#groups-rings-fields

406252 messages · Page 621 of 407

oak grove
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i guess you get uniqueness just by basic number stuff, they can only share the identity

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i guess it's just the original question i asked

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i get why the problem says you have elements of order 5 and order 3

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which seems to be what you said

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but not that you have one of order 15

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im just supposed to see it im sorry i dont bearlain

final oasis
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it is just counting , I leave the rest to you

oak grove
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is it just the original thing i said

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take some element order 5 and some element order 3

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i guess that doesnt make sense

final oasis
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what is the order of each element

oak grove
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in G?

final oasis
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yeah

oak grove
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youll have 2 of order 3, 4 of order 5, 1 of order 1, and the other 8 unknown

final oasis
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you know

oak grove
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i guess you know possibles that theyll have to be one of four orders

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well one of 3 i guess

final oasis
oak grove
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no

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i can try

obsidian sleet
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hello friends

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i am confused about this part in my hw

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so i proved part a I hope correctly

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but part b asks us to deduce (as a result of a, i assume) some stuff

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but idk how to use a to get b

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all i got so far is that if the index is n there are n disjoint left cosets of H in g

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same thing for the right ones of course

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but uh

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that was from an earlier section

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maybe i can use something from part a to get all this

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idk

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pls ping if u have any advice/hints ty :3

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oh dear i guess i should add in the definition of double coset if that helps

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from the previous problem

unknown.png
obsidian sleet
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im kind of struggling with the meaning of this too

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its like

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there would be n elements in g such that we can form left cosets g_i H

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which are disjoint

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and whose union is G

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and then those same elements can form H g_i and these partition G too

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i think

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is what this means

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oh

final oasis
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check this out

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@obsidian sleet

obsidian sleet
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?

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i am confused

tame grove
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So if I am proving the associativity for a binary operation defined on R \ {-1} with the value a*b= a + b + ab, what do I have to prove? I have (a+1)(b+1)=0 to prove that the binary operation exists, am I allowed to introduce c + 1 within R \ {-1} and use it to demonstrate the associative property holds? I am just confused since I am using two different operations (addition and multiplication) defined under the same binary operation *

oak grove
gritty sparrow
# obsidian sleet pls ping if u have any advice/hints ty :3

I think I this should work: so first we will find simultaneous left and right coset representatives for each double cost HgH. To do this we consider the n’ left cosets and the n’ right cosets that are contained in HgH. Now pick a left coset say lH, this will intersect all right cosets, pick any one sat Hr, let a1 be in the intersection. Then a1H=lH and Ha1=Hr. Now pick some other left coset and a different right coset it intersects to get a_2. Continue this way till you get a_{n’}, these will all represent distinct left and right cosets.

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Now just do this for all distinct double cosets

obsidian sleet
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hmm

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yeah i had something like the first part i was thinking to use the intersection proprety to find repreentatives for the left and right cosets

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and i was able to find one but i knew i could find more but i didn't know of an algorithm to find n many

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what is n' though

gritty sparrow
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The number of left (or right) cosets that are a subset of HgH

obsidian sleet
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ah okay

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so each of the n left/right cosets are subsets of the HgH for different g

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as in one HgH may have n' many of the n

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and so we know that there are gonna be n total we can find

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since we just do all the other double cosets

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as well

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we would find n' in each HgH

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and these would sum to n?

gritty sparrow
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Well they have to

obsidian sleet
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right cus there better be exactly n of them

gritty sparrow
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The point is that G is a disjoint union of its double cosets so we can work one double coset at a time

obsidian sleet
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right

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the partition property ensures that we can find all n just in different HgH's

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okay

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makes sense

gritty sparrow
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Does the total proof make sense now?

obsidian sleet
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i think it makes sense i just have to go through it again

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thank you so much

gritty sparrow
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Np

wooden ember
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the set of exercises assumes at the start that R has an identity: would this conclusion be true if R didnt have an identity?

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wooden ember
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i can show that this is finitely generated by nilpotent elements but im not sure how to conclude since the ring isnt necessarily commutative

unknown.png
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i feel there might be some property of p-groups to use (i wont get trapped in terminology and assume it's their nilpotency) but i cant figure out what

sour plume
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anybody who understands spectral sequences: do you feel like "a spectral sequence appears when you have a chain complex whose homology you want to calculate, and it comes equipped with some filtration. the spectral sequence is then the natural product of trying to understand what the filtration does on the level of homology" is sufficiently accurate in most cases

willow mason
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but without 1 ideals would be weird

weak oriole
oak grove
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can i come back to this question blobsweat

unknown.png
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I want to readdress this implying |stab(i)| = 1 for any i coprime with in Aut Z_n

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and why that should be intuitive

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im not sure im 100% following what it would mean for something to be a stabilizer in this group

weak oriole
oak grove
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Are we just saying that like

oak grove
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oh wait

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this might help too blobsweat

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so if |stab(i)| = |G|/phi(n)

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and the next statement is true

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then |stab(i)|=1 is always true?

weak oriole
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Oh, one way to do it it to think about what element the generator <1> can map to

oak grove
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should be able to go to any element coprime with n i figured

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dont cat king blobsweat i dont understand it yet lol

weak oriole
hidden haven
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Well stab(1) = 1, not stab(i) for any i

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See 2 in Z4

oak grove
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you said that before thonk

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i never did figure out what you meant

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OH

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yea i get what you mean

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but i dont see how that means anything here

hidden haven
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Oh right that's where the confusion started KEK

oak grove
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okay so i realize now i was confusing two things

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but its not important

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i get what you mean now

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theres multiple automorphisms that don't move 2

hidden haven
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So 2 is stabilized by the identity automorphism, and the automorphism which does x ↦-x

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Yes

oak grove
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but

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why should that link with |stab(1)|=1

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why cant there be multiple

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why not just not move the generators but swap non generators

hidden haven
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1 being a generator is important

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||a homomorphism is completely determined by what it does to generators||

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But a smaller hint than what I wrote above in case you don't wanna see that

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Is to write any other element in terms of 1 first

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And apply an automorphism to that element

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Assume that that automorphism stabilises 1 and then see what that tells you about the automorphism on other elements

oak grove
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so youre saying that uhh

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Z6 where you only swap 2 and 4 is gonna be a problem

hidden haven
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You can't only swap 2 and 4

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And fix the other stuff

oak grove
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?

weak oriole
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Thinking about automorphisms as random swappings is not a good idea here

oak grove
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oh does that break isomorphism rules

weak oriole
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Since u want something more than a bijection

hidden haven
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Yep

oak grove
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more than a bijection 👀

weak oriole
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A bijection that respects ur operation right?

