groups-rings-fields

[warped wedge]

i tried doing a a to c to b and now b to d and then i have to do d to a but i mkinda lost with how to apply the definitions which is rlly bad

[rocky cloak]

What step are you on? d to a?

[warped wedge]

i did b to d but I don’t think it’s correct

[twilit wraith]

It should be easier to just do a -> b -> c -> d -> a

[warped wedge]

ah ok

hmm i think i can see

im gonna go take a nap

[candid patrol]

Hi everyone, do you think it could be interesting to work on the transfer and its applications in group theory (notably the various theorems of Schur and Burnside) as a topic for a first-year master’s thesis?

[crystal vale]

What does it mean by factors uniquely through P, it is from Ravi Vakil's notes on Sheaves therefore I am asking here

[thorn jay]

exactly as it shows in the diagram

there is a morphism P' → P making the diagram commute, i.e. simultaneously factorising the morphisms ν' and μ'

[crystal vale]

Oh so diagram commutes simultaneously

[thorn jay]

thats quite literally what it means for a diagram to commute

[sudden condor]

im learning elementary group theory (just learned iso thms and now we are doing jordan-holder decomposition) and i find it very boring

are there any good things to read on that make it more interesting

like maybe a further topic which uses this stuff or whatever idk

or just a good book

[karmic moat]

If you care about physics/chemistry, groups show up everywhere

If you like geometry, you can look at algebraic groups

[white oxide]

I agree it is extremely boring

I'm assuming that most of the interesting things lie in applications yeah

[karmic moat]

(In particular, linear algebraic groups, which are GL(n) and its subgroups are a very fruitful class of geometric objects)

[fading acorn]

a lot of symmetry stuffs in physics can be described with groups

in particular lie groups do a big thing

[karmic moat]

If you care about cryptography there’s groups there too

[fading acorn]

yea

ECC

[south patrol]

Down with cryptography

[sudden condor]

its over

[south patrol]

Ruining number theory courses since 1750

[karmic moat]

But the best application of groups is in the fact that every ring is an abelian group with some other stuff and rings are awesome

[sudden condor]

we did dirichlet's theorem instead of cryptography in my intro number theory course

[sudden condor]

alright thank you

[karmic moat]

Imo the fun in commutative algebra starts when you start looking at its applications/intersections with other fields

Representation theory is a big field with a lot of groups

[fading acorn]

literally all algebra stuffs are boring when you learn it, but it's fun when you do other stuffs that use them
e.g. AlgNT, AT, AG, ...

[karmic moat]

Umm let’s see where else have i seen groups

Nah I’m out of ideas

[south patrol]

Gg

[karmic moat]

Oh algebraic topology

[sudden condor]

finite group theorists would be very upset with you right now

[elfin wraith]

What is it with people needing “applications” for their ring theory? Do modules not just bring you joy?

[quiet pelican]

Counterpoint: all three of those are boring (although AT much less than the other two)

[quiet pelican]

I think applying rings to groups is bestest

[elfin wraith]

Group ring moment

[rocky cloak]

In connection with Jordan-Hölder in particular, Galois theory might be good:

An irreducible polynomial has solutions in radicals if and only if all the factors in the composition series of the Galois group are abelian.

In particular this means there are polynomials not solvable in radicals for degrees 5 and above, because the composition factors of Sn are An and C2, and An isn't abelian.

But I guess throwing the question back to you. What things do you find interesting?

[thorn jay]

Jordan-Hölder is all but boring how dare you

[thorn jay]

G-sets appear in the classification of minimal algebras everyone cares about universal algebra cmon

[karmic moat]

Ohhh shit… aita?

[thorn jay]

100 downvotes i cast

[karmic moat]

😢

Now I have to go to reddit bank and get a karma loan

[noble nexus]

do a lot of examples, there are a lot of nifty little groups

there are probably some example focused books out there not sure of any in particular

https://people.maths.bris.ac.uk/~matyd/GroupNames/ I like this page as a starting point

If you want all the groups

[quiet pelican]

The thing I dislike about that page is that it’s all finite groups
Like
I feel one of the worst things about group theory for me for a while was that I didn’t really know any examples of infinite groups
For finite groups, if you know cyclic, dihedral, alternating, symmetric, Quaternion groups (and you know about semi-direct products), you know like
Enough groups for 95% of interesting finite group theory, unless you’re going deep into it

[knotty badger]

wait

what about like

additive groups of integers or rationals or reals

[noble nexus]

True u should know some cool infinite groups

they tend to be harder to work with

[quiet pelican]

I mean like in terms of
Non-abelian ones
And you can contrive such examples, but I don’t know of many natural ones before like surface groups and infinite symmetric groups

[noble nexus]

infinite symmetric groups, SLnZ, infinite wreath products

free groups of course

BS(m,n)

Infinite product of Z/2Z

[quiet pelican]

I learned a really fun theorem recently that the mapping class groups of surfaces of genus at least 3 can’t act properly cocompactly on CAT(0) spaces

[knotty badger]

matrix groups?

[noble nexus]

yeah those ofc

I'm trying to think of non "continuous" groups

since those tend to require more analysis to really study

[quiet pelican]

Mapping class groups are cute, although a pain to compute in non-trivial cases

But also at some point with infinite groups you realise it’s all analysis or analysis-adjacent geo deep down (or at least needs a fair bit of topology) /hj

[noble nexus]

specifically I like the lamplighter group as an example for wreath products

[noble nexus]

cuz it has a fire name

[quiet pelican]

You’re reminding me I have far too many papers to read
In addition to the book I’m reading

although I suspect this is my fate for most the rest of my life at this point

[rocky cloak]

Groups like GLn, SLn, affine transformations should appear pretty early for most people

[quiet pelican]

They’re too big to be groups 🙃
(Yeah idk why I forgot matrix groups
But also linear groups tend to be a little nicer in general)

[rocky cloak]

I mean, is it a problem that groups are nice?

[quiet pelican]

It is if you want counterexamples 🙃

[rocky cloak]

I guess we need a "counterexamples in topology", but for algebra

[tardy hedge]

Examples of infinite groups yeah so true. I think ill always remember Q/Z has being infinite group where every element is finite order lol

[tardy hedge]

I was just thinking about this

Earlier today i googled examples in algebra and didnt find anything

[quiet pelican]

BS(2, 3) gets its own chapter

[elfin wraith]

That’s because we have good definitions in algebra and don’t get random bullshit counterexamples so often that we have to talk about spaces with 16 subtly different connectedness conditions

[tardy hedge]

Lol thats funny

[karmic moat]

Realest shit I’ve heard all week

All month actually

[quiet pelican]

Tbh I feel it’s more because we decide to work with definitions that aren’t just “the minimal condition such that this proof works”
that’s for the UAers

[thorn jay]

its often more interesting the other way around

[tardy hedge]

Probably harder to think about too

[thorn jay]

"when this proof works / conclusion holds in a variety then it satisfies the minimal condition"

for example, if there is some term p(x, y, z) such that it satisfies certain equations, all congruences must commute. But if all congruences commute in a variety, then there must exist such a term

[tardy hedge]

Erm

[thorn jay]

type shit

[thorn jay]

also you try working with operations that arent even guaranteed to satisfy any properties >:(

[tardy hedge]

Universal algebra seems to definitely take a different kind of thinking

[thorn jay]

weird interplay between category theory and the study of generalized monoids

(clones)

[tardy hedge]

Cool

[quiet pelican]

I don’t because I have ε > 0 sanity

[thorn jay]

grrgrghhr4g

i am sane

promise

[quiet pelican]

Promise?

[thorn jay]

pinky

[quiet pelican]

:3

[velvet hull]

let -ε < 0

[tardy hedge]

Upon the Witnessing

[elfin wraith]

No I do joke somewhat, I think the power of topology is that it’s such an incredibly minimal amount of structure that practically everything you care about is a topological space but somehow also rich enough that it’s incredibly useful

But equally that absolutely minimal structure does lead to some pretty frustrating nonsense, like the need for a book of counter examples

This is why I like algtop, you just assume all your spaces are sufficiently nice and everything is better

[knotty badger]

Algebra has good definitions?

