groups-rings-fields

[thorn jay]

yes

[elfin wraith]

Ah nice thank you, seems to be quite a clever proof

[rocky cloak]

It's just 4 elements in the matrix. With u and t on the diagonal

[crystal vale]

I also want to know how do i write any matrix in Smith normal form, if you understand then let me know

[rocky cloak]

Getting the elements along the diagonal is the most important part yes.

Pretty similar to row-echolon form from linalg indeed

[crystal vale]

So in Smith normal form, are we making the matrix into row-echelon form and column echelon form?

[rocky cloak]

Yes, and also you have a relationship between the diagonal entries where they divide each other

[crystal vale]

Source?

[rocky cloak]

I guess in this specific case d can just be a11, as you just take the first column and subtract a multiple of it from the second.

[crystal vale]

Yes

[rocky cloak]

No a11 just stays a11.

But you subtract away a12 in the second column leaving you with 0

Presumably they go on do/ have done some general trick where having a row on that form is helpful

Otherwise a12 would not be a multiple of a11

(so then you wouldn't get 0)

[crystal vale]

Thank you

So this theorem 8.8 is equivalent to fundamental theorem of finitely generated module over PID, right?

[rocky cloak]

How? What do you mean subtract directly?

Multiplying a column by 0 is not invertible, so would not be great

If that was allowed then I guess every normal form would just be the zero-matrix after multiplying every column by 0

[balmy python]

i just learned the concept of an automorphism of a group

do they always map generators to generators?

[tribal moss]

In the sense that if you have some set that generates the group, its image under an automorphism will also generate the group, yes.

[rapid cave]

yes

[crystal vale]

If H is any subgroup of G, then is [G, H ] \normal in G?

[wraith cargo]

Yes

[crystal vale]

How do I show it, I am thinking of characteristic property

Is this subgroup a characteristic subgroup?

[quiet pelican]

No, take any normal non-characteristic H

Then [G, H] = H is not characteristic

[wraith cargo]

Oh wait

Why tf did I think H was normal in your assumption

I can't fucking read

[crystal vale]

If H is normal then [G : H ] \subset H, how do I show the reverse one?

[quiet pelican]

Actually I don’t think the reverse holds
But like that sort of reasoning (ie, consider a normal non-characteristic H) should give you what you want fairly quickly

[crystal vale]

oh if i take G = A5 then H = G\times {1} is normal but not characteristic in K = G \times G. Also, [K, H] = H , because A_5 is perfect i.e., [ A_5, A_5 ] = A_5.
Hence, H is normal in K, but not characterisitic so [K, H] is not characteritic in K
is it correct?

now question is, [G, H] is normal in G for any subgroup H?

[crystal vale]

and if H is normal why [G, H] is normal in G?

[quiet pelican]

Yeah that works

[quiet pelican]

x[g, h]x^{-1} = [xgx^{-1}. xhx^{-1}] \in [G, H] if H is normal

[swift tundra]

I have been thinking more about the structure Zariski Topology, and how it seems very similar to a topology you might put on a general poset (the closed sets are the upwards closed sets). Although I understand this gives perhaps a slightly diferent topology. For example in Spec(Z) (regarded in this moment as just a set), the set containing all nonzero primes is upwards closed, but I believe is not closed in the zariski topology (an integer is only divisible by finitely many primes). But it seems like there is still som sense in which the zariski topology is a more general phenomeon. Like maybe given some Poset (P, <) you could take the smallest topology generated by the closed sets V(x)={y\in P: x < y}.

Appologies if this doesn't make sense. Thinking about closed sets in the zariski topology as upwards closed has helped me understand it more, and I am wondering if it generalizes more.

[quiet pelican]

No
Consider the wreath product Z wreath C_3 where the latter acts by cyclic permutations. Then if we let H = <((1, 0, 0), e)>, [G, H] = <((1, 0, 0), e), ((0, 1, 0), e)> is not normal

[crystal vale]

I don't know about wreath product

[quiet pelican]

Do you know about semidirect products?

If so, my example is Z^3 semidirect C_3, where the latter acts by cyclic permutations on the 3 factors

[crystal vale]

Not much

[delicate orchid]

well if you can't understand this then it's the group <a,b,c,u | [a,b] = [a,c] = [b,c] = u^3 = 1, a^u = b, b^u = c, c^u = a> and they're taking the subgroup generated by a, the only non-trivial commutator is [a,u] = ba^-1 so (as claimed) [G,H] = <a, b> which isn't normal as b^u = c

[crystal vale]

Oh

I will verify the details

Thanks mico and Wew Lads Tbh

[crystal vale]

There is a statement that if S is any subset of a group G, the mapping property defines a homomorphism \phi: F -> G from the free group on S to G.

A family S of elements is said to generate a group G if the map \phi from the free group on S to G is surjective.

And if phi is isomorphism then G is called a free group.

My doubt is what if I take S = G?

[crystal vale]

I got it

[crystal vale]

It is interesting that a free group on two elements contains a free group on three elements

[crystal vale]

can i get more examples where G embedding in K and K is embedding in G, but are they not isomorphic?

[vocal pebble]

Well it also contains the free group on countably many elements by the same construction

[elfin wraith]

How do you do this? I’m imaging some sort of like covering space argument or something like that

[vocal pebble]

I think it's a lot simpler to embed, say F(N) into F_2 by sending n in N to y^nxy^{-n}

[elfin wraith]

Oh yeah that’ll do lol

[rocky cloak]

G = (Z/4)^N and K = G x Z/2

Classic infinite swindle

[azure hull]

[
A = \prod_{i=1}^{\infty} \mathbb{Z}, \quad A \cong A \times A
]

[
G = A, \quad K = A \times \mathbb{Z}
]

There are injective homomorphisms:
[
G \hookrightarrow K, \quad g \mapsto (g, 0)
]
and
[
K = A \times \mathbb{Z} \hookrightarrow A \times A \cong A = G
]
(using the embedding $\mathbb{Z} \hookrightarrow A$).

Thus,
[
G \hookrightarrow K \hookrightarrow G, \quad \text{but } G \not\cong K
]

[cloud walrus]

Yuxuan Wang

[rocky cloak]

G is isomorphic to K here though

They are both just infinite products of Zs

[delicate orchid]

ts pmo to a level never before seen

[azure hull]

if we assume for contradiction, that
G = \prod{i=1}^\infty \mathbb{Z} \cong G \times \mathbb{Z} = K.
then there exists a group isomorphism ( f: G \to K ). Composing ( f ) with the natural projection ( \pi: K \to \mathbb{Z} ) onto the second factor, we obtain a group homomorphism
\pi \circ f: G \to \mathbb{Z}.
this would be a surjective homomorphism from ( G ) onto ( mathbb{Z} ). however if i remember correctly a result that states that any group homomorphism from  G = \prod{i=1}^\infty \mathbb{Z}  to \mathbb{Z}  depends only on finitely many coordinates. in particular, such a homomorphism has image contained in a proper subgroup of ( \mathbb{Z} ), specifically of the form ( n\mathbb{Z} ) for some ( n \geq 0 ). Therefore, no such homomorphism can be surjective.
on the other hand obviously the projection ( \pi: K \to \mathbb{Z} ) is clearly surjective contradiction shows that our initial assumption must be false?

[tribal moss]

"pmo"?

