#groups-rings-fields

406252 messages · Page 446 of 407

little folio
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its not torsion free

faint elm
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god damn it

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you are right

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but how about the first question? @little folio

little folio
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what you said here is correct

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but what thesis?

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what are you trying to prove?

faint elm
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that if $M$ is flat then for every exact sequence $0 \rightarrow X \rightarrow Y \rightarrow M \rightarrow 0$ and for every $N$ module we have that $0 \rightarrow N \otimes X \rightarrow N \otimes Y \rightarrow N \otimes M \rightarrow 0$ is exact

cloud walrusBOT
faint elm
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sinche $N \otimes -$ is right exact we have only to show that $N \otimes X \rightarrow N \otimes Y$ is injective which follows from the snake lemma above

cloud walrusBOT
little folio
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sure

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that works

faint elm
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nice, one down!

chilly ocean
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how do I show $o\left(h^{-1} g h \right) = o\left(g\right)$ when o(g) = infty?

cloud walrusBOT
mild laurel
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Try contradiction

chilly ocean
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ok yeah that was pretty striaght forward lol

chilly ocean
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When one says "x is a group law on set G" does that just mean that (G, x) is a group?

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all that I get from Google is elliptic curves papers lmao

mild laurel
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yes

chilly ocean
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aight thank you!

mild laurel
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Yeah the elliptic curve group composition is pretty complicated

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So they call it a group law idk

chilly ocean
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I saw a presentation about elliptic curves and while I'm not confused as to what they are or how they're useful, the math being done with them is extensive and a little mind-boggling

mild laurel
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Yeah it's honestly crazy

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I'm reading about modular forms right now and how someone related this to elliptic curves is like

chilly ocean
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hahaha also saw a talk about siegel modular forms and was very confused by it

foggy basin
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Is there a proof of the fundamental theorem of algebra that doesn't use topology?

mild laurel
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Not really

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There's a galois theory proof that only uses some easy analytical facts I think

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I think it uses the fact that every poly of an odd degree has a real root, and every quadratic has a root in C

foggy basin
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Yeah I saw that, but I believe it uses IVT for that fact

mild laurel
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Yeah

foggy basin
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wikipedia said that "...any proof must use some form of the analytic completeness of the real numbers..." is that statement formal? Is there a proof that you do in fact need topology?

mild laurel
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No and I'm not sure Wikipedia is super accurate on this

latent anvil
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I think what Wikipedia's trying to get at is that you need to use something about the complex numbers

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Your proof can't work for R, for example

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Or for Fp

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And we know that C is R adjoined a squareroot of -1, and that R is the unique complete ordered field

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Like, if you can find an ordered field F such that F[x]/(x^1+1) is not algebraically closed (and I bet you can) that tells you you need to use completeness of R

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Does that make sense?

little folio
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the proof will work over any field whose algebraic closure is a degree 2 extension

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an algebraic proof can be cooked in that generality

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in fact this follows whenever the algebraic closure is a finite extension

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i'm being imprecise here, let's see if I can find it

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oh I found it

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I guess it's unavoidable that the odd polynomials must be shown to be reducible

misty aspen
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Hey. Anyone here have good intuition for the transfer homomorphism and hall subgroups ? I’m trying to show that the image of the transfer of a group into a Hall subgroup is just the image of the subgroup under the transfer, but I am completely stumped on how to approach this..

barren delta
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I'm having a little bit of trouble understanding the notation of products of transpositions

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we can write (2 5 6) as (2 6)(2 5)

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but the product of those transpositions is confusing me

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I guess I don't understand how to actually read that product

bleak abyss
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Product is composition

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Keep in mind these are functions

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So for example, assume we're in S_6

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(2 5 6) is a function from S_6 to itself

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It takes 2 to 5, 5 to 6, 6 to 2, and everything else to itself

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So for example

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(2 6)(2 5) is a function, let's apply it to 2

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Well, (2 5)(2) = 5

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And then (2 6)(5) = 5

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SO (2 6)(2 5)(2) = 5

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What about 5?

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Well, (2 5)(5) = 2, and then (2 6)(2) = 6

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So (2 6)(2 5)(5) = 6

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And so on

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So that equality is one of functions

latent anvil
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I'm trying to understand double complexes and feeling kind of stupid

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I was trying to do exercise 1.2.6 in weibel, which asks for an example of a second quadrant double complex with exact columns whose total product complex is acyclic but whose total direct sum complex isn't

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And I realized I couldn't think of any examples

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Even just for a second quadrant double complex with no assumptions

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Except for like the maps being trivial/identity or a bunch of zero objects

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Can someone give me some examples of double complexes, and possibly help me with this problem?

woven delta
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So an easy one is a short exact sequence of complexes

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Which I'm sure you've seen before

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There's kind of a trivial double complex you can make from a single complex

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By just writing the complex vertically as well

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Maybe a good exercise is given a chain complex of R modules, you take a free resolution of each R module and turn that into a double complex

latent anvil
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By a short exact sequence of complexes, you mean that that's literally a double complex right?

woven delta
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Yeah

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But they show up pretty often

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So you've probably seen it before

latent anvil
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Yeah

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I have

woven delta
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So that's kind of an easy example

latent anvil
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The last one seems interesting though

fickle brook
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(2 5 6) is a function from S_6 to itself
uh

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did you mean from {1,2,3,4,5,6} to itself? @bleak abyss

woven delta
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Oh lol I know a good example

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You take 2 chain complexes

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And tensor them together

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And each element in the first chain complex

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Gives you a row in the double complex

bleak abyss
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Yeah tru

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I did

woven delta
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I think that gives you a double complex

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Cause of functiorial stuff

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And abelian category stuff

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But I should probably think about it for a couple of minutes

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Yeah this does

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And the tensor product as a chain complex is just the total complex of this

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Wtf

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This is actually very obvious

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But I never realized why the tensor product of 2 chain complexes is the way it is

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Smh

latent anvil
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That sounds important lol

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Im doing this because I tried to read the proof of the acyclic assembly lemma

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And realized I had no idea how double complexes work

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But I think that the application of that lemma involved tensoring two resolutions together

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To get a double complex

woven delta
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And anyway the tensoring gives you a whole bunch of double complexes to think about

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So that's good

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Fuck now I wanna learn more homological Algebra

latent anvil
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It's good!!!!

