groups-rings-fields

[valid fox]

What are the subgroups of $(\mathbb{Z}/p\mathbb{Z})^n$?

[cloud walrus]

Tiessie

[valid fox]

Are they all isomorphic to Z/pZ ?

[quiet pelican]

No, there are ones isomorphic to (Z/pZ)^k for all 0 <= k <= n

(It’s an exercise to prove that that list of isomorphism classes is exhaustive)

[valid fox]

Sorry I meant subgroups generated by one element

Cyclic subgroups

[quiet pelican]

That’s equivalent to every non-identity element having order p, which isn’t hard to prove

[valid fox]

Yeah every non zero component has order p so the whole ordererd tuplet has order p

Makes sense

[white oxide]

This follows from the fact that every subgroup of a cylic group is cyclic right

That is, that it is of order (p - 1)/d

[delicate orchid]

yur, but you still gotta show the set of dth powers is a subgroup

which is a one liner

[white oxide]

yes that's trivial

[delicate orchid]

this entire problem is trivial

[rocky cloak]

I mean, the fact that U(Z/p) is cyclic is somewhat nontrivial

Hmm... how easy is it to do this exercise without this fact?

[delicate orchid]

it follows from this exercise so you would just be reproving it

d = 1

[rocky cloak]

The d=1 case is just saying that U(Z/p) has p-1 elements

The fact that x^d - 1 has exactly d solutions mod p is enough to prove this exercise.

Which is not terribly far from proving that U(Z/p) is cyclic, but still missing the key step.

[delicate orchid]

oh right lol. My brain inserted the word "cyclic" in front of "subgroup"

otherwise why on earth would okey be asking about it being cyclic?

[tardy hedge]

okeyokay

[knotty badger]

Is there a nice group-theoretic interpretation of bezout’s identity

In a sense you can use a pair of coprime integers to obtain a partition of unity

Since $1 = a x + by$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Sounds like something you should be able to interpret in terms of Spec, maybe?

[noble belfry]

suppose there was a non-trivial normal subgroup of A_n

why would it be transitive

I know A_n is transitive ofc

[rocky cloak]

Are you following a proof that An is simple or something?

[noble belfry]

yeah

[rocky cloak]

Alright, take s a nontrivial element.

Then there is some x and y such that s(x) = y.

Now conjugate by something that fixes x, but swaps y with some z (this uses that n>=5), then the conjugate takes x to z.

Hence the group can move x to anything and so is transitive

[noble belfry]

i see yeah

thanks i appreciate the help i see why n >=5 is needed

[rocky cloak]

I guess you can get it down to n>=4 by using a 3-cycle.

But you're only gonna prove simplicity for n>=5 anyway

[velvet cave]

Does anyone know why this cup notation, with the dot, is used?

[rocky cloak]

It means disjoint union

[velvet cave]

ah makes sense

so the same as sqcup

thanks

[mint seal]

in Aluffi, one of the exercises is to obtain Bezout’s identity from the result that the order of [m] in Z/nZ is n/gcd(m,n); in particular if gcd(m,n) = 1, then [m] generates Z/nZ

I guess it goes, if [m] generates Z/nZ, there is some a so that am cong 1 mod n, i.e. am - 1 = bn, and then am - bn = 1

erm, I got a little sloppy and dropped the bracket from the class of m

[balmy python]

i guess you can do this with polynomials

but thats more because they can be uniquely factorised

[torn thorn]

Hello. I'm new to abstract algebra and need help with making a proof more rigorous.

Let G be a cyclic group of order 6. We need to determine how many of its elements generate G.

I've worked out that if, say, we want x^2 to generate G := {1, x, x^2, ... x^5},then x^3 would not possible to be obtained from composing x^2 to itself integer many times. This is because, using something I don't know how to justify, it is equivalent to saying that there exist integers i and j s.t. 2i = 6j + 3; this is not possible, because 6j + 3 = 3(2j + 1) is odd while the LHS is even.

So, presumably I need to show this using the language of abstract algebra, like by defining a new set
S := {k∊ Z | x^k = 1} and doing something with it. But I can't really make that translation. Help

[elfin wraith]

There’s no need to over complicate it. You’ve already defined the elements, and the operation such that x^n * x^m = x^(nm) mod 6

So all there is to do is look at (x^2)^n for different n. You have x^2, x^4, x^6=1, x^2. So there’s no possible way to get the elements x^3 and x^5. A good exercise is to work out what subgroup x^2 does generate.

[rapid cave]

Since G is cyclic, you need to understand the order of the elements of G

The generators will have order 6

[torn thorn]

Hmmm, why are they supposed to have order 6? Sorry if this question is too stupid...

[rapid cave]

The order of an element is the number of elements in its generated subgroup

So for g to generate G it must have 6 elements in the subgroup it generates

[torn thorn]

ahh, that clears out the fog in my mind. Thank you a lot sir 🙏

[tardy hedge]

If R/P is isomorphic to a submodule of M how do I know P is an associated prime of M? So, P = ann(m) for some m. I mean, yes if the isomorphism is given by R->M r -> rm then ok but how would i know the isomorphism has to be that

[rocky cloak]

Do you know of an element of R/P with annihilator P?

[tardy hedge]

Oh, yeah i guess 1. Then sending 1 under the isomorphism you get what you want

Any element of R/P has annihilator P right, but is that only true because P is prime?

[dim wagon]

could anyone hint me what to do here?

Clearly I must construct some SES from the top and apply the tensor functor but im not sure what I can construct at the top.

[lyric lagoon]

Not sure if this is the appropriate channel but is this definition not already inductive? I got a bit confused by exercise 1

[wraith cargo]

Every LES has associated SES to it

Think about what like

An SES looks like

You have a middle term that's general but what do the two terms on the edges look like

So if V_n is you middle term what should the edges be...

[white oxide]

Does anybody know why we can write a and x in this way?

[rocky cloak]

So the odd integers modelo 2^e form a group under multiplication. What is the order of this group?

[white oxide]

Oh is it because we might as well assume that a is a unit? For otherwise (a, 2^e) = 2^l for some 0 <l <= e. But then a is not odd. Therefore a is a unit, and by an earlier proposition, a \equiv (-1)^s 5^t mod 2^e.

[formal laurel]

Is this from the book "A classical introduction to number theory"?

[white oxide]

I'm using this proposition here - if I'm correct, a reduced residue system is just a set of representatives for all units mod 2^e, therefore a can be expressed in that form

Does that work @rocky cloak

[white oxide]

Yes

[formal laurel]

Yes, this is the reason why it works.

[formal laurel]

I had a lot of fun with that book, I love it.

[white oxide]

Yeah it's fun

I'm going to read up until chapter 12 or so then jump into Neukrich

[formal laurel]

Neukirch's book is great, I remember just reading any book on algebraic number theory at the time since for me the theory was too confusing. for that, I fully recomend the book "Primes of the form p=x^2+ny^2".

[white oxide]

Okay yeah I heard it's a bit hard lol

I want to buy a hardcover of Neukrich but it's so fucking expensive 😭😭

[formal laurel]

indeed

[delicate orchid]

I've heard nothing but good things about Primes of the form p = x^2+ny^2

[wraith cargo]

I bought mine for like 20€ when there was some huge sale

[formal laurel]

It is a great book. By all definitions

[tardy hedge]

is it called algebraic number theory by neukrich?

[white oxide]

I believe so yeah

[tardy hedge]

I have a copy of that. I got it for $10 from a professor here who sells his math book collection hehe

[white oxide]

Tf

Sell that shit to me for $12?

So you make profit

[formal laurel]

Damn, math books in where I live cost one liver and a half

[dense raven]

For the group BS(1,2)=< a,b | t a b^{-1}=b^2 > is there like a nice formula for the dehn function

Like obviously it's computable via brute force because it's O(2^l) but is there a nice way to write it

[white oxide]

Does anybody know what congruences are being referenced here?