[azure hull]

piss me off

[tribal moss]

The isomorphism is simply f: Z^N -> Z^N × Z : f(a1,a2,a3,a4,...) = ((a2,a3,a4,...), a1).

pi o f maps (a1,a2,a3,...) to a1, which indeed depends on only finitely many coordinates, but its image is the entire Z.

You can, if you wish, write the image as 1Z, (which is indeed nZ for some n>=0) but that's still not a proper subgroup.

[azure hull]

yea u r right i was misinterpreting theorem to mean that no homomorphism from G to Z could be surjective

should i delete the example ?

[tribal moss]

Just leave it up; otherwise the subsequent conversation becomes confusing.

[azure hull]

okk cool thanks for clearing it up

[supple pecan]

hi new to groups and geometry but how would translations be a bijection onto the same set?
surely some points would fall "outside" the set

[tribal moss]

The set in question is all of R^2.

No matter how far you slide a point that was in R^2 to begin with, it's not going to fall off.

[supple pecan]

but arent we considering a finite subset X of R^2?

we go on to talk about the example of a square and how it has 8 symmetries; but none of them involve translations, how would a translatoin produce a symmetry on a subset of R^2?

[tribal moss]

It doesn't say X is finite.

[supple pecan]

oh so would translations only work on infinite/unbounded sets?

[tribal moss]

Oh, I see how the screenshot could be confusing.

In the first paragraph which defines "isometry of X", X is an arbitrary subset of R^2.
But starting from the line saying "Example 1.1.2" it says "isometries of R^2", which means that in that paragraph X has been chosen to be R^2 in particular.

[supple pecan]

oH im blind tyty

[supple pecan]

so in general translations wont produce isometries for finite sets?

[tribal moss]

Correct.

[supple pecan]

thanksss

[tribal moss]

(This is also the case if you consider bounded sets rather than just finite ones -- which I suspect is what you were actually thinking of).

[supple pecan]

how would a set be bounded but not finite?

[tribal moss]

For example, the unit circle is a bounded subset of R^2 -- every point on it is within a distance of 42 of the point (0,0).
But it is not a finite set -- there are infinitely many different points that lie on the circle.

[crystal vale]

got it

[crystal vale]

got it

[crystal vale]

why?

i have to prove, if p is a prime number and if N is the number of words of length p in a finite set S, then N is divisible by p.

if S = {1,2,3} and p = 2, then N is 9, right?

[elfin wraith]

Assuming your first statement is true then the second one can’t be. I think the number of words is 6 but counting in my head is hard

[crystal vale]

what about 22?

11, 12, 13, 21,22,23,31,32,33

[tribal moss]

So N is |S|^p?

This sounds like a "please show the entire problem as it was given to you" situation.

[crystal vale]

it is in artin

[elfin wraith]

Yeah I only assumed 6 because I guessed we were excluding repeats and I think that makes it work? It does in that case if not in general

[tribal moss]

Weird.

[elfin wraith]

Well yeah unless there’s something in how he defines words you’ve just shown that isn’t true

[rocky cloak]

Are they defining "word" in an unusual way perchance?

[tribal moss]

"words of length p in a finite set S" is a weird wording in the first place.

[crystal vale]

they define word as finite string of elements of S

[tribal moss]

Can't we just choose S to be a set consisting of a single "word of length p", whatever that would mean in the context?

Hmm, but the heading "Generators and Relations" kinda suggests that "word" is not a matter of formal language theory, but instead the special kind of words considered when constructing free groups.

[crystal vale]

not that section is later, after this exercise

[tribal moss]

Yes, but it still gives some information about the general kind of context the exercise is found in.

[vocal pebble]

Well if you take words of length k without repeats, then the total number of such words will be (|S| choose k)•k! Which is always divisible by k even if k isn't prime so you don't need the hypothesis that p is prime, which suggests that it's something else

[tribal moss]

If the sections are the same as in my copy of Artin (which must be 1st edition since it doesn't seem to state a number), then the section immediately before "8. Generators and Relations" is indeed "The Free Group".

Oh, and the exercise is there verbatim in my copy too.

[crystal vale]

yes

[tribal moss]

So I agree your example here shows that the claim cannot be true as stated.

The text doesn't seem to give "word" any special meaning that could salvage it.
(I had wondered whether it could mean "reduced words that may contain inverses of the generators", but that doesn't match how the actual chapter speaks about reduced words anyway).

[crystal vale]

yes

maybe that's a typo, i have to see new edition

oh this problem is missing in new edition

[elfin wraith]

Probably confirms that is just irreparably wrong then

[tardy hedge]

Thinking about stuff in the kernel being about “relations” , is that only a correct perspective in the case S->>M with S a free module on the set M?

Probably not right but right now i feel comfortable in that case but not sure how it generalizes yet i guess?

[thorn jay]

kernels are always relations of some kind

be it as an equivalence relation or not

[tardy hedge]

If you want the kernel to keep track of all relations in M, then you need the free module right?

With the surjection too

[thorn jay]

sure, yeah

given some subset S ⊆ A then theres a unique homomorphism f : F(S) → A where F(S) is the respective free object generated by S such that f(s) = s. The kernel of this homomorphism is exactly the set of algebraic relations between elements of S

[tardy hedge]

Ok yea cause my supervisor i think does things with “syzygies” and this is related

Hilbert syzygy theorem or something

Free resolutions

[thorn jay]

yeah, the nice thing about modules is that kernels are faithfully represented as subobjects

[tardy hedge]

Which is what allows us to do the free resolution of any module yeah?

[crystal vale]

Can I expect something from free groups and free modules, are there any relation between them?

[elfin wraith]

Hardest thing about these is remembering how to spell them

[thorn jay]

its what makes the consept of an exact sequence have any meaning in the first place

[elfin wraith]

Eisenbuds book on them is quite good

What I’ve read of it anyway

[rocky cloak]

They both satisfy the universal property of being free.

Namely, the free group on X is a group F(X) generated by X such that any function from X to a group G extends to a homomorphism from F(X).

The free R-module is exactly the same, but replacing the group G with an R-module. (And replacing homomorphism of groups with homomorphism of R-modules)

[thorn jay]

a free abelian group is a canonical quotient of a free group too

[rocky cloak]

Composition of adjoints is adjoint

[thorn jay]

reducts and subvarieties induce adjoint functors 🔥

[crystal vale]

I see

[tribal moss]

Artin does state the universal property of the free group, but only in the next section (ostensibly about generators and relations) and under the name "mapping property of the free group"...

[crystal vale]

Yes

I want to learn more about free groups

Okay G itself generated G, how do I sure about relations here?

[rapid cave]

{"gh=k" | g,h,k in G}

What I mean is the relations are the different results of the group operation

[crystal vale]

Is there any element in A5 which has order than 1,2,3,5?

[noble nexus]

don't believe so

the only element of order 4 in S5 is a 4 cycle (since you can't get 4 as the lcm of two numbers smaller than 5)

which isn't in A5

[crystal vale]

Or 6, which is also not possible

[noble nexus]

well you can in S5

as the product of a 2 cycle and a transposition

[crystal vale]

But not in A5

[noble nexus]

but you need a transposition yeah

[vapid vale]

great username

[noble nexus]

ya im surprised it wasn't taken

[crystal vale]

5-sylow subgroup is normal in A5, right?

