors? The question is of the first week of the course and we covered just the first two chapters of atiyah

little folio Oct 7, 2019, 4:20 PM

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no, it's not impossible, surely you don't need all the machinery

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but even the exercises on A&M ch 2 assume familiarity with tor

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and it's good for you to learn tor and ext soon rather than later

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what you would have to do is chase the connecting homomorphism Tor_1(M,N) -> NxX -> NxY -> NxM -> 0

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and prove that Tor_1(M,N) = 0

faint elm Oct 7, 2019, 4:27 PM

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I know Tor functor btw

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Ok thank you

little folio Oct 7, 2019, 4:28 PM

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derived functor is a scary name but it's just knowing what Tor is

faint elm Oct 7, 2019, 4:28 PM

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Ah, ok

faint elm Oct 7, 2019, 9:54 PM

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Let $ A = \mathbb{C}[x,y]$. I want to find an $A$ module which is torsion free, finitely generated but is not flat. I think $M = A/(xy-1)$ should be fine, it's torsion free since if $pq \in (xy-1)$ then $xy-1$ is a factor of $p$ or $q$, it is generated by $1, x, y$ and it's not flat bc if you consider $0 \rightarrow (xy-1) \rightarrow A \rightarrow M \rightarrow 0$ then $0 \rightarrow (xy-1) \otimes M \rightarrow A \otimes M \rightarrow M \otimes M \rightarrow 0$ is not exact since $(xy-1) \otimes M = 0$. Do you agree?

cloud walrusBOT Oct 7, 2019, 9:54 PM

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emme:

📎 578598429471801344.png

latent anvil Oct 7, 2019, 10:22 PM

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I don't agree. (xy-1) (×) M -> A (×) M -> M (×) M -> 0 is exact by right exactness of tensor product, so 0 -> A (×) M -> M (×) M -> 0 is exact. You can stick a zero to the left and it stays exact

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I might be wrong

faint elm Oct 7, 2019, 10:46 PM

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Maybe you are right

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Let's see if I can fix it

latent anvil Oct 7, 2019, 11:28 PM

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Is A/(y^2 - x^3) flat?

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I don't actually know the geometrical interpretation of flatness

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I know it's important

little folio Oct 7, 2019, 11:29 PM

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yeah the problem is that your sequence is exact anyway because (xy-1)xM is 0

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so Tor(M,M) never comes into play

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i think A/(y^2 - x^3) is flat

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this might help https://mathoverflow.net/questions/33522/flatness-and-local-freeness

MathOverflow

Flatness and local freeness

The following statement is well-known:

$A$ a commutative Noetherian ring, $M$ a finitely generated $A$-module. Than $M$ is flat if and only if $M_{\mathfrak{p}}$ is free for all $\mathfrak{p}$.

My

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so finitely generated projective = finitely presented flat

little folio Oct 7, 2019, 11:55 PM

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but I believe A/(xy - 1) is flat too

faint elm Oct 8, 2019, 6:12 AM

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Yeah you're right guys

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Maybe there is something more simple

faint elm Oct 8, 2019, 7:59 AM

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Anyway @Flopp, for the previous question: let $N$ be a module, then is the quotient of a free module $L$, say $N = L/K$, so let's consider the short exact sequence $0 \to K \to L \to N \to 0$ and let $0 \to X \to Y \to M \to 0$ be an exact sequence, where $M$ is flat. Now we tensorize this sequence with every factor of the first one.

cloud walrusBOT Oct 8, 2019, 7:59 AM

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emme:

📎 578598429471801344.png

faint elm Oct 8, 2019, 7:59 AM

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So you get something like this

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\begin{tikzcd}
& & & & & 0 & & \
& & & & & \downarrow & & \
& K \otimes X & \rightarrow & K \otimes Y & \rightarrow & K \otimes M & \rightarrow & 0 \
& \downarrow & & \downarrow & & \downarrow & & \
\rightarrow & L \otimes X & \rightarrow & L\otimes Y & \rightarrow & L \otimes M & \rightarrow & \
& \downarrow & & \downarrow & & \downarrow & & \
& N \otimes X & \rightarrow & N \otimes Y & \rightarrow & N \otimes M & & \
& & & & & \downarrow & & \
& & & & & 0 & &
\end{tikzcd}

cloud walrusBOT Oct 8, 2019, 7:59 AM

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emme:

📎 578598429471801344.png

faint elm Oct 8, 2019, 7:59 AM

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First two rows satisfy snake lemma and from that we have the thesis

faint elm Oct 8, 2019, 9:06 AM

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As for as the torsion free f.g. module which is not flat, how about $M = \mathbb{C}[x,y]/(x,y)$, if we take $ p = (x)$ we got $M_p = 0$ and so it's not flat for the result on MO

cloud walrusBOT Oct 8, 2019, 9:06 AM

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emme:

📎 578598429471801344.png

faint elm Oct 8, 2019, 9:07 AM

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Wha do you say?