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I'm trying to work through chapters 2/3/4 of weibel before spring

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So that sheaf cohomology doesn't wreck me

woven delta
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Oh shit

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Let's correspond

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I did chapter 1 in the summer

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But I got sidetracked

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So I wanna do some more

latent anvil
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Saaaame

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Literally exactly that

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For me

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My school had a course on it last spring but I had a conflict

magic owl
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Wait @woven delta i want to read HA as well

mild laurel
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@misty aspen What have you tried?

misty aspen
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Oh I was able to get it afterall. By an order argument you can show that H and the kernel of the transfer generate G to applying the transfer to G=H ker gives the result

mild laurel
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Ah nice

bleak abyss
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Alright time to work through Milne's AG

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I've glanced through a lot of chapter 1 before so hopefully that'll go fairly quickly

bleak abyss
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Okay I ended up playing video games but now for real

uncut girder
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I have that book at home

bleak abyss
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It seems pretty good, and it's apparently closest to what my class is doing

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Actually tbh my working through this will be less a rant and more scratch work

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I'll just create my own workspace server so I'm not flooding this with dumb shit

mild laurel
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I mean, I feel like all the advanced channels aren't used often anyways, and people might be interested in hearing what you say

bleak abyss
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I mean for now a lot of it will just be like

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Glancing through Milne's proofs of commutative algebra and writing down some scratch work to make sure I remember how this one detail he omitted works

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That tends to be less interesting I feel

woven delta
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Can I join your workspace server Dami?

bleak abyss
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If you'd like, yeah, though again a lot of it will just be a storage bunk for the not interesting details of stuff

sullen island
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hi, i came across this while looking at alternating and symmetric squares of tensor products of representations, and im wondering how do we add two tensors like that? i checked on mathstackexchange it seems that addition of two tensors is defined as $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w $ (so the second component is kept constant)

📎 unknown.png
cloud walrusBOT
sullen island
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the circled part is the set of basis elements for the symmetric square

mild laurel
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Elements of the tensor product are just formal sums of v_i tensor v_j

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So that form is the simplest form

sullen island
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im sorry, what's a formal sum?

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im assuming you mean just an abstract sum written that way (which isnt evaluated to anything) ?

mild laurel
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Yeah

sullen island
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ahh okay

mild laurel
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It's kind of like polynomials with variables

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Where like x^2 + x is just that

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There's no way to really combine them

sullen island
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sorry for the silly question, im a year 2 undergrad so im not very good at this XD

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thanks !

mild laurel
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Learning rep theory as a second year is super impressive so don't worry too much

chilly ocean
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if G =Z/nZ, why does ord(a)= n/GCD(n,a)? (a any element in G)

woven delta
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Think about what Z/nZ is

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You're taking Z

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And modding out by nZ

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So given a, when is ka in nZ?

potent lynx
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<a^k> = <a^gcd(n,k)> where |a| =n

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so

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let gcd(n,k) be d

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d divides both n and k

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so n = td and k = qd for some integers q and t

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so

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gcd(n,k) = d ---> there exists integers x and y such that xn+yk = d

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or nah lol

mild laurel
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Uh what

potent lynx
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want to show |a^k| = n/gcd(n,k) where n = |a| ---> (a^k)^n/gcd(n,k) = (a^n)^k/gcd(n,k) = e ---> |a^k| <= n/gcd(n,k) .. suppose there exists i < n/gcd(n,k) such that (a^k)^i = e ... and show contradiction lol

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thats what i wanted to do but i got stuck

thorn delta
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Let r be the smallest such integer such that (a^k)^r = e. Show that r = n/gcd(n, k). I wouldn’t go for a contradiction

potent lynx
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how wouold u do it

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just for knowledge

chilly ocean
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What is next step in solving this equation

woven delta
thorn delta
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@potent lynx Let $r \in \bbZ$ such that $(a^k)^r = a^{kr} = e$. Then $kr$ divides $n$. In other words, there exists $a \in \bbZ$ such that $an = kr$. Let $d = gcd(n, k)$. Can you think of what to do from here?

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wait uhh did i mess something up thonk

cloud walrusBOT
potent lynx
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what do u want to show again?

thorn delta
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ur trying to show that the order of a^k is n/gcd(n,k) lol.

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where n is the order of a

potent lynx
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n=xd and k =yd

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an = axd = kr

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idk lol xd im bad

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ok first i show order is less than or equal n/gcd(n,k)

thorn delta
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wait

potent lynx
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now i suppose there exists a less 1

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what

thorn delta
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wait I should maybe rephrase things. I said some things stupidly and incorrectly. We are trying to find the smallest $r$ such that $a^{kr} = e.$ In other words, the smallest $r$ such that $n$ divides $kr$. Now, let $d = gcd(n,k)$ and ill let you take it from here.

cloud walrusBOT
potent lynx
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n=xd and k =yd

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kr =tn

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i cant im sorry

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fuck

thorn delta
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its kinda tricky tbh. I only know it cuz I've seen it before. But if n divides kr, then n/d divides (k/d)r. Do you see how to conclude now?

potent lynx
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ur nice lol ur lying im jist stupid haha

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yes

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got it now

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tysmm

thorn delta
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np c:

stiff shore
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hey guys, would anyone be willing to look at a problem with me? I've been stuck on it

mild laurel
stiff shore
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I was taking a pic lol

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so these are my thoughts so far

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I is a subset of G but not necessarily a subgroup

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bc the inverses of all elements of a subgroup must also be elements of the subgroup

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I only contains the elements of the isomorphism that get mapped to their own inverses

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and I'm thinking that in the case where | I | = (3/4)|G|, G is finite, therefore I is also finite

mild laurel
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Why doesn't I contain the inverses?

stone fulcrum
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Remember this useful property:
α(x') = α(x)'

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Where ' is inverse

stiff shore
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oh true

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hmmmm

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let me see how I can use that in this case

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@ all rings are noetherian, also I thought that we don't know if it contains the inverses or not

mild laurel
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Ah sure, I thought you concluded that I doesn't always contain the inverses

stiff shore
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lmao going on over three hours working on this question, kms

stiff shore
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for this problem, I'm stuck on figuring out if I is closed. Is α being an isomorphism enough for me to assume I is closed?

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because if x is in I and y is in I, it's not necessarily true that xy will be in I

woven delta
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$g_1, g_2 \in I \implies \alpha(g_1 g_2) = \alpha(g_1)\alpha(g_2) = g_1^{-1} g_2^{-1} = (g_2 g_1)^{-1}$

cloud walrusBOT
woven delta
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So yeah, it's not enough

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I don't see why it being an isomorphism would be enough in fact

stiff shore
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hmmm

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to show G is abelian, I'd need to show xy = yx for x,y in G

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but I'm really stuck on what to do with I here

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how does this particular size of I help

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I guess I'm stuck bc 3/4 sounds like such a particular and specific value

little folio
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that happens cuz it's a number written in a weird way

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it means the center has index 2

stiff shore
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How does it mean that?

onyx lance
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Is there a geometrical way to intuit what the spacetime algebra Cl_{1, 3}(R) is, and how elements of it are supposed to be interpreted?