No

I thought it is pqr

[noble nexus]

sylow 5 subgroup is just cyclic

in a5

[quiet pelican]

A5 is simple, so has no non-trivial normals

[vapid vale]

there is an easy way to say whether a subgroup of A_5 is normal

[crystal vale]

Yes

So can I expect a group such that for any three distinct primes I can get group G such that any element in G has order 1,p,q,r only

I mean for all three distinct primes there exists such group

[delicate orchid]

cite the classification of finite simple groups

[crystal vale]

?

[crystal vale]

[vapid vale]

i dont believe this

[crystal vale]

Counterexample?

[thorn jay]

it just seems like there would really need to be an element of another order

[crystal vale]

If [ G : H ] is finite then [ G : N_G(H) ] finite yes, because we can map [ G: N_G(H) ] -> [G: H] by r N_G(H) -> rH, which is injective

[delicate orchid]

for two primes it works if q|p-1

you take C_p \rtimes C_q

[rocky cloak]

This paper classifies groups where every element has order a prime power
https://projecteuclid.org/journalArticle/Download?urlid=10.1215%2Fijm%2F1258130990

[crystal vale]

I agree

[delicate orchid]

all sylows would be cyclic so it would have to be metacyclic and solvable

that's all I can tell you in general

ahhh not true. Just exponent p or q or r or whatever

[crystal vale]

Thank you jagr ❤️

I don't think I need much for reading this paper, do I need?

[delicate orchid]

you'll need to know what the suzuki groups are I guess

[vapid vale]

https://arxiv.org/pdf/2012.01482

oh jagr beat me

this is different anyway (Theorem 1.7)

[delicate orchid]

oh this is really cool

[crystal vale]

What is the Suzuki group?

I don't know what is locally finite group

I think I have to learn more things

[delicate orchid]

they're a family of 2-groups with weirdo properties

[delicate orchid]

this paper seems better, notknow

[rocky cloak]

If I'm parsing the results correctly, then either just two primes divide the order of G or there are only finitely many examples.

Thus it should not be possible for all triplets of primes

[crystal vale]

But it used graph theory, i don't know graph theory too

[vapid vale]

you dont need to know graph theory

[delicate orchid]

do you know what a graph is

can you visualise points and lines

[crystal vale]

Vertices and edges

[crystal vale]

I can

[delicate orchid]

you're good to go!

[vapid vale]

this is sufficient

xdd

[crystal vale]

Thanks hk

[delicate orchid]

ok this result is really cool, trivially it works for 1 and 2 primes, but for 3 and 4 you need either sporadic or familes of finite simple groups and then it just stops working??? wtf????

am I understanding this right cause that seems surprising

[vapid vale]

im continually shocked that there are just people in the world that can do and like doing this sort of group theory

[delicate orchid]

I just checked PSL(3, 4) and yup it works

[delicate orchid]

I kind of do this type of group theory lol

[vapid vale]

xd

[thorn jay]

this is such a weird thing lol

[elfin wraith]

That theorem looks truly whacky lol

Cool, but crazy stuff

[delicate orchid]

any monster group enjoyers?

[thorn jay]

Theorem 1.7 is actuall crazy

tf

[delicate orchid]

it's pretty standard for these sorts of things to have a random bunch of bullshit and some infinite families

[elfin wraith]

It’s such a random bunch of bullshit though, finite group theory is weird

Super cool, not for me

[thorn jay]

i adore classification results

honestly i want to publish some cool and deep UA classification result in my life

something like the classification of minimal algebras or of abelian algebras

[elfin wraith]

Oh same, I’m pretty sure the opening line to my UG diss was me saying they’re the best thing in maths or something to that effect

[delicate orchid]

I drew it

what is 29 doing

WHAT is going on

[thorn jay]

LOCK TF IN

[kind temple]

he has one friend

[thorn jay]

47 too

[kind temple]

they need to get together fr

[delicate orchid]

didn't bother with the disconnected pieces

the sigma primes 🐺🐺🐺🐺🐺🐺🐺

[thorn jay]

so me fr

[thorn jay]

oh cool lol

what was it about

[delicate orchid]

https://arxiv.org/pdf/1103.2047 Theorem A goes crazy

[slow bison]

prime graph? what group is this?

[elfin wraith]

It was titled betti numbers in algebra and topology, did some combi comalg, ch2 of Hatcher, some TDA, and some homalg

But I kinda used the opening to be like, we don’t have a classification for topological spaces, even making invariants is hard type thing

[delicate orchid]

the monster

[delicate orchid]

from this table

[slow bison]

oops i should have scrolled up thanks

[kind temple]

dawg

[delicate orchid]

I actually use this result in my thesis

[elfin wraith]

How the fuck do you even come up with that

[thorn jay]

what the FUCK

[delicate orchid]

there's big theorems which let you eliminate a shit ton of groups really quickly

[slow bison]

crazy stuff o.o

[thorn jay]

dawg what is group theory even about 💔

[kind temple]

it’s just a competition to come up with the most insane looking theorems that you can

[thorn jay]

what are the primitive relations' intuition / used for?

[delicate orchid]

primitive relations, and I'll explain

[thorn jay]

mm i see thats cool!

[thorn jay]

ive gotta lock in

[elfin wraith]

“So all groups are either the trivial ones of order 2 and 3, in one of these 5 sensible infinite categories, or the fucked group of order 518,151,846”

[thorn jay]

Theorem Z: New NONCYCLIC group of order 3 found

[delicate orchid]

consider the rational representation ring R_Q(G) of a finite group, let B(G) be the burnside ring of G (isomorphism classes of G-sets, addition given by disjoint union, product is given by direct product). Then there is a linearisation map B(G) -> R_Q(G) sending a G-set X to the G-module Q[X] (Q-vector space with X as it's basis set equipped with the inherited action of G). This map has a kernel K(G) of the "Brauer relations" of G. Due to biset functor bullshit, you can induce/inflate between these kernels. The primitive relations are precisely those which do NOT arise as a series of inductions and inflations of relations of subquotients

[delicate orchid]

might want to double check the associaitivity on that one, zunc

rock paper scissors ahh

[thorn jay]

up to isotopy theres only one quasigroup too

hell, up to transformation

[delicate orchid]

yes because it's either C_3 or ROCK PAPER SCISSORS

[thorn jay]

thats not a quasigroup lil bro

[elfin wraith]

“Yeah this is just a virtually semi polycyclic bifinite group that’s naturally the fundamental group of this physics manifold, it disproves the reimann hypothesis”

[delicate orchid]

that's my POINT

[delicate orchid]

the field with one element actually PROVES the riemann hypothesis yunc

[thorn jay]

•
look i stole it

[elfin wraith]

So they say

[delicate orchid]

nice point, where'd you get it? The contractible spaces shop?

[elfin wraith]

But I was more getting at the unit conjecture, it’s a polycyclic bifinite (complete intersection? Something like those words) group that’s the fundamental group of some weird but known manifold and it was the first counterexample found

[thorn jay]

the contractible spaces shop sounds horrible, they only sell one single product

[delicate orchid]

is this how I sound to u mfs

[elfin wraith]

I also know that because it’s the fundamental group of that manifold, apparently K theory would never have worked to disprove the conjecture and I’ve got no fucking clue what that really means

[elfin wraith]

Yes. This is what group theory is.