I don't know if they mean p \equiv -1 (12) and p \equiv -1 (4) or p \equiv 1 (4) and p equiv -1 (4)

[tardy hedge]

hahahaha

[white oxide]

Nvm it's gotta be the latter right, p \equiv 1 (4) and p \equiv -1 (4)

Okay now I'm not even sure if there are primes congruent to 1 mod 4 and -1 mod 4 at the same time

[delicate orchid]

https://tenor.com/view/riddler-paul-dano-new-rockstars-thumbnail-does-he-know-he-doesnt-know-gif-27552850

how can an integer be both 1 mod 4 and 3 mod 4

[white oxide]

Wait a second...

[tardy hedge]

Wew and his reactions 🥰 🥰 🥰

[white oxide]

If that's the case then 1 \equiv -1 mod 4

O.

OHHHH

I get it now

[delicate orchid]

it's two different cases

[elfin wraith]

If someone responded to one of my questions with this id blow my shit smooth off

[chilly ocean]

Well that's not really inductive, that's just taking the smallest subset of G containing X which is closed under inverse and multiplication.

I think an example of an inductive definition could be more like "Define X_0=X, and define X_(n+1) to be the set {a^x+b^y|a, b in X_n, x, y in {-1, 0, 1}}". Then the union of all X_n is the subgroup generated by it(sorry in advance for using + for group operation but also ^ for repeated group operation...)

[kind temple]

lmao a^x b^y doesn't look so bad

[untold hearth]

Define A_n as the set of products of n elements of X (where each one can be put to the power of 1 or -1), so that A_0 is the empty product, i.e. the identity element. Then your generated subgroup is simply the union of all the A_n

For the vector space, you do essentially the same thing (union over n of linear combinations of n vectors)

You can and should write down the details, but that’s the gist of it

[sacred wharf]

Can someone quickly check my proof? I wanna approve that if we have a P silo group which we call big P then if we have a normal P sub group with recall K then K belongs to pee.

So my proof uses the correspondence theorem

Where are the key idea? Is that the group that is the coset of K must have its own P silo group and then we can look at it correspondence and we know that it's correspondence must be a conjugate of P

[rocky cloak]

This is essentially how I would prove it aswell

[rapid cave]

Sylow not silo

[rocky cloak]

I guess you should probably have proven at some point that any p-subgroup sits inside a sylow p-subgroup, so you don't even need to use the correspondence theorem

[sacred wharf]

i know

[sacred wharf]

hmm idk

[sacred wharf]

correspondence is neat isn't it

[rocky cloak]

It is indeed, but it's also neat that Sylow p-subgroups are exactly maximal p-subgroups

[delicate orchid]

the p-core of a group is the intersection of all it's sylows so by maximality of the p-core you are done

[sacred wharf]

wdym isn't that how they defined ?

[tribal moss]

A priori there could be a subgroup of order a smaller power of p that doesn't sit inside a Sylow p-subgroup.

[sacred wharf]

idk what u mean but im trying to do the second part of this question which is proving that every normal p sub is contained in some max normal P subgroup
my dear is to use correspondence again
Because we know every normal P sub group is contained in big P, which is our silo group
If you look at this group, then the centre has to be non-trivial. I can justify this later if you if you think it's important
Of course centre has to be divisible by p so contains an element of order p

let's call it a

We can look at P/<a>

we know this is order p^n-1 hmmmm

maybe dis da wrong approach

[rocky cloak]

If that's how you define them then the correspondence theorem is completely unnecessary.

Then it's just automatic that K is contained in a sylow subgroup

The definition I assumed you were using was that if p^n is the largest power of p dividing |G|, then a p-sylow subgroup is a subgroup of order p^n

[sacred wharf]

ohh u meant maximal in terms. of inclusion

[rocky cloak]

Is there another sense of the word

[sacred wharf]

i was interpreting it in terms of order size

[tribal moss]

Hmm, I think I originally learned it with "Sylow p-subgroup" defined as a subgroup of order the highest power of p that divides |G|, and one of the Sylow theorems was then that every p-subgroup sits within one of those; ergo they are the maximal p-subgroups.
But the presentation in Wikipedia does it the other way around; defines "Sylow p-subgroup" as a maximal p-subgroup and then proves that every such maximal p-subgoup has order the highest power of p that divides |G|.
The net result is the same, of course.

[elfin wraith]

My group theory course took the former approach as well, same difference really

[rocky cloak]

I guess an advantage of the maximal definition is that it also makes sense for infinite groups.

Though I guess sylow subgroups of infinite groups aren't particularly well behaved in general, so it's not a big win.

[sacred wharf]

how do u have sylow group of infinite group?

[rocky cloak]

"maximal p-subgroup"

[tribal moss]

(Which may or may not exist in the infinite case -- e.g. Q/Z has none).

[rocky cloak]

They always exist, and Q/Z is in fact the direct sum of its p-sylow subgroups

But in general you can't talk about "the sylow p-subgroup" because they might not be conjugate in the infinite case

[tribal moss]

Really? Then I must have internalized a wrong definition of "p-subgroup".

[rocky cloak]

Subgroup that is a p-group, i.e. ever element has order p^k for some k

[tribal moss]

Yeah, I thought of it as "subgroup whose order is some power of p"

[rocky cloak]

That would be equivalent in the finite case yeah

[tribal moss]

Okay, with that definition Zorn's lemma applies.

[sacred wharf]

what theorem is it for?

[formal laurel]

don't you need some profinite definition? like using supernatural numbers or something like that?

[tribal moss]

Maximal p-subgroups exist in every group.

[sacred wharf]

i mean in general

[tribal moss]

I don't understand what you're asking.

[sacred wharf]

Arent lemmas used to build up to a theorem ?

[wraith cargo]

it's for a lot of different theorems

[tribal moss]

It's not an ironclad rule. Zorn's lemma can be used for a whole truckload of different purposes.
I supposed it's named like that because Zorn had some particular purpose in mind in his original publication. Perhaps it was the well-ordering theorem.

[sacred wharf]

should be a theorem itself atp fr

[elfin wraith]

It may upset you to learn that the first isomorphism theorem is really a definition

Something I learned after losing my mind trying to write lean for far too long

[sacred wharf]

?

[tribal moss]

Hmm, looks like Zorn himself actually proposed it as an axiom, and it was only Tukey who several years later started calling it "lemma".

[sacred wharf]

ig it is equivalent to aoc?

so it is kind of an axiom

[tribal moss]

Yes, given ZF.

[elfin wraith]

Because we dont just care that its true. The actual content of the "theorem" is the definition of a specific isomorphism

Words are all meaningless

[sacred wharf]

well its proving the homorphism is well defined and bijective no?

[tribal moss]

There are several different claims that are called "first isomorphism theorem" by different authors. Which one are we talking about here?

[sacred wharf]

G/ker is iso to image of bla bla i think

[tribal moss]

(https://en.wikipedia.org/wiki/Isomorphism_theorems#Note_on_numbers_and_names)

[elfin wraith]

Yeah this is what I still dont really get but im absolutley assured by the lean community that its a definition because it carries this computational aspect

Everything about doing this is making me realise how much I appriciate informal maths and do not want to do anything formal

[tribal moss]

I suppose "absolutley assured by the lean community" is a step up from "GPT told me".

[elfin wraith]

Dont worry, Kevin Buzzard gave me a nice snippy response to my confusion about this, truly infailable

But I have also had some people also in type theory try to explain this to me and apparently there is some actual distincition.

Very much looking forward to going back to simply saying things and have people know what I mean

[karmic moat]

Do graph theory

[elfin wraith]

Im trying to minimise my suffering

[south patrol]

One problem is how you define the image lol

One is that the image of f: G -> H is the kernel of the cokernel

And with that it does not seem to be a definition

Ig the first iso theorem is saying cokernel of kernel is kernel of cokernel

[glad osprey]

I dunno if this is what Kevin Buzzard meant, but I think the point is that there isn't just an isomorphism between G/ker(f) and im(f), but there's a canonical map. And sometimes it matters what the exact map is from G/ker(f) to im(f) - like, you can't just prove that it's canonical, so you have to define it

[knotty badger]

Doesn’t this only work for abelian groups

[wraith cargo]

do u mean abelian categories?