[thorn jay]

lwk i do think so

[tardy hedge]

Stuff does work out nicely in algebra doesnt it

[thorn jay]

thats probably just because y'all work with the nice algebraic structures

[karmic moat]

Anyway I always just pretend my rings are noetherian 😋

[tardy hedge]

I was actually thinking earlier whats the significance of noetherian rings in practice. I feel like im aware somewhat now but not totally

Of course ideals being finitely generated is a nice property

I know of that fact f: M->N map of modules and if injective then surjective if theyre noetherian or something

That one seems nice

[velvet hull]

it is a surprisingly powerful property

one that I know of is that in Noetherian rings, every ideal has a reduced primary decomposition

[tardy hedge]

Oh yea

I heard that fact but didnt study it yet

[knotty badger]

Idk…

[karmic moat]

If youre willing to accept some algebraic geometry, you have that Spec A is noetherian (as a topological space) iff A is noetherian (as a ring)

Noetherian topological spaces are nice as hell

[tardy hedge]

I kinda forget the topology on Spec A

[karmic moat]

Zariski topology

[thorn jay]

closed sets are the sets of prime ideals containing some ideal

[tardy hedge]

In point set topology i think we defined Zariski on R^2 for example to have closed sets as R^2 minus finitely many points

[karmic moat]

Most rings one encounters naturally are noetherian

[tardy hedge]

I know theres the connection to the prime ideals thing but i forget atm

[thorn jay]

prime ideals act like "points" in affine space

[elfin wraith]

It’s essentially just a finiteness condition, it can be hard to work with rings without it because they can just be so ridiculously large that you can’t really get a handle on them

And thankfully a lot of interesting rings are

[tardy hedge]

I got confused by prime ideals in like multi variables now

[thorn jay]

this is because points in k^n correspond to k-algebra homomorphisms k[x1, ..., xn] → k, so one can approximate them using ideals. These are the maximal ideals of the polynomial ring. Then, as simply taking maximal ideals doesnt work and you want your stuff to be a coherent condition you take all prime ideals

[karmic moat]

This sounds weird

If not wrong

[tardy hedge]

Cuz i know over one variable its kinda like field theory stuff

[thorn jay]

it is

that is the cofinite topology

[karmic moat]

Okay good I’m not crazy

Yeah

[tardy hedge]

Hmm ill look then

That example should be in Janich

[thorn jay]

decidedly different from Zariski (f.e. consider the complement of V(x^2 - y) )

[tardy hedge]

Wtf is what im saying then

Isnt there some connection

[thorn jay]

cofinite topology is certainly included in Zariski on the reals

[elfin wraith]

My general topology exam had a massive question about the zariski topology but didn’t assume anyone knew any ring theory (for some unknown reason) so everything was described set theoretically and it was the worst fucking question I’ve ever had to do in my life

[tardy hedge]

Lol

[thorn jay]

elaborate please that sounds so funny

[elfin wraith]

I took algebraic geometry and still made a mess of that problem because just parsing it was neigh on impossible

I was also like dying with some unknown illness at the time which didn’t help

[tardy hedge]

The open sets in my example were the finite collection of points i think

[tardy hedge]

Closed sets were R^2-p

[elfin wraith]

Truly just terrible to parse, I honestly think having done alggeo made this question harder

I got an 80 in alggeo and a 45 in that exam

[tardy hedge]

It kinda looks fun though

[elfin wraith]

Ill be honest in an exam setting while youre also extremly ill... wasnt

[tardy hedge]

Yeah i could see that

A lot of random shit to parse

[elfin wraith]

But it does mean that I have As in 2 algebraic topology modules but I have D in both metric spaces and topology which is just amusing

[tardy hedge]

Lol

[karmic moat]

Just notational jibber jabber

[thorn jay]

EVERY FINITE SET??? NO WAY NEW THEOREM JUST DROPPED

[elfin wraith]

But yeah I think after much struggling with notation I did everything up to half of e and just gave up and moved on

[thorn jay]

The class of finite sets is not proper!!

[karmic moat]

Took me a second to read

[elfin wraith]

People were not happy about that question lol, especially the people who hadnt taken AG because we never did any zariski stuff in class and it was almost 50% of the exam

[quiet pelican]

This seems very easy if you’ve done AG, and incredibly horrible if you’ve not
Like
this should not be on an exam for anything other than AG

[elfin wraith]

That entire course was a disaster becase the course organiser, the guy who wrote the exam, (he does FA) didnt teach any of the class due to illness, so we were taught by a homotopy theorist who doesnt belive in exams or lectures really, from notes he fundamentally disagreed with

none of it was good

[elfin wraith]

Yeah this was the general sentiment among students. I honeslty think having done AG was helpful but I think it was still doable if not. In any case it was ass because the notation was just incomprehensible

[quiet pelican]

The green highlighted part is everything in this question you wouldn’t have memorised if you did our AG exams

And that’s not to say it’s hard

[thorn jay]

why did he fundamentally disagree with it lol

[elfin wraith]

Kinda just everything about it from the notation and words used to the things that were given attention and the order they were done in. Like theres an entire chapter proving Browers fixed point theorem just using pointset (its essentially stokes theorem but painfully unwrapped) and he just went yeah thats dumb were not doing that, take algtop next semester and do it in 2 lines

[thorn jay]

thats valid though

[elfin wraith]

He just kinda waffled about schemes a lot, he had a lecture once where he turned off the recording and ranted about the algebra at this uni being shit because no one knew what a free group was and no one does any category theory

[quiet pelican]

I feel like in a modern context there’s little reason to do a pure pointset proof of Brower’s

[elfin wraith]

No it fully was, I wouldve loved to have taken his topology course, he taught it for like years before that I mean hes a pretty big name in topology

[thorn jay]

i like him but this does not seem good for students' learning lollll

[elfin wraith]

The only reason he did it is because he taught it for the first time the year before and finished the course weeks early, and also realised all the content is incredibly easy

[tardy hedge]

Can someone elaborate on b)

The green part

[quiet pelican]

The identity map from Euclidean to Zariski is cts

[elfin wraith]

Its a mixed bag lol, I wouldve been upset had it not been an topology course in my penultimate semester of UG after a required metric spaces course

Though when I audited his higher cat theory course it didnt seem much more organised

[tardy hedge]

Oh yea ok cool but why do we care about that?

[elfin wraith]

He told everyone they were going to use some textbook, then in the first lecture said actually nah im writing my own notes, but theyre not done and I don really have a plan but ill upload them when I can, they might not be in order

[quiet pelican]

I don’t know of any uses for that specifically, but being able to transport properties between Euclidean and Zariski topologies I’ve been told is useful when you can do it

[knotty badger]

Algebra 😭

[glad osprey]

Algebra 🔥

[tardy hedge]

Im over it but i have nothing else to do

Atm

[thorn jay]

algebra

[tardy hedge]

Also maybe more generally too

Who knows

[quiet pelican]

Algebra is cute
Fill brain with all the algebra!

[tardy hedge]

Lol

[thorn jay]

i might've found a connection between tame congruence theory and universal algebraic geometry

owo

[tardy hedge]

Breh

[thorn jay]

might be worth studying

[knotty badger]

I don’t even know how to feel about algebra anymore

[tardy hedge]

Why

you know more than me and i still like it so what does that make me then

Yk

Just enjoy what u can

Or idk i mean tbh i dont really know on what level youre trying to understand algebra

[rocky cloak]

It's exactly the condition needed for the category of finitely generated (or finitely presented) modules to be abelian.

[tardy hedge]

Thats cool, so for morphisms to have kernels and u can “add” morphisms or something?

Idk if i remember the definitions

[rocky cloak]

Yeah. Specifically having kernels is what Noetherianess grants you here

[elfin wraith]

Oh that’s sick I didn’t know that

[tardy hedge]

Do i need to understand what kernels are category theoretically for this

I havent learned that yet

[rocky cloak]

It's sort of obvious if you think about it. An ideal I is the kernel of
R -> R/I
which is a map of finitely generated modules.

So you need all ideals to be finitely generated to get an abelian category

[rocky cloak]

Not really.

Though learning the universal property of kernels is probably a good use of your time

[tardy hedge]

U need ideals to be fg because kernel needs to be an object in the category or?

[rocky cloak]

Yeah, otherwise you don't have kernels in the category

[elfin wraith]

Oh yeah I guess haha, I just never put that together before

[tardy hedge]

Ohh lol

Whats an example of a non abelian category?

[south patrol]

Rings

[tardy hedge]

Oh ya

[elfin wraith]

The category of Hilbert spaces isn’t, neither is Ring

[south patrol]

But also just lol

Much less algebraic categories than that

[tardy hedge]

Yeah

[south patrol]

I just say this because it catches some out initially

[tardy hedge]

Yeah

[south patrol]

Like topological spaces or Set for example would work lmao

[elfin wraith]

I think infinite dimensions make things go badly generally

[south patrol]

Hilbert spaces is another nice thing

[rocky cloak]

A fun example is the category of topological abelian groups, which is preabelian, but not abelian

[south patrol]

Condensed abelian groups tho

[south patrol]

This is cool actually like lol

[rocky cloak]

(the first isomorphism theorem fails)

[tardy hedge]

In non abelian categories there still can be a notion of kernel but that object isnt in the category, or there also just couldnt be any notion of kernel at all? I guess if we’re not talking about algebraic things

[south patrol]

Have you seen these quasi-abelian cats

[south patrol]

No I mean you can still have kernels and stuff meaningfully, just the notions don't quite interact as in an abelian category

[tardy hedge]

Cause kernel can be developed in cat theory ?

[south patrol]

Ig that is what Jagr's example is for

[rocky cloak]

Kernels can be defined in pointed categories

[south patrol]

Yeah I mean kernel is defined by a pullback

[tardy hedge]

f-1(0)

[south patrol]

Indeed

[tardy hedge]

Lol

[rocky cloak]

In general you have kernel-pairs or whatever it's called

[south patrol]

Ye

[tardy hedge]

In abelian categories you can also add morphisms or something?

