groups-rings-fields

[coral shale]

<S' u R | R>

ah and then statement 4, is kindof what we wanted for 'simplifying the relations'

math is cool

<A u B> = <A><B>

damn, think itd be nice to write out all the iso thms for free groups

[chilly ocean]

good theorems indeed

[feral timber]

Can someone explain me how to work out $\mathbb{Z}*_\mathbb{Z} \mathbb{Z}$

[cloud walrus]

Tardis

[feral timber]

I need the presentation

[coral shale]

what product is that

semidirect?

[feral timber]

Amalgamated free product

[coral shale]

ah that thing

count me out tho

[feral timber]

I never understood Amalgamated free products

[coral shale]

alg top and van kampen

probably best context to understand them

im like struggling to reget what this means purely within gt

[feral timber]

Lol I needed to work out the presentation of fundamental group of two mobius bands glued at their boundary, and van kampen says this is the fundamental group

I have no clue how to work out a presentation for this

[coral shale]

u might have better luck in #point-set-topology

but someone in here should know anyways

[cloud walrus]

grothendieckfan1 (Ryx)

[feral timber]

This is the pushout diagram I got using van kampen

[coral shale]

i mean in terms of the presentation

this thing will be

(G * H)/N

and u just gotta figure out what N is right

N will be your relations

And G * H is...

the disjoint union of the generating sets right

with the disjoint union of the relations

[feral timber]

Won't the presentation of Z have only trivial relation?

[coral shale]

Im stating all of the above without having clocked what N is here, yet

But <S1 | R1> * <S2 | R2> = <S1 u S2 | R1 u R2> (those unions being disjoint)

right

[crystal turtle]

So what you do, is you take the free product Z * Z, let's say <a,b> and quotient your relations ab^{-1} I think?
Since you want to identify the image of each z in Z (the top left one) under both identity maps

[feral timber]

I guess that is the free product when they are disjoint, yes

[coral shale]

then you figure out what the amalgamation is and stick those in your relations also

[feral timber]

Why do I quotient out the relations by ab^{-1}

Wait

[coral shale]

I think I agree

[coral shale]

maybe its a bit clearer if we label the different Z's involved nvm

I recall this being a good read for myself to understand this https://unapologetic.wordpress.com/2007/03/03/amalgamated-free-products/

[crystal turtle]

Yeah it's clearer when everything doesn't look the same

[coral shale]

well as long as ur aware the top left is subgroup of the things its pointing to in most contexts

[crystal turtle]

But essentially, to take the pushout of $g: A \to G$ and $h: A \to H$, you take first the free product $G * H$, and then you quotient by identifying $g(a) = h(a)$ for all $a \in A$. So your relations will be ${g(a) *h(a)^{-1} : a \in A}$.\
Now here, since $\mathbb{Z}$ is the free group on a single generator, $\mathbb{Z} * \mathbb{Z}$ is the free group on two generators $\langle a,b\rangle$. And when we consider maps from $\mathbb{Z}$, they are uniquely determined by what $1$ is sent to, so we only need to quotient by $g(1) *h(1)^{-1}$. Here, the top map is $h = id$, so $h(1) = a$ [one of the generators], and the left map is $g = id$, so $g(1) = b$ [the other generator]. Hence why we quotient by $ab^{-1}$.

hopefully I did not miss anything

[coral shale]

what do your square brackets mean

[crystal turtle]

nothing

Just hopefully(?) clearer than g(a)h(a)^{-1}

[summer path]

[[]]

[coral shale]

i find the latter better . Anyways, I follow.

[crystal turtle]

But maybe not

[cloud walrus]

grothendieckfan1 (Ryx)

[feral timber]

In the diagram, we will consider inclusion from G to G*H, and same from H, right?

[coral shale]

I think G*H is one of the things that can be pushed out?

[crystal turtle]

For the bottom and right maps? Yes, it's g |--> g and h |--> h, regarded as single length words

or like whatever equivalence class they're in I guess

[feral timber]

So I get a diagram like this?

[crystal turtle]

That doesn't commute unless you quotient by ab^{-1}

Left then bottom sends 1 to b, top then right sends 1 to a

[feral timber]

Yeah, for the diagram to commute, I need a and b to be sent to same places

So this?

[crystal turtle]

Well that's the pushout, but it doesn't only commute for the pushout (for example, it clearly commutes taking the bottom right group to be {0}). Just that this is the universal square for which it commutes

[feral timber]

How do I show that this is universal?

Like how do I know the square won't commute for any other group H and any two other homomorphism j_1 and j_2 from Z to H at the bottom right when H is not isomorphic to <a,b|ab^-1>

Can I say that if it does, then I have the square commuting with two different groups at bottom left which are not isomorphic, so since the actual pushout must be unique upto isomorphims, it cannot be isomorphic to both H and this group, hence violating the universal property of pushouts?

Also, does the existence of one group (and associated homomorphism) with which the square commutes show the existence of a pushout?

[feral timber]

Okay, I think here I can say this as Van Kampen theorem proves that the diagram we got is a pushout diagram, so if we got the diagram to commute with another group, that must be isomorphic to the fundamental group required.

Am I right?

[crystal turtle]

I think you're a bit confused on what a pushout is. It's not the pair j_1, j_2 homomorphsims to H at bottom right for which the square commutes, it's a pair j_1,j_2 which satisfies that universal property.
Also, you can have a pair j_1, j_2 which "fills out" a square with prescribed top/left maps, even if there is no pushout of that diagram. This cannot happen in categories like Grp, since Grp has all (small) colimits, but it can happen in other categories

[feral timber]

Commuting of the square is fine ig. But how do I know that the universal property will hold if the square commutes for some H at the bottom right? Like if the square commutes for some other H' with associated homomorphisms j_1, j_2, how do I know that there is a homomorphism from H to H' (and when do I have one)?

Like I say a diagram is a pushout diagram when it commutes and the universal property holds, right?

[crystal turtle]

Show that there is always a unique one that fits in the diagram

[rocky cloak]

The pushout of a morphism along the identity is itself, so the pushout here is Z. But I think you may have set of the up the wrong maps.

[crystal turtle]

that is a very good point that I forgor about

[rocky cloak]

The inclusion of the boundary into a Mobius strip is multiplication by 2 on fundamental groups

[feral timber]

What I did was I took M_1 and M_2 to be the mobius bands and intersection to be S_1. All the fundamental groups are Z. But yes! The induced homomorphism is not identity, right?

[rocky cloak]

Yeah, it should be multiplication by 2 I think

[feral timber]

So it's a going to a^2 and b going to b^2, hence we need to quotient by a^2=b^2, right?

[rocky cloak]

Yeah

[feral timber]

Okay, now I need to see why this is true. Gimme a minute

[crystal turtle]

Anyways, to finish this, if you want to show that this process generally describes the pushout, then for any other commutative square with B <-- A --> C on the top left (let's say the bottom right is D),

1. You need to construct a map from the bottom right B *_A C --> D. The fact that you have a commutative square should imply that there is a canonical choice for what you map an equivalence class to (or in other words, commutativity means it interacts well with your relations)
2. You need to show that it is the only square which fits in the diagram. This should just be that imB U imC generates B *_A C I guess

[cloud walrus]

grothendieckfan1 (Ryx)

[crystal turtle]

This being the diagram

[feral timber]

Hmm, I understand a bit now

Would I have needed some nice exposure to commutative algebra before venturing on to algebraic Topology?

[crystal turtle]

No

[feral timber]

As you can see, not good at algebra, hence I am understandably struggling with alg topo.

What do you think the adequate course for an undergrad course in alg topo should be?