[knotty badger]

I like to view this as the distinction between knowing that two things are iso, and knowing how two things are iso

[knotty badger]

Yes

[glad osprey]

and another point is that in Lean, you can't look into the proof and see what map did I actually use to prove the isomorphism, since Lean is proof irrelevant

[wraith cargo]

lol kevin can uh

be a handful

he's probably very knowledgable

but talking to him usually makes me want to kill myself

anyways

[tribal moss]

"Such and such map is an isomorphism" still sounds like a theorem to me, rather than a definition.

[knotty badger]

For example, there is a bijection between the set of continuous solutions to Cauchy’s functional equation and the real numbers

[wraith cargo]

well you can define the map as the definition of an isomorphism perhaps

[knotty badger]

But it’s helpful to know that this iso takes the form “evaluate at 1”

[tribal moss]

I'm resonably happy with the definition of "isomorphism" I already have.

[wraith cargo]

but the issue is that when you do formal computations that definition can be very difficult to work with

[knotty badger]

Wait which def of iso are you working with

[wraith cargo]

like you can say "finite sets are very easy to work with" but in lean they're some of the most difficult objects to grasp

[knotty badger]

I complained about the way Yoneda is presented like this a couple months ago

The type 1 statement says how two things are iso, whereas the type 2 statement says that two things are iso

[south patrol]

Tbh I didn't think of groups and thought of modules but good point

[sacred wharf]

modules are groups tho

[karmic moat]

other way around

abelian groups are modules

[knotty badger]

Yes but the definition of kernel/cokernel will give you different results

[wraith cargo]

well modules are also abelian groups with extra structure

[karmic moat]

ewwwww

[sacred wharf]

maybe ur way is correct but how is this way wrong ?

[karmic moat]

technically it's not wrong because like irony said it's an abelian group with more structure

but that's a gross way of thinking about it 🤢

[wraith cargo]

I genuinely don't know why that's weird to think about because that is quite literally the definition

[karmic moat]

oh wait yea under addition lol

i'm thinking about multiplication

oopsie

i thought the underlying abelian group we were talking about was that of the underlying ring

[sacred wharf]

cokernel =

[wraith cargo]

there is no underlying ring to a module

[sacred wharf]

?

[wraith cargo]

unless you mean the ring of coefficients

[karmic moat]

yeah

[sacred wharf]

i think they mean in the case where we define the module to be 'the ring'

[wraith cargo]

idk what this means

a module that's also a ring is called an algebra

[sacred wharf]

Like if R is our ring, we can define the module to be all the elements of R, with addition of R, since R can also be seen as a group

[wraith cargo]

you don't need R to be abelian

[sacred wharf]

yeah just realized

[sacred wharf]

do we ever look at 'rings' where addition is not abelian

?

is that even possible

[karmic moat]

codomain mod image

[sacred wharf]

so if surjective, cokernel is nothing

or 0

[karmic moat]

yeah

[karmic moat]

https://en.wikipedia.org/wiki/Near-ring

[elfin wraith]

Fun exercise. Take the usual definition of a ring and drop the word abelian from its additive group. Prove that addition is commutative

[sacred wharf]

it's from distribution law ig ?

[elfin wraith]

Yuh

[sacred wharf]

so if u remove it ig that's the near ring

?

[elfin wraith]

I’ve only just heard of them, but from a very brief skim of the Wikipedia yeah it looks like you just define it to be one sided

Point is you need to drop more than the assumption that the additive ring is abelian

[copper kestrel]

quick question about number 5, is this just asking to find n when (26 + <12>)^n = <12>?

sorry if im interrupting!

[wraith cargo]

yes

[copper kestrel]

sick

[wraith cargo]

except just remember you aren't literally exponentiating

you're multiplying by n

[rocky cloak]

I don't know of such a definition, but "maximal p-subgroup" works fine

[copper kestrel]

yeah exactly :]

i thiiiiiink the order is 30

[wraith cargo]

err that's a bit too high

[copper kestrel]

rats

[wraith cargo]

Z_60/<12> has order what

[copper kestrel]

great question

would it be 60 / o(<12>)?

i dont really know i apologize

[wraith cargo]

yes indeed

[copper kestrel]

oh does it have order 6 maybe? bc i think o(<12>) = 10?

[wraith cargo]

no

what's 60/12

[copper kestrel]

5

oh

lmao

[wraith cargo]

yeeesssss

good so the order of Z_60/<12> is 12 :)

[copper kestrel]

wait what

[wraith cargo]

what's 60/5

[copper kestrel]

12

[wraith cargo]

and we conclude from this that

[copper kestrel]

but i thought you meant that o(<12>) was 12

[karmic moat]

<12> is {12, 24, 36, 48, 60=0}

[wraith cargo]

oh no

I mean 5*12 = 60, so it's 0 mod 60

[copper kestrel]

truth nuke

i apologize

[wraith cargo]

lol nothing to apologize for

[karmic moat]

https://tenor.com/view/true-truth-nuke-super-truth-nova-truth-meme-gif-16889273424352737553

all good

[wraith cargo]

anyways

[copper kestrel]

so then 26 + <12> has to divide the order of 12

[wraith cargo]

yes so it has order a divisor of 12

[copper kestrel]

so its order must be of order 1, 2, 3, 4, 6, 12

[sacred wharf]

if you divide you just get 2 or smth idk?

[wraith cargo]

a hint would be that 26+<12> = 2+<12> :3

[copper kestrel]

how-

oh wait i can see it

[sacred wharf]

these are cosets ye

[copper kestrel]

bc 26 = 24 + 2 + <12> = 2 + <12>

[sacred wharf]

yessir

[wraith cargo]

yes exactly

[copper kestrel]

so then the order must be 6(?)

[wraith cargo]

yep

https://tenor.com/view/fast-cat-cat-excited-jumping-gif-5054843775616689213

[copper kestrel]

yippeeeee

i will be back to ask more questions

[sacred wharf]

can we define gcd in groups or is that ring only

hmm

[tardy hedge]

Why are there finitely many minimal primes over I an ideal if R is noetherian?

[karmic moat]

for gcd to make sense you need some ordering

[wraith cargo]

uh

well okay this is a kind of subtle question

[sacred wharf]

i guess you kinda could just looking at the definition ? idk

[karmic moat]

Spec R is noetherian, minimal primes correspond to irreducible components, and noetherian topological spaces have finitely many irreducible components

[wraith cargo]

you can prove this with prime avoidance I think

in the ring R/I

[tardy hedge]

Ok. I was hoping using a theorem about Ass(M) when M is fg R mod

[wraith cargo]

because it's still noetherian

[tardy hedge]

Cuz this is from Eisenbud and he mentioned that fact after

[wraith cargo]

this is exercise 1.2 right?

[tardy hedge]

It was actually just from the intro discussion in the chapter about associated primes

[wraith cargo]

yeah you can def use some result like

Ass(R/I) is finite or something

[sacred wharf]

m is gcd of a,b if:
mk = a for some k in G
mv = b for some v in G
(this is defined as dividing a,b)
and for any z that divides a,b
zl = m for some l in G

[wraith cargo]

let me think for a sec

[tardy hedge]

I wouldve liked to say ann(R/I) = I but thats not true if I is not prime

[wraith cargo]

that is still true

even if I isn't prime

ann(R/I) isn't the set of zero divisors
It's the elements that annihilate the whole ring

[tardy hedge]

This is why i mentioned it

Ha

I said something the other day that made me think thats true but maybe i interpreted it wrong

[elfin wraith]

The proof in my noncom class showed did it by showing first there’s primes P_1,…,P_n containing I such that P_1\cdotsP_n id contained in I, and then some argument with that

[wraith cargo]

the easiest argument is via a contradiction

and this is spelled out in exercise 1.2 in eisenbud

[wraith cargo]

so the proof can go something like this

let R be noetherian

[tardy hedge]

Oh so anything in ann(R/I) has to be in I because it needs to kill 1

[wraith cargo]

You have to use the primary decomposition here I think

but the idea is that like

minimal primes in Ass(R/I) are the minimal primes lying over I because these are the isolated primes in the primary decomposition

so if you can prove R/I is a f.g. R-modules you're done because Ass(R/I) is finite

but to prove R/I is f.g. you only have to prove every submodule is f.g. but this is easy

[tardy hedge]

am i silly or isnt R/I just generated by 1 ?