[delicate orchid]

I guess K-theory of manifolds is easy because they're just R^n

[thorn jay]

i see, so primitive relations are anything that's really "the group's own", cool

wouldnt know how to use that in any way but im sure its useful

[delicate orchid]

it just basically solves how this kernel behaves completely

see Theorem B

which means you can use group actions to essentially deduce almost all of the rational representation theory of G

[elfin wraith]

This is true, but I never worked out what that means really, Giles Gardham said it though and I trust the man

[delicate orchid]

I say almost all cause unless G is a p-group there's a tiny cokernel

[elfin wraith]

But hey maybe I’m becoming a K theorist this year, maybe it’ll all become clear

[delicate orchid]

excellent. Then I can vomit spectra bullshit at you

[elfin wraith]

I did see spectra and Mackay functors in some of the papers I skimmed and I thought of you

[delicate orchid]

actually are you doing algebraic K-theory or topological K-theory

how wholesome

[thorn jay]

paper SOLVES representation theory, leaves group theorists BAFFLED

[delicate orchid]

over positive characteristic it was solved in the 1950s

[elfin wraith]

No clue, a potential advisor said he had something in mind with some nonsense using Milnor K theory, he drew a horrible diagram and wanted me to find when something was an isomorphism or something like that

I am no where near actually understanding K theory yet

[delicate orchid]

good project. Do it.

[elfin wraith]

The other option I have is to do rep theory in monoidal categories which also seems very cool

[delicate orchid]

better project. Do that

[elfin wraith]

I’m kinda between those but I’ve got another person to talk to

[thorn jay]

sleeper agent activated upon hearing "representation"

[delicate orchid]

how would this go?

[elfin wraith]

Yeah I’m leaning more towards that, but like, shit sounds fucked, professor is scary

[elfin wraith]

I forget the exact specifics of it, but it’s to do with Tannakian reconstruction in Verlinde categories, and the prof has some conjecture that’s been shown up to like char 5 and he wants me to try to generalise it

But honestly right now I can only almost understand the definitions of the category

[delicate orchid]

I thought you said "shown in char 5 [only]" and I nearly fell out of my chair

[elfin wraith]

And my rep theory is borderline nonexistent

[elfin wraith]

It possibly is, he did say it’s pretty fucked and like it’s not that clear how to move it on from where it is

He kinda just roasted me for 30 minutes prior to this so I wasn’t the most focused

[delicate orchid]

I only really know Tannakian duality for G-sets icl

[delicate orchid]

hmm not the best vibes

[thorn jay]

how hard can the more general case be

[elfin wraith]

I think it’s just a Russian prof diff tbh, I don’t know that it was intentional but yes that is the concern in my mind that’s pushing me towards the K theory stuff

[delicate orchid]

https://tenor.com/view/cat-osu-greece-greek-cat-osu-cat-greek-gif-2694212203346871017

[thorn jay]

Russian profs are crazy

weve got one here and im nervous just being around him that ill do something stupid lmao

[tardy hedge]

I feel like even if its not intentional it'd still make u feel uneasy when u got to meet him every week

which probably wont be too fun

[elfin wraith]

Yeah that’s kinda where I’m at

[delicate orchid]

the russians are built different I concur

if you go with that guy you will learn LOADS

[elfin wraith]

I’ve heard he’s lovely but the project sounds really hard and I’m concerned that I’ll just disappoint him

[thorn jay]

though he did put a convo with him and chatgpt in the recent lecture notes showing him getting increasingly annoyed that chatgpt couldnt write a proper poem about math

[elfin wraith]

but if I pull through I’ll be cracked

3 months to prove a conjecture in full generality that experts have managed 3 cases of? Easy stuff

[thorn jay]

i think even one extra case is great

[elfin wraith]

Yeah I think honestly even understanding the existing cases is plenty for a masters thesis, like there’s not even any courses at this uni that would theoretically prepare you to even do this project

It’s only because of the ring theory and quantum programming I did in my UG that I’m vaguely aware of what he’s on about

[delicate orchid]

you will not do ts in 3 months

I hate to be a hater

[elfin wraith]

No im well aware

[delicate orchid]

I thought this was a PhD level project when you said it

[elfin wraith]

No this is my masters thesis

Bros just cracked and seemed to be disappointed I’m not too

[thorn jay]

thats fucking crazy for a masters thesis 💀

[elfin wraith]

I mean even the K theory one the guy wants me to work on some conjecture that was made like 5 years ago, bro this uni doesn’t even cover Hatcher ch4

[thorn jay]

your uni has hatcher?

[delicate orchid]

I never did alg top lol

[crystal vale]

Which uni are you in?

[crystal vale]

How is this possible?

[elfin wraith]

We cover up to ch3, which is nice at least

[elfin wraith]

Warwick

[thorn jay]

i think our uni uses Bredon for algtop courses if they even exist

[delicate orchid]

it's very possible

[thorn jay]

which yeah they prolly do ig

[thorn jay]

i never did alg top either :P

[elfin wraith]

Is that topology and groupoids? I think that’s what our cat theory course uses

[delicate orchid]

pointset was a third year course and then they just didn't have any more

you have a cat theory course?

[thorn jay]

Topology & Geometry

[crystal vale]

But haven't you done in your graduation?

[delicate orchid]

I graduated in 2022

I don't know what you're asking

[elfin wraith]

Yeah, somewhat disappointed I didn’t do it at Edi with Tom Leinster though, but also somewhat not disappointed…

Actually main reason would be that if I had I could’ve took Sheafs and Higher Cat theory with Clark Barwick

[delicate orchid]

I didn't do alg top but I can tell you how to set up a grid based fluid simulator using Lax-Friedrich's

[crystal vale]

I mean, in my country, first we have to do undergraduate, then graduation then PhD, so in my graduation I have alg top in semester III

[delicate orchid]

yes I also did that, I went to an applied focused uni

[elfin wraith]

I belive wew does now know algtop yes

[delicate orchid]

wtf is a nerve 💔

[elfin wraith]

The preimage of uhhhh

[thorn jay]

simplicial set smt smt

[elfin wraith]

Something like that

[delicate orchid]

WRONG

IT IS A TRIANGLE

[elfin wraith]

Is it not? Thought it was something about coverings and like preimages of the projection

[karmic moat]

I was just wondering how to do that how could you tell

[thorn jay]

(this gives a description of Cech homology in terms of the homology of some simplicial object)

[elfin wraith]

Genuinely don’t remember, not something I’ve needed

[elfin wraith]

Ah, for Cech homology you just plug into Pegasus

[rocky cloak]

It is true, yes

[rapid cave]

(inner) direct sum is about uniqueness

its not yeah

if A1=A2=A3...

then the sum is just A1 while the (outer) direct sum will be A1^n

[cloud walrus]

mq

[rapid cave]

no mistake

what are you confused about

what is the first step and what is the second step

also what are you proving?

an infinite sum might be different then all of the finite sums

[balmy python]

what're some good tips to doing well in group theory questions

and like finding counterexamples to certain statements

or trying to prove certain statements that don't seem that straightforward

i've never heard of these 😭

[rapid cave]

then ignore them for now

but eventually you should learn about them

[balmy python]

well i've heard of them

but

never came across them

my course doesn't mention them tbh

[balmy python]

like right now i've done up to the first isomorphism theorem and i know the proofs and stuff but when it comes to doing questions it's a lot different

not yet

not yet

went through that today acc

[rapid cave]

if you have a specific question, you are welcome to ask here or in the help channels

[balmy python]

map from an image of a homomorphism to G/Ker(f)

i think i struggle more with coming up with counterexamples

[rapid cave]

yes

||hint: binomial theorem||

[balmy python]

oooh i get to do rings in 3 weeks time

😭

[rapid cave]

its fun

[balmy python]

my course is quite rapid

but smth like gt takes time for me to marinate in my head

[rapid cave]

hmmm

fields are the best

[balmy python]

like don't spoil it but with a question like this, I don't know whether i'm right or wrong 😭

[rapid cave]

yeah I was typing too fast

[balmy python]

and i don't know if im being stupid and the answer is right there

[thorn jay]

module theory is NOT group theory

it's closer to category theory

[rapid cave]

real

[thorn jay]

they're abelian groups with a ring acting on them

you wouldn't call group actions sets, would you?