[rocky cloak]

You may want to ask in #point-set-topology , but something like homotopy groups, singular homology, CW complexes would make sense

[feral timber]

Ah, right

Anyways, thanks everyone for helping a lot and sticking with me for a long time!

[rocky cloak]

I don't think you'll see much commutative algebra in AT, it's mostly homological algebra

[crystal turtle]

I am always happy to procrastinate doing my own work

[sonic coral]

how can i argue by orders that S4 / H is isomorphic to S3, where H is the normal subgroup containing the disjoint two cycles

[elfin prairie]

hey guys

I was watching those harvard lectures and it introduced smth called an automorphism

which the prof said is the same as bijection

I am kind of confused

[crystal turtle]

Automorphism = isomorphism from a group to itself. So for example, the map n |--> -n is an automorphism of (Z, +)

[elfin prairie]

okay let's re-trace our steps a bit

what is informally a "insert prefix"morphism

I am at the point where I have seen examples of groups, subgroups (of S_n and GL_n(R)) and things like that

and I know what an order of an element of the group is

[topaz solar]

[prefix] is usually an adjective describing the type of morphism it is

Automorphism is an isomorphism from an object to itself

[coral shale]

do you know what a homomorphism is...

[topaz solar]

“Auto-“ meaning self

[warm badge]

uwu

[elfin prairie]

um vaguely

we haven't gotten to that point yet

[coral shale]

🤨

[elfin prairie]

just 2 lectures in

[topaz solar]

Ok then automorphism is irrelevant to you for now

[coral shale]

well, theres no point considering anything else before knowing what a homomorphism is

[topaz solar]

gg

[coral shale]

and no, its not something hard, but you need to know the formal definition

one of the pins here explains the idea... interestingly

[elfin prairie]

oh ill look into it

can you link to it?

[crystal turtle]

Do you agree that S3 is the only nonabelian group of order 6?

[sonic coral]

i know i should argue that there are no events of order 6 so it must be isomorphic to S3

yeah the other one is cyclic

[warm badge]

ryx, can i ask you a question, it's not related to this really

[crystal turtle]

Uhh sure

[warm badge]

what are the exact prereq's for hartshorne

lmao

say artin alg/munkres top/atiyah macdonald?

[south patrol]

willingness to suffer

[crystal turtle]

i have not started reading it But what I will say is that I took a look at it this summer and thought "damn I don't know enough comm alg"

[warm badge]

i have that in spades

[summer path]

a lot of comm alg

[crystal turtle]

But from what I hear, Atiyah-Macdonald should be enough? And then just review stuff you don't know as it comes up?

[topaz solar]

[crystal turtle]

Exactly. So you want to show 1) that S4 / H has order 6, and 2) that it's not cyclic, or it's not abelian. So for 1), that just amounts to showing that H has order 4, and for 2) uhh trial and error?

[warm badge]

i’m looking to get into algebraic geometry, and i haven’t found the explicit prerequisites, i read somewhere that differential geometry was required but i’m not sure ab that

and i saw you talking about it yesterday

[sonic coral]

i have two of the cosets, is it enough to just take two elements out of each and show those don’t give anything in either one of the cosets

[crystal turtle]

Yeah that's next on my reading list; some combination of AG books but I think I'll try Hartshorne first

Well you'd want to pick two elements from distinct cosets, say a and b, and check that ab and ba belong to different cosets

Since that would mean multiplication S4/H is not commutative

[sonic coral]

that’s kinda what i was thinking. i’ve shown that like (124)(24 = (14) and (24)(124) = (12) and (14) and (12) are not in either of the two cosets i have computed

so that would be sufficient? or would i need to do all the other pair wise stuff

[rocky cloak]

Differential geometry motivates a lot of stuff in algebraic geometry, but I don't think you really need anything from it.

AG uses a lot of commutative algebra and homological algebra, but I'm not sure you really need to know it before hand, as you can pick up stuff as you go. You probably want to know the basics of rings and modules though.

Could also ask in #algebraic-geometry

[topaz solar]

If those are in distinct cosets, that says S4/H is noncommutative

That’s a big problem with being cyclic

[warm badge]

will do

done

[open sluice]

oh hartshorne
I only know him for his axiomatic geometry

[crystal turtle]

[sonic coral]

isn’t this sufficient

since the two groups of order 6 are either abelian or not abelian

[topaz solar]

If you already know there’s only 2 order 6 groups yep

[sonic coral]

gotcha thanks

[topaz solar]

It might not feel like enough but it is

[crystal turtle]

More generally, you only really need that S_3 is the only nonabelian group of order 6

[sonic coral]

yeah i figured i’d need to do the whole cosets multiplication

yeah we know there are only two groups of order 6

and that one is Z/6Z and the other is S3

[topaz solar]

So gg

[south patrol]

What is the coolest proof that that is the case lol

I remember trying to learn funne proofs for smol groups

[crystal turtle]

Well fundamental theorem of finitely generate abelian groups gives you that Z/6 is the only abelian group of that order (but is like taking a sledgehammer to the problem)

Uhh something something playing around with order of generators? Not very cool tho

[sonic coral]

he recommended an order argument for this but i thought it would be easier to show abelian vs no abelian

[crystal turtle]

You need the order argument to also know that the quotient has order 6.

[sonic coral]

well we have been given an H

[rocky cloak]

Not a very cool proof, but: if G has order 6, then by sylow theory it has Z/3 as a normal subgroup and Z/2 as a subgroup. Since Z/2 can't intersect Z/3, G is a semidirect product of Z/3 and Z/2.

So the number of such groups is the number of homomorphisms Z/2 -> Aut(Z/3)

[south patrol]

Cool

[south patrol]

Hm, I was wondering - let V be a module over a commutative ring R, S an R-algebra and q: V -> R a quadratic form. Does q always extend to a quadratic form on M (x)_R S sensibly? If, say, R and S are fields of characteristic not 2, then this follows easily by working with the associated bilinear form instead

One would have to define it by like s (x) v |-> s^2 q(v) but you can't extend by linearity because it isn't linear lol

[chilly ocean]

could anyone explain what this question means?

[next obsidian]

@chilly ocean it says if you have {v1,…,vk} linearly independent you can find w_j so that {v_1,…,v_k,w_k+1,…,w_n} is linearly independent

[chilly ocean]

hmmm

so v1,...vk is the linearly independent subset of Rn?

[next obsidian]

Yes

[chilly ocean]

how should I start this problem? I feel like I still don't fully understand what I'm supposed to do

I guess more specificall

I don't understand how these proceeding questions are in the same context

[terse crystal]

Hints:

1. omitted
2. consider S^-1 preserves direct sum
3. S^-1R otimes M is isomorphic to S^-1M

[chilly ocean]

oh, I've done 1-3 already

I'm trying to do 4

I don't get how 4 is related to 1-3

[terse crystal]

Can’t see 4

[chilly ocean]

this one

[terse crystal]

S^-1R is a field where S=R-{0}

So M is this module generated by k linear independent elements, consider S^-1M ( a vector space of dimension k)

S^-1M contained in S^-1N thing

[chilly ocean]

What is N?

[terse crystal]

That’s what you need to find out

S^-1N=F^n where F=S^-1R

Hint:

S^-1M is isomorphic to F^k= direct sum of Fei for 1<=i<=k, which is contained in F^n=direct sum of Fei for 1<=i<=n (for some e_j: k+1<=j<=n)
||there exists xi from M si from S such that xi/si=ei||
Take ||N=ΣRxi|| and prove those ||xi|| are linear independent

[chilly ocean]

sorry some of those notation is confusing me

[terse crystal]

My bad, just R^n

[chilly ocean]

What is the direct sum of Fei

[terse crystal]

Look M is a submodule of N=R^n right

[chilly ocean]

yea

[terse crystal]

So S^-1M is a dimension k subspace of S^-1N=F^n

[chilly ocean]

can you explain why?