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Infact, try to characterize all homomorphisms f : Z/nZ to Z/nZ

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You'll see that f takes a very restricted form

oak grove
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we havent seen those yet

hidden haven
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Not seen homomorphisms yet? stare

oak grove
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no

hidden haven
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oof

oak grove
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we havent seen a lot of the stuff you guys are talking about lol

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but its not important

weak oriole
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What is your definition of automorphism then?

oak grove
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automorphism is an isomorphism that maps from a group to a group

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same group

weak oriole
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Yes, and what's a isomorphism

oak grove
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a mapping from some group to some group

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with conditions

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im not seeing how my suggested mapping breaks anything

weak oriole
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Which conditions

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Elaborating them will probably help

oak grove
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oh

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well i didnt see it through the conditions

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but youre right i guess

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sorry i gotta remember to breathe and think

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i see what you mean im gonna write it down

weak oriole
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Cool cool

oak grove
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no wait

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idk i dont understand but i cant not finish this assignment today im just gonna write stuff down KEK

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so im guessing induction isnt the method here thonk

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actually itd probably be easier to just show it directly than going through stab

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well i assume

hidden haven
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Replace homomorphism with automorphism

oak grove
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oh bearlain

weak oriole
oak grove
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i will when i have time catthumbsup

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i still have to do that other question thonk

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yea im still not seeing this one either blobsweat

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i think im close? or im on the right track

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i have a bad feeling bchaotic completely explained it and i cant tell

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youll need to use |G|/|H| = # distinct left and right cosets or number of ways to partition up G here

oak grove
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how can we know the order of the rest of the elements?

willow mason
weak oriole
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that is too much of a overkill

silent osprey
# oak grove

subgroups are prime order so they must be cyclic take the generator of subgroup of order 3 to be 'a' and generator of subgroup of order to be 'b' then 'ab' must have order 15

urban acorn
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not saying this isn't a good direction

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but it's not a complete proof

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but I have an idea for a different way to prove this

urban acorn
# oak grove

if these subgroups are the unique subgroups of their order, then they must be normal since any conjugates would have the same order

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and therefore you can use the characterization of internal direct products to show that G is the direct product of both subgroups

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and then - because of primality the subgroups are cyclic - and since 3 and 5 are coprime their direct product is cyclic

silent osprey
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Yeah it wasn't meant to be a proof

urban acorn
weak oriole
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You can probably just do order bounds
number of elements of order n= (number of cyclic subgroups of order n)*phi(n)

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In particular, a nice generalization of the problem would be to show the following :
Given a finite group such that x^n=1 has atmost n solutions for each n, show that G is cyclic

urban acorn
crisp badge
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Given a particular element g \in G, consider <g>. The size of <g> is either 1, 3, 5 or 15 - and there is only 1 subgroup of size 3 and 1 of size 5.

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From this, show that |<g>| is size 15 for some g

urban acorn
weak oriole
urban acorn
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well genie's way seems like the right way to go

weak oriole
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Ya they are basically the same

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Mine and genie's I mean

urban acorn
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re-reading what you were saying, I see how you had a similar thing to genie's argument in mind, but I couldn't figure it out from what you were saying

weak oriole
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Ah, I meant the map $x \to <x>$ from set of elements of order $n$ to set of cyclic subgroups of order $n$ is $\phi(n)$ - to - $1$

cloud walrusBOT
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Noob666

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willow mason
crisp badge
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But this sort of question is a beginning group theory question

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Oh wait, that was a joke 🤦

willow mason
weak oriole
willow mason
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if we dont have 1

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we need to change the definition of homomorphisms and ideals slightly to deal with it

urban acorn
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I have certain concepts about rings already ingrained in my mind, and I typically assume rings have a 1, so when you ask me about whether something holds without a 1 I can never be 100% sure the argument I come up with doesn't somehow rely on the existence of a 1 in a way I'm not thinking of

urban acorn
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and I thought: this property (that every a has n s.t. a^n = a) easily carries over to quotient rings

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so we just need to show that integral domains with this property must be fields

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and that's easy (cancel both sides of the equation by a)

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but I can't actually answer the question because I have no idea if I implicitly used the presence of identity here

willow mason
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how would a field without 1 work in the usual definition

hidden haven
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you can't define inverses without 1

willow mason
wooden ember
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Is there a particular reason we take polynomial rings over commutative rings?

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Like what’s uninteresting about say a polynomial ring over M_n(R)

willow mason
wooden ember
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But surely it would be interesting to study non commutative polynomial rings ?

willow mason
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people probably have

wooden ember
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True

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But I’m wondering what makes them less interesting that regular polynomial rings (since they aren’t included in d&f for example)

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Or less standard let’s say

willow mason
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just like semi-groups and groups

wooden ember
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Seems a bit simple of an answer

willow mason
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why not study semi groups

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cuz weak definition

wooden ember
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Like we study group rings and they don’t require group commutativity

willow mason
wooden ember
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Fair enough lmao

urban acorn
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because in the polynomials themselves we assume the elements commute

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in Z[x, y] we have xy = yx

wooden ember
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fair, x commutes with stuff

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but x is a formal object it doesnt necessarily have to be an analogue to an indeterminate in our ring

urban acorn
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the real motivation for R[x] isn't a vague idea that "polynomials" in the sense of elementary functions you learn about in high school are important so let's make rings out of them

wooden ember
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in the same way in a group ring the elements of the ring commute with those of the group

urban acorn
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the real motivation for R[x] is that in the category of commutative rings, a map out of R[x] is precisely the same thing as a map out of R and a free choice where x goes

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and same thing for R[x1, ..., xn], or in any arbitrary set of indeterminates

wooden ember
urban acorn
wooden ember
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huh so we have a sort of universal property then

urban acorn
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yes

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you can do a construction equivalent to this in the category of non-commutative rings

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and then for example you get a version of ℤ[x, y] where xy ≠ yx

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and they're called non-commutative polynomials

wooden ember
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i see

urban acorn
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however, here's another reason why we specifically care about commutative polynomials:

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polynomial rings are for algebraic geometry

wooden ember
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so (R[x])[x] also satisfies thuis universal property right?

urban acorn
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and algebraic geometry is over commutative rings

wooden ember
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cause then if you pick the same y as for R[x] the x in R[x] and the x in R[x][x] both get mapped to y hmmCat

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funky

urban acorn
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"R[x]" actually conveys more information than "polynomial ring with coefficients in R"