[wraith cargo]

yeah ig this is true lol

welllll

okay this is a bit subtle maybe hm

I think when R isn't noetherian R/I can be pretty weird

but maybe I'm missign something here

oh right yeah

it has to be a f.g. module over a noetherian ring

so yeah you're right it doesn't matter

[tardy hedge]

If M fg R-mod then Ass(M) is finite and includes all minimal primes over ann(M). so since you said ann(R/I) = I then thats it

[wraith cargo]

R/I is always f.g.

yeah you just need the ring to be noetherian for these lemmas about Ass(M) to work

[karmic moat]

okay so the ordering is divisibility, I guess my only concern is uniqueness

[tardy hedge]

in the intro eisenbud mentioned that theorem and then just said there are finitely many minimal primes over I as a corollary

the only thing i missed is that ann(R/I) really is I

[sacred wharf]

do u need uniqueness ? for example even in standard numbers isn't both -2 and 2 gcd of 4 ?

[wraith cargo]

?????

4 and what

[next obsidian]

4

67

[sacred wharf]

oh wait i'm confusing smth

haha

[next obsidian]

More like amanono

[sacred wharf]

bro had to sneak it in

[wraith cargo]

it must be unique up to unit

[karmic moat]

chmonkey i hope we can bury the hatchet and i hope you have a wonderful holiday season

[wraith cargo]

there was drama??????????

[next obsidian]

There’s no hatchet

[karmic moat]

well chmonkey wont admit it because they dont wanna lose honorable

[next obsidian]

I just think amanono is fun to say

[tardy hedge]

ive seen this same interaction like 4 times now

[karmic moat]

yea im just kidding

[next obsidian]

More like kiand67

[tardy hedge]

true

more like amanono

[karmic moat]

hmm

idk there's just something about the way chmonkey says it

[next obsidian]

More like amanono

Maybe it’s the capital m

[karmic moat]

maybe

could also be the chmonkey pfp with the pigtails and foot border

[next obsidian]

https://tenor.com/view/juri-juri-face-juri-pout-street-fighter-gif-26491263

[karmic moat]

what is that from

kpop demon hunters?

[next obsidian]

Street fighter

[karmic moat]

ohhhh shit

i look stupid asf now

[next obsidian]

It’s Juri

https://tenor.com/view/juri-street-fighter-6-juri-han-face-reaction-gif-11401483181172890420

[karmic moat]

i see

[tardy hedge]

it looks like it

[karmic moat]

well my pfp is my cat JJ and he is 16 years old

[next obsidian]

He’s old

[karmic moat]

indeed

[next obsidian]

My car js

Umm

Maybe around that old

[karmic moat]

damn

[next obsidian]

I think he was 1 when I was in like… 4th grade?

And I’m in uh

21st grade?

[karmic moat]

i also have two younger cats they're both 2 or 3 i can't remember i think 2

and they're twin sisters

[next obsidian]

Twinsies

[karmic moat]

well actually originally it was 3 but someone adopted 1 so we adopted the other two

yea

[karmic moat]

damnnn

super super deluxe senior

[sacred wharf]

i'm trying to prove every normal p group is contained in a maximal normal p group i've proven that every normal p group belongs to the sylow p group P. How do i start ?

[wraith cargo]

more like nochmoney

[next obsidian]

Apparently my cat is 16

[karmic moat]

do we have the same cat

[next obsidian]

Shut up shut up shut up

[next obsidian]

Maybe

[karmic moat]

wtf...

do u live in my walls or something

[next obsidian]

Is this ur cat

[karmic moat]

conjugate up to pattern change

[next obsidian]

Wow

[karmic moat]

your cat smiles mine has RBF

[spark veldt]

Hi so this might be something simple but I can't work it idk so i have $G=H\times K$ and let $\overline{H}={(h,1):h\in H}$ and $\overline{K}={(1,k):k\in K}$ So apparently $\overline{K}$ centralizes $\overline{H}$, so I know this means $C_G(\overline{H})=\overline{K}$, but i can't really show this idk.. Ik
$$C_G(\overline{H})={(g_1,g_2)\in G:(g_1,g_2)(h,1)=(h,1)(g_1,g_2) \text{ for all } h\in H}$$
I got $g_1h=hg_1$ and $g_2=g_2$ which... idk how does $\overline{K}$ appear here...

[cloud walrus]

bluepianist

[wraith cargo]

does H here have trivial center?

[spark veldt]

H and K can be any subgroup

[wraith cargo]

then this isn't true I think
You can take g_2 to be any element of K and g_1 to be any element of the center of H and get that C_G(H-bar) = C(H) x K so the claim isn't correct

[vapid vale]

the statement that K centralizes H doesn’t mean K is the entire centralizer

[candid patrol]

(g1, g2)(h,1)(g1,g2)^(-1) = (g1hg1^-1, 1)

[wraith cargo]

Yeah it just means K-bar is a subset of the centralizer of H-bar

[candid patrol]

If H is abelian, then C(H bar) = G

[vapid vale]

have you verified by computation that K bar lies in the centralizer? since that’s all you need to show

[spark veldt]

ohhh so it's not the entire centralizer i see. yeah i can see why Kbar now is in the centralizer of Hbar, because (1,k)(h,1)=(h,k)=(h,1)(1,k)

thanks for clearing that up yall

[sacred wharf]

something isn't right here...

isn't AB just B?

[next obsidian]

no

[sacred wharf]

or i gues snot

but a subset of B ig ?

[next obsidian]

yeah

[sacred wharf]

i had a similar proof idea to them cn u verify if mine achieves same result?
if 1 = a + m
then (1-m)b belongs to AB belongs to M
b -mb belongs to M
b belongs to M

[copper kestrel]

im a bit lost on how im supposed to get an element of G from this :/

like im so close to proving it

[hidden haven]

(ab)^(-1) is b^(-1)a^(-1). The order changes.

[copper kestrel]

WAIT YOURE RIGHT OMG

YAYYYY

[sacred wharf]

ayo bro is trying to fight pseudo

[hidden haven]

[rocky cloak]

Pseudo and Moldi might just be equivalent up to unique isomorphism

[copper kestrel]

i feel like this is just a dumb computation problem bc theres no theorem that helps us

[next obsidian]

Moldi

There’s 6 elements

[copper kestrel]

ye

[next obsidian]

You can do it in liek 30 secs

[copper kestrel]

so im gonna need to compute ;wil

[copper kestrel]

im lazy :(

(i dont want to)

but sigh.. i suppose i shall

[next obsidian]

https://tenor.com/view/funny-sad-emoji-getting-disintegrated-into-dust-denigrating-denigrating-face-emoji-meme-lol-gif-27508363

[rapid cave]

its not even a subgroup

unless \iota is the identity

[copper kestrel]

it is

[rapid cave]

and then its weird notation

[copper kestrel]

it is i agree

my book only does that for dihedral groups

everything else is e

[elfin wraith]

There’s only 5 elements you actually need to care about 🥀

[copper kestrel]

i just got e and \mu that work

the rest give some other element other than mu

[knotty badger]

ew no

[rocky cloak]

I guess being #1 fan is not a universal property after all

[copper kestrel]

i did this so damn wrong ignore this LMFAOOOO

[rocky cloak]

As it's the index of the normalizer it must be something that divides 6/2 = 3.

So you just need to rule out 1

[copper kestrel]

normalizer?

and rule out 1 reflection/rotation?

[sacred wharf]

isn't nil trivially nilpotent ?

nvm it's opposite way around

[tardy hedge]

Said ew lol

[sacred wharf]

?

[tardy hedge]

Pseudo in the chat

[sacred wharf]

ohh

[wraith cargo]

where are you stuck?

[marsh fulcrum]

Showing I_0=I at every localisation at M maximal and why it means they are equal

[wraith cargo]

what happens at the M_i and what happens at all the other M's?