[terse crystal]

So a basis { e_1 , … , e_k } of S^-1M can be expanded to a basis { e_1 , … e_n } of S^-1N

[chilly ocean]

can you explain why this statement is true?

[terse crystal]

? M->N injective

Taking S^-1 is exact, (or in other words, S^-1R is flat)

S^-1M -> S^-1N is still injective

[chilly ocean]

what does flat mean ahah

sorry, I feel like I shouldn't be in this class 😔

[terse crystal]

This property you can leave to someday in the future. Which is

Short exact sequence of modules 0->M’->M->M”->0 over a commutative ring R, S is a multiplication closed subset of R, then 0->S^-1M’->S^-1M->S^-1M”->0 is still exact

This property must be contained in your textbook, you can skip this step and find proof of it in your textbook later

[chilly ocean]

ah I think that's on our next weeks homework actually

[terse crystal]

Okay so for now it’s just S^-1M -> S^-1N being injective we need here

[terse crystal]

So this

[chilly ocean]

right

[chilly ocean]

wait this is still just confusing me, what does S^N mean?

[terse crystal]

Sorry typo, S^-1N

[chilly ocean]

oooh

ok that makes more sense haha

ok so you want to show that the basis can be expanded to the bigger basis

[terse crystal]

Yes.
M=Σ R x_i for linear independent elements x_i
e_i = x_i /1 form a basis of S^-1M, which can be extended to a basis { e_1 , … , e_n } of S^-1N
||You can find some x_j / s = e_j for x_j in N, s in S, k+1<=j<=n||
I think you can see now
how x_i 1<=i<=k is extended to R-linear independent elements ||x_i 1<=i<=n||

Jagr is the best person here constructing counterexamples. If you ask him he will provide you a counterexample for when R is not integral faster

[chilly ocean]

ugh, I'm still kinda confused about this construction

I've been trying to wrap my head around it

so

M and N are both submodules right of R right?

[terse crystal]

No M is a submodule of N, and N is R^n as a R module

[chilly ocean]

ah ok

so M can be formed by a basis of (e_1,..., e_k) and you can use that to form a basis of S^{-1}M

[terse crystal]

No
R-linear independent elements x_i
M=ΣR x_i then

e_i = x_i /1 in S^-1M for a basis of S^-1M

[chilly ocean]

ok bare with me if I'm just saying stupid stuff, I'm still working through the definitions. Would M be considered the span of these x_i then? what is the R used for here

[terse crystal]

Span…

I deleted the counterexample part when R is not integral, I need to think more about it

[chilly ocean]

my mind is still boggled, how does the basis expanding work?

ah man

[terse crystal]

e_i form a basis of S^-1M

[chilly ocean]

how do I get better at algebra, I feel like none of it makes sense to me

[terse crystal]

Which can be extended to a basis e_1 … e_n of S^-1N

[chilly ocean]

somehow managed to get an A in algebra 1 but now I'm dying in algebra 2

[terse crystal]

For some e_j k+1<=j<=n in S^-1N

[chilly ocean]

hmmm

ok

[ivory trail]

i personally did commutative algebra with algebraic geometry visualizations and intuition

[chilly ocean]

did that help in understanding it?

I feel like I just don't have a good grasp on algebra in general rn

[ivory trail]

yeah like going up and going down for instance

and localization and quotients

[terse crystal]

I don’t think anyone can walk in that higher path

[chilly ocean]

like algebra 1 I was able to just fraud myself through by memorizing definitions, but now I feel like I need to apply a lot of those definitions and I just can't

[ivory trail]

it is not easy if you don't know any AG

[chilly ocean]

maybe it was just a bad algebra 1 professor

[terse crystal]

Always focus on yourself

[chilly ocean]

yea

[terse crystal]

Improve oneself , find no excuse

[chilly ocean]

that's valid

I definitely wasn't a good algebra student

[ivory trail]

it's important to build good intuitions for everything

it's how you remember stuff

[chilly ocean]

that's true

ok so

You can find some x_j / s = e_j for x_j in N, s in S, k+1<=j<=n

lemme think about this

sorry, I don't see why this matters? Isn't e_j already just being used as basis elements for S^-1N? What does x_j/s have to do with this?

[terse crystal]

by definition e_j = x_j / s_j since e_j are elements of S^-1N

I took common s

So there exist some new x_j in N such that x_j /s= e_j

[chilly ocean]

ah

I see

ok, can you kinda sketch out the proof outline? I feel like I understand these smaller steps but now I don't really understand how they apply to the bigger picture

like just a big picture kinda thing I guess

like I don't get how this extensino so S^-1M and S^-1N apply to what we're doing directly

[terse crystal]

That is not a math question I think

Like how and why we do or don’t do something

A proof is valid if it’s logically true, irrelevant with what we think or feel

[chilly ocean]

no I get that

I don't get how what you're trying to do proves what the question is asking for

is what I'm tryign to say

[terse crystal]

I see

Prove x_i for 1<=i<=n is linear independent

Original x_i 1<=i<=k along with new x_j in N for k+1<=j<=n

(Remember e_i = x_i /1 , e_j = x_j /s form a basis of F^n)

[chilly ocean]

ahhhh, none of it is connecting haha

[terse crystal]

?

[chilly ocean]

so x_i \in M?

[terse crystal]

Yes

[chilly ocean]

e_i is a basis for S^-1M?

[terse crystal]

Yes 1<=i<=k

[chilly ocean]

and e_i = x_i/1

so e_i essentially equals x_i

kinda

[terse crystal]

It’s not rigorous to say that, but okay if you like

[chilly ocean]

kinda just tyring to intuitively understand the idea of the proof

so N = R^n, and it has a basis of e_i from 1 to n

[terse crystal]

I use i for index between 1 and k
j for index between k+1 and n

[chilly ocean]

and S^-1N has basis of x_j/s

where s is the commond denominator? So why can't we again say that e_j = x_j/1?

[terse crystal]

we don’t know that

[chilly ocean]

but how do we know that for S^-1M?

[terse crystal]

Let us go through it again:

[chilly ocean]

ok

[terse crystal]

Along with this agreement

[chilly ocean]

right

[terse crystal]

linear independent elements x_i in R^n

Let M=ΣR x_i, which is a submodule of N=R^n

Let e_i = x_i /1 in S^-1M

They form a basis of S^-1M

S^-1M is a subspace of S^-1N=F^n where F=S^-1R

So there exist … e_j … such that e_i extend to a basis
… e_i … … e_j … of S^-1N

e_j = x_j /s for some x_j in N, s in S

Now … e_i = x_i /1 … … e_j = x_j /s form a basis of S^-1N=F^n

Prove … x_i … … x_j … are R-linear independent

(Σ r_i x_i + Σ r_j x_j =0 in N
Σ r_i /s x_i /1 + Σ r_j /1 x_j /s =0 in S^-1N
r_i /s =0/1= r_j /1 in S^-1R, use the fact that R is integral and s is non-zero)

[chilly ocean]

$\frac{r_i}{s} = \frac{0}{1} = \frac{r_j}{1}$

that?

[cloud walrus]

Pancaker

[terse crystal]

Yes

[chilly ocean]

Why are we trying to show this?
Σ r_i x_i + Σ r_j x_j =0 in N

I thought we were trying to show linear independence?

or is this for contradiction?

or are we showing that this is only true trivially?

when r_i and r_j = 0?