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it also tells you what to name the indeterminate: x

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so when you take another polynomial ring, this time with coefficients in R[x]

wooden ember
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fair enough so i shouldve said R[x][y] or smth

urban acorn
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you're going to want to name the determinate something different

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yeah

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and R[x][y] is (naturally, in fact) isomorphic to R[x, y]

wooden ember
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makes sense

urban acorn
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one way to think of it is like this: every polynomial in the indeterminates x, y can be rearranged around the powers of y in order to look like a polynomial in y with coefficients that are polynomials in x

wooden ember
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wait a second wouldnt R[[x]] also satisfy the previous universal property?

urban acorn
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no

wooden ember
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why not

urban acorn
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where do you send 1 + x + x^2 + x^3 + ...?

wooden ember
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oh fair enough lol

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cant be having infinite sums in our target space

urban acorn
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yeah

urban acorn
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i.e. a map R[[x]] -> S conveys more information, because you have more choices as to were to send the infinite elements

wooden ember
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yeah

urban acorn
wooden ember
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that makes me think, in R[[x]] is the only ideal that is all of the ring (1)?

urban acorn
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I guess technically polynomial rings are also highly relevant in algebraic number theory though

wooden ember
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is there no other representative?

urban acorn
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"all of the ring" completely specifies an ideal

wooden ember
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no i know

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i meant as representing it as a principal ideal

urban acorn
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so you're asking if 1 is the only element that generates the entire ring as an ideal?

wooden ember
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i dont know how to formally say it (is the terminology generating set for your ideal?)

wooden ember
urban acorn
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well, consider (2) in ℚ[[x]]

wooden ember
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fair enough

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what if R isnt a division ring though

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well a field since R is commutative

urban acorn
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Okay, so bear in mind that my intuitions are only for commutative rings with 1

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but

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x ∈ R satisfies (x) = R if and only if x is a unit in R

wooden ember
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yeah so it comes down to classifying units in R[[x]] and im pretty sure every power series starting with a unit from R is a unit in R[[x]]

urban acorn
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so you're asking what the units are in R[[x]]. clearly every unit of R is also a unit of R[[x]], but I think it actually turns out that things that really shouldn't be units, are units

urban acorn
wooden ember
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i think ive proved this gimme a sec

urban acorn
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so, as an example (1 + x) feels like it shouldn't be a unit, but consider (1 + x)(1 - x + x^2 - x^3 + x^4 - x^5 + ...)

wooden ember
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yeah ive seen thjis before

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yeah thought so

urban acorn
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yeah I think it's an exercise in dummit and foote

wooden ember
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a power series is a unit in R[[x]] iff a_0 is a unit in R

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i did prove it

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gotta head out but thanks for your time

urban acorn
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alright, nice to talk to you

oak grove
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I reached out to the teacher and she got back

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all she wanted us to use was all elements orders divide 15 -> look at 8 elements not in our two specific subgroups -> assume they have order 3 or 5 -> show that implies they're not one of the 8 remaining elements -> they must all be generators

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so basically what bchaotic said bearlain

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sorry to clog up this channel with such a silly question, thanks for your help yall 🙇‍♂️ ill study harder before the next one so my questions arent as silly

urban acorn
unreal portal
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In class today, my professor wanted to prove that a maximal ideal $\mathfrak{m}\subset R$ was unique. He did this by showing for all $x\in (R\setminus\mathfrak{m})$, $x$ was invertible. Could someone elaborate on how this shows the maximal ideal is unique?

cloud walrusBOT
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cgodfrey

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wooden ember
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what do you mean by M is unique?

unreal portal
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I believe to show that it's the only maximal ideal

chilly ocean
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every ring is local pepeomegaworry

wooden ember
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is did you mean R/m or R\m, cause if R is commutative every element of R/m will be invertible no matter what

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whether it's unique or not

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ah no you meant R\M of course

unreal portal
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oh yeah I meant R\m

wooden ember
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well suppose there is another maximal ideal M'. Then it contains a unit x that is not in M. But then M' is the whole ring and so M' is not a maximal ideal

unreal portal
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ooh that makes sense

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as a follow up question, is it always true that the noninvertible elements of a ring form a maximal ideal?

wooden ember
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i just realized this idea comes up in 3 exercises lol

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no because then R is necessarily local

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and not every ring is local

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im definitely not the most qualified on local rings since i havent even gotten to the exercises on them yet but someone else might be able to give you good examples and intuition

unreal portal
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I'm not familiar with what a local ring is (although I should be since I'm taking algebraic geometry lmao), I'll look more into that

wooden ember
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a local ring is a ring with a unique maximal ideal

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i have a question of my own if that's all you had to ask: im not seeing at all how to proceed with this one

unknown.png
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the previous exercises were on radical ideals and the jacobson ideals so maybe those play a role but im not sure how

unreal portal
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is that the atiyah-macdonald book?

wooden ember
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nah this is just d&f

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i dont have the prereqs on rings and modules to get into AM yet

chilly ocean
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a&m is a fun problem book

maiden ocean
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and then we'll be done by maximality

wooden ember
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yeah hold on i think im onto something

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sorry though i appreciate the answer haha

maiden ocean
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no problem

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ill leave a hint if you get stuck

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||you want to show that the set of common zeroes of M is non-empty. compactness of the interval is key||

wooden ember
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aight i got it

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it's a very neat proof

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topology is so cool

maiden ocean
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nice nozoomi

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This result leads into algebraic geometry

wooden ember
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if i hadnt been looking at compactness the other day i would have never thought of this monkaS

wooden ember
maiden ocean
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Well i mean you dont need it to do any AG or anything but this idea of looking at zero sets of polynomials is the key

wooden ember
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stone cech compactification or smth

maiden ocean
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not sure what u mean by that exactly but theres probably context im missing

wooden ember
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i dont know i havent read the comment yet

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but right after the exercise there's a comment about stone cech

maiden ocean
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Oh yes the stone cech compactification will recover MaxSpec in this case

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filters or something

wooden ember
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what's MaxSpec?

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the maximal spectrum?

maiden ocean
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Yes

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maximal ideals

wooden ember
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c) and d) would be true if we were considering C^1 functions right?

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nvm

maiden ocean
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Uh

wooden ember
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theyd only be true for C^infty functions i think

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by true i mean (x-c) = M_c (and d) follows)

urban acorn
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it's a maximal ideal that doesn't correspond to a point in your space

sinful mirage
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so far what I understand is, that a representation induces a special left action(given by the rep). But why is every left-action a representation?