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and, if the vector space R^4 with the Minkowski metric is supposed to be included in it, how do we find the representation of a given vector via the gamma matrices?

hot lake
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?

meager flint
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Hey all, I'm asked to decomposed the set $\mathbb{C}^{2\times 2}$ of complex matrices into orbits for the operations of left multiplication and conjugation of $GL_2(\mathbb{C})$

cloud walrusBOT
meager flint
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I guess I don't exactly understand what the question even is or how to approach it

meager flint
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help?:(

fading wagon
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ah

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Basically, an orbit is an equivalence class of where the group action sends the matrix

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@meager flint

meager flint
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yeah, that makes sense, but doesnt $\mathbb{C}^{2\times 2}$ just get split into $GL_2(\mathbb{C})$ and the non-invertible matrices?

cloud walrusBOT
fading wagon
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no

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you are talking about left multiplication?

meager flint
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Yes

fading wagon
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iirc there are 3 orbits

meager flint
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Oh wait

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{I} and the set of echelon forms would be two of them no?

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{I} and the set of echelon forms would be two of them no?

fading wagon
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?

meager flint
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Yeah no nvm I have no idea of what im talking about I have to think about this lol

fading wagon
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neither do I

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all I know is some group theory

latent anvil
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You can use left multiplication to do row operations

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Two matrices are conjugate by an invertible matrix (in the same orbit of the second action) iff they are similar

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Just by unwrapping definitions

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Do you know any techniques for classifying similar matrices?

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@meager flint

smoky cypress
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If a permutation is written as the product of cycles that are not necessarily disjoint, how do I composite them to get the permutation?

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Do I simply go from right to left like function decomposition? (Also considering function composition is associative I don’t really need to start from the most right cycle)

sharp sonnet
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a cycle is just a map and a product of cycles is a composition of cycles

smoky cypress
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K

bleak abyss
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Alright time to understand Zariski's lemma

cloud walrusBOT
latent anvil
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it's good to see I'm not the only person trying to understand commutative algebra at like 11pm on a Sunday

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v reassuring

cloud walrusBOT
bleak abyss
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But there are infinitely many irreducible polynomials

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So we're done

cloud walrusBOT
bleak abyss
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And then Rabinowitsch trick, which I'll do out but in terms of localization

cloud walrusBOT
languid moss
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Need help with how to start this

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suppose that R is a ring with no zero divisors, and S is a subring of R. Show that S has no zero divisors

mild laurel
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Contradiction

languid moss
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okay

steep hull
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Don’t overthink it

bleak abyss
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Time to do some more notes

cloud walrusBOT
little folio
weary terrace
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p-adic analysis:
Anyone knows good reference for the proof: $1+2\mathbb Z_2\cong \mathbb Z/2\mathbb Z\times\mathbb Z_2$?

Or better, has a quick proof in his/her sleeve?

cloud walrusBOT
cobalt pilot
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I was watching a lecture and he assumed that a maximal ideal m + an ideal generated by a non-unit x, where x \notin m will always be the whole ring. Is this true, and how do you prove it?

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(commutative ring with identity)

mild laurel
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What would happen if it weren't the whole ring?

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What would that say about the maximality of m?

cobalt pilot
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There can be co-maximal ideals though

mild laurel
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Sure but this isn't the case

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One ideal strictly contains the other

cobalt pilot
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and how do we know this?

mild laurel
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Oh I misunderstood what you meant by plus

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Well not really

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If you take two ideals I, J, the ideal I + J contains both I and J

cobalt pilot
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Yes, but how de we know that I+J = A when I is maximal and J=(x) for a non-unit x \notin I

mild laurel
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I + J contains I

stone fulcrum
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A lecture on YouTube?

mild laurel
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I is maximal

cobalt pilot
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I guess I could look at A/I

mild laurel
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What does an ideal being maximal mean

cobalt pilot
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@stone fulcrum Yeah, there is a Indian university that had published lectures on commutative algebra that I look at to freshen up my memory. I have not had com alg for 10 years

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That there are no porper ideals that contains it

mild laurel
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But I + J contains I

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which is maximal

woven delta
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And it properly contains it

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Cause it contains J

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Which has an element not in I

cobalt pilot
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Let's say we are in Z. Then 3Z is a maximal ideal. 10 is a non-unit, not in 3Z and (10) = 10Z is an ideal in Z. So 3Z +(10) = Z ?

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But (10) is not contained in 3Z

bleak abyss
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Yup

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And I mean, 10-3(3)=1

woven delta
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That's why lol

mild laurel
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but (3) + (10) contains (3)

cobalt pilot
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OK, so it works in this example and I think I understand it, but how would I prove it?

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nvm, I guess I can just look at A/I

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if I is maxiaml

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Then (x) is not an ideal in A/I

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at least not proer

woven delta
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Yes that works too

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Consider the image of (x) in A/I, it is non trivial

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So it must be the whole A/I

cobalt pilot
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Thanks. It just bugged me a but that he used this in a prof without proving it first, although it's probably quite trivial for him.

woven delta
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It's quite trivial in general

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You just got to get used to it

cobalt pilot
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yes. Sorry for digging so much, but I hope that once I can grasp things like this better then it might help understanding the whole theory better as well

stone fulcrum
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Those lectures follow mcdonald-atiyah to the T

cobalt pilot
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They do

stone fulcrum
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If you're ever unsure, the pdf likely has some clarification

cobalt pilot
#

And that's the book we used 10 years ago as well, so it fits me well

stone fulcrum
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I was thinking this one, but they could be similar

cobalt pilot
#

Oh, this is actually just the book?

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Would have saved me $100 10 years ago XD

stone fulcrum
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Rofl, who knows if it was online then. At least now it's free to download

cobalt pilot
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Probably the most I paid per page of a book ever

golden pasture
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theres libgen now most books you can get for free

cobalt pilot
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If we have a sequence of nested ideals containing each other with a maximal element, would it be correct to say that the union is isomorphic to the maximal ideal?

mild laurel
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The union is just equal to the maximal ideal

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No isomorphism required

cobalt pilot
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ok, thanks

lime basin
#

Hey guys!

I have a certain function f(x,y,z) belonging to a spectrum of some operator O, that is f \in E\lambda(O). Then, I take the 2D fourier transform with respect to x and y of f. f is periodic, so I obtain f= sum over n,m, of h{n,m}(z) exp(blah blah). What do you call the space of these functions? What symbol would you use to write h_{m,n}(z) belongs to ... ?

languid moss
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Quesstion

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if m is odd, find the inverse of [2]_m in Z/mZ

latent anvil
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Suppose [a]_m is an inverse of [2]_m. What can you say about 2a?

languid moss
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2a is even

latent anvil
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Does that fact that 2 and a are inverses mod m give us any more information?

languid moss
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no

latent anvil
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It does

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What does it mean for x and y to be inverses?

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Actually hang on, does inverse mean multiplicative or additive inverse? I assumed multiplicative

languid moss
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multiplicative

latent anvil
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Okay, cool

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So what does it mean for [x]_m and [y]_m to be inverses?

languid moss
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I'm not sure. I'm confuse

latent anvil
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Before you try and solve a problem, you should understand what it's asking

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it means that [x]_m * [y]_m = [1]_m

languid moss
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Let me review

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my stuff

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I will be back

languid moss
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For the order

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Do i take account the smallest order?