“Add”

[south patrol]

That needs less than abelian

This is being additive (or less even)

[tardy hedge]

Ok

[rocky cloak]

Technically preadditive

[south patrol]

Yeah

[tardy hedge]

Is it just some binary operation on morphisms

[south patrol]

The fact I still feel is underrated is how the addition in a (say, additive) category can be constructed from just universal properties

[rocky cloak]

Each Hom-set should be an abelian group and composition should be bilinear

[south patrol]

Like given maps f, g: a -> b you can form like a -> a (+) a -> b (+) b -> b

[tardy hedge]

Cool im not totally sure what the bilinear condition is for but it makes sense to have that

[rocky cloak]

Yeah, it's fun like
preadditivity is a structure you put on a category, but additivity is a property

[south patrol]

3(f + g) = 3f + 3g etc

[south patrol]

Ye

[tardy hedge]

Yeah but the significance of it ig

[south patrol]

Which seems reasonable

[rocky cloak]

(which implies a unique preadditive structure)

[tardy hedge]

Also 3? I thought more like (f+g) o h = foh+goh right

[south patrol]

Maths is broken for me

Wait yeah ignore me

I guess what I wrote is fine if you interpret 3 as 3 id

Etc

[rocky cloak]

Well a consequence is that it makes Hom(X, X) into a ring

[tardy hedge]

Nice

[rocky cloak]

Indeed preadditive categories are usually called "rings with several objects"

[thorn jay]

any category of algebras that isn't a category of abelian algebras I'm gonna guess tbh

(because that would just be equivalent to a module category)

[tardy hedge]

Enpeace says stuff i dont understand lowkey

[thorn jay]

basically any category of algebraic structures that aren't ridiculously nice

[tardy hedge]

an abelian category needs to be a (pre)?additive category?

[thorn jay]

both

[south patrol]

Should we Google the definition for you

Jk lmao

I say that playfully

[tardy hedge]

Lmao

learning is more joyous this wa

way

[south patrol]

Yes

[tardy hedge]

especially when im just sitting here bored on a friday night

[thorn jay]

(ridiculously nice when looking at their equational logic, as opposed to how ridiculously nice and convenient e.g. commutative rings are)

[rocky cloak]

Anyway, the hierarchy is
additive category = category with finite biproducts
example category of projective modules

preabelian = additive category with all kernels and cokernels
Example, topological abelian groups

abelian = preabelian and satisfies first isomorphism theorem
Examples, module categories, sheaves of abelian groups, finite length modules

[south patrol]

An abelian category is what you get when you prove the snake lemma in RMod and write down everything categorical you use in the proof

[thorn jay]

actual generalised elements

[vapid vale]

that’s the other nice thing about (small) abelian categories, there is a fully faithful exact functor into R-mod for some ring

so you can do element chasing

[crystal vale]

if R is an integral domain and M is an free R-module. Say B = { e_i | i in I } is a basis for M, how do i show any maximal linearly independent set of M has same cardinality as B?

[rocky cloak]

A hint could be that the statement is true when R is a field. So if you can somehow reduce to the case...

[crystal vale]

you mean i have to consider its ring of fraction

[rocky cloak]

There's an idea

[crystal vale]

does we need to use scalar extension here?

tensor product not introduce yet

[velvet hull]

I mean, this is a pretty good idea...

[crystal vale]

If M is module over PID, then is it true that if M = N_1 + N_2 then rank M = rank N_1 + rank N_2 - rank (N_1 \cap N2), where M has finite rank?

Because I am not sure that I can extend the basis of N_1 to basis of M

[rocky cloak]

Well, I guess that's technically what it is, but really you just need to notice that R^n is a subset of K^n, where K is the field of fractions

[rocky cloak]

It is true, even for integral domains more generally

[crystal vale]

You mean I don't need PID there?

I assumed to pid to make sure rank of submodules exists

How can I show this?

I am trying to extend the basis but I don't think that's a way to show this result

[rocky cloak]

Well, here extensions of scalars would be useful

[crystal vale]

Again extension of scalars

[rocky cloak]

But you're talking about basis, is M free?

[crystal vale]

Yes M is a free module of rank n

[rocky cloak]

Well, then you can use the PID condition to simplify I guess.

Like you have a short exact sequence
0 -> N1nN2 -> N1(+)N2 -> N1+N2 -> 0

Then because N1+N2 is a submodule of M, and R is a PID this is also free. Hence the sequence splits.

So you can extend a basis for N1nN2 to a basis for N1(+)N2

[crystal vale]

I see

[rocky cloak]

But turning your attention back to this:

Are you able to show that a maximally linearly independent set in R^n also is maximally linearly independent in K^n ?

(hence is a basis of size n)

[crystal vale]

To show maximal linearly independent first I will show they are linearly independent.

Claim: { r_1,..,r_k} are LI in R^n then they are LI in K^n.

Say, \sum k_i r_i = 0. But k_i are a_i/b_i, where a_i, b_i in R, b_i ≠0.

So \sum (a_i Π(b_j)_j≠i r_i) = 0.

Since a_i Π(b_j) in R, implies a_i Π(b_j) = 0. Since Π(b_j) ≠ 0 in R otherwise they are zero in K, so it implies a_i = 0.

Hence k_i = 0.

This is my first step, is it correct?

Yes I can show maximally independent set in R^n is also is maximally linearly independent in K^n

So in finite case we are done

What about infinite case?

I got it

I am dumb

Thanks @rocky cloak

[rocky cloak]

The proof is exactly the same in the infinite case

[crystal vale]

Yes

[thorn jay]

i have looked around and genuinely NOTHING, not even ai can convince itself that such a connection exists

[tardy hedge]

So what

Like the point is you have a good idea or something?

[elfin wraith]

50/50 biggest UA discovery in years or enpeace has finally succumbed to full on crankery

[thorn jay]

haha who knows

the intersection between people who think about tame congruence theory and people who think about universal algebraic geometry is the singleton set containing me

[white oxide]

Can I have a hint on how to show $\sqrt{(4, t)} \subseteq (2, t)$? Here we're in $\mathbb{Z}[t]$

[cloud walrus]

okeyokay

[thorn jay]

(2, t) is a prime ideal

[white oxide]

Wait does this have something to do with the Nullstellensatz lol I took a course in algebraic geometry but I forgot most of everything

[elfin wraith]

Z isn’t a field so you’d struggle to apply that

[thorn jay]

radical of I is the intersection of prime ideals containing it

thats the fact relevant here

that is probably how id define it in the first place too but thats too based of an opinion

[white oxide]

Oh I see

thanks

[elfin wraith]

Recall as well that over a commutative ring semiprime and radical ideals are the same thing, if that gives you a push in the right direction

[white oxide]

I've never heard of a semiprime ideal negl

[elfin wraith]

Smh, kids these days don’t even study noncom rings

[elfin wraith]

Yeah that’s fine, I wouldn’t worry about it then lol, my point was just to remind you that all prime ideals are semi prime but like all semi prime ideals over a commutative ring are just prime so it doesn’t actually matter

[white oxide]

who the FUCK would study non commutative things

[karmic moat]

noncommutative algebraic geometry is popular

[thorn jay]

our uni has that as a masters course

[south patrol]

does noncommutative algebraic geometry really use noncommutative rings though

[thorn jay]

seems too analysis pilled tbh for ke

me

[south patrol]

i joke cause it does a bit but not nearly as much as one might expect

[south patrol]

depends what you do lol i mean a lot doesn't have analysis

[karmic moat]

i have a friend who does stuff with quivers or something like that and he uses noncommutative rings

[thorn jay]

could universlw algebraic geometry be clasified under noncomm lol

[south patrol]

Sure

[karmic moat]

oh yeah i didn't mean that as a counterpoint to you just some general comment

[elfin wraith]

There’s non commutative geometry which is just functional analysis but non com AG is very much algebraic. It’s a bunch of motives nonsense. Theres also something about universal enveloping algebras of Lie algebras I think but I don’t really know how that ties together

[thorn jay]

ooo thats awesomeee

[elfin wraith]

I may possibly try to do some NCAG stuff for my masters, there’s a pretty big name in it here

[thorn jay]

whats a motive again

[white oxide]

Can somebody help me see why n^c = n'^c, I know we have n^c \subseteq n'^c since q \subseteq q'

[tardy hedge]

I feel like this was the part i got stuck on when i looked at this chapter earlier in the year

One of these little lemmas lol

[elfin wraith]

Some alggeo nonsense to do with generalising cohomology, I’ve got no idea about the specifics

[thorn jay]

draw a diagram and see what's happening

youve got this commutative diagram of rings, which implies that, if I ⊆ B is a prime ideal, then first contracting it to A and then passing to A_p is the same as first passing to B_p and then contracting to A_p

[tardy hedge]

Omg real life enpeace handwriting 🤯

Guys this is crazy

[thorn jay]

lmao

[fading acorn]

lol

[tardy hedge]

Handwriting Reveal Handwriting Reveal !!!