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or is the statement even true?

urban acorn
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the statement is absolutely true

urban acorn
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it needs to preserve the vector space structure, i.e. v → gv is linear for every g

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then that's the homomorphism you get from G → GL(V), you just send g ∈ G to the map v → gv which is in GL(V)

sinful mirage
cloud walrusBOT
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Goose on a Moose
Compile Error! Click the errors reaction for more information.
(You may edit your message to recompile.)

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snow cedar
tawny pine
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<@&681260374879633482>

shell brook
sly nexus
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yes
PepeLaugh
each combination ideally

tawny pine
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ok done for you; for some reason can't get perms to work

raven merlin
latent anvil
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Yo

narrow marsh
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owo

sly nexus
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maybe you can cut it a bit by some amount, but thats an exercise on itself

narrow marsh
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associativity is a pain to prove in general

snow cedar
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yeah I figured as much, I know there is a shortcut with abelians groups, if the cayley table is symmetrical about the diagonal than it is commutative, I was wondering if there is something similar with associativity.

narrow marsh
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not that i know of

nova cliff
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one way to do this is to realize Z2 as functions on Z2: 0 is the do-nothing function, and 1 is the swapping function
then associativity follows because function composition is associative

snow cedar
nova cliff
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this is more "pure thought" but it [might not be] strictly less work

jagged geyser
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I was taged

sly nexus
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yah
its 8

jagged geyser
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Does anybody need help with mathematics today

snow cedar
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All good, my question was answered. Thanks for the help kind peeps.

native sand
# jagged geyser Does anybody need help with mathematics today

Are there any simple examples of non-linear groups encoding symmetry transformations that a layperson would understand pretty easily, as like an everyday example? (Like how 3d rotations are a simple counterexample showing composition doesn’t have to be commutative)

jagged geyser
native sand
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Fair enough, I can’t think of any either. I’m trying to motivate group axioms to myself by just considering transformations that encode symmetry, can put a check mark on all the axioms, except, every example I can possibly think of is linear, googling things, some esoteric lie group stuff pops up, but that’s way out there compared to first starting a study of groups

maiden ocean
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I dont know what your group theory background is so im not sure if any of them will be particularly insightful

native sand
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alright ty, for motivation then I’m just gonna go with the following justification:
When considering a group, we just want to focus on one operation, and linearity, while nice, is a red herring that involves introducing potentially two more operations

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not technically correct, as I’ll see later, but, it’s nice and intuitive

maiden ocean
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Is this for educational purposes

native sand
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I know the axioms of a group, and some basic examples, and I’ve worked through the first couple chapters of Pinter

maiden ocean
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Oh wait what do you mean by "linear"

native sand
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I’m not an expert or anything on it though

maiden ocean
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I assumed you meant embedding into GL_n(K)

native sand
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linear means a function that obeys two properties
L(ax) = aL(x) for some constant a, vector x
L(x + y) = L(x) + L(y) for vectors x,y

maiden ocean
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What is the function here, an element of a group G acting on some set?

native sand
#

yes, like rotations for example

#

or addition on integers, multiplication on reals (with no 0) and so on, all the examples I know off the top of my head are linear

native sand
# maiden ocean Is this for educational purposes

I’m learning CS and physics and group theory is useful for both so, I’m learning it, as with all of math though, I like to have some intuitive way to think about it to make everything obvious

maiden ocean
#

and you want this specifically on a vector space V?

native sand
#

I want to say it doesn’t have to be, but tbh idk any examples of a linear function of interest acting on something that isn’t a vector

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so yes

maiden ocean
#

Well i can do something like take V = R and then let Z act on R by making n(x) = x + n. now this is definitely gonna be associative, and 0(x) = x + 0 = x, and obviously (-n)(n)(x) = x + n - n = x, so the set of transformations i just defined forms a group. but its not compatible with the vector space structure on R. n(0 * x) = n(0) = n, while 0 * n(x) = 0.

#

@native sand makes sense?

native sand
#

ah right because addition is affine not linear

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yeah I can’t believe that slipped my mind lol, thanks

maiden ocean
#

no problem nozoomi

restive kiln
#

If you’re interested in actions of groups and want to think about ones that don’t arise from linear algebra, the canonical example is the action of a group on itself via conjugation.

Each conjugation map is a bijective homomorphism from the group to itself and so is a symmetry of the group!

Another sort of fun thing to think about is the symmetric group acting on strings in three dimensional space. This gives you something called the braid group and it is very, very cool.

latent anvil
#

Yeah so you can't always embed lie groups smoothtly/continuously into some GL_n but every group can be embedded into the automorphism group of some vector space. This is essentially cayley's theorem, given G make a vector space k[G] whose basis is the elements of your group and have your group act on that space by permuting it

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@maiden ocean exercise: decode the above so the person who originally asked cna understand it

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Your example with addition should be embedding n in Z into GL_2(R) by sending n to [[1 n] [0 1]] I think

#

I think your original definition of linear was actually correct here, they're interested in saying groups can be thought of as linear transformations and they're right

gritty sparrow
#

I think the embedding of the permutation group into the endomorphism group of a vector space doesn’t work for infinite groups

latent anvil
#

Why do you think that

gritty sparrow
#

Because vectors are a finite linear combination of basis elements

latent anvil
#

yes, but endomorphisms on vector spaces can do infinite permutations of the basis vectors

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This is like how V* will be uncountable if V is countable dimensionsal over Q

gritty sparrow
#

I see

latent anvil
#

consider the endomorphism of V, a space with basis {v0,v1,...} defined by σ(v_i) = v_{i+1}

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This isn't invertible but it gives you the idea

gritty sparrow
#

Yeah I got how its done

#

I was just being a dumbass for a sec

latent anvil
# native sand ah right because addition is affine not linear

See above, as a summary: addition is actually the composition of linear transformations, namely adding n and m is like multiplying the matrices [[1 n] [0 1]]. You can always find a group like this, even for those lie groups you were talking about, it's just that the vector space you're acting on will be infinite dimensional

#

This idea of studying a group by looking at its elements as linear transformations is called representation theory, but in that general setting we don't usually ask that each element necessarily acts like a different matrix

chilly ocean
#

would someone be available to help me understand frobenius automorphisms

#

I'm studying galois theory

#

so like this construction you take a monic integer polynomial that is irreducible over Q and if it's separable mod p you reduce it mod p

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so the quotients commute and you get F_p/(f(x)) where f(x) is the original polynomial

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next step: factor f over F_p into irreducibles. since it's separable each is distinct, so the quotient F_p/(f(x)) ends up being the direct product of finite fields F_p/(f_i(x))