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[4]^3 and [4]^6

latent anvil
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What do you mean?

languid moss
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sorry

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I thought I send picture

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4

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part b

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I got it

tribal pasture
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Is there any relation between the cosets of a centralizer of x and the conjugacy class of x (aside the fact that the number of cosets will equal the order of conjugacy class)?

wind steeple
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cosets are conjugates

tribal pasture
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I mean over here we can see that the conjugates are very different from the cosets but there should be some relation between them

📎 unknown.png
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I was thinking maybe conjugacy classes of a point x is just some element from the distinct cosets of the centralizer (eg. C(y) = {y,xy,yx} and each one of them is in a distinct coset) but that hypothesis does not hold true for C(x) (since x and x^2 are in the same coset)

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This is ofcourse an attempt to visualize is there any relation in how do cosets partition and how conjugates partition a finite group

wind steeple
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G/Stab(x) = g^-1(G/Stab(y))g for x= g^-1yg

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Or something like this

cunning geyser
#

this server has been mean at times

latent anvil
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?

ripe basalt
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@Sameer Varma Jalebi Gulabjamun#9911 I think the people on the server can be more mean than the server itself

urban acorn
#

Cool group theory exercise: Prove there are infinitely many groups unique to their order. (Hint: ||use Lagrange's theorem||)

bleak abyss
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Unique to their order meaning what exactly? Like, show that there are infinitely many n such that there is only one group of order n up to isomorphism?

urban acorn
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Yes.

bleak abyss
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Z/p

urban acorn
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indeed, that's the solution

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at least the one I thought of

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would be better if we put solutions in spoilers though

plush vector
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wtf i was solving this

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you just ruined my answer

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ugh

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just kidding

bleak abyss
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I'll say that groups of order p being cyclic is seen as a bit of a standard fact in group theory

wind steeple
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Btw is the result true for any cardinal ?

latent anvil
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I doubt it but don't know any set theory

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Is |Q^\oplus κ| = κ?

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I'm thinking you can look at vector spaces over Q versus like F2

bleak abyss
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Wait which result?

latent anvil
#

That there are infinitely many distinct groups of order κ

#

My idea was to look at vector spaces of dimension κ over countable fields with varying characteristic

#

I think it works, as long as the cardinality of k^(\oplus α) is α for any countably infinite field k and a cardinal α

urban acorn
#

@wind steeple if you mean the result that there are infinitely many groups unique to their order, then obviously no (if the generalisation to "every cardinal" you mean is that for any cardinal, there's just one group with that order up to isomorphism). I'm pretty sure the first counterexample is the klein four group and integers modulo 4 being of the same order but distinct isomorphism types.

wind steeple
#

No I was asking if for any cardinal, there exists a group of this cardinal

woven delta
#

Yes

#

Take the free abelian group on that cardinal many generators if it's an infinite cardinal

#

Otherwise use Z/nZ

#

If you want to be indirect, the lowenheim skolem theorem from model theory tells you that there are infinitely many groups for each infinite cardinality lol

wind steeple
#

Thanks

#

Nice theorem xD

fringe nexus
#

How do we do euclidean domain in gaussian integers exactly?

#

like how would i find the gcd of 85, 1+13i?

mild laurel
#

Technically the euclidean domain only gives the existence of a "remainder"

#

So you just use the existence of this to keep reducing down

fringe nexus
#

Could you just give example of the first step?

#

like i find the norm of (1+13i) and the norm of (85) and i divide them right

#

and i round down?

mild laurel
#

That's what I'm saying

#

Technically there's no actual way to get the remainder just using the euclidean domain stuff

#

E.g. you know the existence of this remainder

#

But the definition doesn't tell you how to find it

#

In this case you can though I think

#

If you're dividing 85 by 1 + 13i

fringe nexus
#

oh wait i multiply the denominator by the conjugate of 1+13i

#

and the numerator

mild laurel
#

The problem is that won't give you a gaussian integer

#

And rounding to the nearest gaussian integer won't really work either because the remainder isn't a gaussian integer

fringe nexus
#

why won't it give me a gaussian integer if i just round down?

#

like element wise

mild laurel
#

Hm okay maybe I'm dumb

#

The remainder will be a gaussian integer

#

You'd have to prove that this always works

#

You'd have to show that the norm of the remainder will be less than the norm of what you're dividing by

#

So I think you'd have to round to closest and not just round down

fringe nexus
#

oh right oops

#

closest

mild laurel
#

I think the other way to do this is by considering the lattice created by 1 + 13i times all elements of Z[i]

#

Then 85 will be in some parallelogram

fringe nexus
#

okie thanks

mild laurel
#

And the idea is that you can find the place where 85 is in the first parallelogram

#

Which is just 85 - some multiple of 1 + 13i

cloud walrusBOT
woven delta
#

So given a factorization in Q[x], you can make this into a factorization in Z[x] by clearing denominators

#

And then you mod out by (7), making a factorization in Z/(7)[x]

#

But in Z/(7)[x] this thing is just -1

#

So due to it being a Euclidean domain, it can't have a factorization

#

I guess

#

I'm super rusty on this stuff tbh

#

@chilly ocean does this look right?

covert vector
#

can't u just use rational root theorem

#

and reduce it to 4 negative roots

woven delta
#

Is what I said right though?

covert vector
#

oh

woven delta
#

It's been a while since I thought about this shit

#

So it's possible this is just garbage

potent lynx
#

totally unrelated question

#

when do u get to polynomials

#

in algebra?

latent anvil
#

It depends on what you mean by algebra

potent lynx
#

abstract algebra

somber bramble
#

depends on your prof it was pretty much the first topic for us and we kept returning to them in different contexts

#

they’re a natural example for rings and we started abstract algebra with ring theory

potent lynx
#

oh ok

wind steeple
#

polynomials are omnipresent in algebraic geometry

uncut girder
#

Hey guys

#

I'm trying to prove that if p is a prime that is 1 (mod 4) then there exists some integer n such that n^2=-1 (mod p)

#

Here's my proof, please verify if it works

#

So U(Z_p) multiplicative group of units mod p, is cyclic of order p-1 so there exists some element in there, call it m, with 4 diving the order of m.