[thorn jay]

also funny typo, we should be passing to B_p^e, not B_p

im pretty sure

or idk what they mean specifically by B_p

maybe they do still mean B localised by A\p?

yeah probably, ignore what i said

[white oxide]

Wait so is it a typo

Or is it true

[tardy hedge]

Holay Molay

[thorn jay]

i dont know how do they define B_p

[tardy hedge]

Ok i gotta get off this server ive been here for like 24hrs

[thorn jay]

because p is an ideal of A

[white oxide]

Localization of b

[thorn jay]

by what? can't be p itself, that contains 0

oh wait nvm the complement of p in A can't intersect n or n' we're all good

[thorn jay]

this is because q and q' correspond uniquely to their extensions in B_p

[thorn jay]

its true

[vapid vale]

kind of yeah? (i know it was joke) but like from my limited knowledge you are working with homological/"representational" properties of the ring

so instead of rings you more talk about their module category, instead of schemes you talk about their coherent sheaves

so like if you want to talk about the noncommutative equivalent of coherent sheaves on Proj R, you can talk about the quotient of the module category. by the finite lenght modules

[rocky cloak]

Isn't non-commutative algebraic geometry, just taking questions from alggeo, formulating them as questions about the derived category of quasi-coherent sheaves, then asking the same question about arbitrary dg-categories?

In particular noncommutative dg-algebras

[rapid cave]

What is dg?

[rocky cloak]

Differentially graded

https://en.m.wikipedia.org/wiki/Differential_graded_algebra

[rocky cloak]

Seems maybe ncag is wider than this

This would be more derived ncag

[vapid vale]

thats all i understand about it lol

[south patrol]

this is what i view it as more lol

My poitn anyway was more to emphasise that non-commutative AG doesn't (usually) use non-commutative rings in the way one might expect from its name

[south patrol]

agreed

[crystal vale]

I can say there is always a unique ring homomorphism Z to any unitial ring [ I am allowing that 1 maps to 1 ], right?

[crystal vale]

ring of End(Z\times Z) is isomorphic to M(2, Z), right?

so Aut(Z \times Z) will be isomorphic multiplicative group of M(2,Z)

how it looks like? I know matrix should have determinant +- 1 here

[limber sequoia]

quite a lot... for every a,d you choose you can have
a ad-1
-1 d

[rocky cloak]

Famously if you quotient out by ±1 it should look like the free product of C2 and C3

But otherwise it looks like matrices with determinant ±1

[crystal vale]

what is the free product?

[velvet hull]

Coproduct in the category of groups

[rocky cloak]

Basically the elements are alternating sequence of group elements from each group, and the multiplication is concatenation where you multiply neighboring elements from the same group and remove the element any time you get the identity.

[crystal vale]

i see

[rocky cloak]

Or in terms of group presentations, if you have a presentation for each group you just combine them

So C2 = <x | x^2> and C3 = < y | y^3 > and their free product is
<x, y | x^2, y^3 >

[crystal vale]

got it

[rocky cloak]

The presentation for SL(2, Z) is fairly similar. It should be
< x, y | x^4, y^6, x^2=y^3 >

Here x^2 corresponds to -1, which you can then mod out to get the free product

All of this is just matrices with determinant 1

Then you can throw in a semidirect product with C2 to get negative determinants

[crystal vale]

how can i find this presentation by myself?

[delicate orchid]

very painfully

[rocky cloak]

I mean finding elements of finite order that generate the group is not too hard.

Think about elementary row operations.

But showing that there aren't any more relations ones you have the right generating set is probably pretty hard

[delicate orchid]

you use something called the ping-pong lemma for the last step iirc

[rocky cloak]

https://chiasme.wordpress.com/2015/03/08/an-elementary-application-of-ping-pong-lemma/

[crystal vale]

oaky

thanks jagr

can there be finitely generated module but it has no basis, i think any finite abelian group will work

[elfin wraith]

The difference between free and fg really tripped me up when I first learned about it

[rocky cloak]

That's kinda weird, since they're not really related at all

[elfin wraith]

Yeah I’m not sure, I think I just got tripped up with how something could have a basis and yet not be finitely generated, and sorta vice versa

Like I don’t think I appreciated that the basis could be infinite and then also just forgetting that bases need to be linearly independent

[crystal vale]

okay, I have to prove that, Let M a left Q-module. Show that the given action of Q is the only one which can be used to make M a left Q-module.

So action of Q induce the ring homomorphism \phi: Q -> End(M).

Here now i am trying to give the module structure to End(M) as (q,f) -> qf given by (qf)(x) = q(fx).

Then we can see \phi( r ) = r\phi(1), but since \phi(1) is identity in End(M).

so it shows that \phi( r ) : M -> M given by m -> rm.

So if other action exists so it induce the same ring homorphism which shows that there only one possible way to make a M left Q-module, is it correct?

[rocky cloak]

I'm not sure I follow your argument.

Surely relying on End(M) also having a unique Q-action is just pushing the problem...

Though thinking about the value of phi(r) for simple rational numbers is a good idea.

phi(1) and phi(0) you should have

[crystal vale]

what's the missing part in my argument?

[rocky cloak]

Well, I don't understand what the parts are.

First you say you try to define a module structure on End(M), which you don't justify and I don't see the relevance of.

Then phi(r) = rphi(1) comes out of nowhere with no justification.

[crystal vale]

okay, we have to say phi ( r ) = rphi (1), we need to define what is rphi(1), because phi (1) in End(M) and r in R.

Therefore i defined the R-module on End(M).

Then we have phi ( r + s ) = phi( r ) + phi ( s ), [ i used the same technique here as cauchy functional equation, which says that if f(x+y) = f(x) + f(y) then f(x) = xf(1), domain is Q ]

[rocky cloak]

Okay, I see. You have essentially proven it before and are just supressing the proof

[crystal vale]

just verifying that there is no problem in my argument

if M is R-module of rank n, if some subset of M spans M and cardnilaity of that subset is n, then is that basis set?

[rocky cloak]

No problem, but yeah the missing part would be "by the same proof as the Cauchy functional equation"

[crystal vale]

[rocky cloak]

It is if R is commutative or Noetherian (or more generally IBN)

[crystal vale]

how is it true when R is commutative?

say S is the subset of M which spans M, can i reduce S to basis set?

oh no, we can't

actually i have to show if R is commutative, then any surjective endomorphism of M[ has rank n ] is bijective

[rocky cloak]

A useful thing is that homomorphisms of free modules can be described as matrices.

[crystal vale]

yes

[rocky cloak]

If R is commutative, then you have things like determinants and adjugate matrix that are very helpful

[crystal vale]

yeah, but if i can show this then we are done

i am trying to avoid matrix computations here

[rocky cloak]

Alright, can you use that a surjection onto a free module splits?

[crystal vale]

no, i am following Jacobson so it not introduced yet

but you can tell

[rocky cloak]

Hmm, so what do you know?

[crystal vale]

i think according to Jacobson textbook, matrix approach is needed

[rocky cloak]

Wait, did you assume M to be free or just a module?

And do you define rank as smallest linearly independent subset?

[crystal vale]

in question he mention R^n instead of M

[rocky cloak]

Alright, then no problem

[crystal vale]

so i think if M has rank n so it is isomorphic R^n, right?

[rocky cloak]

Depends on your definition of rank, but let's say yes

[crystal vale]

yes there is confusion, yesterday i read about structure theorem of fg modules over pid, and then author define rank according to the decomposition

[rocky cloak]

Anyhow.

You have a surjective map R^n -> R^n

Then in particular each of the standard basis vectors is in the image. So you can define a map in the other direction by sending each basis vector to one of its preimages.

This then means that R^n is isomorphic to K (+) R^n where K is the kernel of the original map.

From there you can finish using Nakayamas lemma if you have that...

[crystal vale]

i am trying to see this

do i have to use splitting lemma here?

[wraith cargo]

Yes

But you can construct the isomorphism coordinate wise

[rocky cloak]

You can, I guess you can also just show that K and the image don't intersect and so on, but I guess that would be reproving most of the splitting lemma

[tardy hedge]

me too

[elfin wraith]

Yeah

[rocky cloak]

If HL = {hl | h in H and l in L}, then you usually wouldn't have HL be a subgroup

[elfin wraith]

I’m think this is true but I’m not 100% sure off the top of my head, I would need to check. I know it’s certainly true if they intersect trivially but I’d need to think about non trivial intersections

Second iso assumes normality of one of the groups

[rocky cloak]

That never happens if they're sylow subgroups

Unless H=L

Alright, then sure. If H normalizes L, then
HL/L = H/HnL is an isomorphism of groups

So yeah HL will have size |L| times a power of p

[cloud walrus]

mq

[rocky cloak]

Looks good

[crystal vale]

man, i took 1.5 hr to prove splitting lemma

[knotty badger]

ooh this looks interesting

[crystal vale]

crazy

[knotty badger]

so f is monic and g is epic, right?

[crystal vale]

yes

[knotty badger]

and also ker(g) = im(f)

[crystal vale]

should i send my arguments to verify?

[knotty badger]

mhm, sure

[crystal vale]

yes

[crystal vale]

then i have to write a lot

i will send it later

i will try to write in latex

[tardy hedge]

yea i quite enjoyed learning about split exact sequences

its nice

[thorn jay]

i looove split exact sequences

[tardy hedge]

🥵

When the sequence is exact … 😲

When the sequence is split … 😍😍😍😍

[gilded hatch]

Oh, sorry I did mistake actually b,c are from G',

[crystal vale]

what is your doubt ?