[terse crystal]

Yes

[terse crystal]

This gives you r_i =0 , r_j =0

[chilly ocean]

Σ r_i x_i + Σ r_j x_j =0 in N

Shoudl these be e_i and e_j? I don't get how you turn \sum r_ix_i into \sum r_i/s x_i/1

[terse crystal]

Something =0 in N

Something /s =0/1 in S^-1N

[chilly ocean]

ah fair

[rotund aurora]

Given groups G and H it is not always true that the sequence 0->G->X->H->0 always splits. I'm interested in how bad can this be, ie, how many groups X could satisfy 0->G->X->H->0 ?

[coral spindle]

Well this is fairly hard to work out in general, but there is a very very nice characterisation of these things in the case of Abelian groups (and modules more generally). The answer is: pretty much as many as you like, but you have to think carefully about when two extensions are isomorphic as mere groups, not extensions

The way to measure the number of extensions is the Ext functor.

I don't believe the group extension problem has really been characterised in any nice way in the case of general groups.

[rocky cloak]

Like Boytjie says, if G, X and H are all abelian then such extensions are classified by the Ext functor. But keep in mind that two extensions can be non-isomorphic even if the middle terms are isomorphic.

For example the extensions

3Z/9 -> Z/9 -> Z/3

And

3Z/9 -> Z/9 -2-> Z/3

Are different extensions, so if you're only interested in the middle term then the Ext functor will overcount a bit.

Other types of extensions that have a classification would be the semidirect products are classified by homomorphism H -> Aut(G) (up to isomorphisms)

And when G is abelian, central extensions are classified by group cohomology H^2(H; G).

In general finding group extensions is quite complicated.

[rotund aurora]

thanks

[wet zodiac]

[topaz solar]

Uhhhh show it’s divisible by 2

Then apply one of the many results of a certain French analyst I think

[south patrol]

Who that lol

French?

[solar vessel]

Cauchy

[south patrol]

Oh

Yeah fair lol now I have read it properly that seems more clear

[wet zodiac]

what

[topaz solar]

That one theorem of Cauchy that says something like

p | |G| means you have an element of order p?

[wet zodiac]

yeah

[delicate orchid]

sharp what the fuck are you doing

[wet zodiac]

OH

i see what we're trying to do

if 2 | |G| then we have an element of order 2

and the only way possible is if g^2=e

[delicate orchid]

chat please chat

[rotund aurora]

why would you do that

[delicate orchid]

chat please just fucking put k = -1 in chat

I'm losing my marbles

[wet zodiac]

the question is how do i prove that order of U(n) is divisible by 2 now

[rotund aurora]

EULEREULEREULEREULER

but

bruh you dont need Cauchy's theorem, its overkill imo

or well, now you would have to prove Cauchy

[wet zodiac]

we proved cauchy in class i think

[delicate orchid]

:thecosmoshumswithatunemostsour:

[night onyx]

hint: (n-1)^2 = n^2 - 2n + 1

[rotund aurora]

ok but I was thinking of using CRT, so that it is enough to prove it for n a prime power, and now consider x^((p-1)/2)

[delicate orchid]

hint: MINUS 1

[rotund aurora]

bruh wtf

the cauchy thing distracted me xDD

[wet zodiac]

oh wait

eulers totient function is multiplicative

[rotund aurora]

0

like what is this exercise

[delicate orchid]

units of F_p is C_p-1

p-1 is even unless p = 2

[hushed dock]

ah yea indeed

[topaz solar]

The flames of Babylon

This has been an elite level troll

[delicate orchid]

trolled

[topaz solar]

But also yeah please don’t go this route

[crystal turtle]

oh hello again wew

[topaz solar]

Literally what’s (n-1)^2

[delicate orchid]

I don't understand how this question single handedly assasinated the hippocampus of every chatter under the sun

[crystal turtle]

oh i guess that says sour, not sweet

[delicate orchid]

yeah it's the converse

[topaz solar]

Someone should give wew nitro just for the ascended emotes

[topaz solar]

Im not the dragonslayer for nothing fr

[crystal turtle]

this one did?

bruh

[delicate orchid]

yeah, read the chat lol

[topaz solar]

Ring units don’t exist

-1 isn’t a unit

[crystal turtle]

Oh did Sharp forget about -1

[delicate orchid]

anyway it exists because Z_{2^n}^\times \cong C_2 x C_{2^{n-2}} and Z_{p} \cong C_{p-1} for p odd so apply chinese remainder theorem nerds

[topaz solar]

Wew why do your emotes have those names like

:thefifthgateopens:

[delicate orchid]

because I find it amusing

[crystal turtle]

:thesixthcircleofhellunleashed:

[delicate orchid]

:theeyeofthestormbeckonstheawoken:

[topaz solar]

Is that one real

[delicate orchid]

not yet

[coral spindle]

Welcome back, sorry this is the first thing u had to see

[delicate orchid]

it's ok - it's like the one in a billion occasion where it's not ME making the catastrophic blunder

[coral spindle]

Tbh I liked Sharp's idea, cute general approach

you're VALID sharpie

xoxo

[topaz solar]

Here wew as a treat I have a representation question

[coral spindle]

:O

[delicate orchid]

father in heaven. Hallowed be thy name. Thy kingdom come, thy shall be done on earth as you art in heaven.

[coral spindle]

Are u talking to Brauer or sth

he's not ur dad

probably

[topaz solar]

Smh “Thy will”

[warped fable]

Hi @delicate orchid. How's health?

[coral spindle]

Hurry up Sharp I want to see some reps!!!!!!!!!

[topaz solar]

ok so, I’m tryna look at unitary dual of a locally compact group

tf is containment supposed to look like on the representation side

[coral spindle]

nvm bye

[delicate orchid]

locally compact group
🚪 🚶‍♂️

[coral spindle]

SAME BABES

[topaz solar]

Because I only saw it as “1_G \subset pi” as like

[delicate orchid]

ngl pretty fuckin shit but I'm not anemic and I ate 6 apples today

[topaz solar]

|pi(x) xi - xi| = 0

[warped fable]

Good!

[coral spindle]

It's likely to be that there is a rep isomorphic to 1_G in pi

If those are reps

[delicate orchid]

wait uh

hold on

[coral spindle]

not sure tbh. Usually people write chi_1 | chi_2 for chi_1 being a constituent of chi_2.

[delicate orchid]

there's a thing with characters where the trivial is always a constituent of \phi\overline{\phi}

[warped fable]

I hope you get well soon

[delicate orchid]

I imagine the proof is exactly the same but just with integrals

[topaz solar]

Idk this was all done in terms of like, functions of positive type

[coral spindle]

then sharpie, it truly beats me

wew might have an inkling

[topaz solar]

I have no idea what this should look like in terms of those

[delicate orchid]

ty. I put details in a spoiler in you know where about what actually happened if you care

[delicate orchid]

what the fuck is a positive type

like we talking root systems here?

[topaz solar]

Uhhh

[delicate orchid]

lemme check this - the proof might work for every class function?

I know it works for brauer characters too so

[topaz solar]

$x\mapsto\langle \pi(x) \xi,\xi\rangle$

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

[delicate orchid]

right with something in here positive definitive

?

[solar vessel]

no

[topaz solar]

It’s a bilinear form so yes

[solar vessel]

anything of this form is of positive type

[delicate orchid]

daybroken I'm passing the relay baton to u

[topaz solar]

But it’s not always positive no

[solar vessel]

what was the question

[topaz solar]

None of this was described in terms of modules

[topaz solar]

.

[solar vessel]

unitary dual?

[topaz solar]

Yeah, unitary irreps of your LC group

[solar vessel]

so the dual group

[topaz solar]

It’s not got a group operation afaik?