#

Then there is an element in the subgroup of U(Z_p), generated by m, with order 4

#

Call this element n. Then n^2 has order 2

#

Now we prove that any element of order 2 in U(Z_p) is -1

#

(I'm wondering if this part is true:)

#

So let x be such an element
Then x^2 =1
So 0 = x^2 -1 =(x+1)(x-1)
So x=1 or x=-1
But 1 has order 1 in U(Z_p) so x=-1

#

[End of proof]

delicate bloom
#

which step in particular are you doubtful of?

uncut girder
#

Everything seems fine to me

mild laurel
#

It is fine

urban acorn
#

This proof is correct. I've actually seen it elsewhere before.

carmine sentinel
#

can someone explain the first answer

#

how is Fqn a subgroup of GL(m,Fq)

velvet zinc
#

the diagonal

carmine sentinel
#

like choosing n elements along the diagonal?

velvet zinc
#

n elements of F_q*

carmine sentinel
#

F*q^n has q^n-1 elements though

velvet zinc
#

oh im wrong

#

you're looking for F*_{q^n}

#

not (F*_q)^n

#

so then what you want to do is this

#

notice that F_{q^n} is a degree n extension of F_q

#

therefore each nonzero element acts as a matrix of GL_n(F_q) on F_{q^n} by multiplication, since it's an invertible F_q linear transformation on (F_q)^n

#

this identification embeds F_{q^n}* into GL_n(F_q)

carmine sentinel
#

how do i construct a matrix for an element?

#

like what basis should i choose

velvet zinc
#

1, x, ..., x^{n-1}

#

is the natural choice

#

interpreting F_{q^n} = F_q[x] / p(x)

clear obsidian
#

Is this all correct (I’m pretty sure it is, just need affirmation)

📎 image0.jpg
mild laurel
#

Uh

#

What's Z_p

clear obsidian
#

Cyclic subgroup of Z with addition mod p, where p is prime

#

<0,1, ... , p-1>

mild laurel
#

Then yeah this is fine

clear obsidian
#

Ok thanks!

ripe basalt
#

it's the p-adic integers mang

delicate bloom
#

tell everyone you know F_p is the new Z_p and fight anyone who dare oppose you; Z_p are p-adic integers,

urban acorn
#

F_p doesn't work if you want to extend it to non-primes because the motivation for F is field and it's not a field with non-primes

sharp sonnet
#

there is a finite field for every prime power, not just every prime though

somber bramble
#

yea, though I’ve usually seen it notated e.g. $\mathbb{F}{2^4}$ rather than $\mathbb{F}{16}$

cloud walrusBOT
somber bramble
#

one could argue that’s the same thing ofc

#

also $\Z/p\Z$ or if you’re lazy, $\Z/(p)$ or even $\Z/p$

cloud walrusBOT
somber bramble
#

taking modding out by an element to mean modding out by the ideal generated by it

sharp sonnet
#

$\mathbb{Z}_p$ is common notation especially in introductory classes, because you do not need the concept of a quotient ring/group to introduce it

delicate bloom
#

@urban acorn that's when you should use Z/nZ

cloud walrusBOT
sharp sonnet
#

but yeah, it is notation also commonly used for p-adic integers

delicate bloom
#

why do we need Z/nZ and Z_n to mean the same thing, just write the left, how often are people working in Z/nZ really

somber bramble
#

I mean you also don’t need to know why it’s written Z/nZ to introduce the notation

sharp sonnet
#

it's a good example for a first algebra class

delicate bloom
#

if you're not working in F_p you should be and then go back to Z/nZ with the chinese remainder theorem

sharp sonnet
#

yeah, but the notation might be confusing in the beginning

#

i.e. "why use such cumbersome notation"

somber bramble
#

I mean you can just say two sentences about it

#

“the /nZ basically means that we make all numbers that are divisible by n into 0”

#

“ask me after class if you want to hear more or just wait like five weeks and it’ll make sense”

sharp sonnet
#

i guess

somber bramble
#

could even make a very informal analogy with division. intuitively, Z is "n times bigger" than nZ, so when you divide Z by nZ you should end up with n elements

#

this is of course wrong on so many levels, but it works out :P

sharp sonnet
#

i think most students encountered quotients before at least in the context of equivalence classes

urban acorn
#

@somber bramble actually, the "n times bigger" becomes formally justified if you use density

somber bramble
#

yea sure

urban acorn
#

What's your favourite algebraic structure of each major type?

delicate bloom
#

Z/1Z, Z_1, F_1

urban acorn
#

Do you hate fun?

#

Mine are the klein 4 group, integers, and rationals.

delicate bloom
#

haha no fun zone

urban acorn
#

btw i was thinking about how much i like the klein 4 group

#

and i realized i can visualise it because the even dihedral groups have it as a subgroup (up to isomorphism of course)

#

Check out {1, s, r^(n/2), sr^(n/2)}

#

okay i thought of an amazing proof there's no group of order 0

#

Suppose G is a group of order 0.

#

G acts on itself via left multiplication.

#

This action is faithful.

#

Thus there's an injective homomorphism from G to S_0.

#

Therefore, G is isomorphic to a subgroup of S_0.

#

So S_0 has a subgroup of order 0, but S_0 is of order 0!=1, and 0 doesn't divide 1, and so according to Lagrange's theorem we arrive at a contradiction.

#

Q. E. D.

#

god damnit we should add a memes channel

ripe basalt
#

looks dum

urban acorn
#

you think?

urban acorn
#

Let Q be a field.
Let A be the ring of integers of Q except 0. Let B contain precisely the multiplicative inverses of the elements of A.
Let the elements of Q be uniquely identified each in the form ab with a in A, b in B.

wind steeple
#

Is it uniquely ?

golden pasture
#

If we just use rationals doesnt seem like?
2(1/4) = 1(1/2)

wind steeple
#

If you take b = 1 and b = another thing you'll not have uniqueness

#

The ring of integers of rational numbers is Z @golden pasture

somber bramble
#

What is the ring of integers? Is it just the subring generated by 1?

fading wagon
#

depends, it might be the ring of algebraic integers in a number field

#

but I think without any further context, it's probably just $\bbZ$

cloud walrusBOT
chilly ocean
#

agreed

fading wagon
#

@somber bramble

chilly ocean
#

though there's a more general setting

#

https://en.wikipedia.org/wiki/Ring_of_integers

Ring of integers

In mathematics, the ring of integers of an algebraic number field K is the ring of all integral elements contained in K. An integral element is a root of a monic polynomial with integer coefficients, xn + cn−1xn−1 + … + c0 . This ring is often denoted by OK or

...

somber bramble
#

also to answer the above, SO(n) is a good group; ℂ[X] is a good ring and 𝔽₂ is a good field

#

(honorable mention to quaternions)

chilly ocean
#

@somber bramble lmao so good! my thesis subject is about bad groups 😭

somber bramble
#

what's your thesis about?

urban acorn
#

@somber bramble I've seen a pretty cool technical application of F_2 in a computer related problem, and since you're in the discord api discord i assume u know a bit about computers, wanna hear it?

somber bramble
#

since you're in the discord api discord
I wasn’t even aware of that, am I?