[wraith cargo]

also they're equal because I_0 \subseteq I and I/I_0 = 0 iff all the localizations of the quotients are 0

Follows from this

[copper kestrel]

im thinking question 4 is just a counter example bc i cant possibly think it can be right

now comes the task of finding a counter example :(

[wraith cargo]

why don't you think it's right?

oh I see why

well

why don't you think it's right actually :)

[copper kestrel]

bc theres no way cosets are that well behaved

[wraith cargo]

what's (aH)^-1

[copper kestrel]

a^{-1} H

[wraith cargo]

are you sure?

it's elements of the form ah

so (ah)^-1 = h^-1 a^-1

so it'd be H(a^-1)

[copper kestrel]

oh wait yeah

rats

[wraith cargo]

hehe

[copper kestrel]

im not good at algebra 💀

but i dont see how it can be right

like for all groups

[wraith cargo]

equality is a very powerful thing

[copper kestrel]

nonabelian and abelian

etc

[wraith cargo]

lol you literally just started so dw

[copper kestrel]

truth nuke

[wraith cargo]

you'll make it there sooner than later you'll see

[copper kestrel]

would the counter example be in the dihedral group

unless you were hinting that its right

[wraith cargo]

I mean there isn't a counter example because it's a true statement

[copper kestrel]

and i did not pick that up

RATS

[wraith cargo]

do you want a proof or do you wanna think about it with the hint I gave

[wraith cargo]

tbh I get why you think that because you're usually thinking about normal subgroups where (aH)^-1 = a^-1H but for non-normal subgroups that might not be true and you get that a^-1 goes on the RHS

[copper kestrel]

i mean i think its going to be similar to

aH = bH
(aH)^{-1} = (bH)^{-1}? then go from there?

[wraith cargo]

yessss

[copper kestrel]

yeah im used to normal subgroups being well behaved :(

[wraith cargo]

hehe yeah I get it

[copper kestrel]

question 3 is kinda confusing me too

like we know ghg^-1 in G for all g in G and h in H

then consider the left coset gH = {gh : h in H}

and i want to say then do gHg^-1 but that doesnt seem right

[wraith cargo]

they meant ghg^-1 is in H not in G

I think

[copper kestrel]

did i write the question down wrong let me check

oh i did and thats not even the question i need to solve

lord have mercy

i think i need a nap

the question i actually need to solve is if aH = bH then Ha = Hb

proof/counter example

[kind kindle]

Can someone help me here gap is being stupid

gap> A := ElementaryAbelianGroup(8);
<pc group of size 8 with 3 generators>
gap> C2 := CyclicGroup(2);
<pc group of size 2 with 1 generator>
gap> AllHomomorphism
AllHomomorphismClasses AllHomomorphisms
gap> phi := AllHomomorphisms(C2, AutomorphismGroup(A));
[ [ f1 ] -> [ IdentityMapping( <pc group of size 8 with 3 generators> ) ], [ f1 ] -> [ Pcgs([ f1, f2, f3 ]) -> [ f1*f2*f3, f2, f3 ] ], [ f1 ] -> [ Pcgs([ f1, f2, f3 ]) -> [ f1, f1*f2*f3, f3 ] ],
.....
gap> B := SemidirectProduct(A, phi[2], C2);
Error, the families of the element or collection <elm> and Source(<map>) don't match, maybe <elm> is not contained in Source(<map>) or is not a homogeneous list or collection at /usr/share/gap/lib/mapping.gi:148 called from
Image( aut, pcgsG[i] ) at /usr/share/gap/lib/grppcext.gi:1220 called from
SplitExtension( G, aut, N ) at /usr/share/gap/lib/gprdpc.gi:140 called from
<function "SemidirectProduct generic method for pc groups">( <arguments> )
called from read-eval loop at stdin:4
type 'quit;' to quit to outer loop
brk>

nvm

[torpid radish]

$$\begin{array}{rl}
c * a &= c * b \
c^{-1} * (c * a) &= c^{-1} * (c * b) \
(c^{-1} * c) * a &= (c^{-1} * c) * b \
e * a &= e * b \
a &= b \
\end{array}
\quad\quad
\begin{array}{rl}
a * c &= b * c \
(a * c) * c^{-1} &= (b * c) * c^{-1} \
a * (c * c^{-1}) &= b * (c * c^{-1}) \
a * e &= b * e \
a &= b \
\end{array}$$

[cloud walrus]

Abiria

[torpid radish]

is this right?

seems like i need to use associativity axiom as well to prove that every group is cancellable...

[kind temple]

here is the one line proof version:
a = 1 a = (c^-1 c) a = c^-1 (c a) = c^-1 (c b) = (c^-1 c) b = 1 b = b
i like this style of proof if i can use it

[elfin wraith]

It’s a good proof to know for the “basic” proofs like that, the ideas similar to show the identity or inverses are unique or whatever

[knotty badger]

If I have a ring R and an idempotent e, can I form a “subring” with identity element e, consisting of all elements r such that er = r?

Say the ring is commutative for now

[topaz solar]

Its got a working multiplication and addition, an identity, a 0

[knotty badger]

I’m trying to understand in what sense you can decompose rings using idempotents

[topaz solar]

consider eR (+) (1-e)R

or write a times or your favorite choice of symbol

[knotty badger]

Hm I see

[topaz solar]

key bit there is orthogonality, i.e. ef = 0 for idempotent e, f

[knotty badger]

So idempotents allow you to write a ring as a direct sum?

[topaz solar]

its a projection to a subspace basically

just imagine things like orthogonal projections

[knotty badger]

Mhm

[topaz solar]

then check your intuition works

[knotty badger]

I don’t see how this’d work for non-commutative rings though..

[topaz solar]

once again, consider orthogonal projections

what works there?

B(H) is non commutative after all

how about block decompositions e.g.

[knotty badger]

In this case Re wouldn’t be a ring I think..

[topaz solar]

like p1Ap1, p1Ap2, p2Ap1. p2Ap2

[knotty badger]

Oh you mean that

Hm

If I take eRe

[topaz solar]

note that eAe = eA in commutative

[knotty badger]

Right right

[knotty badger]

This should form a subring

[topaz solar]

and e and (1-e) aren't the two pieces, since you have the diagonal elements

[knotty badger]

Well, “subring”

With eee = e as identity

[topaz solar]

but you need ef = fe = 0

[knotty badger]

So is it like

[topaz solar]

I think amplifications are a keyword here?

[knotty badger]

eRe + eR(1-e) + (1-e)Re + (1-e)R(1-e)

[topaz solar]

but this is literally just block decompositions of matrices

[topaz solar]

prove it works

[knotty badger]

So any element can be written as a sum of elements in those

eRe and (1-e)R(1-e) are subrings

Not sure about eR(1-e) though

I can’t see what an identity element would be

Anything in the ring squares to 0 after all

But it might have nonzero elements

[topaz solar]

e(1-e) = 0

so multiplication is always 0

[knotty badger]

Mhm

[topaz solar]

but yes you have nonzero elements at time

[knotty badger]

I don’t think that’s allowed for rings

[topaz solar]

again, literally block decomposition of matrices

[knotty badger]

Yeah

So now what

[topaz solar]

why would you be multiplying the off diagonal rectangles

but if you want a decomp of matrices you need those

so thats what you get

[knotty badger]

So what actually is eRe + eR(1-e) + (1-e)Re + (1-e)R(1-e) ring-theoretically

[topaz solar]

R

[knotty badger]

Does the decomposition mean anything

[topaz solar]

just look up keywords like orthogonal idempotents and such

i think the wiki page for idempotents has the e, 1-e thing even in the commutative case

and keywords around it are easier to look for

but its just block decomp of matrices

you have projections down so here you are

[knotty badger]

Hm so

It talks about central idempotents

[topaz solar]

(I dunno, you can consider rings as nice functions on a space, interpret it in that light if you like that pov)

[topaz solar]

in which case the off diagonal disappears

it tells you about when you can fall apart into a product

so those idempotents are right where it happens

and general idempotents are where you get a product up to slapping on some rng bits

[rocky cloak]

The ring eRe is called an idempotent subring, and has some pretty nice properties.