#

this is the part where I sort of lose it

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I understand why the quotient is isomorphic to the product of finite fields by the chinese remainder theorem, but I am unsure what the original roots of f are supposed to look like in this product

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so then when you take automorphisms of the ring Z[x]/(f(x)) (or maybe it's Q[x]/(f(x))) I'm not sure how they corespond to automorphisms of F_p/(f(x))

latent anvil
#

It's Z[x]/(f(x))

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This is an interesting question

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So this ring has a root of f, namely x

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Right?

chilly ocean
#

so like f_i(x) are the factors of f in F_p, and the finite fields in the product are given by F_p/(f_i(x)). I understand that automorphisms will permute these fields, but apparently any automorphism that maps one of these fields in the product to itself must act no trivially on that field unless it's the identity

chilly ocean
latent anvil
#

Right, and then f factors mod p into a product of monic irreducible polynomials

#

So f = p1*...*pn

chilly ocean
chilly ocean
latent anvil
#

ah sorry

#

So let α = x + (f(x)) in Z[x]/(p, f(x))

#

Then f(α) = 0, but we don't actually know pi(α) = 0 for any i because the ring isn't an integral domain

chilly ocean
#

it's not 😢

latent anvil
#

The Chinese remainder theorem tells us our ring is isomorphic to Fp[y1]/(p1(y1)) ×... × Fp[yn]/(pn(yn)), right?

#

Do you know what the map from our ring to that product is, explicitly?

chilly ocean
#

I think (1,1,1,1,1,1) maps to 1

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that's just the way I rememberwd ir

latent anvil
#

Sure

chilly ocean
#

like generators

latent anvil
#

So here's how I think of it

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There's always a map R -> R/I1 ×... × R/In

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Just reduce mod I_j in the jth component of the output

chilly ocean
#

yup

latent anvil
#

and then the CRT gives conditions under which we can make this map pass through the quotient and be an isomorphism

#

so when we map Z[x]/(p, f(x)) into this product of quotients of Fp, we're just going to send the residue class of x to the residue class of x

#

We're just reducing mod (pi(x))

#

Does that make sense?

chilly ocean
#

what's z

latent anvil
#

Sorry, x

chilly ocean
#

oh haha

latent anvil
#

So specifically it'll be (x, x,...,x)

chilly ocean
#

yeahhhhhhhhhhhh

latent anvil
#

That's where our original (formal) root x goes

chilly ocean
#

omfg you're right it seems like so easy

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here's the example that tripped me up though

latent anvil
#

It's not!!!

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I got confused writing this and I should really know what I'm talking about lmao

chilly ocean
#

let's take x²+1 over F_5. this splits as (x-2)(x-3), so the product of fields is just (F_5)²

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now

latent anvil
#

Makes sense

chilly ocean
#

in this example I can clearly see that the 2 roots 3 and 2 map to (3,3) and (2,2) respectively

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however

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(2,3) also satisfies the relation x²+1=0

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so I'm like

latent anvil
#

Yep

chilly ocean
#

mystery root

latent anvil
#

So here's the trick

#

it's that F5[x]/(f(x)) is not an integral domain

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Let's try taking f(x) = x^2 - x

chilly ocean
#

ohhhhhh

#

you're right

#

it's never an internal domain if f splits over F_p!!!!!!

latent anvil
#

Ah okay I'll skip the example then

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(unless f is linear)

chilly ocean
#

yeah stupid linear

latent anvil
#

(or constant lol)

chilly ocean
#

🐣

latent anvil
#

:peep:

#

Anyways yeah worth thinking about the number of solutions to x^2 = x in F2 × F2

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Imo

chilly ocean
#

ok so now I see we have to give special treatment to the elements (3,3) and (2,2). they are the imahes of the original roots

chilly ocean
latent anvil
#

Solutions to this equation actually identify the number of ways to break up your ring into a product of two rings in general

chilly ocean
#

no way

latent anvil
#

yep!

chilly ocean
#

how general are you talking

latent anvil
#

Let me find the exercise I'm thinking of so I don't misstate it

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But it's pretty general

#

For all unital commutative rings general (maybe not even commutative is necessary?)

#

If you don't know what Spec A is don't worry, just ignore part i

chilly ocean
#

yeah I don't 😿

latent anvil
#

So the specific way this works is if you have e1, e2 you can define A1 = (e1) and A2 = (e2)

chilly ocean
#

I thought so

latent anvil
#

These don't seem like unital rings, they don't contain 1

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But!

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e1 is a unit of (e1)

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Try to prove this!

chilly ocean
#

hmm

#

it's late but I'm thinking

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e1²=e1 so like

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idk I feel like that should be enough 😢

latent anvil
# chilly ocean yeah I don't 😿

(also don't worry about that, this is from an algebraic geometry textbook. This all says geometric things but it's not important what they are for the ring theory)

#

Hmm, sort of

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How do you show something is an identity?

chilly ocean
#

give an inverse

latent anvil
#

Not quite, 2 has the inverse 1/2 in the ring Q but it isn't the identity 1

#

You're thinking of being a unit

chilly ocean
#

yeah that's what I thought you said oops

latent anvil
#

hahaha

#

I mean how do you know that something is 1?

chilly ocean
#

wait but aren't we proving that e1 is just a unit

latent anvil
#

What property does 1 have to satisfy?

#

Nope

chilly ocean
#

I mean 1x=x

latent anvil
#

e1 isn't a unit!

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Consider (1, 0)

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Say in Z×Z

#

This satisfies the equation but is not invertible

latent anvil
#

Does that make sense?

chilly ocean
#

oh of e1

latent anvil
#

Omfg

#

No

#

I'm so sorry

#

Ive been saying unit

chilly ocean
#

ok so like anything in e1 has the form e1x

latent anvil
#

And meaning the *multiplicative unit*

chilly ocean
#

so like multiply

latent anvil
#

Right, exactly!

chilly ocean
#

e1²x=e1x so boon

latent anvil
#

Sorry I just felt like this is sort of subtle so I wanted to talk about it

#

The having different multiplicative identities

chilly ocean
#

not that's fine it's interesting

latent anvil
#

Anyways yeah

#

So this goes the other way

#

If A = A1 × A2, take e1 = (1,0) and e2 = (0,1)

#

= is really isomorphic here

#

I'm just lazy

#

So we have a correspondence between ways of breaking A up into a product of rings and idempotent (meaning x^2 = x) elements of A

chilly ocean
#

yeah I get that now

#

that's cool

latent anvil
#

yeah!!!

chilly ocean
#

cause like (1,0) is sort of a half identity

latent anvil
#

Commutative algebra & algebraic geometry are very fun

chilly ocean
#

I can feel it

latent anvil
#

Do you know any topology at all?

chilly ocean
latent anvil
#

Once again, not expecting you to necessarily and no shame if not

chilly ocean
latent anvil
#

Okay cool

#

So the geometric side of things is that a product of rings is like a disjoint union of spaces

#

Say you have a disjoint union of spaces

chilly ocean
#

if you are about to mix commalg and topology...

latent anvil
#

What do continuous functions on that space look like?