#

so I am

#

but sure

urban acorn
#

I checked our mutual servers to see if you know about computers since a lot of people here have computer realted mutual servers with me

#

aight so basically there was this security CTF competition

#

and there was a particular problem where you needed to pwn a router (i think) by making it accept a fake firmware update

#

and the way it checked the firmware update

#

is by taking a sha256 of all the files

#

bitwise xor-ing them all together

#

and then checking that up against what it should be in a legit update

#

and with a bit of linear algebra involving $F_2$ the solution involved selecting of a random list of outputs of sha256 which ones to include in order for the xor of them all to be a particular value

cloud walrusBOT
urban acorn
#

and essentially it goes like this

#

xor is equivalent to addition in $F_2$

cloud walrusBOT
urban acorn
#

and so bitwise xor-ing the (256 bit) output of the sha256 algorithm is equivalent to adding vectors in ${F_2} ^{256}$

cloud walrusBOT
urban acorn
#

and choosing which ones to leave in is equivalent to which coefficients to set to 1 as opposed to 0 in a linear combination of them

#

so you just repeatedly take sha256 of stuff, make sure that adding it to your list would preserve linear independence, and then adding it (or leaving it alone and iterating until you get something you can add without making your list linearly independent)

#

until you get 256 different ones

#

and then you have a list of 256 linearly independent vectors in the 256 dimensional vector space, so it must be a basis

#

so then to figure out the coefficients you just have to solve a system of 256 linear equations in $F_2$

cloud walrusBOT
somber bramble
#

neat

urban acorn
#

boom, pwn

chilly ocean
#

@somber bramble bad groups 😂

somber bramble
#

oh is that a technical term?

#

of course it is

chilly ocean
#

yes xD

somber bramble
#

nonsoluble connected groups of finite Morley rank all of whose proper connected definable subgroups are nilpotent
this?

chilly ocean
#

😮 NOIIIICE

somber bramble
#

(I understand like half the words in that sentence)

chilly ocean
#

remove nonsoluble and you get the most popular definition

somber bramble
#

nonsoluble \✔
connected ✘ no clue what it means in the context of groups
group \✔
finite Morley rank ✘
proper \✔
definable ✘
subgroup \✔
nilpotent ✘ again not sure what it means in the context of groups

chilly ocean
#

when one says connected

#

the typical algebraist would think about connectedness of algebraic groups

#

but in this context it's the connectedness of stable groups

somber bramble
#

btw is this a master’s thesis or a phd thesis or?

chilly ocean
#

it's a model-theoretic notion

#

phd

#

starting it

somber bramble
#

nice

chilly ocean
#

1st year

somber bramble
#

so you’re gonna be mostly on your own doing research?

chilly ocean
#

😭

#

no one in the lab knows model theory except for my advisor 😂

somber bramble
#

what exactly is a lab in the context of math anyway

chilly ocean
#

aah the research? nah under the guidance of advisor

#

a group of mathematicians 😂

#

and their desks

#

and an assignment function

#

that assigns to each mathematician a desk

#

it may not be injective though

somber bramble
#

is the group abelian? does it have any itneresting properties?

chilly ocean
#

it has 4 cosets

#

Algebraic geometers

#

(and people on Langland stuff)

#

Probabilists and statisticians

#

Applied PDEs people

#

and forgot 4th 😅

somber bramble
#

(I’mma interpret “Langland stuff” as some kind of drug) so which coset contains the identity?

chilly ocean
#

this is not a politically correct question 😂

#

but now the director of the lab is from the algebraic geometry team

somber bramble
#

I assume the group multiplication has sth to do with proofreading and calling the other’s work garbage right?

chilly ocean
#

also my advisor and I are assigned to that team as it's the "closest" to our topic

somber bramble
#

oh, no I know

#

it’s probably redirection, right?

chilly ocean
#

😂

somber bramble
#

a*b=c means “when a presents their work to b, b says ‘talk to c instead‘”

chilly ocean
#

this happened to me twice 😂

somber bramble
#

scientifica² is thus the person you’d first go to to ask for help

chilly ocean
#

scientifica²=AoE2

#

😂

somber bramble
#

and the identity is therefore whoever can always convince people to read their work

chilly ocean
#

🤔

somber bramble
#

(but interestingly they will also always read everyone’s stuff)

chilly ocean
#

:o*

#

😮

somber bramble
#

this seems to point to it being the director

chilly ocean
#

anyway got to go

#

hungry roopopcorn

somber bramble
#

bon apetit

urban acorn
#

I'm so regretful I didn't witness this conversation when it occured.

#

the meeeeemmeeeessss

#

I'll keep studying the labs group

#

I proved there's an even amount of mathematicians in the lab

somber bramble
#

I can one-up that

#

the number of mathematicians is divisible by 4

urban acorn
#

but every number is of mathematicians megathink

#

I'm currently working on a stronger result.

#

btw what's your proof?

#

my proof that it's even is quite elegant

#

Scientifica mentioned that he and the advisor are the only model theorists.

#

So we have the subgroup {Advisor, Scientifica}.

#

And the order of every subgroup of a finite group divides the order of the group.

#

By the way this proves Scientifica² is the advisor.

somber bramble
#

my proof is that he said they had there were four cosets

urban acorn
#

oh fuck

somber bramble
#

(of some unspecified subgroup)

urban acorn
#

Anyway, at least we know who scientifica constantly harrases now since he sucks at maths.

cloud walrusBOT
urban acorn
#

I'm afraid you're a bit behind.

#

your results aren't quite as interesting as mine

#

and they're trivial anyway

#

(aka your work is garbage)

somber bramble
#

hey now my results also included deducing the identity of the identity element

urban acorn
#

that can also be easily obtained using the ideas I thought of

somber bramble
#

yea but I thought of it first

urban acorn
#

but my proofs are more elegant

somber bramble
#

also it’s not clear that {x | x is a model theorist} is closed under proofreading

urban acorn
#

of course it is

#

and also inverses

#

No moron model theorists stupid enough to let an algebraic geometer who doesn't know any model theory proofread a model theory proof by a fellow model theorist will be doing a thesis in maths.

#

btw I just got stronger versions of my results

#

@somber bramble HAHA IN YOUR FACE

#

I just proved the number of mathematicians in the lab is divisible by 638.

chilly ocean
#

👀

somber bramble
#

that seems unlikely

urban acorn
#

So he told me on DMs there are 29 algebraic geometers and 11 applied PDE people.

somber bramble
#

(for one it could be infinite)

#

oh

#

that’s unfair

urban acorn
#

lcm(2, 29, 11) = 638

#

in your face

chilly ocean
#

sry man was out and when connected back saw his DMs before this discussion here

#

hey

#

Intel

urban acorn
#

yeah?

chilly ocean
#

the advisor and me are included in the 29...

urban acorn
#

mathematicians proved 2 divides 29

chilly ocean
urban acorn
#

btw my result also explains why the function associating mathematicians to desks isn't injective

chilly ocean
#

oh dear...