You get a functor Mod R -> Mod eRe given by M |-> eM
and this functor has adjoints on both sides and has as kernel the embedding of Mod R/ReR, which also has adjoints forming a recollement

https://ncatlab.org/nlab/show/recollement

[knotty badger]

oh dear

[elfin wraith]

The appropriate response to being sent an nlab article lol

[knotty badger]

No usually I like them

This one is too algebraic for my tastes

Since I’m still not really a fan of algebra

[rocky cloak]

I just meant it as a link to the definition

So I didn't have to type it all out

It's many functors that are adjoint to each other is the thing

[knotty badger]

yeah I’m sure algebraists would find that interesting

[rocky cloak]

Yes, this decomposition is a matrix decomposition.

If you think of R as a right R module, then
R = eR (+) (1-e)R

Now R = End(R) can be written as a matrix ring
[ End(eR), Hom((1-e)R ]
[ Hom(eR, (1-e)R) , eR), End((1-e)R) ]

[ eRe, eR(1-e) ]
[ (1-e)Re, (1-e)R(1-e) ]

It's not really interesting just useful... Not sure why you're being so weirdly dismissal, but you don't have to learn about them if you don't want to.

[knotty badger]

Just got reminded how much I dislike algebra recently is all

[elfin wraith]

I think I may have done an exercise along these lines before actually

I’m very much looking forward to next semester and having some actual algebra to do, Lie algebras should be fun

[sacred wharf]

lie algebras when they meet truth algebras

[elfin wraith]

Boolean moment

[lyric lagoon]

What you've written is also an inductive defintion but the one given in the image also fits the framework for an inductive definition I was given in my book

The general inductive framework is given by a triple (U, B, F).
In the subgroup case, it seems we can take:
U = G
B = X
F = { e (0-ary), inverse (1-ary), o (2-ary) }
Then the inductively generated set is:
C(B, F) = [X]_G
In this setup:
The first condition of the inductive framework (B is contained in C) corresponds to condition 2 of the subgroup definition (X is contained in [X]_G).

The second condition (C is closed under all operations in F) corresponds to subgroup conditions 1, 3, and 4:
• condition 1 holds because the 0-ary operation e guarantees e is in [X]_G
• condition 3 holds because of the unary inverse operation
• condition 4 holds because of the binary operation o

[noble nexus]

This seems highly convoluted unless you're doing like model theory

or something similar

[elfin wraith]

I think im just being really dumb here, but why is cd=0?

I assume this is like immediate and I’m just being a fool

(The footnote just says “don’t expand c and d”)

[rapid cave]

"since {1,\alpha} is a basis for K(alpha)/K(beta)"

[quiet pelican]

The lhs has no alpha component

It’s a - β

Not a fan of using a and α simultaneously

[elfin wraith]

Ahh yeah obviously, I knew it was something silly lol

[elfin wraith]

Yeah I think this is why I was getting confused

[rapid cave]

too much symbols to keep track of

[quiet pelican]

It’s more that, at a glance, it’s very easy to confuse $\alpha, a$

[cloud walrus]

micoi the group things (8/1/6)

[knotty badger]

-# cdeeznuts

[copper kestrel]

am i overthinking this, so i have that an element is of order 6, and then im supposed to consider the cyclic group generated by that element of order 6, the cyclic group generated by that group also has order 6 right?

like it seems to me yes but i want to make sure im not going crazy

[karmic moat]

wdym cyclic group generated by that group?

[copper kestrel]

sorry "by that element"

[karmic moat]

oh

[copper kestrel]

i mistyped

[karmic moat]

yea it'll be order 6

[copper kestrel]

SICK

[karmic moat]

i think that's one of the (equivalent) definitions of order of an element actually

[copper kestrel]

that would make sense

so then the index of this would be |S_6|/|\mu| = 120?

since S_6 is finite

[karmic moat]

yeah, by lagrange's theorem

[copper kestrel]

HECK yeah

[tardy hedge]

Hell yeaaaaaaaaaaaaaaaaaaa

[tardy hedge]

why does torsion free module over a PID imply flat module

[topaz solar]

Block matrix decomposition strikes once again, I am vindicated

[topaz solar]

It’s talking about formulas so

[south patrol]

Here's how I'd do it: firstly, note that for the finitely generated case, torsionfree modules are even free (in particular flat). Now any module is a filtered colimit of its finitely generated submodules, and hence your module is a filtered colimit of flat modules. You can now check that this implies it is still flat

I am sure there are other ways of doing this but this is a like "structural" way to do it I guess

You don't quite need PID for your statement (it is enough for each fg ideal to principal), which I guess is probably enough to prove that bit about torsionfree + fg => flat

Ok sure in general this statement holding for arbitrary modules is a characterisation of Prüfer domains

[copper kestrel]

so true

[tardy hedge]

Oh ok, huh more complicated than i expected

[rocky cloak]

A different proof than potato suggested could be:

Tor(M, R/r) = the r-torsion of M as you can see from the projective resolution
R -r-> R -> R/r

So if M is torsion free Tor(M, N) = 0 for all fg modules N. Tor commutes with direct limits and every module is the direct limit of its fg submodules, so Tor(M, N) = 0.

[south patrol]

Ah okay so similar proof but using the different sides of Tor lol

[south patrol]

(I guess also then the remark that higher tor groups vanish as we are over a pid)

[rocky cloak]

It's not really necessary. First tor group vanishing already implies flat

[south patrol]

It sort of seems the natural way because the answer is "easy" in the fg case and you want to leverage that

[south patrol]

True lol

I was just wondering where you explicitly used PID but I guess it is cause you are using the classification in the penultimate sentence

[rocky cloak]

I'm using that finitely generated = product of R/r thingys

[south patrol]

And to be honest i had in my mind the classification of flat modules as filtered colimits of finite free things (Lazard's theorem)

[rocky cloak]

fg projective no? Or does it also work with free?

[south patrol]

This is how i've always seen it stated

[rocky cloak]

I guess an fg projective is the direct limit of
R^n -e-> R^n -e-> ...
for an idempotent e

[south patrol]

https://stacks.math.columbia.edu/tag/058G

Oh true lol

[rocky cloak]

This theorem is like the one I example I have of filtered colimits being useful instead of just direct limits.

In that M is flat iff the over category
(fg free)/M is filtered

[south patrol]

Ah nice lol

[tardy hedge]

kind of silly question but how can you technically say that (like, in k[a,b,c,d]), b+d not in (a+c)? Can you just argue by degree?

Multidegree, like in Z^4 or something

[south patrol]

Yes by degree works well. Like note that the usual properties still work for multiple variables

You can also reduce to the one variable case by writing this as k[b,c,d][a]. Like you have a constant poly vs a nonconstant

[tardy hedge]

b+d = f*(a+c) then deg f has some negative components which isnt possible i guess

[south patrol]

Ye same argument as 1 var

[tardy hedge]

I only recently started working with graded objects but didnt realize that actually is helpful

[south patrol]

Graded objects r great

[tardy hedge]

But actually does this make sense for Z^4 grading, because b+d wouldnt be homogenous in it so deg(b+d) doesnt make sense?

[south patrol]

You can e.g. lexicographically order to get a Z^4 grading

[tardy hedge]

Whats the degree of b+d in that?

[south patrol]

Uhh maybe I am silly tbh lol

No I mean I think you can just lexicographically order monomials and then pick the biggest monomial

Ig this isn't quite a grading rip

I am silly

[tardy hedge]

This makes sense to me but at first i tried doing something with Z^4

[south patrol]

But ye this ordering is still all you need for your question

[tardy hedge]

Lexicographic thing?

[south patrol]

Ye

[next obsidian]

Going back to the PID flatness thing,

I think the lowest tech proof is as follows.

To show M is flat we only need to show it preserves the injections of ideals into R. This does not require Tor to prove (but it’s a pain in the ass).

Any ideal I is of the form (r) and so any tensor in M (x) (r) is of the form m (x) r, they’re all simple tensors.