#

C(X disjoint union Y) =...

chilly ocean
latent anvil
#

it's a good field :)

#

anyways

chilly ocean
latent anvil
#

How do you define a function on a disjoint union of spaces

#

Right!

#

Which means

#

C(X×Y) = C(X) × C(Y)

#

Where C(S) = ring of continuous maps S -> \R

chilly ocean
#

omg

latent anvil
#

So in algebraic geometry you like to think of rings as the ring of functions on some space

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and so breaking up a ring into a product is like taking a disjoint union of spaces

#

And the idempotents are the functions which are identically 1 on a connected component and 0 elsewhere

chilly ocean
#

😋😋😋

chilly ocean
#

ok but frobenius automorphisms 😳

latent anvil
#

Oh right

#

Sorry lmao

chilly ocean
#

why is it that any automorph8sm that fixes a field F_p/(f_i(x)) acts nontriviallu on that field

#

if it isn't the identity*

#

like can't I have an automorphism tbag leaves one of my fields in the product alone and let's the others go wild

latent anvil
#

umm

#

Okay so

chilly ocean
latent anvil
#

What constraints are there on the automorphism?

#

Like, this is false for just any ring automorphisms

#

Take f(x) = x^2 - 1 and you can just swpa the copies of Fp

chilly ocean
#

it's the projection of an element of the galois group of f over Q

#

I think

latent anvil
#

hmm

#

Does this make sense to me

#

I think so

#

This descends to Z

#

er

#

No

#

It doesn't make sense to me

#

How do we descend that, sorry?

#

Like when you say projection

#

What's the projection Gal((splitting field of f)/Q) -> Aut(Fp[x]/(f(x)))?

chilly ocean
#

I just mean like the galois group elements can be defined as permutations of the roots of f. the roots in Z[x]/(f(x)) project in the quotient by (p) to roots in F_p/(f(x))

#

so you take the same permutation but of the projected roofs

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roots

latent anvil
#

hmm

#

hmm

#

I think I see it

#

if you adjoin a single root

#

You might also adjoin more roots

#

right?

chilly ocean
#

yes

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by accident

#

they tag along

latent anvil
#

These roots will be in the same orbit of the galois group action

#

eight

#

*right

chilly ocean
#

the galois group acts transitively on the roots

#

so that happens anyway

latent anvil
#

So we can make sense of the galois group as acting on the roots of Z[x]/(f(x))

#

not quite

#

Well

#

Yes

#

I'm thinking of something definitely

#

No you're right, I'm thinking of something different about than them being in the same orbit

#

I'm thinking that is β is in Q(α) then σ(β) is in Q(σ(α))

#

Yeah?

#

So yeah I'm def confused

#

Okay let me try to explain why I'm confused

chilly ocean
#

makes sense

latent anvil
#

You say we let the galois group act on the roots of f in Z[x]/(f(x))

#

This doesn't quite make sense, because the galois group acts on the roots of f in the splitting field of f

#

Right?

chilly ocean
#

yes

latent anvil
#

It's not even clear to me that Z[x]/(f(x)) has that many roots of f

#

like

#

An element of the galois group can permute a root out of an intermediate field

#

Right?

chilly ocean
#

ohhhh omg I'm sorry I forgot to give a totally crucial condition omg

latent anvil
#

lmfao

#

Sick I love it

#

dw I do this to my friend all the time

chilly ocean
#

you have a galois extension of Q and since all extensions of Q are simple, you just take f to have as a root any single element that gives the whole extension

#

so like you know Q[x]/(f(x)) is the splitting field

latent anvil
#

Also, I'm pretty sure the answer in the end will be that permuting the roots of f via an element of the galois group keeps you as a root of the same irreducible factor of f, and the irred factors of f correspond to the rings in the product at the end

latent anvil
#

Omg

chilly ocean
#

it's irreducible over Q

latent anvil
#

You were assuming it's irreducible over Q all along

#

I totally missed that

#

Okay haha

chilly ocean
#

but it can splitt over F_p

latent anvil
#

Yep

chilly ocean
#

sorry 😿

latent anvil
#

I thought you were just asking for it to split separably over Fp

#

No haha it's fine

#

Okay

heavy fractal
#

guys can u give me some questions from algebra

latent anvil
#

Not in the middle of a conversation, no

chilly ocean
#

question 1: classify all groups

#

come back when you solved it

heavy fractal
latent anvil
#

Okay, so you're saying Q[x]/f(x) is the splitting field of Q

#

Very cool

chilly ocean
#

yes

latent anvil
#

Okay

chilly ocean
#

splitting field of x

#

of f*

latent anvil
#

I fully understand how the galois group descends now

#

The roots of f are just polynomials

#

I guess there might be a concern about the coefficients being non integer?

#

But I don't think so

chilly ocean
#

polynomials in a given one of them

#

f is monic knteger

#

if x is one root all other roots are polynomials in x

latent anvil
#

Could you say that again

latent anvil
chilly ocean
latent anvil
#

This is the whole f being the min poly of the generator of the extension thing

#

okay so

#

It makes sense that this descends

#

but yeah it's not clear to me why you can't stabilize a field hmm

#

So how does f factoring actually work

#

hmm

#

so what do we want to know in the end? I think we want to know that if α is a root of some pi then σ(α) is a root of the same pi, where σ is a descended galois automorphism

#

Does that sound right?

chilly ocean
#

not necessarily

#

you can have automorphisms that permute the fields

#

but the thing is saying if sigma is not the identity and sigma fixes the field pi then it acts nontrivially on that fiekd

latent anvil
#

sorry, my phone just crashed

#

ah

#

hm

chilly ocean
#

ohhhhh

#

is it because

latent anvil
#

so

chilly ocean
#

all roots are polynoia8s in the root x

#

so if a given root is fixed

#

all roots are fixed?

latent anvil
#

Hm, I don't think I see why that is

chilly ocean
#

if the roots are x and 3x²+1

#

and if x is fixed

#

then 3x²+1 always has to be sent to 3x²+1

#

like and all roots can be expressed as polynomials in any given root

latent anvil
#

hm, I guess I was concerned because I was thinking of the roots as elements of Z[x]/(f(x)), but the relation will descend all the way down to Fp[x]/(pi(x))

#

I think

chilly ocean
#

so the galois group acting on roots has no elements being fixed unless you're mr. identity

latent anvil
#

you just reduce stuff mod (pi)

#

this makes sense to me

#

I was going to say that the automorphisms of Fp[x]/(pi(x)) are cyclic of order deg pi

chilly ocean
#

I mean stuff should reduce just from the quotient being well defined

latent anvil
#

they're generated by the map a -> a^p

latent anvil
chilly ocean
#

this whole thing is supposed to be about using the easy structure of finite fields to learn more about galois groups over Q

latent anvil
#

ahh neat!