[balmy python]

could i have a hint for this?

not a bait one

[thorn jay]

exactly

[rapid cave]

do you know that every subgroup of index 2 is normal?

[balmy python]

what does index mean 😭

[thorn jay]

the index of a subgroup H < G is the amount of left-cosets of H in G

[balmy python]

Also if someone told me to describe a group would that mean just state properties like cyclic, abelian etc

[thorn jay]

denoted [G : H]

[balmy python]

ohhh yeah where lagranges theorem comes from

[thorn jay]

yeah

|G| = [G:H] * |H|

[balmy python]

i did not know ever group of index 2 is normal 😭

[rapid cave]

describe the entire structure of the group, using known groups and ways to combine them

[thorn jay]

give the Jordan-Holder decomposition of the group

[rapid cave]

or using generators and relations

actually there are a lot of things this could mean

do you have a picture?

[balmy python]

yeah 😭

yep

don't spoil though i want to try ofc

gxg^-1?

i've been staring at that for time 😭

[thorn jay]

yeah probably give an explicit description of the elements

[balmy python]

really?

[rocky cloak]

They might mean, which group we have introduced a notation for are these isomorphic to

[balmy python]

yeah 😭

i think you can have elements other than squares

[balmy python]

ohhh okkk

otherwise it's quite trivial i think

[rocky cloak]

That would at least count for a good description in my book

They could also mean list out the elements and they're combinations...

Both are doable

[balmy python]

for a group of homomorphisms? 😭

sounds long

[rocky cloak]

Not really very long in these cases

[karmic moat]

aren't minimal ideals defined to be non-zero

[thorn jay]

you mean the fact that any set of ideals has a minimal element?

well, if your set contains the zero ideal then that's automatically the minimal element

well yeah obviously, the definition would be useless otherwise lol

[karmic moat]

yeah so then this is true just by definition, dont need artinian or anything

[balmy python]

anyone able to give a hint to this?

[rapid cave]

take an element of gHg^-1

show its in H

[balmy python]

hmm yeah ive been trying that

[thorn jay]

no

[balmy python]

the hard part is that if x E H then x may not necessarily be a square

hmmm

[thorn jay]

you can also use \in instead of E, everyone will know what you're saying :>

[balmy python]

oh yeah sorry 😭

should use latex

$x \in H$

[cloud walrus]

pentium

[thorn jay]

I think it's quicker to just write out the latex for this kinda stuff

like in text

looks ugly

and is way less readable

[rapid cave]

btw, if you know H is normal iff G/H is a group then you can solve this easily

@balmy python

[thorn jay]

I don't see how tbh

[rapid cave]

||G/H will be a group... really inverses are the only thing preventing that. And every coset will be its own inverse||

[balmy python]

i got that gh^2g^-1 \in H and g(gh^2g^-1)g^-1 \in H which might be able to lead somewhere

[balmy python]

hmmm

i need to memorise definitions properly 😭

[thorn jay]

||no the thing preventing it being a group is well-definedness of the operation||

[rapid cave]

Oh yeah xD

[rapid cave]

if you want to show ghg^-1 in H, look at (gh)^2

[balmy python]

aw man that was easier than i thought itd be 😭

ghgh

then apply h^-1 on the right

then (g^-1)^2

boom its in H

[thorn jay]

oh that's clean

[balmy python]

how did you even find that i was just staring for a while 😭

[rapid cave]

Tried random things and that works

[balmy python]

that was too big a spoiler

next time make it harder for me

[rapid cave]

Dont know how to make that easier

Except making it a puzzle for you

Which sounds annoying more then helping

[south patrol]

Groups.

[tardy hedge]

He’s right^

[south patrol]

Me?

[tardy hedge]

Finitely generated module M isn’t necessarily projective because you dont actually have complete control over where you can send generators of M right

[tardy hedge]

Sorry btw, i forgot you use they/them pronouns

[south patrol]

Indeed. But also just think of the fact projectives are summands of free things

[south patrol]

No dw lol I was just unsure

[south patrol]

So ljke

Z/2 as a Z-module for example

[thorn jay]

well usually potatoes can't talk

[tardy hedge]

So a fg module that is not free cannot be a direct summand of a free module?

[south patrol]

No this is not true

(Though happens to be true for Z)

[thorn jay]

PID ahh

[tardy hedge]

Lol

[south patrol]

My point was more like

[thorn jay]

(free modules cannot have torsion elements, so by the structure theorem every fg submodule of free module must be free)

[south patrol]

The presence of torsion here

[south patrol]

Well don't even need fg here

Just any submodule of a free module over a PID is free

[thorn jay]

right right

zorn n allat

[south patrol]

Real

[tardy hedge]

Every fg module that is not free has torsion element?

[south patrol]

Well-ordering jumpscare

[thorn jay]

(over PID)

[south patrol]

Not generally no

Again like good to just have any basic examples of projective non free stuff

A nice example is k^n as a left M_n(k)-module

[tardy hedge]

Yea earlier today actually this started with me thinking that i dont even really know examples of projective modules that arent free

[south patrol]

Ah fair

[thorn jay]

honestly same lol

gotta up my game

[south patrol]

Ig issue is that stuff is usually good over PIDs or local rings lol

[thorn jay]

can you blame me for preferring to think over PIDs and local rings ;w;

[south patrol]

Lol why

[thorn jay]

at least noncomm rings have nice properties

well tbf I basically work with monoid-like objects

[south patrol]

Like rings

[thorn jay]

clones; monoids where the multiplication is "elongated" like n-ary compositions rather than binary composition

[south patrol]

Real

Yeee lol

No u

[tardy hedge]

So then back here this logic is only relevant if the ring is a PID or w/e right

[south patrol]

Tbh that was just to generate examples of non-projective fg stuff but being a PID isn't really needed. For example if you consider a module over a domain and you start getting torsion then things have gone wrong

And similar things in non-domain settings

[tardy hedge]

Ok so to summarize this is what I'm going with:

In general, I can think of a module with a generating set doesn't imply projective bc u cant always control where generators go. If R is a domain, you can also think of how M might have torsion so then it cannot be a direct summand a free R-module

[chilly ocean]

Is there any useful(or at least somewhat interesting) generalization of a ring in a groupoid-like direction? Say, the underlying additive or multiplicative(or both) structure is a groupoid, not in general a group, but still works as expected with addition when defined, e.g. a•(b+c)=a•b+a•c and so on?

[thorn jay]

thats basically an additive category right?

a ring is an Ab-enriched monoid, so the oidification is an Ab-enriched category

ah, a preadditive category

[chilly ocean]

I guess a categorical definition is probably the best I can hope for, but is there any purely algebraic definition(e.g. defining it as a set with two potentially partial operations such that […]) like a groupoid has?

[twilit wraith]

How is possible that theres a non normal subgroup of Q8 x Z4

Apparently there is one but I have no clue how

[tough raven]

Yes.