The map M (x) (r) -> M (x) R is, after identifying M (x) R with M via multiplication, the map
m (x) r -> mr. If this is zero then m was zero to begin with as M is torsion free so m (x) r is zero.

There’s a lot of things that’s been said so far that’s more general, as other noted this proof (or the one with Tor) will show that if R is a domain such that every finitely generated ideal is principal, torsionfree and flat are the same. Just do a direct limit on the ideal to reduce to finitely generated ideals.

This says in particular that torsionfree modules over valuation rings are flat which is useful I guess

[sacred wharf]

tensor products r cool

thats all i want to say

good night

[next obsidian]

Tru

[sacred wharf]

Formal definition of free abelian group?

[tardy hedge]

Is that first statement something like Baer’s

[next obsidian]

In spirit yes

[velvet hull]

formally it's defined with a universal property

but in the case of finite rank it's just Z^n

[next obsidian]

You can take a few, here’s one.

Isomorphic to Z^kappa for some cardinal kappa

Where this is a direct sum

[sacred wharf]

[next obsidian]

Because it’s a domain (I think) you can just say it has a basis as well

When there’s zero divisors the word “basis” is tough

[sacred wharf]

i only learnt domain in context of rings

[south patrol]

This this is nice, I like this

[next obsidian]

I meant Z is a domain

Because if it wasn’t even the ring R itself has some “torsion”

But actually I think basis is fine even not over a domain

But whatever

[south patrol]

S of cardinality k being a basis for an R-module M is equivalent to saying the corresponding map R^k -> M is an isomorphism regardless of R

[next obsidian]

Yeah

[south patrol]

I worried about this too but it's just the fact you cannot kill 1 in R by any nonzero element ig

[next obsidian]

Yeah

I thought about it between stops

[sacred wharf]

wdym cardinal so k is like any real number as well as like different kinds of infinities

[next obsidian]

Which prompted this

[next obsidian]

Yes

But lik

[south patrol]

Not a real number

[next obsidian]

It’s the size of a set

Not a real number

[sacred wharf]

oh ye i mesn pos ints

[south patrol]

It just means it is of the form R^(S) for some set S

Where this is the S-fold direct sum

[next obsidian]

Except that isn’t right lol

You need finite support

Now it’s right

[south patrol]

Ye me being dumb there dw

[next obsidian]

By redefining what notation means

[south patrol]

Ig your original thing had slightly off notation too lol

[next obsidian]

Yeah

Lol

[south patrol]

Hm

Chmonkey I want to compute like coho of Gm for a projective variety

[next obsidian]

Idk how to do that shit

[south patrol]

Do you know if there is any funny trick for this like with the Kähler differentials etc

Valid

[sacred wharf]

fight fight fight

[south patrol]

The thing I know is like you can compute torsion by using various SES and stuff but eh

Well also I have seen it for curves but lol

[next obsidian]

Idk why do you think this would come in play?

If you have an idea I can think about it

Also do u mean étale coho?

[south patrol]

Ye

[next obsidian]

I don’t know how to compute anything in étale

Lol

[south patrol]

Valid

[next obsidian]

What’s ur variety defined over

[south patrol]

F3 bar lol

Well defined even over F9 but yes

[next obsidian]

Lmfao

[sacred wharf]

étale the type of word to make me suffer 200 hours of studying just so i can say it

[next obsidian]

How deep in the dumpster did you have to dig to get this variety?

[south patrol]

What makes it dumpster lol

[next obsidian]

The number 3

And 9

[south patrol]

Basically it is an interesting variety in a paper which seems to warrant more study

Like provides some counterexamples to conjectures

I think basically the analogues at bigger primes cease to be interesting

[wraith cargo]

isn't H^2 like the brauer group

I feel like for a general projective variety getting a concrete description of it is fucked

[south patrol]

Its torsion at least yes

[south patrol]

Yeah that sounds reasonable lol

[wraith cargo]

https://link.springer.com/article/10.1007/BF02765009

you can get a description but it seems hard

[south patrol]

Thank

Yeah lol

Ask chatgpt to create an algorithm

[wraith cargo]

mfw in a meeting with my advisor and he doesn't know how to prove something so he asks chatGPT

[tardy hedge]

Lol

[sacred wharf]

shouldve used gemini

[foggy cedar]

$\mathbf{Proposition:}$
Let $G$ be a group with $H \leq G$ and suppose $a,b \in G$, then either $aH=bH$ or $aH \cap bH= \emptyset$.

$\mathbf{Proof:}$
Suppose $aH \cap bH \neq \emptyset$.
...
Then, $aH=bH$, which completes the proof. Q.E.D.

[cloud walrus]

Brandon

[foggy cedar]

Question
This proposition require that only one statement is true and cannot be both true.
WLOG, if A is true, then B is false. By contraposition, if B is true, then A is false.
The proof has shown only one direction. Is it true that we should also prove if A is false, then B is true? Then by contraposition, we have if B is false, then A is true.

[tall igloo]

i think they are just omitting the other direction, since it is clear that if aH \cap bH = \emptyset then we cannot have aH = bH

[foggy cedar]

I guess ignoring the other direction makes sense, since its obvious.... I guess I'm just frustrated that the lecture notes didn't point that out.

[tardy hedge]

b+c not in I = (a+d, e+f). can I show this by having a map k[a,b,c,d,e,f]->k[b,c,d,f] where I is in kernel but b+c is not

[rocky cloak]

Sure

[sacred wharf]

does existence of a left R module M guarantee existence of right R module M

?

[quiet pelican]

Do you mean modules on the same underlying abelian group, or?

[sacred wharf]

Suppose we have a left R module M

Does this guarantee right R module M?

[quiet pelican]

There always exists some left and right R module (namely, R)

[sacred wharf]

No I mean like

if we have a certain module

M

it is a left R module

can we also have M as a right R module ?

[delicate orchid]

U can have it as a right R^op-module

[sacred wharf]

nice

[sacred wharf]

in tensor product

wait

what is the lateX for tensor product ?

$\cross$

[cloud walrus]

mq

[sacred wharf]

ok i'll pretend this is correct

$M \cross N$

[cloud walrus]

mq

[sacred wharf]

Suppose we have this

wait no

that doesn't work

hm

[lusty marlin]

\otimes

[delicate orchid]

It’s otimes

[sacred wharf]

$M \otimes N$

[cloud walrus]

mq

[sacred wharf]

ok suppose we have this

M is our left R module

and N is our right R module yea ?

[delicate orchid]

So is this another instance of not asking what you actually want to ask lol

[delicate orchid]

I think one of them needs to be a bimodule

[sacred wharf]

bimodule ?

[sacred wharf]

what if i want to ask both ?

[rocky cloak]

What does it mean to "have M as a right module"?

[sacred wharf]

That M satisfies the axioms of being a right module of R

[delicate orchid]

Can you not just always impose the trivial module structure then

I thought you wanted something more canonical than that

[rocky cloak]

So a right R-module is an abelian group M together with a map MxR -> M.

You haven't mentioned any map, so it doesn't make sense to fulfill the axioms

[sacred wharf]

when you say R^op that just means r_1 * r_2 = r_2 * r_1 right or..

[sacred wharf]

hm yea

[delicate orchid]

Yes

[sacred wharf]

can we define rm = mr

[rocky cloak]

You cannot in general

[sacred wharf]

y

[rocky cloak]

But
rm = mr
defines a right R^op-module structure on M

[sacred wharf]

oh yea that's what i meant

[rocky cloak]

So if R is commutative R=R^op and this would work

[sacred wharf]

why we call it R^op ?

isn't R^op just for elements of R

[fading acorn]

opposite category

[delicate orchid]

Because it’s the opposite ring

[fading acorn]

i.e. arrows reversed

[rocky cloak]

Because it's R with the opposite multiplication

[sacred wharf]

but m isn't necessarily an element in R

[rocky cloak]

(multiplication happens in the opposite order)

[sacred wharf]

i agree here

but i thought R^op is multiplication happens in opposite order only for elements in R

[rocky cloak]

m very much is not in R, it's in M

[delicate orchid]

So?