#

I don't really know much number theory, this is cool

chilly ocean
#

ok so I see why no elements are fixed

#

nowwwwww

#

the next thing is

latent anvil
#

wait, I think I actually still have a concern

chilly ocean
#

ok

latent anvil
#

we know that the roots are fixed

#

of our automorphism

chilly ocean
#

aren't*

latent anvil
#

what?

latent anvil
chilly ocean
#

fpe the identity map yes

#

I'm sayijg if 1 root is fixed

latent anvil
#

nevermind, I understand now

chilly ocean
#

then all roots are fixed

latent anvil
#

x generates Fp[x]/(pi(x))

chilly ocean
#

and I get that now

latent anvil
#

neat!

chilly ocean
#

and x generates Fp[x]/f too

#

cause x is actually the element (x,x,x,x,...)

latent anvil
#

many such cases

chilly ocean
#

this follows because x generates the splitting field in Q so you just carry it down to the quotient

#

ok ready for next step?

#

this is actually getting so much clearer for me

#

if two distinct automorphisms a and b map the field pi to the map pj then they have to have restrict to different homomorphisms from pi to pj

#

this is easy because otherwise ab^-1 is the identity

latent anvil
#

I don't understand this

chilly ocean
#

oh

latent anvil
#

If the automorphisms are distinct

#

then they're different homomorphisms?

chilly ocean
#

so the automorphisms permure the fields

latent anvil
#

er, hang on

#

automorphisms pi -> pj?

#

oh

#

automorphisms of the product of fields?

chilly ocean
#

automorphisms of the products of fiekds

latent anvil
#

which map the ith field to the jth field

#

sure

#

this makes sense

chilly ocean
#

yes

#

yes

latent anvil
#

otherwise ab^{-1} fixes but acts nontrivally

chilly ocean
#

yes

latent anvil
#

cool

chilly ocean
#

which is nono

#

ok here's the cool part

#

we count the maximum number of automorphisms using what we just learned

latent anvil
#

ooh

#

this is getting spicy

chilly ocean
#

so first of all

#

we know that the number of maps that send p1 to pi is at most the number of homomorphisms from p1 to pi

latent anvil
#

yes

#

and we know how many of those there are

#

bc classification of finite fields

chilly ocean
#

also this is at most the order of Gal(pi/fp)

latent anvil
#

What is that galois group of?

chilly ocean
#

and the order of the WHOLE automorphism group is at most the sum over i of the number of maps from p1 to pi

latent anvil
#

Fp[x]/(pi(x)) over what?

chilly ocean
latent anvil
#

Ah sure

chilly ocean
#

which has galois group C_(degree of pi)

latent anvil
#

the ways of embedding it into something bigger are the ways of permuting the roots

#

yep

latent anvil
#

or Fp[x]/(f(x))

#

like f_p

chilly ocean
latent anvil
#

yep

chilly ocean
#

which is the degree of f

latent anvil
#

ah okay

chilly ocean
#

so the number of maps is at most the degree of f

#

butttttt

#

the group is galois

#

sooooo

#

the galois group of Q[x]/f is given by all automorphisms of the product of fields!!

latent anvil
#

neat!

chilly ocean
#

in particular, all pi have the same degree

#

because then there actually are the maximal number of homomorphisms from each pi to each pj

#

and that's where frobenius comes in cause we know those automorphisms are generated by sigma(x)=x^p

latent anvil
#

Oh neat

chilly ocean
#

so now you know for example all polynomials of prime degree are either irreducible over F_p or they split completely

#

which is a sexy fact

#

it reminds me of the fact that the polynomial x^p-x-a always either splits completely or it is irreducible whenever the field has characteristic p

latent anvil
#

Very cool

chilly ocean
#

something special about polynomials of prime dehree

#

ok I have to sleep

chilly ocean
#

fun problem

latent anvil
#

Maybe? It's been a while

#

Hmm

#

So say it has a root α

chilly ocean
#

I know the answer in advance so honestly idk if I could figure it out

latent anvil
#

So (α + k)^p - (α + k) - a = k^p - k

#

Right

#

So take k to range over Fp

#

then these give p distinct roots

chilly ocean
#

yessss

#

woop woop

latent anvil
#

So that does it I think?

chilly ocean
#

yup

latent anvil
#

Neat

#

Good problem

chilly ocean
#

and you can do similar things where the galois group of x^(p^n)-x-a over F_p is the additive group in F_(p^n)

latent anvil
#

Right

chilly ocean
#

there's lots of sexy math behind these things

latent anvil
#

it is

#

Number theory

chilly ocean
#

these kinds of extensions

#

are quite the sex

latent anvil
#

nomber theory

chilly ocean
#

ok sleepy times

modest zenith
#

Hey. Shamrock if ya around. can ya help with a little something?

latent anvil
#

Sure

#

I am less sober than optimal for doing math problems but I'm 2 for 2 so far

modest zenith
#

Like, the relation question really doesn't make sense to me

latent anvil
#

Sure

modest zenith
#

The range I understand, and the third I can guess from the second

latent anvil
#

Which of the three sub parts?

modest zenith
#

The first one

latent anvil
#

Ah okay

#

What's the definition of a function?

modest zenith
#

a group of variables

latent anvil
#

Not quite

#

in set theory, a function is a certain kind of subset of A×B, ie a certain kind of relation

#

It's a way of giving you a choice of element of B for every element of A, right?

#

So you want the subset F of A×B to have the property that for every a in A, there's some b in B such that (a, b) in F

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But you also want to know that giving the input a can only give one output, ie for all a in A if b, b' in B both satisfy (a, b) in F and (a, b') in F then b = b'

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You can think of the association between functions and relation like every function f defines a relation aRb iff b = f(a)

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Does that last bit makes sense?