#

on thing for sure

#

it is irregular

somber bramble
#

because no sensible lab is that big?

chilly ocean
#

like

#

there's a PhD students desk with 5 people

#

another one with 2

#

as for mine

#

we're 3

#

but the 2 others are almost never here

#

cuz one of them (statistician) has a desk in a hospital

#

the other guy is finishing soon and is teaching in schools

somber bramble
#

that’s not an ambiguous statement at all

chilly ocean
#

because no sensible lab is that big?
agreed

#

proof by realism

#

🔳

urban acorn
#

better than proof by intimidation

#

@chilly ocean how many chairs are there?

chilly ocean
#

my favorite is proof by inexistent reference

somber bramble
#

proof by personal communication tho

chilly ocean
#

how many chairs are there?
oh dear.....

somber bramble
#

also proof by cosmology (it would be an affront to the notion of a beautiful universe if it was false)

chilly ocean
#

got to go

#

got some bad groups to tame

somber bramble
#

and proof by funding (“how could three different agencies all be wrong?”)

urban acorn
chilly ocean
#

😂

somber bramble
#

and my algtopo prof’s favourite:
“I don’t feel like finishing this. Fix it □”

chilly ocean
#

😂

urban acorn
#

proof by cumbersome notation is just a special case of proof by intimidation

chilly ocean
#

😂 I still don't think so though

#

like

somber bramble
#

I still believe physics would’ve been better off if it used meaningful Hanzi for its units and constants

chilly ocean
#

go read the proof of the classification of the finite simple groups

#

or the proof of the Feit-Thompson theorem

somber bramble
#

instead of using the same five latin letters twenty times

urban acorn
#

read the proof for the classification of the finite simple groups

somber bramble
#

quick! list five meanings of e in math&physics

urban acorn
#

Are you stupid?

#

I'm mortal.

chilly ocean
#

😂

urban acorn
#

I'm gonna die at some point in my life.

somber bramble
#

mortals wrote it

#

so mortals can read it too

urban acorn
#

many of them

chilly ocean
#

250x mortals wrote it

somber bramble
#

so what writing takes longer than reading

#

:P

#

it’s only what

#

how many thousand pages?

chilly ocean
#

proof by "250 people believe it's right"

#

10000 pages...

somber bramble
#

see

chilly ocean
#

no fake

somber bramble
#

that’s only 100 big books

#

you can totally read that in a lifetime

urban acorn
#

not with my attention span

somber bramble
#

wait no

#

10 big books

golden pasture
#

@空间都是紧空，我答应我们都应该用汉字，有更多的选择啊！

somber bramble
#

or 100 small ones

#

just because I know the word “hanzi” doesn’t mean I speak chinese

#

if that was directed at me

golden pasture
#

(was joking a bit)

chilly ocean
#

BTW, what other theorems follow from Feit-Thompson ?

#

@golden pasture 但是60亿其他人呢？

golden pasture
#

甭管他们吧

urban acorn
#

@somber bramble 我想你会说中文

somber bramble
#

ni hao to you too

chilly ocean
#

@chilly ocean dunno :/

golden pasture
#

all spaces are compact ~ 空间都是紧空 lol

#

imagine in a proof just start using chinese characters for variables

urban acorn
#

I know that proof.

chilly ocean
#

امسك الشاي

urban acorn
#

rotation rhymes with notation

golden pasture
#

Consider $\frac{\mathbb{R}[假]}{假^2+1}$ it is trivial ...

cloud walrusBOT
golden pasture
#

wew

#

that looks

#

pain

chilly ocean
#

😂

#

😂

#

😂

urban acorn
#

By the way.

chilly ocean
#

you don't know the real pain

urban acorn
#

excuse me what the fuck?

golden pasture
#

LOL

somber bramble
#

that just looks like bad kerning
and is it just me or does the font change

golden pasture
#

i thought reading landau in chinese was hardcore enuf

urban acorn
#

if someone said this was a beautiful elementary one line proof of fermat's last theorem i'd just take their word for it and not bother

chilly ocean
#

excuse my ignorance, what's landau?

urban acorn
#

by the way

somber bramble
#

the theoretical physics books?

urban acorn
#

fun fact

golden pasture
#

yea

#

10 books covering like most of say pre-1950 theoretical physics

urban acorn
#

merosity's favourite algebraic structures are trivial and left as an exercise for the reader

chilly ocean
#

😂

urban acorn
#

literally

#

trivial

chilly ocean
golden pasture
#

||I would argue F_1 doesnt exist||

chilly ocean
#

oh no

#

||you don't want to bring the 0=1 debate ||

delicate bloom
#

||lmao||

somber bramble
#

I believe that 0=1 but not 0=2

chilly ocean
#

0=1 => 0=2

somber bramble
#

I refuse

golden pasture
#

what is 2 tho

chilly ocean
#

1+1

somber bramble
#

well, roughly

chilly ocean
#

have a nice day y'all

urban acorn
#

@golden pasture i would argue that too, i'm not a mathematical platonist

#

that basically means i believe mathematicians are like people in love with doki doki literature club characters

#

that is we spend all our time thinking about things that don't even exist

#

it took me ten seconds to realize why the statements
||F_1 doesn't exist||; and
||you don't wanna bring back the 0 = 1 debate|| are related

#

lmfao now whenever someone says something like "but that simplifies to 5 = 4 which is obviously false" in a proof I'll be like "no in F_1 it isn't"

velvet zinc
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F_1, the field of one element, isn't a field and doesn't have one element

urban acorn
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isn't a field

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What field axiom does it not satisfy?

velvet zinc
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maybe you don't know what F_1 is

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just google "field with one element"

delicate bloom
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lol

somber bramble
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the naive approach of just taking the field axioms but saying it has one element fails at the axiom that 0 and 1 are two distinct elements, but from what I gather F₁ isn’t seen/constructed/whatevered as such anyway

urban acorn
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ohhhh

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right

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you're right

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I forgot

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doesn't mean it doesn't exist, just that it's not a field

somber bramble
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(and if you drop that axiom you just get the 0 ring)

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(which isn’t interesting)

urban acorn
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I guess you could drop that axiom, yeah

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it isn't interesting, but neither is the {0} vector space over any field, and the trivial group, yet no one denies their being groups

somber bramble
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yea but apparently F₁ actually is interesting

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whatever exactly it is

urban acorn
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it's just a matter of when you say "field" is it more convinient to include or exclude the trivial field

somber bramble
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the wiki page has some information but it’s mostly gibberish to me

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I mean one pragmatic reason for saying the 0 ring is not a field is because linear algebra over it doesn’t work very well

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e.g. try defining dimension over it

urban acorn
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scalar multiplication by it would just be doing nothing

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so a linear combination is just a sum

velvet zinc
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the general linear group GL_n(F_1) isn't linear

urban acorn
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so the span of any list of vectors consists of one element, and must thus just be 0

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so the sum of any list of vectors is 0

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so the only vector space is the trivial space

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so here's how I define dimension

somber bramble
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for every other field you have that Fⁿ is an n-dimensional vector space over F

urban acorn
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"Let V be a vector space over F_1. dim(V) = 0"

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sue me

somber bramble
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and that Fⁿ and Fᵐ are not isomorphic for n≠m

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over the 0-ring, I agree with what you said

velvet zinc
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the vector space F_1^n isn't a vector space

urban acorn
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yeah, these are sum good points

somber bramble
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I’m not talking about F₁ but about the 0-ring

urban acorn
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@velvet zinc yeah it is

somber bramble
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(cause I don’t really know what F₁ even is)

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a vector space needs a field and the 0-ring is not a field (and neither is F₁, whatever it is exactly)

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and the fact that you get inconsistencies like above justify why it’s not a field

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it’s a module tho

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a pretty boring one

urban acorn
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F_1 is the 3-tuple ({x}, +, *) with + : {x} -> {x} : x -> x, * : {x} -> {x} : x -> x

velvet zinc
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why don't you just read what F_1 is instead of making false things up

urban acorn
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is that not true?