Since it’s not abelian

[solar vessel]

cringe

ok so containment of what

[wet zodiac]

uh

i shall send you all to the containment of #advanced-algebra

(i think idk)

[topaz solar]

$1_G \subset \pi$ meaning something like, $\Vert \pi(x)\xi -\xi\Vert=0$ for all x, some xi

[cloud walrus]

Dragonslayer Sharp

[solar vessel]

so pi(x)\xi=\xi

[crystal turtle]

spell failed to cast

[topaz solar]

Would I just want like

Some lil subspace where two reps are equal?

[delicate orchid]

it's alright homeslice just chat yo shit irregardless of these nerds

[solar vessel]

wait so

is xi nonzero

[topaz solar]

Yes

[solar vessel]

as in pi contains the trivial representation

[topaz solar]

Yes

This is the definition I have for weak containment though

[solar vessel]

so are you asking like

what this corresponds to when it isn't weak

[topaz solar]

I have no idea how containment should look in this sense ye

Because everything here is done in terms of these things

Do I just want that sum on all of G to be 0 or like

But that’s for every xi, whereas 1 \subset pi seems a bit different?

[solar vessel]

so you want something that looks similar to this but gives actual honest containment?

[topaz solar]

Yep

[solar vessel]

sure

ok well

if pi is contained in rho

then every function positive type associated to pi is also a function of positive type associated to rho

[topaz solar]

I see

Yeah ok that’s not too bad

Dats it for my peabrain intrusion, wew can get back to witnessing the flames upon Babylon and the spheres

[solar vessel]

I should probably say as well

if pi is irreducible then the converse holds

[rotund aurora]

was Z^N={ functions from the naturals to the integers } free as a Z-module?

[dim widget]

No

[rotund aurora]

I asked this almost a year ago

[coral spindle]

It's relatively hard to show this, btw

[rotund aurora]

but Im still intrigued by these questions

[coral spindle]

sigh.... time to go back to Rotman

[delicate orchid]

pretty nifty that it's not free

[coral spindle]

Theorem 4.17 in Rotman's Homological Algebra

[rotund aurora]

dayum its a theorem

[delicate orchid]

lemme think

[dim widget]

The reason it’s true is surprisingly tricky I guess

[dim widget]

So look at the compactly supported functions inside of the regular functions

[coral spindle]

Footnote

[rotund aurora]

idk why I vaguely remembered someone saying something about counting, but that must have been another thing

[rotund aurora]

spoiler 😭

[chilly ocean]

#feet-note

[coral spindle]

Yeah you WOULD say that Terra

you WOULD

[delicate orchid]

S/pS would be uncountable
based?

cool proof

[coral spindle]

Ikr, it's sick

[delicate orchid]

also is Nielsen-Screier a fuckin corollary

[coral spindle]

hurried googling ensues

[delicate orchid]

Cor 4.15

[coral spindle]

Oh, ig so but that's the Corollary 4.15 that Rotman cites

Yea

But it's abeeeeeeeeeeeeelian

[delicate orchid]

Corollary 11.4: The classification of finite simple groups

[coral spindle]

It's a consequence of some hereditary ring stuff

apparently

I don't remember this

[dim widget]

So if it were free you could take a surjection \phi: Z^{\oplus S} \to Z^N and there is some smallest subset K in S such that Z^K maps onto the direct sum inside of your direct product. You can see that K will have to be countable. Then this means that the quotient Z^N/Z^{\oplus N} is the direct sum of a free group Z^{S - K} and a countable group. But now you can show that there is an uncountable amount of divisible elements in the quotient by constructing it explicitly

[delicate orchid]

I think it's free because you have vectors (1,0,...),(0,1,....) so it is free 👍

[coral spindle]

Dear liberals: if Q^N is free, how come Z^N isn't? Curious.

[delicate orchid]

upon the witnessing

ohhh wait a minute does this generalise nicely

replace "p" with a prime ideal perhaps ;3

[wraith cargo]

this directly contradicts my religious views and as such I will ignore it

[delicate orchid]

@coral spindle trolled by discord API

[coral spindle]

I'll call that a success

[rotund aurora]

mmh I think if you refresh it fixes it

[delicate orchid]

... browser user detected?

[rotund aurora]

yeah

[delicate orchid]

I mainly want to know if my idea works out or not

[rotund aurora]

I have google opened almost all the time, so its a window less

[delicate orchid]

but it's a tab more

[rotund aurora]

window sucks more than tab

[coral spindle]

Let us all come together to mock the browser user for no good reason

[rotund aurora]

like what is a nonfield ring R such that R^N is free

[delicate orchid]

zero ring

[elder wave]

browser user

[coral spindle]

interesting question

[elder wave]

is there such a ring

[rotund aurora]

idk

I asked this also a while back

I asked it for arbitrary cardinals

[coral spindle]

I would like to try doing a similar argument to the one Rotman gave but it's too specific

Ig it works for PIDs

and presumably UFDs as well

[elder wave]

hmm

[coral spindle]

Anyway my guess is that there exists such a ring, because the argument showing that they're always non-free seems hard

Oh well actually now that I mention it, isn't R = Z^N an example?

Unital ring too

[elder wave]

[delicate orchid]

well if the ring is less nice then R^N must be less nice. This is rigorous it's called "model theory"

[coral spindle]

I actually love you

Marry me?

But like, in minecraft

[rotund aurora]

Z^K for infinite K contains Z^N which is nonfree, therefore its nonfree

no?

[coral spindle]

No

[rotund aurora]

mmh how did that work

[coral spindle]

Z^N is free over Z^N!

[rotund aurora]

ah

mb

lol yeah this is stupid

wait

no its not nvm

[coral spindle]

And in general, as long as I haven't cocked up, if you choose any cardinal a then R = Z^a should be an example

Now is there a non-field for which R^a is free for any cardinal? Non skew field too ofc.

[delicate orchid]

upon the witnessing

[coral spindle]

Ofc there are still details to consider

[rotund aurora]

I dont see why R=Z^N works honestly

[coral spindle]

I'm suddenly not convinced yeah

[delicate orchid]

wait so are we saying that Z^N^N works?

[coral spindle]

The module structure is fucky

[delicate orchid]

curry it down to get like ZxN -> N maybe?

this doesn't seem free

[coral spindle]

Aaaaaaaaagh does IBN get us?

[rotund aurora]

like I think your (heuristic) reasoning was (Z^N)^N=Z^N as sets, no? but as modules, (Z^N)^N over Z^N is infinite dimensional lmao

[delicate orchid]

ah wait, Z^N-modules

[coral spindle]

Dimension? Wym dimension

[rotund aurora]

maximal independent set

[delicate orchid]

it has dimension because it is free :)

[rotund aurora]

yeah I goofied with the term, but you get it

[coral spindle]

Why should there exist a maximal independent set

anyway

[rotund aurora]

it exists

no?

idk

how did module theory work

[coral spindle]

Why

Anyway, yeah in any case I don't think it works

[rotund aurora]

mmh

I think it should be an app of zorn's lemma

but like in Q over Z, its 1

[coral spindle]

it's easy over domains, but this isn't a domain

I think in any case, invariant basis number fucks us over

That generalises to cardinals right?

So I would expect that no commutative ring would work.

[rotund aurora]

wow

crazy

[delicate orchid]

called it

[coral spindle]

Nvm, I don't know if IBN generalises to cardinals

this is a tough question

[delicate orchid]

Z is inital... I won?