[gilded hatch]

Idk How to prove it . Even How start

[thorn jay]

that remark is so ahh

First prove this:
φ(a) = φ(b) <=> φ(ab^-1) = e

[gilded hatch]

If is showing that kernel is trivial subgroup iff HOM... IS INJECTION

[thorn jay]

?

yes i got that

[thorn jay]

this is a standard fact

look at it closely

read it carefully

[gilded hatch]

Yeah next question is about showing that phi is isomorphism if oder Domain group is prime number or target is trivial group

I did mistake in typo on Q. b,c are from Target not domain

So now I should start with onto on if |G| is prime , they already said it is injection, but idk how to show |G| being prime number does led to PHI IS ISO... or G' as just indentity

[crystal vale]

i am not used to Nakayamas lemma, can you tell me how can i use it here, and there are many version of it, i don't know how to use it

[rocky cloak]

Say you take some prime p then localizing at p you have
(R_p)^n = K_p (+) (R_p)^n
so moding out by p you get that
K_p/(pK_p) = 0
then Nakayamas lemma say K_p = 0.

Then K_p = 0 for all primes means K=0

[karmic moat]

when in doubt, nakayama's always works

[elfin wraith]

One of the 50 versions will apply

[rocky cloak]

I feel like I've only really seen / used two versions.

The one that also works in the noncommutative case, and the one you need to reach for when proving something for commutative rings that's not true in the noncommutative case

[crystal vale]

ah localization, i have to learn about it

and this exercise also show that any subset of M which spans M and it has cardinality as rank of M then that subset is basis, when R is commutative

[tardy hedge]

you've been focusing more on group theory for a while right

[crystal vale]

yes

[tardy hedge]

ive been the opposite

focused on commutative algebra and now i kinda barely know group theory unfortunately

i havent had to think about groups much for a while

i never took a proper class in group theory

[crystal vale]

that's funny , now i want to understand module theory and i think i need commutative algebra too

[tardy hedge]

i like modules

yeah ring theory is important there

[rocky cloak]

Can always look at G-modules if you're missing some groups in your life

[tardy hedge]

Oh yeah

I kinda wanted to do that because of Galois cohomology

[rocky cloak]

(you're life being modules)

[tardy hedge]

idk why i barely ever seen artinian rings jagr

[crystal vale]

Yes

[tardy hedge]

not p^8Z tho

just kidding

lol

its a field

so what are its ideals

[crystal vale]

Kiand, what book do you used for commutative algebra?

[tardy hedge]

atiyah macdonald is very good

think about what could happen if you had some ideal in a field K

yeah

yeah good job

because any ideal would have 1

btw this also implies something about any homomorphism of fields f: K -> F

yeah

would be the same definition anyway because there is the same underlying operations u are working with

consider the kernel of f

im just asking you to think about the kernel if you had a homomorphism f: K -> F

there is something u can observe about it

[rapid cave]

If A is a 2-(krull)dim k-algebra generated by two elements is A=k[x,y] (polynomial ring)?

[tardy hedge]

based on u saying that a field can only have 0 or itself as ideals

[tardy hedge]

the fact im talking about is also pretty easy to see directly tho fyi

ideals map to ideals in the image

[fading acorn]

[tardy hedge]

yeah nice

so what does that mean about the kinds of maps you could have?

yeah

thats kinda neat right

np

[rapid cave]

Bump

[fading acorn]

some recs:
atiyah-macdonald(standard recommendation)
altman (it has solutions)

eisenbud
(but this book is too long and i dont think it is not good for doing introductory stuffs)

[rocky cloak]

Any commutative k-algebra generated by two things is a quotient of k[x, y]. So yes

[tardy hedge]

how do u see that jagr

[rapid cave]

How does it follow from that?

Does a quotient must reduce the dimension?

[tardy hedge]

probably correspondence theorem for ideals idk

[rocky cloak]

The dimension is just the length of a chain of prime ideals

If you mod out by a non-zero ideal, then you can no longer have (0) in that chain, so it won't be maximal

[rapid cave]

Oh nice

[tardy hedge]

i dont get it

[rocky cloak]

The primes of R/I are the primes of R that contain I.

Now say I <= p0 < p1 < p2 is such a chain, then
(0) < p0 < p1 < p2 would be a chain of length 3

Which doesn't happen if R = k[x, y] has dimension 2

[crystal vale]

Thanks

[rapid cave]

Ring inside Ring

[crystal vale]

What about Matsumura, pardon me if I wrote wrong spelling

[tardy hedge]

I dont think thats a good intro book personally

[rocky cloak]

You also want 1 in the subring to be the same as 1 in the superring

[tardy hedge]

Matrices with entries in Q?

[rocky cloak]

You could classify the ideals...

[tardy hedge]

I should try that question too tbh

[rocky cloak]

There is an easier way though. One that uses linear algebra.

||You can show that it's a finite dimensional vector space over Q, and that||
||any ideal is also a vector subspace||

[crystal vale]

Can't we define determinants for matrices over non-commutative rings?

[tardy hedge]

Probably

[rocky cloak]

You might be able to define them, but they don't really have any nice properties. Won't be multiplicative and so on

[karmic moat]

Strictly speaking I don't see why not (since the determinant is just a polynomial), but I'm not sure if the nice properties would carry over

Damn sniped

[rocky cloak]

Defining polynomials over noncommutative rings is already pretty fishy

[karmic moat]

Interesting

Why can't you just define it the normal way as you would for commutative rings?

[rocky cloak]

Well what is "the normal way"?

[karmic moat]

I mean for commutative rings you just write them as sums of the monomials

[tardy hedge]

I was thinking maybe something with evaluation map goes bad

[rocky cloak]

Yeah, so if your base is a commutative ring (I guess Z would be natural here) you can define the free algebra
Z<x1, x2, ...>
which would be a non-commutative polynomial ring

[karmic moat]

Ohh I see

[tardy hedge]

Someone was just asking about that earlier today lol

[rocky cloak]

But then it begs the question, what is the determinant of
[a, b]
[c, d]
?

Is it ad - bc, or da - cb, or maybe ad - cb, or ad + ab - ba + cd - dc - cb, or something else

[karmic moat]

I was thinking one could just fix an order, say ad-bc

[rocky cloak]

Yeah you can do that. Like that is a well defined function

[elfin wraith]

Noo-ter (roughly)

[rocky cloak]

Closer to the i in bird, than the oo in zoo I'd say

[elfin wraith]

Yeah this is why I went with roughly I’m not sure exactly how I’d transcribe it using only English sounds (and I have a pretty strong accent which kinda makes it harder for me to transcribe phonetically in the first place)

Most people in English tend to just call her No-ther or No-ter in my experience though, but that is strictly speaking wrong

[rocky cloak]

Either way, best is to just look up a sound clip. But it's fun to come up with sounds that sound like other sounds

[elfin wraith]

Or learn the IPA, everyone needs a hobby

[spark veldt]

My book uses the notation x^g to mean conjugation by g, that is g-1 x g. Not quite sure where to start here for (a).

Something like this perhaps?

[wraith cargo]

x^sigma here means sigma(x)

[spark veldt]

oh, so not the conjugation then?

[wraith cargo]

no because sigma isn't an element of G

it's an automorphism of G

so it's a function acting on elements of G

[spark veldt]

oh right okay

If thats the case why does the author write it as a power and not just sigma(x)

just a matter of style?

[wraith cargo]

it's more compact

idk it's a thing that's popular in algebra for some reason

[wraith cargo]

anyways yes this is a good starting point

just remember that if xyx^-1y^-1 = e for every x,y then G is abelian

this is basically the solution lol

[elfin wraith]

We need some sort of international group theory convention to decide on official notation because why does everyone and their mother do something differently

[rocky cloak]

Probably so it's consistent with writing x^g for conjugation

[tribal moss]

obXkcd: https://xkcd.com/927

[spark veldt]

thanks guys

[thorn jay]

subset of a ring that contains 0, 1 and is closed under all the operations

[elfin wraith]

Are there any nice results which use subrings (under the correct definition of a ring) other than the noncommutative hilberts basis theorem? I don’t think I know any others off the top of my head

Like subrings just dont seem to come up much (unless you’re a heathen)

[tardy hedge]

I should review the proof for hilbert basis thm

Was thinking abt that the other day

Ik u gotta knock down degrees or smth

[rocky cloak]

What's the nc hilbert basis theorem?

That an Ore extension is Noetherian?

Does that use subrings in any specific way?

[elfin wraith]

That’s the one I’m referring to anyway, we called it the NC basis theorem in my course but maybe it’s not generally known as that

[rocky cloak]

Well, the name makes sense I guess

[elfin wraith]

I was thinking about this today too, my comalg is rather rusty, I’m kinda vague on the proofs of the NSS and basis theorem which is not ideal

[crystal vale]

how can i find B?

and isn't f_i in R^n ?

[elfin wraith]

It’s a nice result, a few things to check but rather handy. But I think that’s the only theorem that, off the top of my head, I remember using subrings

[rocky cloak]

But anyway, something like if S is an integral extension of R, then they have the same Krull dimension.

Does that count as using subrings?