[rocky cloak]

That is correct yes

[delicate orchid]

It does

[rocky cloak]

Notice
if rm = mr then
(st)m = s(tm) = s(mt) = mts = m(ts)

The multiplication of s and t has been reversed

[sacred wharf]

yes

i see

ok

thanks

[delicate orchid]

pmo(ts)

[sacred wharf]

pmots ?

[chilly ocean]

Is there any interesting stuff or work in chain complexes and homology with groupoids instead of groups?

[delicate orchid]

There is a lot of theory on the cohomology of categories. Not sure about groupoids specifically cause each connected component should be homotopic to a group iirc

Maybe it’s a Kan complex thing actually I take that back

[sly crescent]

stomp backwards

[sacred wharf]

ohh

[delicate orchid]

https://tenor.com/view/biden-joe-biden-microphone-interview-walk-gif-2478024445377708296

[sacred wharf]

i want to show a

can someone check my work ?

if a is finite abelian group

[quiet pelican]

Sure

[sacred wharf]

any element a, there is n such that na = 0

n being in the integers

take some element in Q \otimes A, say q \otimes a

q = q/n * n

q otimes a
n(q/n otimes a)
q/n otimes an)

[quiet pelican]

That isn’t a generic element of Q \otimes A

[sacred wharf]

= 0

[quiet pelican]

Your proof is basically right, but you need more justification here (ie why elements of the form q \otimes a being zero implies all elements are)

[sacred wharf]

a generic element is the sum ig of those elements right

and since each one is 0

0 + 0 + 0 = 0

let's gooo i got the catyes emoji

ok but i'm not done yet

since i have part b

i use associativity property

so we got Q otimes Q/Z

any generic element is a sum of elements of the form
a/b \otimes n/d (n/d notably being in between 0 and 1)
d(a/bd otimes n/d(
a/bd otimes n
= 0

is this good ?

[quiet pelican]

Yes

[sacred wharf]

thank you micoi

[south patrol]

do you means katseye

[sacred wharf]

wym

[south patrol]

Dw lol

[sacred wharf]

what 😭

yo can someone check my work ?

My idea is to use orbit stabilizer theorem

if we take x from each orbit and sum up from the above equation

we must have things adding up to 11

As in

Orb(x_1) + Orb(x_2) ... = 11

where each x are in distinct orbits

Orb(x_1) must be either 1,7,5 by the above

So we must partition 11 with 1s,7s,5s

there is no partition without 1

So Orb(x) = 1

so the stabilizer of x is G

we are done!

[quiet pelican]

Or, in 2 lines:
All orbits are of size 1, 5, 7 by O-S
There’s no way to make 11 without a 1.

[sacred wharf]

ye but i want to justify my work

wait

aren't they just telling you to assume what you want to prove ?

[quiet pelican]

No

[sacred wharf]

ah hah

[quiet pelican]

168 = 7*3*8

[sacred wharf]

i see

i got it because the normal group can be 2 for instance

but the p sylow group would be 8

Alright

this is trivial if the group is order 3 or 7 we are done

so we assume it is of order 2

so i would assume, maybe it's not the path forward, that we want to show p_8 is our sylow normal group

we know the number of p_8 sylow groups must be 1,3,7

if 1 we are done

we know the normal group is contained in the center

[rocky cloak]

I don't know if you want a hint, but ||I would think about how many 7-sylow subgroups there can be||

[sacred wharf]

i think i got Z(P_8) = 2

[sacred wharf]

8

i don't see why that's helpful

[rocky cloak]

8 or 1.

And what would you need for it to be 8?

[sacred wharf]

I'm not sure what you mean

[rocky cloak]

So you know that G acts transitively on the sylow 7-subgroups by conjugation right?

[sacred wharf]

Yes

Second sylow I believe ?

[sacred wharf]

this is just a fancy way of stating every sylow 7 group is conjugate

[rocky cloak]

Right, so you want your sylow 7 subgroup to have 8 conjugates

[sacred wharf]

Correct I don't see what the path is from here

[rocky cloak]

So by orbit stabilizer what's the size of the normalizer of the sylow 7 subgroup?

[sacred wharf]

168/8

21

[rocky cloak]

Indeed. So as 21 is odd this group can't have any element of order 2

But then I ask you. Does it have an element of order 2?

[sacred wharf]

Are you trying to say that if assume a normal group of order 2, it has an element of order 2 and the normal property must put in normalizer

[rocky cloak]

Well, is that true?

[sacred wharf]

i think so

quickly checking

take any element in this group of order 7

suppose n is in N that is order 2

gng^-1 = n
g = ngn^-1

[sacred wharf]

but this property that normal group must. be in normalizer only holds in this very specific case since the orde ris 2 right ?

that is so hyper specific I am so destroyed in the exam, how do you think of like 10 unique ideas

[rocky cloak]

Yeah. The key idea is really that a normal subgroup of order 2 is in the center

[rocky cloak]

I guess maybe it's good to have some familiarity with examples. Like if you cooked up
C7 x C2 x C2 x S3
you would see that the 2-sylow subgroup need not be normal

[sacred wharf]

wdym ?

how

[rocky cloak]

Like that's an example of a group of order 168 with a normal subgroup of order 2 where the 2-sylow subgroup is not normal.

So that tells you you should not try to prove the 2-sylow subgroup is normal

Then once you're onto 7 you just need to exclude there being 8 of them, and in such situations it's always good to think about the normalizer

But yeah it's hard. Just gotta practice a lot of old exams and hope there are some recurring ideas you can remember

[sacred wharf]

i have one day to do exams

[rocky cloak]

Then my tip to you is to get enough sleep

[sacred wharf]

if i'm asking you for help at 1 am, you can guess i probably won't sleep well

anyway for this question, can someone check my work!

For part a) trivial or so i thought nvm
let's show it is a group, ofc it has 0
suppose x-y are in Iso_M(N)
rx-ry
N_1 - N_2 = N_3 done!
now to show it is closed suppose m is in Iso. but rm isn't, so rm isn't in N but we have r_1m is in iso and therefore in N. then simply look at r_1rm
we have commutativity so r_1rm = rr_1m = rN = n_4 so clearly it is closed

b)
suppose an element in that group isn't torsion free then we can just show it has to be in iso_M(N) so then it is zero

for non trivial submodule that is equal to its isolator just choose a submodule over a field

so just standard vector base ig

R^n

where R is Q the field

[rocky cloak]

For x-y note that a priori it might be a different r that takes x and y into N

[sacred wharf]

ah hah so first have to show closed under multiplication in R

is everything else correct ?

[rocky cloak]

As far as I can tell

[tough raven]

Let Q be a finite acyclic quiver and k a field. The "vertexwise dual" contravariant functor D maps Rep(Q) (finite-dimensional representations of Q over k) to Rep(Q^op) by mapping each vector space of a representation to its dual and each map of the representation to its transpose. In terms of A-Mod and A^op-Mod = Mod-A, where A is the path algebra of Q, that D is representable as Hom_A(-, X) for some (A,A)-bimodule X?

[rocky cloak]

Yes, X = DA

The same works for any finite dimensional algebra

You can see this just from Hom-tensor adjunction.

Hom_A( - , DA) = Hom_A( - , Hom_k(A, k)) = Hom_k( A(x)- , k) = D(-)

[tough raven]

Is it true in general that DA is the direct product of injective envelopes of all simple modules (which exist for an Artinian ring IIRC)?

In particular, is this why Hom(-, DA) is the well-behaved functor rather than Hom(-, A) (while Hom(A, -) (which of course is just the identity) is well-behaved rather than Hom(DA, -) as A is the direct sum of all projective covers)?

[rocky cloak]

Yes

And injective envelopes exists for arbitrary rings.

So the nice thing here is really that End(DA) = A

[rocky cloak]

I guess another way to think about this is if F is any left exact contravariant functor left-mod R -> right-mod S. Then F(R) is an R-S-bimodule and F is equal to Hom(-, F(R))

So if you have a duality D it must be off the form Hom(-,DA)

[sacred wharf]

did you pull an allnighter