[rotund aurora]

no

[coral spindle]

Might be a good question for mathoverflow, idk

[warm saffron]

Had a question. I proved in a hw problem that if $K$ is a group and $k^2=e$ for al $k\in K$, then $K$ is abelian. Wondering if anyone agrees with my logic on why I think I can invoke it on the problem: Let $G$ be a group and define $H={g\in G:g^2=e}\cup{e}\subseteq G$. Show $H$ is a subgroup of $G$. I know $H$ contains the identity and that every element of $H$ is it's on inverse. For closure, if $a,b\in H$, I think I can invoke that hw problem statement and say that $ab=ba$. That problem was for groups $K$. But I think we can use it here since $H$ has the identity, has inverses, and since it is a subset of $G$ the law of composition is associative. So $H$ is a group.

[cloud walrus]

logician

[coral spindle]

No, you cannot use the homework problem, as part of the assumptions there state that K must be a group, but that is exactly what you need to prove.

[rotund aurora]

@coral spindle Have you heard about slender groups?

[coral spindle]

Instead, you should use the same argument as you used in the proof of the problem. It will work

I've not heard of slender groups, no – what are they?

[rotund aurora]

idk

[chilly ocean]

slenderman group

[coral spindle]

[rotund aurora]

and Z is a slender group apparently

[delicate orchid]

what is this dog water

[warm saffron]

I'm saying H is a group itself I just need to show it's closed.

[delicate orchid]

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

[coral spindle]

This definition is fucked up

[rotund aurora]

Its late now, so Im not reading new definitions

[warm saffron]

I proved it has the identity and inverses and I'm given that the law of composition is associativ soo it seems like I can invoke the theorem as H=K

[delicate orchid]

ah so that's why we care

[coral spindle]

Yes, and you cannot assume it is closed in order to show it's closed. That is circular logic.

[rotund aurora]

like this doesn't seem at all right. Here Z_i=Z

[delicate orchid]

hmm

[coral spindle]

The homework result as stated requires closure to be invoked, so it is not valid to use here. I am repeating myself.

[warm saffron]

I see what you mean. I forgot the implicit assumption about groups being closed in the first place.

Thanks

[south patrol]

Slender group

Lol

[delicate orchid]

Hom_{ab}(Z,A) = Hom_{ab}({1}, U(A))

what happens

hmmmmmmm

[coral spindle]

I'm stumbling on this but yeah, Hom does funny things with products

[delicate orchid]

damn they're the wrong way around

[coral spindle]

Yeah they are the wrong way around

I don't know if there's a nice description of Hom(Prod_i X_i, Y)

But this is the right set of isomorphisms

[delicate orchid]

that's pretty insane

wait

or is it

[coral spindle]

It's nice, imo it makes sense like

[delicate orchid]

is it a preserves limits/colimits thing (or sends colimits to limits)

[coral spindle]

Idk maybe

I suppose so

[delicate orchid]

hmmm

[rotund aurora]

yeah this does make sense

[safe meteor]

why does only a map to x+<p(x)>

?

[rotund aurora]

is this from your notes? 👀

[safe meteor]

couldnt all zeros of p(x) map to it? (so how is it a isomorphism or are we assumign p(x) has just one zero?) i think i got it nvm since we only include the zero a in F(a)?

[coral spindle]

The author really wanted you to notice the text so they put it in bold and also they put it in italics and also they put it in a blue box

[coral spindle]

This is from my master's thesis

[rotund aurora]

[coral spindle]

It was mostly a lit review

[rotund aurora]

can i red it

[coral spindle]

I don't want to dox myself here yet

plus it's not really worth reading. Just read rotman tbh. My only major improvement was a handful of better proofs from elsewhere and nicer typesetting lol

[delicate orchid]

u mfs will pull my masters thesis from my cold dead hands

[delicate orchid]

what's the full name of this so called "r"otman

[coral spindle]

Joseph J Rotman I think

Rotman, Introduction to Homological Algebra

[rotund aurora]

Muhammad Rotman

[coral spindle]

^ statistically correct

[barren sierra]

Ye

[coral spindle]

It's a pretty good book. There are some annoying typos early on.

[rotund aurora]

I spotted another one today

[summer path]

we just need to get wew drunk again, then convince him that we just want a peek of his master's thesis /j

[delicate orchid]

and it's garbage!

[coral spindle]

Lol yeah that is an annoying feature, but most of those exercises are quite reasonable I assure you

[warm saffron]

I'm starting to think that this statement isn't true: If $G$ is a group and $H={g\in G:g^2=e}\cup{e}$, then $H$ is a subgroup of $G$. What I have so far for my proof is: Since $e\in H$, we must check closure and existence of inverses. Given $h\in H$, $hh=e$, so $h^{-1}=h\in H$. Now let $a,b\in H$. We must show $(ab)(ab)=e$. I can't figure out how to do that....

[coral spindle]

It's not a reference, it's an intro!

[cloud walrus]

logician

[delicate orchid]

are you even able to comprehend how incompetent i am

[coral spindle]

Wew you're awesome

[crystal turtle]

e^2 = e so there's no purpose to U {e}

And it's not true unless it's abelian

[warm saffron]

my prof technically wrote this statement in words for H he wrote that H is the subset containing elements of G with order 2...yea my set notation is redundant

[coral spindle]

Then I suppose you should try working on a counterexample!

[warm saffron]

working on it lolll

[delicate orchid]

I thought of one!!! pick me!!!

[coral spindle]

Hmmm can someone ELSE answer a question? You at the back?

[delicate orchid]

simmer down in the back

[warm saffron]

I realize all I need to come up with is a set that isn't closed but is a subset of a group and contains inverses and the identity

and that set must contain elements with order 2 and e

hmmmm

[delicate orchid]

@coral spindle we need words

[coral spindle]

It's an introoooooooooooo

[delicate orchid]

this is an algebra book. do NOT give me a line integral on page 1 I WILL skip the entire chapter

[coral spindle]

He doesn't do any actual analytic stuff, it's just to exposit homology

Ugh just noticed a typo in my thesis

[delicate orchid]

my thesis is nothing but typos

my entire field of study is held up by the fact I wrote \chi' instead of \chi 13 months ago

[coral spindle]

Clearly I did a dodgy search-and-replace somewhere

[rotund aurora]

isnt sheafs and sheaf cohomology very close to anal?

he does talk about sheafs frequently and states the Riemann-Roch theorem, idk if he proves it

[coral spindle]

Idk man I don't know shit about sheaves

maybe in a few years I'll do character sheaves and Deligne–Lusztig stuff idk

[rotund aurora]

I will probably read the sheaf parts, once I get the algebra part down, specially because of Riemann-Roch

[warm saffron]

I got it I'm thinking of D_3. And then the set of all elements of D_3 that has order 2 along with the identity isn't closed.

[delicate orchid]

yeah

(12)(23) = (123) is the example I had in mind

[south patrol]

D_3, symmetries of the 1.5-gon

:)

[warm saffron]

LMFAO

[warm saffron]

this made me laugh so hard thank u for that ahahahaha

[south patrol]

Lol I am glad

because there was a high probability somebody would just say I should allow for other conventions and get annoyed at me

[barren sierra]

There's a joke paper in here somewhere I think

[south patrol]

Yes, by me, to appear in AMS next month.

[warm saffron]

lol lemme add it to my cart real quick

[delicate orchid]

this is TOTALLY unreasonable... I may cry...

[south patrol]

No

[crystal turtle]

Literally impossible

[elder wave]

Rotman is a great expositor

his exercises tend to be a bit too simple though

[crystal turtle]

his exercis

[devout radish]

Why are the integers not a complete ordered field.