[elfin wraith]

Oh that’s true actually I kinda forgot about integral extensions and stuff being about subrings

And I guess in a similar vein, localisation and stuff actually cares somewhat about subrings

[tardy hedge]

Why localization? I guess localization is exact so that says something about how it interacts with subrings

[rocky cloak]

If you localize at a regular set, then R is a subring of S^-1R

[tardy hedge]

One that has no zero divisors?

[brisk tendon]

sorry to butt in to these but i am interested @rocky cloak what localization r you thinking about ? i know it from a geometric sense like equi localization on some rings

but in this part thm i confused

[rocky cloak]

If S is a multiplicatively closed set in R, then S^-1R has as elements r/s with r in R and s in S, such that rt/st = r/s for t in S

https://en.m.wikipedia.org/wiki/Localization_(commutative_algebra)

[elfin wraith]

I guess when you’re localising you’re really dealing with over rings as opposed to sub rings but that’s just a matter of perspective

[tardy hedge]

Meant to say modules over rings?

[elfin wraith]

No, if A is a subring of B then B is an overring of A

[tribal moss]

No, "over-" as the opposite as "sub-". In analogy to sets we might call them super-rings (but that comes with its own risks of ambiguity). It's most usual to speak of them in the language of unital algebras, I think.

[tardy hedge]

Oh ok

Not sure why u brought that up

[brisk tendon]

i just have a diff def of localization in my head fr derived sheave on CY complex @rocky cloak but i think i need to learn more algebra 🙂

[tardy hedge]

How u learning about sheaves without algebra

[brisk tendon]

@tardy hedge i know algebra lol

[tardy hedge]

My bad

[brisk tendon]

i was wondering about the discussion about localization and the nc theorem

[tribal moss]

The point was that there's a difference in perspective between starting with a larger object and asking what its subobjects are (as is common, e.g. for groups and modules), and starting with the smaller object and asking how to make a larger object that it can embed in.

[elfin wraith]

If you have some ring (it may have to be a domain I’ve only studied localisation for domains) A and some multiplicative set S then a left ring of fractions for A is an over ring B with some conditions so that it looks like a ring of fractions

[tribal moss]

The latter is the perspective of localization.

[tardy hedge]

I see, thanks

In practice tho localization is more used to exploit the properties of its ideals right

How does one know if property P is a local property?

Or how would you intuit it

[elfin wraith]

I’ll be honest I learned both commutative algebra and algebraic geometry from non com ring theorists, I actually know more about non com localisation than I do about “normal” localisation and I have pretty much no intuition for it

That’s roughly the extent of my understanding of it as well

[tardy hedge]

For A is 0 iff Ap is 0 i guess there you notice that the property of an element being 0 is reflected in an ideal ann(x)

Bc if ann(x) is R then x must be 0

Yeah i guess if any property is somehow reflected in ideals its good

Because a prime ideal always contains such and such

[elfin wraith]

I mention this because the comalg class did essentially nothing with it because, and I quote “I don’t really like geometry or commutative rings, and this is really only useful there”

[tardy hedge]

Lol

[elfin wraith]

That lecturer was insane, lovely and incredibly talented, but truly terrible to take a class was

[thorn jay]

tangentially related to this topic, subalgebras behave in general a lot less nice than quotients

for example, every collection of algebras can be some quotient of a bigger algebra, but not every collection of algebras can be simultaneously embedded into another algebra

congruences are in general also just way nicer objects than subalgebras

[tardy hedge]

Does UA have any utility in other areas of math

[thorn jay]

its big in constraint satisfaction problems

[crystal vale]

I have a doubt

[thorn jay]

other than that you can convince yourself operads are an instance of UA

[crystal vale]

I don't see how det( delta \phi - a_ij) annihilates each x_i, I am not sure what is det( delta \phi - a_ij)

[south patrol]

An infamous passage

[tribal moss]

Hmmm, thinking aloud here: Formally it looks a lot like the entries of tI-A from linear algebra, whose determinant is (up to sign conventions) the characteristic polynomial. Instead of t we have the endomorphism phi, and the a_ij are also the coefficients of phi in matrix form ... hmm, this smells like some Cayley-Hamilton analogue.

[south patrol]

Indeed it is precisely a form of Cayley-Hamilton

Though actually I would view this more as a nice proof of Cayley-Hamilton

[karmic moat]

it's probably more obvious if you write it out as a vector lol, as in you have
[ (\phi\text{Id}n - A) \begin{pmatrix} v_1 \ \vdots \ v_n \end{pmatrix} = 0]
where $A = [a{ij}]$

[cloud walrus]

anamono

[tribal moss]

As for "what is the determinant": It must be a determinant of a matrix over the subring of the endomorphism ring generated by phi . Subrings with a single generator are and multiples of I, which is always commutative, so speaking of determinants is meaningful and (delta_ij phi - a_ij) is the expression for entry i,j of the matrix whose determinant you're taking.

[south patrol]

The point is this: firstly, given a matrix $M$ over any commutative ring $R$, there's a matrix $\mathrm{adj}(M)$ such that $\mathrm{adj}(M) M = \det(M) I$. Now the point here is that given a a module $M$ over $A$ with an endomorphism $\phi$, you can view $M$ as a module over $A[t]$ (via like $t.m \coloneq \phi(m)$. Now you can make sense of the matrix $(\delta_{ij} t - a_{ij})$ with coefficients in $A[t]$ and run the previous sentence for that matrix

Equivalently you can just do subring as Troposphere says. I find this a little more universal

[cloud walrus]

Prismatic Potato

[tribal moss]

I think what I said might be nonsense, since the subring in question must be generated not just by phi, but by phi and \mathfrak{a}, and I'm not immediately sure how obvious it is that this would be commutative.

[south patrol]

phi is A-linear so it should be fine

[tribal moss]

Oh, wait, \mathfrak{a} here means scalar multiples of I (which are central anyway), so it's okay.

[south patrol]

(I have seen this written up precisely as you said btw troposphere, to give you confidence aha)

This seems a sort of famous passage cause (1) it's quite early on in Atiyah-Macdonald, (2) it's used in the proof of Nakayama, which is super important, and (3) it is quite terse lol

[crystal vale]

Sorry but I don't get it, i understand that we can make M as A[t]-module.

What I am thinking, if B is matrix representation of \phi, then if we can show det(tI-B) = 0 then we are done.

[crystal vale]

How (I\phi -A) is it valid here?

Oh here we are taking entries in matrix from A[t]

[karmic moat]

i'm just writing out what they're doing as a matrix

which will hopefully make it a little more clear

so you have
[ (\text{Id}_n\phi - A) \begin{pmatrix} v_1 \ \vdots \ v_n \end{pmatrix} = \begin{pmatrix} 0 \ \vdots \ 0 \end{pmatrix}, ]
multiplying both sides by the adjugate matrix of $(\text{Id}_n\phi - A$), as potato mentioned above, gives you
[ \det(\text{Id}_n\phi - A) \begin{pmatrix} v_1 \ \vdots \ v_n \end{pmatrix} = \begin{pmatrix} 0 \ \vdots \ 0 \end{pmatrix} ]

[cloud walrus]

anamono

[karmic moat]

Expanding the determinant, you'll get the desired form (you can try it out for n=2 if you'd like)

(To be precise, multiplying by the adjugate matrix gives you the n x n diagonal matrix whose entries are det(Id \phi - A), from which you get det(Id \phi - A)x_i = 0 for each i. But that gets you the same thing as above)

Oh sorry replace v_i with x_i

[crystal vale]

Now i got it

So, take A[t] then (It - A) acts on M^n, and then (It-A) is a matrix in A[t] such that (It-A)x = 0.

So det(It-A)I x = 0.

But det(It-A)I x gives det(It-A)x_i = 0 for all i.

Therefore when we expand det(It-A) and act on x_i it gives 0.

Is it correct?

[crystal vale]

If there is any mistake then please let me know

[karmic moat]

Yeah looks right

[earnest delta]

How can i solve it?

[wraith cargo]

look at where x and y are sent via the homomorphism

tbh it's a good idea to count how many elements the target ring has

and then look at where x and y can be sent

[rapid cave]

no

a direct sum would be a new module of tuples where the operations/actions are element wise

[fading acorn]

I think mq is referring to internal direct sums

[rapid cave]

these would be isomorphic

[fading acorn]

yes they should but i mean they have different representitives

[elfin wraith]

Nigerian rings 🇳🇬

You only need to prove it’s true for the sum of 2 rings if that helps

I can help you more if you’re still stuck but it’s generally true that if R = S+T and I is an ideal of R then I is of the form J+K for J an ideal of S and K an ideal of T

It is straightforward to show that btw, you don’t need to come up with too much

[azure hull]

Let $R = R_1 \times R_2 \times \cdots \times R_n$ be a finite direct product of rings, and let $A \subseteq R$ be a left ideal. For each $i$, define $e^{(i)} = (0, \ldots, 1, \ldots, 0) \in R$, where $1$ is in the $i$-th component and zeros elsewhere these $e^{(i)}$ are idempotent elements that act as projections so for any $a = (a_1, \ldots, a_n) \in A$, we have $e^{(i)} \cdot a = (0, \ldots, a_i, \ldots, 0) \in A$ by ideal closure under left multiplication this shows you that $A$ contains each component vector of any of its elements and Since $A$ is also closed under addition it must contain all finite sums of such vectors, i.e., $A = A_1 \times \cdots \times A_n$, where $A_i = { a_i \in R_i \mid \exists a \in A \text{ with } a = (a_1, \ldots, a_n) }$. One can check that each $A_i$ is a left ideal in $R_i$, so every left ideal of $R$ is a product of left ideals in the $R_i$

[cloud walrus]

Yuxuan Wang (Trusted By Mods ®)

[elfin wraith]

This is true but I think harder than it needs to be, if you just prove it for R+S the rest follows by induction and the proof is much more clear

I guess, yuxan has gave the proof above

[rocky cloak]

I guess worth noting that a submodule of M1(+)M2 need not be of the form N1(+)N2 with Ni a submodule of Mi.