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because that's what I'd expect

velvet zinc
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all you need to do is google "field with one element"

urban acorn
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I read the beginning of the wiki page

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It says it's the ring of the characteristic 1, where the characteristic of a ring is said to be the smallest number of times you need to add 1 to get 0, and thus 1 = 0

somber bramble
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F₁ is some wierd thing that kiinda behaves like a field with characteristic 1

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but it’s not a field

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the notation is suggestive but misleading

urban acorn
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yes, it's not a field because fields are taken here (as they are pretty much everywhere, I assume) to include the axiom that 1 ≠ 0

somber bramble
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no

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you’re still equating F₁ = 0-ring

urban acorn
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"ring of characteristic 1" seems like the 0 ring to me

somber bramble
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we’re saying F₁ is not that (though I’m not stating what it actually is, because I don’t really understand it)

urban acorn
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According to what I came up with when searching that, it does seem like F₁ is that

somber bramble
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the wiki page in fact says it’s not a ring either

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and in fact

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Instead, most proposed theories of F1 replace abstract algebra entirely

urban acorn
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nvm

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yeah

delicate bloom
urban acorn
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i didn't read up to that part when i finished paragraph 1

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it's related enough to abstract algebra

delicate bloom
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MODS

urban acorn
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MODS BAN MEROSITY HIS FAVOURITE GROUP IS Z/1Z

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AND HIS FAVOURITE FIELD... ISN'T EVEN A FIELD!

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horrified face

chilly ocean
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How do I find all homomorphisms from Z8 to Z12?

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What Im thinking is if f(0)=0 then f(1+1+...+1)=8f(1)=0 which implies 8f(1) needs to be in form 12k?

hot lake
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that's a very strange use of if ... then ...

chilly ocean
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so f(1)=0 (then f=0), f(1)=3, f(1)=6 and f(1)=9?

hot lake
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something like that yes

somber bramble
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you start with a generator of the first group

chilly ocean
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those are additive yes

somber bramble
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and then you have to map it to something whose order divides its order

chilly ocean
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so the thing I described above is what you meant by starting with a generator?

somber bramble
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yea so basically f(0) = 0 is fixed right? now the key insight is that if you know f(1) you know f on all values

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so you just need to find out which values it can get mapped to

chilly ocean
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yep yep

somber bramble
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and you know that 1 has order 8 in the first group

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so f(1) added to itself 8 times must give 0

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so the order of f(1) must divide 8

chilly ocean
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And how would I show there are only 4 homomoprhisms from Z8 to Z12?

somber bramble
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show that there are only 4 elements in Z/12Z whose order divides 8

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there’s only 12 elements in Z/12Z

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so you can just list them out by order

chilly ocean
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oh I see

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Thanks

somber bramble
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(technically you also have to show that for each of these you actually do get a group homomorphism if you try to map 1 to it, but I think it always works with these groups)

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(I’m not quite sure how to show that tho)

bleak abyss
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@somber bramble so let's say G is any group and x\in G has order dividing n. You have <x>, which is isomorphic to Z/d, and the inclusion of <x> into G

woven delta
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Suppose your (commutative) ring has a nontrivial ideal. Is the only object that is both injective and projective in R-mod 0?

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Obviously such an object can't be free, but that's all I was able to show

velvet zinc
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I believe k[x]/(x^2) is injective over itself?

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im a bit rushed so I cant verify but check it

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otherwise k[x]/(x^2 - 1)

woven delta
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Yeah that works thanks

meager flint
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Is $SO_n$ isomorphic to $O_n\times{\pm I}$?

cloud walrusBOT
woven delta
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No

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Maybe you're thinking the opposite

bleak abyss
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-I has det (-1)^n

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So shouldn't be

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Since in even dimensions -I \in SO(n)

meager flint
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Oh lmao no I meant is $O_n$ isomorphic to $SO_n\times{\pm I}

woven delta
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Don't you need more

meager flint
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I know it depends on the parity of n

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they are isomorphic for n odd and not when n is even, but I just don't understand why

woven delta
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Where did you see this?

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Oh cool, it actually is a thing

meager flint
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Artin problem 5.1.3

woven delta
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Okay let's see if I can reason through this

meager flint
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Please

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Thanks

woven delta
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Okay so the determinant is a group homomorphism to R for both (assuming you're working over R)

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No that probably won't help

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Oh this is stupid

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You have -I in SO(n) like Dami said

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So you will have a bunch of square roots of 1 in SO(n)x \pm I

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But only 2 in O(n)

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Cause you have -IxI, -Ix-I, IxI, and Ix-I

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But only I and -I in O(n)

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So that's a way of telling them apart

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Okay, so you now want to construct an isomorphism in the odd case

meager flint
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The product map works I think

woven delta
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Why?

meager flint
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It's a homomorphism because I and -I commute with everything

woven delta
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Why is it onto?

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Also it can't work

meager flint
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I'm not sure tbh it was just a guess lol

woven delta
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Cause of the determinant

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The determinant of the product will be -1

meager flint
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Wait how does the fact that we have more than to sqrt1 tell us that we can't construct an isomorphism between the sets?

woven delta
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Wait am I being silly

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I am

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Yeah it might work

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I thought orthogonal meant something else for a second lol

bleak abyss
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Okay sos

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In the even case I'm pretty sure what I said just kills it

woven delta
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Yeah

bleak abyss
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In the odd case

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\pm I is normal

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And SO(n) is normal

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So yes

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Something something if you have two normal subgroups with trivial intersection then G = H\times K

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Err you'd need HK = G

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But that's trivial

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Since multiply something with det = -1 by -I

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Now it's in SO(n)

woven delta
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Yeah

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I guess it is trivial

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The product map that @meager flint described works

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It is exactly the same thing that Dami described

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For some reason I forgot that every matrix had determinant \pm 1

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Smh

meager flint
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Okay yeah that makes sense I'm just not following what happens in the even case

bleak abyss
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The issue with the even case is that, at least concretely (when you think of {+/- I} and SO(n) as subgroups of O(n)), your issue is that the determinant of negative identity is already 1