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and then the question is asking you if there's a function which makes the relation f have thus form

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nope

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we're given two sets A, B

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And a relation between them

modest zenith
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Gack. I hate set theories.

latent anvil
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Right?

modest zenith
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Yup

latent anvil
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You want to know if there's a function F : A -> B such that b = F(a) if and only if (a, b) in f

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Does that make sense?

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Set theorists actually define F to be this function f. They say that functions are certain types of relations

modest zenith
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How would I write it in set theory form?

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Well. I meant

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As a proof

latent anvil
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You would say that for every a in A, there is a unique b in B such that (a, b) in f

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this says every "input" has exactly one "output"

modest zenith
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Since I believe that is what the professor wants everything to be in

latent anvil
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Sure

modest zenith
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Wouldn't that be something to note?

latent anvil
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Yep

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This is like how y = x^2 will miss negative numbers

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If we think of it as a function R -> R

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How would you state this in the case of squaring?

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It doesn't contradict my set theory definition of a function, because I only said that for every element of A, there's an element of B. I didn't say anything about every element of B having an associated element of A!

modest zenith
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Ok I see

latent anvil
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Right!

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Think about functions you know, like squaring or sine

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The graph has to pass the vertical line test but not the horizontal line test

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Nope

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What's your understanding of the range of a function in general?

modest zenith
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I think I misunderstood what is ranged when I took my note in class

latent anvil
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So the range is all of the outputs of a function

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For squaring this would be all nonnegative numbers, [0, infinity)

modest zenith
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Yup, that part I understand

latent anvil
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For the sine function it would be all numbers between -1 and 1, [-1,1]

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So the thing to keep in mind is that f has a direction

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The first component of a pair is the input, and the second is the output

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When we say (a, i) in f, we mean that f sends a to i

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Right?

modest zenith
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Yup

latent anvil
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So when you say o is in the range, that means f contains some pair (x, o)

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Right?

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I'm not sure what hit mean

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Elements can't be duplicated in a set, but they can in a pair

modest zenith
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Oooo ok.

latent anvil
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Sorry, I meant o not a

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(a, a) is a pair, but it's not in f

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No

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You want only the outputs of f

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This is the set of input-output pairs

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Say your function was the height of a person, it's domain was the set of your classmates, and the codomain was the real numbers

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Probably nobody is a negative height, and nobody is 100 feet tall

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The outputs here would be some numbers likely between 4 feet abs 7 feet tall

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In your example, what are the outputs of f?

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No, that's a pair of an input to f and its output on that input

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Here the output would be i and the input would be a

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In the height example, we would have a pair (shamrock, 6 foot 1) in our function

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Right!

modest zenith
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Right. I'm just an idiot then.

latent anvil
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Hahaha

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It's okay

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But really, the first step should be to review definitions

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Yup, that's right!

modest zenith
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Professor didn't really give us any definition. Legit. We heard about it one day, given the homework, and then move on to the next

latent anvil
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This is what you were noting before, btw

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That the range doesn't contain o or u, but does contain everything else in B

latent anvil
modest zenith
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Pretty rapid fire when it comes to our lessons

latent anvil
modest zenith
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Noted, thank you

latent anvil
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It's not a huge deal but now you know

modest zenith
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Gotcha. I'll make a note of it

latent anvil
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Wikipedia is also a good source if you want to review definitions like this, but the notation might not be what you're used to

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It's weirdly good as a reference for higher math stuff

modest zenith
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Yeah no. This is not something I am familiar with, and getting shotgun'ed these daily is rough

latent anvil
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Shotgunned?

modest zenith
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Just one after the other, one after the other, without breaks.

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We cover like 2+ chapters worth of new notes daily

latent anvil
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Rip

modest zenith
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The 10-15 min breaks we do get, it's all swarming the professors asking question so... yeah. Rough

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Thank you though, that helped alot 👍

latent anvil
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Sounds like a shit class

modest zenith
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It's one of those higher level math class that promises to be "in bite size and easy for student who is interested in math"

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Turns out, bite size means bite size lessons with minimal examples

deft wave
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Ping?

urban acorn
modest zenith
# urban acorn ouch

The professor somehow managed to both be SUPER quick with his lessons yet drawn it out for 4 hours

The only thing I got perfect was mod

edgy vault
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@tawny pine don't @ abstragebra again

ionic ingot
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heya! this may be a really dumb question but, i'm trying to write up addition and multiplication tables for a finite field. i know how to do it when it is just Fn, but right now i'm looking at an example that's squared. in that case, how does one go about it? does it make any difference to the tables? for example F₃²

chilly ocean
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do you know field extensjons

tawny pine
edgy vault
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@tawny pine oh.. so from being helped to being helper

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kool

viscid pewter
wraith obsidian
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Is left identity & left inverse enough to axiomatize groups? I.e. if we consider the signature (e, i, ◌, ) as a universal algebra, do associativity, i(x)◌x≈e and e◌x≈x suffice for the rest?

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This was in an exercise but it feels very wrong because we don't even get uniqueness of e due to lack of being a right identity, and I also don't see how „i is involution” would follow

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Naturally tho, I don't know any non-commutative semigroups which are nasty enough to give a counterexample

tawny pine
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indeed, associativity+left inv+left e implies group

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group just requires existence of e, if you already know group implies uniqueness of e then you don’t need to worry about it during the exercise

hidden haven
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Starts a couple messages before that one

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It is tricky to prove and not true when you change conditions to assoc + left id + right inv

plucky flicker
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Hii! Can someone give me an example of a self-normalizing proper subgroup in an infinite 2-group?

wraith obsidian
urban acorn
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I can't even think of any non-abelian infinite 2-groups on top of my head rn lmao

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I guess you can take any infinite 2-group

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and then do a semidirect product

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with Z2

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and I think that's an example

urban acorn
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so take any abelian infinite 2-group

waxen hedge
cloud walrusBOT
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Adrien

280791755300012032.png
hidden haven
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They said operation tables of finite fields so F₉ presumably

plucky flicker
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If we have a H a subgroup of G, then is it true that the frattini subgroup of H is a subgroup of the Frattini subgroup of G? I guess it is true, but i don't see it clearly

unique berry
plucky flicker
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You mean my question?

plucky flicker
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Ohh nope

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the normalizer is the whole group G in that case

hidden haven
cloud walrusBOT
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Yeetus

842836522029678612.png
chilly ocean
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f(X) = 17?