While for rings the ideals of RxS are of the form IxJ.

[thorn jay]

funny that they use the notation of direct sums for this

[delicate orchid]

Ts? Pmo.

[south patrol]

It's bad fr

[elfin wraith]

It’s not technically a direct product, the co product in ring is the tensor product

[thorn jay]

it also doesn't really "sum" anything

[elfin wraith]

It’s actually just a direct sum of abelian groups with a suitable ring multiplication iirc

But uh don’t quote me on that part

[thorn jay]

yeah sure but algebraically we care about the fact that its a product

[thorn jay]

i mean yeah lol, but thats just because Ab has biproducts

for nonabelian groups i can still somewhat stand by the notation to denote the group of sequences of elements with finite support, i.e. finite nunber of nonzero elements, but this does not form a subring in the infinite case

[elfin wraith]

I personally am of the opinion that if it walks like a direct sum and quacks like a direct sum slap an \oplus on it so I’m happy enough to use that notation but yeah I guess it is worth bearing in mind that it’s not quite the same thing

[thorn jay]

but it doesnt walk or quack like a direct sum, there is no (guaranteed) embedding of R into R × S

[tribal moss]

(... when rings are unital, at least)

[alpine plank]

Yeah I was about to suggest sandwiching it in an exact sequence before realizing we weren't in R-Mod so I think it's genuinely bad notation

[elfin wraith]

Ok yeah that’s fair, but as far as the context in which it was being used, I.e. for looking at its ideals, it’s close enough that I’m not personally bothered, but I do see that it’s not great notation

[elfin wraith]

(So always…)

[thorn jay]

in the context of ideals we are treating them as rings

[tribal moss]

The most convincing argument might be that is does walk and quack like a product, so why not talk about them like that.

[delicate orchid]

Ts? Pmo. It ain’t a biproduct it’s a product. Use \times

[karmic moat]

I literally just went to a pond and saw it. It was honking like a goose

[elfin wraith]

Alright fair enough, I’ll concede it’s just irredeemable notation

[thorn jay]

🤝

[spark veldt]

So I'm trying to show this right, and I thought of a solution already and written it. But I thought of another solution like this, but kinda skeptical about it. Does this idea perhaps work?

So essentially I'm trying to create like this homomorphism (of course I'll have to show that pi is a homomorphism) then use the 1st isomorphism theorem.

[thorn jay]

H isn't required to be normal

so G/H might not be a group

[spark veldt]

oh rightt

crap

thanks

[kind temple]

hmm

[kind temple]

on the level of sets this almost works

the only thing that you need is that equivalence classes of ker pi are all the same size

[velvet hull]

you could still invoke the first iso theorem actually, just a little differently

[thorn jay]

first iso theorem for sets? sure

maybe you could do some G-set stuff lolol

[kind temple]

yea, this is what i am alluding to

but like

the issue is that equivalence classes in the kernel may not be the same size

so the division statement doesn’t follow immediately

[thorn jay]

hmmm

it shouldn't be too hard to prove that, if f : X -> Y is a map of G-sets, then G permutes the fibres of f

i.e. all the fibres of elements of connected components in Y must be the same size

and, as phi(G)/phi(H) is acted on transitively by G (using the action induced by that on G/H), then every fibre of pi, and thus equivalence class of ker pi, must be the same size

seems a little involved though lol

[kind temple]

connected components in Y?

[thorn jay]

orbits*

I like calling them connected components though

[kind temple]

i can see the reasoning there lol

[thorn jay]

Im a geometer at heart..

[kind temple]

would not have guessed it haha

[thorn jay]

can't tell if that's ironic lol

[kind temple]

i just meant because we only really interact in this channel and the advanced algebra channel lol

[thorn jay]

well I'm learning about AG rn and have done some work in universal AG and generally enjoyed that a lot more than just the dry algebra

also tame congruence theory has somewhat geometric interpretations

[kind temple]

sweet

[thorn jay]

and ofc quandles

[south patrol]

AG.

[thorn jay]

do you mean: equational logic of classes closed under embeddings?

[glad osprey]

I hate to break it to you, but nothing you do is actually geometry. You can't just say "oh this latin square is actually a curve in quandle space" and call it geometry

[kind temple]

i have been reading it as quandale this whole time

[thorn jay]

😔 I've been got

I should pack it up tbh

[spark veldt]

How do we know there's a guarantee of existence of maximal subgroup K here?

[south patrol]

Dingle

[south patrol]

G is finite

So you can literally pick a (proper) subgroup of largest order

[thorn jay]

bingle bongle dingla dangle

yickety do yickety da tippy tappy too ta

[spark veldt]

ooh i see okay thanks

[glad osprey]

Am I the only one disliking < here to mean strict subgroup? So confusing, I read it as "let H be a subgroup of G"

[south patrol]

No I mean then you would do <= innit

[thorn jay]

I guess it is already subsumed in the "maximal subgroup" part

[south patrol]

Could just use subset symbols though smh

[glad osprey]

Or just "if H \neq G"

[south patrol]

No need to even introduce H either smh

[thorn jay]

yeah

[noble nexus]

on the contrary, I hate when people use $\subset$ or < to denote non strict inclusion

[cloud walrus]

Blake

[thorn jay]

< for strict comparison is alright imo, the rest is eh

[noble nexus]

Using $\subseteq$ for subset and $\subset$ for strict subset is so natural

[cloud walrus]

Blake

[chilly ocean]

They are precisely the connected components of the action groupoid, so that fits!

[noble nexus]

nah that's even more demonic arguably, because < is always strict with numbers

[thorn jay]

🙏

[quiet pelican]

Unfortunately because some people don’t realise this is objectively correct, I usually use $\subseteq, \subsetneq$

[cloud walrus]

mico

[noble nexus]

Yeah that's why it sucks

having to use \subsetneq for clarity

[twilit wraith]

I used to do \subseteq but then I noticed no one cares

[thorn jay]

I honestly don't really like how \subset looks often in combination with letters such as C and then also using \cup

everything is too round

[karmic moat]

\subset sucks because one author will mean strict containment and another won't

[thorn jay]

I just changed all instances of \subset to \subseteq in what im working on and it looks much better

[tardy hedge]

[tough raven]

What is the endomorphism ring of ℤ/pℤ (+) ℤ/pℤ (+) ℤ/p^2ℤ (or more generally any finite abelian p-group)?

[tribal moss]

Hmm, one general strategy would be to consider the group as a subgroup of (Z/nZ)^m where m is the number of factors, and n is the highest power of p that appears among the orders -- namely, in each factor allow only elements from the subgroup of the appropriate summand. Then an endomorphism can be expressed as a an m×m matrix over Z/nZ with some conditions of the off-diagonal elements to ensure that each generator maps to something that's actually allowed in the other summands.

In the concrete example, I think we'd be looking at 3×3 matrices with entires in Z/p²Z
( a b c )
( d e f )
( g h i )
with the condition that c and f must be multiples of p.

Hmm, but then we need to quotient out matrices where a,b,d,e,g,h differ by a multiple of p too. Oof, that's somewhat messy.

(At least "a,b,d,e,g,h are multiples of p, and c,f,i are zero" is actually a two-sided ideal of "c,f multiples of 3", so it makes sense to quotient them out).

[tough raven]

OK, I got a pretty picture for this (or rather the case of nilpotent matrices, which replaces ℤ, p with k[x], x for k a field) which implies that the endomorphism ring has length 1^2 + 1^2 + 2^2 (and in general sum of squares of exponents of factors) as a ℤ-module, i.e., it has cardinality p^{that sum of squares}.

[tough raven]

Yeah, I think this kind of thing is the way to go. Specifically I can see 3 methods: let M be the group with r factors, the largest exponent of p being k; then (i) view the group M as a subgroup of L := (ℤ/p^kℤ)^r (IG this rs a bit messy because there's no guarantee of endomorphism of the subgroup extending to the larger group) (ii) view M as a quotient of L, so that endomorphisms of M are endomorphisms of L preserving the kernel, up to homomorphisms from L into M (iii) basically specify by mapping ℤ/p^kℤ summands into M, ℤ/p^{k-1}ℤ summands into M[p^{k-1}] = sum of <=p^{k-1} summands, extending that ... and so on until we've extended to the whole thing. (The last is what I did.)

That was a long rambly paragraph but here's the picture of approach 3.

(This is where there are c summands with exponent 3, b-c summands with exponent 2, and a-b summands with exponent 1.)