[tough raven]

If they have a common subgroup H (up to isomorphism) which is not central in Q8, you could consider the diagonal copy of H in the product. Conjugation by an element of Q8 will preserve the fact that the left components of the elements of the subgroup form H but will shuffle them around, resulting in a different subgroup of the product.

[twilit wraith]

Oh I got u

[tough raven]

Explicitly, H = {1, i, -1, -i} \cong Z4 will do. We can take as a subgroup {(1, 0), (i, 1), (-1, 2), (-i, 3)} and conjugation by (j, 0) will turn it into {(1, 0), (-i, 1), (-1, 2), (i, 3)} (since jij^{-1} = -jij = -i).

[tough raven]

Fun problem BTW; thanks for it.

[halcyon peak]

S -> R is a ring morphism (not necessarily commutative), M a left R-module, then i see some places (like nlab https://ncatlab.org/nlab/show/coextension+of+scalars) define the left S-action on Hom_R(S,M) to be
(sf)(s') = f(s's) , but i dont see why the associativity holds. are the definition just wrong ? i personally think (sf)(s') = f(s's^{-1}) works but a book Im refering to also use the above definition and it makes me confused

[rocky cloak]

So s^-1 doesn't necessarily exist, but also your suggestion would fail associativity.

As for why it holds in the correct formula

(t(sf))(x) = (sf)(xt) = f(xts) = ((ts)f)(x)

[rocky cloak]

Some good examples can be for a product ring RxS, the factors, e.g. Rx{0} are projective.

More generally a left ideal generated by an idempotent is projective, so
Mn(k)E11 = k^n is projective over the matrix ring.

Another example is that over Dedekind domains, all ideals are projective, so for example
(2, 1+sqrt(-5)) is a projective Z[sqrt(-5)] module.

[crystal vale]

I have to show if F1 and F2 are free groups of finite rank. Prove that F1 isomorphic to F2 if and only if they have the same rank.

One direction is easy, how do I show F1 isomorphic to F2 then they have the same rank?

Give me an hint

[rocky cloak]

A hint could be to exploit the universal property of each, i.e. compare Hom(F1, G) and Hom(F2, G)

[elfin wraith]

I think you can also argue via their abelianisations which could be nice

[crystal vale]

How?

[crystal vale]

I don't get it, how do I compare Hom(F1, G) and Hom(F2, G) i think I can say Hon(F1, G) is bijective to Hom(F2, G)

[rocky cloak]

Yes, if they where isomorphic there would be a bijection between those homsets. Now pick a group G and compute the homsets, are they in bijection?

[crystal vale]

Oh, say G is finite set then hom(F1, G) and Hom(F2, G) has different cardinality if there rank is not same

Right?

[knotty badger]

Yeah

[crystal vale]

Thanks @rocky cloak

[crystal vale]

And what can I say when free groups on infinite set?

[rocky cloak]

If you assume the generalized continuum hypothesis, then the same argument works without change.

Otherwise it's possible to beef up the argument with some more category theory, or you can follow Nopes suggestion of looking at the abelianization.

[knotty badger]

Do you actually need cat theory for this part

[crystal vale]

Abelianzation refers to considering the quotient group of G by its commutator?

[knotty badger]

Mhm

I think it’s not too hard to show that, if X is infinite, the free group on X has the same cardinality as X? So long as you have choice

[rapid cave]

Do we really need choice?

[halcyon peak]

thanks, thought it would be a stupid miss but cannot work it out for some while 🙏

[rocky cloak]

Yes, this would also be a viable argument

[rocky cloak]

It appears you do
https://mathoverflow.net/a/450084/157483

[crystal vale]

How?

I am not good at infinite cardinality stuffs

How do I think F[X], it is larger than a family of finite subsets of X, but when I am saying larger maybe that's not the correct phrase to say

So it is safe to say, F[X] is a family of all finite list of X

So it is a countable union of finite Cartesian products of X, right?

Why the commutator of a free group on {a,b} is not finitely generated, hint?

[knotty badger]

Yep

[knotty badger]

Yeah it turns out that A being isomorphic to A^2 for infinite A is equivalent to choice lol

Let alone finite families

Cardinality is a significantly less useful notion without choice

[rapid cave]

Ok

[iron arrow]

Do you guys have any examples of very interesting comm rings?

[rapid cave]

There are plenty

Power series rings, rings of integers

[iron arrow]

Yeah but like, weirder rings

Not more "unknown" but not as basic (in the sense of introductory to an algebraic structures class)

[rapid cave]

Rings of integers are not that introductory, and I do not mean Z, I mean the algebraic integers in some number field

You have also I-adic completions

[iron arrow]

Like Z[sqrt[(d)] where d is a non square complex?

[rapid cave]

Thats one example yes. But d would be an integer

[iron arrow]

Oh yeah sorry that's what I meant

[iron arrow]

What is that? I've heard p-adic for primes but I'm not sure about which rings are those?

[rocky cloak]

I guess you also need d not 1 mod 4 if you want it to be a ring of integers

[rapid cave]

Yeah, ofc

[rapid cave]

Same concept, a bit more general

[iron arrow]

Yeah yeah

[tardy hedge]

Why is that so obvious to everyone

[iron arrow]

Is it for prime ideals?

Well I can search

[iron arrow]

The d not 1 mod 4? It is a kind of basic exercise

At least how I did it

[tardy hedge]

Yeah it is an exercise involved there right

[rocky cloak]

Works for any ideal.

The vauge idea is kinda "power series evaluated at elements of I".

And indeed the power series ring is exactly the (x)-adic completion of k[x]

[iron arrow]

Interesting

[wise wagon]

Hello peps

I have a question

[rapid cave]

Seen it in class and used it countless times

[rocky cloak]

R = k[x, y, z]/(xy - z^2) has a prime ideal p = (x, z) such that p^2 isn't primary. That's pretty interesting I think.

It is also isomorphic to the fixed ring of C2 acting on k[u, v] by multiplying u and v by -1

Then there's Nagatas example of a Noetherian ring with infinite Krull dimension
https://math.stackexchange.com/a/1837164/306319

The ring of algebraic integers is a Bezout domain, meaning every finitely generated ideal is principal, but it also has infinitely generated ideals

[iron arrow]

Yeah as I was searching I found that one example of nagata

[iron arrow]

Hm

[rocky cloak]

The endomorphism ring of Q/Z is the profinite integers, which is equal to the product of the p-adic integers for all p

[rapid cave]

Learned about this recently from neukirch

[rocky cloak]

The prime ideals of R^N correspond to ultrafilters on N. The quotient is R for principal ultrafilters and a model of the hyperreal numbers otherwise.

(R real numbers, N natural numbers)

[thorn jay]

well i wouldnt say groupoids are algebraically defined either :P

[iron arrow]

That's really cool lol

[chilly ocean]

You can define them quite algebraically(see e.g. https://www.matem.unam.mx/~omar/groupoids/day1.html), in fact they were defined as Brandt groupoids in like 1927, which afaik was before “category” was even defined, so I was wondering if there was a similar purely algebraic definition of a ringoid. I guess you could just unravel the definition in very high detail of an Ab-enriched category, but this feels somehow different? Idk. Maybe just cause it is more complicated.

[rocky cloak]

I mean, once you allow partial operations you can do:

There is a set of elements 1_x such that 1_x * 1_x = 1_x and whenever the products are defined a * 1_x = a and 1_x * b = b.