This should be obvious but I was just thinking it’s complete and ordered

[chilly ocean]

well, uh, they're not a field

[devout radish]

Ohh

Lmao

Yah right

[south patrol]

They are complete though I suppose lol

[tender wharf]

Ordered too ig

[hollow mica]

complete ordered ring

[warm saffron]

Had a random conjecture....I know that Lagranges theorem guarantees that if G is a group with order n and if H is a subgroup of G, then |H| divides |G|.... I was wondering if the set {|H|:H is a subgroup of G}={m : m divides |G|, m>0}.

I know {|H|:H is a subgroup of G} is a subset of {m : m divides |G|, m>0} from Lagranges theorem, but wondering if {m : m divides |G|, m>0} is a subset of {|H|:H is a subgroup of G}

if not true in the general case wondering which groups specifically satisfy this

[south patrol]

No, indeed A4 has no subgroup of order 6

You can find stuff on the converse to Lagrange's theorem like https://bearworks.missouristate.edu/cgi/viewcontent.cgi?article=4484&context=theses#:~:text=The converse to Lagrange's theorem is that for a finite,≤ G of order d.

[warm saffron]

ah okay thanks. Oh yea I suppose I was asking about the converse

[open sluice]

central limit theorem group

[rocky cloak]

Any Artinian ring works.

In fact this is true iff R is left perfect and every finitely generated right ideal is finitely presented.

[rocky cloak]

So in general modules over perfect rings satisfies flat iff projective, and the right ideal condition is just ensure that product of flat modules is flat.

Also I guess you asked about free not just projective, but for a perfect ring a projective is determined by its top, and R^N / J = (R/J)^N (at least if R is artinian (?)) so R^N is free.

[delicate bloom]

if you want to send it over I'll try working out a sed or awk command to fix it for you

[terse crystal]

Where can I find a reference of this?

[rocky cloak]

https://www.ams.org/journals/tran/1960-097-03/S0002-9947-1960-0120260-3/S0002-9947-1960-0120260-3.pdf

[terse crystal]

Thank you☺️

[rocky cloak]

(its thereom 3.3)

[terse crystal]

Oh I see.

[terse crystal]

Wait is finitely related and finitely presented the same thing there?

[rocky cloak]

Yeah, I guess it's just an oldtimy word

They define it on page 459

[terse crystal]

This is part of a book?

[terse crystal]

I see, thank you

[rocky cloak]

Part of a journal issue, I think

[terse crystal]

I see

[ivory trail]

isn't finitely related "exists epimorphism R^I -> M with f.g. kernel"

and then finitely presented is finitely generated and related

[rocky cloak]

Maybe it usually means that, but not in the linked article

[ivory trail]

oh i see, weird

[terse crystal]

I see, indeed. Finitely related there is definition of finitely presented on wiki

[rocky cloak]

I've never seen the term finitely related used outside this article anyway, so wouldn't know what it "usually" means

But yeah here it means finitely presented

[chilly ocean]

can someone help with 2 and 3? I think I did 1 already

[glossy crag]

I've been struck by a fit of stupidity: if G is a group, X is a subset, and g is an element of G s.t. gXg^{-1}\subset X, why does it follow that gXg^{-1}=X?

[void cosmos]

do both directly

2 is already given for you out

and for 3 define a map to the direct product M* x N

and by universla property u get a map out of the tensor product

construct an inverse and ur done

[void cosmos]

gXg^-1 is the set of all elements of the form gxg^-1 for x in X correct?

this being a subset of X would imply that all elements of this form are elements of X

[glossy crag]

Why would that imply this, that's the entire question

[void cosmos]

what does

the statement " every element in gXg^-1 is an element of X " mean

[glossy crag]

It means that for every x there is a y with gxg'=y, my point is how does the statement with reversed quantifiers follow (for every y there is an x with gxg'=y). You're confusing what I'm asking with the equivalent definitions for normal subgroups (the fact that gNg'<=N for all g iff gNg'=N for all g).

[void cosmos]

yes

u are right

mb

are there no anymore conditions

on X

[glossy crag]

Nope.

[void cosmos]

if X is closed then this is a subgroup

and ur done

right?

yea idk either

so maybe both of us got struck by a fit of stupidity

what context did u get this

[glossy crag]

Maybe.

[glossy crag]

I was trying to recall the definition of a normaliser of a subset and I couldn't remember if you have to require that gXg'<=X or gXg'=X. I thought the two conditions must be equivalent, but I don't see how to show the normaliser is closed under inversion in the first case.

[void cosmos]

huh

[rocky cloak]

It's not true

[void cosmos]

i think u require gXg' = X

iirc

yea you do

N_G(S) = {g | gSg^-1 = S}

[glossy crag]

Yeah, I figured.

[void cosmos]

yeah its because the normalizer is just the stabilizer of the conjugacy action

of a certain set

[glossy crag]

True, but for whatever reason I had the recollection that these conditions are equivalent. Probably same false logic as you had.

[void cosmos]

yea

everythings clear now should be

so we are not the stupidest after all

(optional)

[rocky cloak]

If X is finite it's true

[glossy crag]

Sure. That's another reason why I had this false recollection.

[feral timber]

If I have a group with presentation <a,b|a^2=b^2>, how do I find a subgroup of this which is isomorphic to Z×Z?

[rocky cloak]

One place to start is that if you want x and y to commute, then you have to involve this relation a^2 = b^2 somehow.

So x and y should be such that a^2 or b^2 apears in xy somehow.

What would be the easiest way to make that happen?

[feral timber]

Last letter of x and first letter of y being same (x and y being words)?

Like x being something like x=a^m b^n a and y being something like y=a^pb^q?

[rocky cloak]

Something like that yeah

Even the simplest possible choice should work here

(i.e. ||x = ab, y = ba||)

[feral timber]

Okay, so I can say that a subgroup generated by ||<ab,ba>|| is a subgroup that commutes?

Just because generators commute?

Okay, so when I consider this subgroup, I define the map f from Z×Z to this subgroup as ||(m,n)->(ab)^m(ba)^n||. This is clearly injective (because I can write every element of subgroup as ||(ab)^m(ba)^n for some m,n, ince it is commutative)||, and it is also clearly surjective. Then ||(m_1,n_1)+(m_2,n_2)->(ab)^m_1(ba)^n_1(ab)^m_2(ba)^n_2=(ab)^m_1+m_2(ba)^n_1+n_2=f(m_1+m_2,n_1,n_2)||, hence it is a homomorphism, thus f is an isomorphism

Would this be correct?

But ||(ab)^m(ba)^n|| is not a reduced word. Do I need to consider only reduced words?

[delicate orchid]

The word “clearly” is very dangerous

abbaba = ba so (1,2) and (0,1) map to the same element no?

I would just show that <x, y> satisfies the same realtors as the ones in the presentation for ZxZ

[delicate orchid]

Never mind I misread the realtors in the original group!

I think your thing works now

[slate forge]

Trivial.

[void cosmos]

literally my whole exam last wednesday is full of "clearly" "obvious"

"it is clear"

lmfao

bro thinks he is him

[rocky cloak]

I think it should work yeah, but whether it "clearly" works is another question

[terse crystal]

I didn’t understand the answer. I thought <ab, ba> is simply <ab> which is Z, not Z times Z…

Nvm I misread the relators in G too…

[delicate orchid]

the phantom "=1" claims another victim

[primal tusk]

Anyone know how to show that for a polynomial with degree d>0 there exists an irreducible polynomial of at most degree d/2 that divides the polynomial?