[spark veldt]

Can I get some help on this? I think orbit-stabilizer theorem is involved somehow I’m not really sure

I’ve written down the definitions but I’m still kinda struggling to make the connection

[velvet hull]

you haven't used the fact that the action is transitive

hint: ||if (x,y) is in some orbit, the transitivtiy guarantees that we can WLOG set x=alpha||

[spark veldt]

Oh i see so like basically the first term of the ( , ) can be kept constant and we’re free to make some kind of bijection from the second term to the orbits on Omega by the stabilizer of alpha?

[velvet hull]

something like that, you sound like youre on the right track

[spark veldt]

got it thanks

[karmic moat]

You have a ring morphism R -> R_i which gives each R_i an R-module structure

[tough raven]

R_i is a two-sided ideal of R_1 (+) ... (+) R_n, so yes.

or that

(both are the same module structure)

[karmic moat]

🤝

[tough raven]

I think this is the way to go - it seems easier to get at the ring structure this way.

By multiplying and seeing.

What's your definition of the ring R_1 (+) ... (+) R_n?

Better find that out first, then.

[rapid cave]

here R is an outer direct sum

which is isomorphic to your definition

you can view each R_i as a submodule of R using an embedding

R is tuples of the form (r1, r2, ..., rn) where r_i in R_i

so you have an embedding r --> (0,....,0,r,0,....0)

[karmic moat]

yes multiplication and addition in R is done component-wise

[thorn jay]

wow it's almost as if.. they should've used direct product notation...

no way...

[delicate orchid]

oh. my. science. it's just like le heckin vector spaces

[karmic moat]

#uponthewitnessing

[delicate orchid]

what are we talking about

[karmic moat]

mq is trying to understand the structure of a direct sum of rings

i think

[tough raven]

Are you sure R1 (+) ... (+) Rn has not been equipped with a ring structure?

[thorn jay]

well, there's really no other way to make sense of the "Noetherian" bit right?

[tough raven]

TBH that should be specified by the question (or previous definitions), not by us.

[thorn jay]

surely it's not Noetherian as an abelian group, and it carries no natural module structure from any of its direct factors

[elfin wraith]

I thought we solved this yesterday

[karmic moat]

yeah but it is what it is

[thorn jay]

we did lol

[tough raven]

Correct, so if there's no ring structure given, the question is underspecified.

[thorn jay]

again I pledge, they shouldn't've fucking used the direct sum notation 😭

wdym "ideal method", this is a question about ideals

how

[elfin wraith]

This was how my non com class did this

At least for the ideal structure stuff, I share this because my class also used the illegal notation

[thorn jay]

??? genuinely confused why they call it the direct sum

[elfin wraith]

It is particularly confusing because that lecturers work is quite categorical so she definitely knows lol

I guess she maybe just wanted to avoid any in-depth discussion of products etc

[thorn jay]

but its called a product for groups

:(

[karmic moat]

why are you always cooked before exams do you have exams like every other week

[noble nexus]

diabolical to call that a direct sum when the natural inclusion maps are not unital homomorphisms

[thorn jay]

bro's slowly cooking at low temperature for the midterms in a month

[karmic moat]

bro getting sous vided

[thorn jay]

yeah that's my main gripe lol

[delicate orchid]

"ermmmm the unit must be preserved because uhhh ermm uhh"

[thorn jay]

because else it wouldn't be a ring homomorphism, it would be a rng homomorphism

[delicate orchid]

you said ring twice

you mean it would be a ring homomorphism, not a unital ring homomorphism

[thorn jay]

do not ever talk to me again 💔

[noble nexus]

even if you only care about nonunital rings isn't the coproduct still the tensor product

[delicate orchid]

it is lol

or is it

[noble nexus]

actually maybe u don't have one

[thorn jay]

for commutative rings I can imagine the direct sum still being the free product

[thorn jay]

no all categories of algebraic structures are complete and cocomplete

[karmic moat]

ah okay

[rapid cave]

enpeace 🎵

[south patrol]

Or like nonunital rings are equivalently augmented rings ig so it is all chill

[elfin wraith]

My intro to ring theory class didn’t require ring homomorphisms to be unital, much to the annoyance of literally everyone involved

[rapid cave]

all rings are unital

[thorn jay]

I'll find out what our convention here is next quarter lol

[elfin wraith]

I agree, we also used that definition in the course, they just didn’t make homomorphisms map 1 to 1

[thorn jay]

what's even the point then 😭

[elfin wraith]

Edinburgh could be good, I think you’ll be in the new curriculum as well which will be even better

[elfin wraith]

Genuinely no one knew, the lecturer didn’t write the notes and he literally said it’s incredibly stupid please just ignore that and take care when they talk about homomorphisms, I will always require them to be unital

It’s not that cold, Edinburgh is very mild relative to the rest of Scotland, you’re in Sweden are you not? That’ll be colder

But yeah it’s not cheap

Very good

It was 16th in the world when I started, I think it’s slipped a bit now but it’s still top 50 for sure

For maths specifically it’s also very good, lots of very good algebraists and people in PDEs

The system for bachelors thesis at Edinburgh is a little odd but you will get a supervisor assuming you get to do one

[south patrol]

big up edinburgh

[elfin wraith]

If it’s required and they’re paired them I assume you are allowed but the system is weird

You either choose to do a 20 ECTS (but credits at Edinburgh are essentially meaningless) thesis on your own at a masters level OR you do a 10 credit UG level thesis in a group of up to 5

You also, unless you organise something in advance specially for you, choose 5 options from a big list of pre proposed projects and then it goes to a ranked choice lottery

Most people get their first choice, but like one or two people get one that wasn’t even in their options

[karmic moat]

group thesis is interesting

[elfin wraith]

Nah edi is genuinely one of the like 3 places worth living in the UK

[elfin wraith]

Yeah it’s an odd one, but you can all do entirely separate stuff and just stitch it together, that’s what I did

Yeah I mean you’ll need to speak to your people at your uni. As I said credits at Edinburgh are literally meaningless, but you can do a 10 or a 20 credit final project

I would recommend doing the 20 credit

Manchester and London

[delicate orchid]

WRONG

WROOONNGGG

[elfin wraith]

Certainly not fucking Aberdeen

Place should be nuked

[delicate orchid]

NEWCASTLE UPONNN TYNEEEE NUMBER 111

[karmic moat]

liver pool beatles

manchester oasis

[elfin wraith]

Ooh look at us we have 17 bridges and shit beer

[karmic moat]

wonderwaaallllll

[delicate orchid]

the beer is dog shit but the city is great

saying "london" is absolutely fuckin mental btw

[elfin wraith]

You’re watching too much right wing news, London is the most gentrified place in the world, it’s chill

[karmic moat]

sounds like you just said the city is shit

[delicate orchid]

it's half a step away from birmingham

[karmic moat]

if the beer isn't good what am i supposed to drink?

[delicate orchid]

technically it's ale not beer

[karmic moat]

unlivable

[elfin wraith]

Isn’t there a Greggs pub now, that’s sorta redeeming

[delicate orchid]

yeah there is

[karmic moat]

oh wow ale so classy omg

[rapid cave]

people in tokyo

[elfin wraith]

London is huge though, like it’s not all that dense relatively speaking

[elfin wraith]

Which is a style of beer, smh you’re not making the case of Newcastle

[rapid cave]

yep xD

[delicate orchid]

We’ve got like 90 Wetherspoons per capita just simmer down

[karmic moat]

reese?

[delicate orchid]

You make me sick

[karmic moat]

https://tenor.com/view/uncanny-cat-robert-helpmann-gif-11858302410808955060

[elfin wraith]

That’s actually an incredible stat, that does kinda sell me on Newcastle a bit

Leicester is really the odd one out there

Yeah I mean oxfords just kinda a town, no real reason to be there if not for the uni

[delicate orchid]

Leicester is a dump

[elfin wraith]

It’s in the midlands, that goes without saying

[south patrol]

thanks

[elfin wraith]

Warwick could offer me 50k a year to stay and do my PhD and I’d turn it down

[south patrol]

Bruh

If they offered me 50k i'd transfer very quickly

[delicate orchid]

Ts is a lie

[elfin wraith]

Have you been to the West Midlands

[south patrol]

Many times

My partner is from there

[delicate orchid]

Warwick is uncomfortably close to Birmingham but it’s just far enough

[elfin wraith]

And they’re no longer there for a good reason

[south patrol]

Offend potato speedrun jk

[elfin wraith]

Place is wank

Hate it here

I stand by Edinburgh Manchester London as being the only places worth living in the UK

Maybe Cambridge if you’re at uni

[delicate orchid]

U gotta stop saying London unc

No

[elfin wraith]

Actually revolting take

[delicate orchid]

Leicester is a bottom tier dump

I lived there for a year

[elfin wraith]

London is class im sorry, it’s maybe massively over priced but it’s not at all bad

Everything you could want there, and top tier public transport

[delicate orchid]

Huh wonder if that’s because they’ve been siphoning money from the rest of the country since 1947

[elfin wraith]

I didn’t say the reasons for it being good are also good, the UK being entirely reliant on one city is like a super not good idea, but unfortunately it does mean it’s significantly less shit than the rest of this dump