For each a there exists a 1 on either side, and + is defined only between elements for which 1_x * a * 1_y is defined.

[thorn jay]

yeah sure but i believe its more fruitful to think of it as a category than as a partial algebra

mainly because partial algebras arent that nice imo

idk maybe im just a purist, they are used in a nice proof of some theorem

[rocky cloak]

I guess the fact that the image of a groupoid isn't necessarily a subgroupoid of the codomain is kinda wack

[thorn jay]

yeah, it sucks

and partial algebras dont have a nice interpretation as the representation theory of certain objects either

[chilly ocean]

I see, thanks!

[rocky cloak]

How do you mean?

[thorn jay]

universal algebra is the representation theory of clones (generalised monoids; they are to small Lawvere theories what monoids are to categories with one object)

and Mal'cev conditions essentially give the existence of certain elements in a clone in terms of their representation theory (for example how the congruence lattice behaves of the algebras)

but partial algebras dont have this

its precisely because of this that algebras are so nice (all the isomorphism theorems and such come from the fact that they hold in Set)

[chilly ocean]

Does the usage of representation theory here relate to like, representing a group as automorphisms of a vector space? Or do you mean representation theory differently?

[thorn jay]

thats related!

[chilly ocean]

[thorn jay]

a G-set is exactly the special case when your clone consists of only invertible unary operations and projections (they model the natural projection maps A^n → A)

and so you might want to ask "what if we want to represent in more general categories" and thats totally fine! topological algebraic structures are precisely clone representations in Top, but you can also go in reps of schemes, manifolds, whatever

[chilly ocean]

That’s pretty neat!

[thorn jay]

though if we want something like rings as a representation in Ab of the monoid clone, then we run into a problem being that we cannot turn the product into any monoidal product of a monoidal category

because, as discussed in #category-theory they might not have a diagonal map Δ : X → X ⊗ X that acts the way youd want it to

this is, however, remedied using operads, where such a thing isnt necessary. (you can think of it as being defined by equations where each side uses a variable at most once)

[delicate orchid]

I was gonna ask if your clones aren't just multicategories

[tardy hedge]

Rx{0} is projective as an R^2-module right, and (Rx{0})x({0}xR) is a free R2-module? (So its a direct summand of one)

[thorn jay]

theyre Lawvere theories

just seen as some generalised monoid rather than a category

[delicate orchid]

hmm I'm not seeing it

is the n in n-ary multiplication fixed?

[thorn jay]

it looks like multicategories are an oidification of operads

rather than clones

[delicate orchid]

yes they are

that doesn't mean they can't be the same thing

[thorn jay]

hmm, well no, clones are not multicategories

the reason they look similar is because operads are a restricted form of clones (although the restriction allows for a much richer representation theory), and multicategories are the oidification of operads

wait huh

[thorn jay]

wait yeah nvm this is correct

clones are different in that they require duplication maps x → x × x to exist, so identities like f(x, ..., x) = x can be formed

this is not possible in multicategories with a single object. You have to do this in the way that just gives you back an algebraic theory

[thorn jay]

okay heres the difference: in algebraic theory, if youve got terms t1(x1, ..., xn), ..., tm(x1, ..., xn) and an m-ary operation f, then we have the composition f(t1, ..., tm)(x1, ..., xn), i.e. the variables are in that sense "duplicated". However, in multicategories / operads, one would necessarily have f(t1, ..., tm)(x11, ..., x1n, ..., xm1, ..., xmn), i.e. each variable automatically is a new instance

[delicate orchid]

ah I see

[thorn jay]

this is, for example, why there is no group operad

[tardy hedge]

Free implies torsion free only over a domain right

But even over a non domain, if everything in the module is a torsion element then it cannot be free right

[thorn jay]

yes, because 1 cannot have torsion

[tardy hedge]

1 in ring?

[thorn jay]

ye

free module over R is a direct sum of copies of R

[tardy hedge]

I was thinking every m in M has some nonzero r in R so rm=0

Not 1m = 0

[thorn jay]

and, in particular, contains a copy of R

[tidal schooner]

I think usually the definition of "torsion-free" for modules over general rings is that 0 is the only thing annihilated by a non-zero divisor. So with that definition, then actually free always implies torsion-free

[thorn jay]

so because R cannot be a torsion module (containing 1), no free module can be a torsion module

[tribal moss]

Hmm, that doesn't sound right. Shouldn't any R be free as a module over itself (since {1} is a basis)?

[thorn jay]

thats what i said

[tribal moss]

Ah, sorry, looks like I thought of the wrong kind of torsion.

[thorn jay]

oh lol

[tardy hedge]

Erm basic question but every unital ring has units right? I think yes and the way i thought of it was cause u always have a Z->R map

[alpine plank]

Yes

[tardy hedge]

Ty

[tribal moss]

But more elementarily it's because 1 is always unit (since 1·1=1 by definition).

[alpine plank]

Wait yeah how do you even define units in a non unital ring

[tardy hedge]

Yea i guess units besides 1

[alpine plank]

Then no, consider Z/2Z

[thorn jay]

(or any boolean ring for that matter)

[tardy hedge]

In a local ring R-m are all units. In other settings do we know where units are “located”?

[thorn jay]

not really, but the jacobson radical gives some of them

namely 1 + x for x ∈ J(R)

[tardy hedge]

Oh yeah

[next obsidian]

R \ U m where m runs over all maximal or equivalently prime or equivalently non-trivial ideals

[thorn jay]

that works lol

[tardy hedge]

Is it just take away jacobson radical

[tough raven]

If a group G acts simply transitively on a set X then the orbits of X ⨯ X are in bijection with G: for any x in X, {(x, gx) : g in G} is a transversal. But the bijection ψ_x: G → G\(X ⨯ X) depends on the choice of an element x in X (we have ψ_{gx}(h) = ψ_x(g^{-1}hg). Is there an object (whose expression may involve G and X but not quotients by G-actions) which is canonically in bijection with G\(X ⨯ X)?

[tardy hedge]

Ok ya if an element is not in any maximal ideal it has to be a unit yeah

[tough raven]

No: J(ℤ) = 0 but ℤ^⨯ is not ℤ{0}. J = intersection of all maximal ideals, while {non-units} = union of all maximal ideals.

[next obsidian]

No that’s taking away the intersection of the m

[tardy hedge]

This was the point tho right

[tough raven]

Maybe Hom(Hom(X, G), X)? (If this is actually the answer I will give up on caring about this.)

[thorn jay]

what are we taking the Hom of? homomorphisms of G-sets?

[thorn jay]

does Hom(X, G) come equipped with the action f |-> (x -> f(x)g^-1) ?

[tough raven]

OK, maybe what I actuallly want is the decompositon X ⨯ X = ∪_g {(x, gx) : x in X} even though this is different from the orbit decomposition when G is not abelian. So it's OK. (I think this is the orbit decomposition (X ⨯ X)/H, where H = Hom(X, X) is non-canonically isomorphic to G.)

[tough raven]

Actually I have no idea what I meant, but this is probably the only correct way to do it.