Feel like this should be super easy but im not seeing how to prove it lol

[rocky cloak]

I mean the polynomial could be irreducible

[primal tusk]

i think you can assume its not

[rocky cloak]

So say f = gh, what can you say about the degree of g and h

[primal tusk]

i see

this reminds me of a similar thing with prime factorization

[coral shale]

it likely wants the polynomial reducible and to not consider constant polynomials when dividing

[primal tusk]

but cant degree g be d-1 and degree h = 1 ?

[rocky cloak]

Exactly and 1 is less than d/2

[primal tusk]

but what about d-1?

[rocky cloak]

What about it?

[primal tusk]

that could be greater than d/2 right?

[rocky cloak]

Indeed it can

[primal tusk]

so is this a false statement?

[rocky cloak]

No, the statement just said it was divisible by an irreducible of degree less than d/2

[primal tusk]

that seems too easy... my freind told me he had this question on a quiz this morning and didnt know how to do it. Maybe he told me the wrong thing

but also maybe its just not supposed to be that bad

I was interpereting the question as "all irreducible factors are less that degree d/2" for some reason

[rocky cloak]

Sound like a reasonable question, like if you want to check if a polynomial is irreducible by brute force, you just have to check its not divisible by any polynomial of degree < d/2

So the statement is at least a little bit useful

But yeah, not that hard

[primal tusk]

ohh yea i see now that is pretty useful

[drowsy oxide]

i was looking at the direct product isomorphism theorem for groups and was looking at a specific step where it is sufficient that every element from one subgroup commutes with every element from the other. many authors assume normality of both of these subgroups among the hypotheses. i tried to find an equivalence between these two properties of the pair of subgroups but couldnt prove the reverse direction. is this true?

[rocky cloak]

Consider

x = mnm^-n^-

If N is normal, then mnm^- is in N, so x is in N. If M is normal then nm^-n^- is in M, so x is in M. Hence if both are normal x is e, and so mn = nm

[drowsy oxide]

that is the direction i was able to prove

im just looking for a counterexample or a proof to the converse

[rocky cloak]

Okay, but the other direction is trivial. If m and n commute, then nmn^- = m

[drowsy oxide]

im trying to get at N is normal in G, not in M

besides, we dont know N is contained in M

[rocky cloak]

So g = mn and h is in N, then

ghg^- = mn h n^- m^- = n h n^-

[rocky cloak]

G is generated by M and N

[drowsy oxide]

oh my god

[drowsy oxide]

i kept thinking you needed normality for that

thank you

[hidden wind]

i was signed up for a course in abstract algebra some two years ago, but instead of trying to learn the material i obsessively generated images on my computer that i don't even remember what is supposed to represent

this one is quite simple

this one less so

i have them tagged as Zm36 and Zm2520, whatever that means

[delicate orchid]

cayley tables for the groups of units of the rings Z/36Z and Z/2520Z

[hidden wind]

[coral spindle]

Zm36 means "Z mod 36" clearly

[hidden wind]

i think, making these i just wanted to see some nice colours

as opposed to the current blue in my nick

[delicate orchid]

you can get rid of it I think

yeah in id:customize

[summer path]

not very ppl

the light blue is a lot nicer to look at than the active blue

[delicate orchid]

yeah

[summer path]

so just type more and your name will be readable again

[delicate orchid]

I'll bully the mods to fix it

[hidden wind]

i am content to be not very ppl

[wooden ember]

couldn't really find this anywhere by looking it up: are there any other compact fields than than finite fields with the discrete topology (or any field with the trivial top)?

[coral shale]

cofinite topology ?

not sure if that ever gets you a topological field if thats necessary

[wooden ember]

yeah im looking for topological fields here

[coral shale]

https://math.stackexchange.com/questions/1560871/nontrivial-examples-of-compact-topological-fields

here u go

some progress

[wooden ember]

thanks a bunch

oh yeah i guess compact Hausdorff fields are in particular locally compact so theyve gotta be finite dimensional, so it's easy to home in onto finite fields

[chilly ocean]

So I'm getting the idea of a map where it's like f(phi, y) maps to \phi(m)y, but I don't really get how to apply the universal property, and why the fact that it's finitely generated matters

[void cosmos]

the universal property of the tensor product would give you a map from the tensor product to the Hom

yeah this is showingf that the tensor product is isommoprphic to the Hom

not that the Hom is the tensor product

ty tropo

find a g and then u will get a homomoprhism u

find an inverse to that homomorphism and ur don

done*

[chilly ocean]

to the homomorphism that I had in mind?

does the homomorphism that I suggested work?

[void cosmos]

try it out queen

is it a homomoprhism to the Hom

if so then yes

if not then no

maybe this is where f.g matters

[chilly ocean]

ok, I think I'm getting tripped up about how to in general show something is isomorphic to homomorphisms.

So I'm mapping an element of M*xN to I'm not really sure what, because if I'm taking f(phi,y) to be phi(m)y does that uniquely define a homomorphism?

[void cosmos]

what does phi(m)y mean

this is an element in N no?

[chilly ocean]

yea

[void cosmos]

so this doesnt really make much sense right?

[chilly ocean]

yea

so what does it mean to map and element to a homomorphism?

[void cosmos]

remember

R is a field

and M and N are f.g R-modules

[chilly ocean]

Why exactly does that matter?

[void cosmos]

those are basically vector spaces

[chilly ocean]

right, but I don't see how it relates to finding a map between the direct product and homomorphisms

[void cosmos]

in general they might be infiinte dimensional

unless R is artinian

but since they are f.g then ig that implies finite dimensional

and R is artinan anyways

so meh

[chilly ocean]

ugh, but I still don't get why that's relevant

why does the dimension matter?

[mossy lintel]

how do i prove a polynomial is irreducible over a field, for example i want to show x^2 + 1 is irreducible over Z_11, should i show it cannot be factorized to non-units explicitly, or should i prove (x^2+1) is a maximal ideal of Z_11[x], or should i show that Z_11[x]/(x^2+1) is a field?

[void cosmos]

it will matter

when trying to show its an isomoprhism

[void cosmos]

.

try to make more sense of this map

like

g: M* x N --> Hom(M,N)

g( (f,y) ) = what

[chilly ocean]

phi(m)y is an element of N, how does that uniquely define a homomorphism? Can't there be multiple that result in that?

[void cosmos]

ig it would be better to say that

x --> y*f(x)

where y is fixed in N

this is in Hom(M,N) right?

where f is fixed too

f is in M*

the assignment that is

not y*f(x)

[chilly ocean]

sorry, you're going to need to be even more specific. I don't get why that's in Hom(M,N)

[void cosmos]

y*f(x) is an element in N

[chilly ocean]

right

[void cosmos]

the assignment x --> y*f(x)

is a bilinear map from M to N

[chilly ocean]

yes

[void cosmos]

so now by varying over the y and

f

u get ur map to the Hom from the direct product

that is sending the pair (f,y) --> to that assignment

[crystal turtle]

So for lower degrees over a field it's easier. Any degree one poly over a field is irreducible and (exercise!) A poly in F[x] of degree 2 or 3 is irreducible iff it has no roots in F.

[void cosmos]

do u get it @panckaer

[chilly ocean]

I'm getting a bit confused about this

[void cosmos]

okay

its my bad im not being clear

okay

we want to find a bilinear map from the direct product to the Hom

[chilly ocean]

right

[void cosmos]

so we get our first homomoprhism from the tensor product by the universal property

and then we want to find an inverse

or show that its a bijection whatever

[chilly ocean]

right right, I'm with you on that

[void cosmos]

so with these things the thing that comes to mind first almost always work lmfao

so

f: M* x N --> Hom(M,N)

we want to define this

right?

[chilly ocean]

yes

[void cosmos]

what does this f take

and what should it spit

[chilly ocean]

a homomorphism and an element in N

oh