[thorn jay]

lol

that seems like a kinda "hacky" action

[tough raven]

I think the correct answer is Hom(X, X) anyway.

psi(f) = {(x, f(x)) : x in X}

[thorn jay]

i see, yeah

and because X is free this will give every orbit

(i.e. Hom(X, X) ≈ Hom(G, G) ≈ G)

and the non-canonical nature of the isomorphism of G\(X × X) with G is because the isomorphism X ≈ G is not canonical

funny how that works

[delicate orchid]

not funny at all, sets with a G-action are basically modules

[thorn jay]

except the union of invariant subsets is invariant

[tough raven]

O(sum) 🤷

[rocky cloak]

I would say one of three things are true
free always implies torsion free,
You only define "torsion free" for domains,
You have a bad definition of torsionfree

[delicate orchid]

that's a quotient of the coproduct I don't get ur point

that's literally the (non-direct) sum of two modules

it's kind of cool how close they are, you can tensor them and as a functor it's adjoint to an internal hom

[vapid vale]

thought this was gonna be a limerick

[tardy hedge]

Honestly, i remember not fully understanding this before, which is why its coming up again. I think a module over any ring is torsion free if for any nonzero m in M, rm = 0 implies r = 0

[delicate orchid]

just like modules wooaaahh

[tardy hedge]

Kinda reads like one lol

[tough raven]

I think torsion-free should be defined wrt a given multiplicative set.

[rocky cloak]

That would be option 3: bad definition

[tough raven]

Torsion-free over a domain = wrt the multiplicative set of non-zero elements.

[thorn jay]

wow no way the join of two subalgebras is their pushout over A ∩ B

[delicate orchid]

https://tenor.com/view/bad-ass-uh-oh-ndt-neil-neil-degrass-tyson-gif-14802463180959376587

[vapid vale]

thjis should at least be zerodivisor, not 0

[thorn jay]

you mean tensor the G-sets?

that is kinda awesome

[delicate orchid]

yus, write down the definition you'd expect and it's right

ok I should say, only a right G-set with a left G-set

they're kind of woke in how they're one sided

[thorn jay]

so the set of (x,y) quotiented out by (xg, y) = (x, gy)?

[rocky cloak]

The definition usually used in my field is that a module T is torsion if
Hom(T, R) = 0
and torsionfree if
Hom(T, F) = 0 for all torsion T.

[tardy hedge]

Because otherwise the set of torsion elements (those m with rm = 0 imply r = 0) would not be a submodule ?

Is that why its a bad defn if u leave out the zero divisor part

[delicate orchid]

yur

typically you'd do this with bisets but you can do it with just G-sets

[tardy hedge]

Thank u random dummit and foote exercise

[thorn jay]

bisets are with relation to a pair of groups (G, H) right?

left-rep of G and right-rep of H

[delicate orchid]

"rep" is a stretch

they're like bimodules but it's sets and groups not modules and rings

[thorn jay]

a rep is a rep

[delicate orchid]

"any group homomorphism is a rep" ahh

[thorn jay]

any homomorphism is an interpretation or generalised element depending on what youre doing

[thorn jay]

(dw its just copium so i can call universal algebra a form of rep theory)

[delicate orchid]

no it's an arrow I draw between two letters

[thorn jay]

some guy wrote a book on general rep theory for UA and it explored like a single interesting concept right at the end and didnt even do anything cool with it

[delicate orchid]

my thesis

[thorn jay]

its a published book

[tardy hedge]

Youre probably swimming with possibilities in ua

[vapid vale]

i once watched a twitch streamer who was an universal algebra professor

they'd just do math on stream

[thorn jay]

sure but its hard to motivate why you should care about certain stuff, even in universal algebra

[thorn jay]

K-theory ahh

(the youtuber)

[delicate orchid]

do you do algebraic K-theory or youtube K-theory

[thorn jay]

no way there still exist UA professors

thats crazy

[thorn jay]

SO WHATSUP GUYS I JUST PROVED THAT THIS NEW K-L-M-N-O-P THEORY ACTUALLY SOLVES THE UNIVERSE, so stay watching because YOU WONT BELIEVE WHAT HAPPENS NEXT

[delicate orchid]

what are we doing here 💔

[thorn jay]

an attempt at comedy

[rapid cave]

1$ MATHMATICIAN VS $1000000 MATHMATICIAN

[thorn jay]

student vs phd vs professor

[tardy hedge]

K-theory thats about me right 😎

[thorn jay]

k for kiand123

[tardy hedge]

k for kian my name

Psa: my name is kian not kiand

[delicate orchid]

your name is kiand123

[thorn jay]

i know

[rocky cloak]

I feel like there's a pickup line in here somewhere

[tardy hedge]

teehee

[delicate orchid]

take my kiand in marriage

[tardy hedge]

wow

Look who’s typing

[karmic moat]

jagr is a poet and a freak

wow

[delicate orchid]

you've gotta kiand it to the guy, he knows how to write a post

[rocky cloak]

Someone had to put that d in there

[karmic moat]

i kiand handle your shit anymore

[rapid cave]

what can we say about Hom(A \times B, C) in relation to Hom(A,C) and Hom(B,C)?

for algebras A,B,C

[karmic moat]

there's hom tensor adjunction if you want that

[tardy hedge]

Chat got the giggles today

[delicate orchid]

booooo change it back

[rapid cave]

\oplus not tensor

mb

[karmic moat]

Though hom tensor doesn't give you something about hom(A, C)

oh

[delicate orchid]

ts is the cuh cuh classic big matrix of hom spaces

[thorn jay]

actually crazy I love it

[delicate orchid]

wait are these modules or algebras

[thorn jay]

there is NO toothfairy, NO santa claus, and NO DIRECT SUM OF ALGEBRAS

[rapid cave]

fine

how do i call it then

\times?

[thorn jay]

YES

[karmic moat]

bullshit explain the $5 left under my pillow last night

[rapid cave]

fixed ❤️

[delicate orchid]

modules: coproducts are biproducts, Hom(-, C) sends it to a product

[thorn jay]

thank you ❤️

[rapid cave]

from the local crackhead

[delicate orchid]

algebras big matrix YEAahhhHH

[karmic moat]

hmm

[thorn jay]

this is just stand universal property of products, Hom(-, C) commutes with products

[rocky cloak]

(let's say first that everything is commutative)

If C is connected, then it's then
Hom(AxB, C) is the disjoint union of Hom(A, C) and Hom(B, C)

[rapid cave]

connected?

[delicate orchid]

yeah what does that mean

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

[rocky cloak]

Only idempotents are 0 and 1

[thorn jay]

it means that it has only a single connected component guys, cmon

[delicate orchid]

ZAMN!

[rocky cloak]

Or C not the product of two rings

[rapid cave]

I think in my case that is true

but why do we care?

[rocky cloak]

Or spectrum is a connected space

About what?

[thorn jay]

lets bring AG into this ofc

[rapid cave]

about connecteness

[thorn jay]

your new name

[rapid cave]

this always has been about AG

[karmic moat]

rich coming from u mr "one can explain [basic concept] with universal algebra..."

[rocky cloak]

If C is not connected then you can first split C into a product of rings and use
Hom(X, CxC') = Hom(X, C)xHom(X, C')

[thorn jay]

wait i got my stuff the wrong way around lol

[rocky cloak]

And if C cannot be written as a product of connected rings, then I will throw in the towel

[delicate orchid]

it works cause ts a biproduct yunc

[rocky cloak]

Because it gives you the answer to your question I guess...

[thorn jay]

hm, if C is finite dimensional then products will always reduce the dimension, so this method works

[crystal vale]

Can I ask what is the connected ring?

[rapid cave]

I asked like why is connectedness is related to this

[thorn jay]

A, B, C are algebras

[rapid cave]

but you answered it already

[delicate orchid]

we're cooked

[thorn jay]

we are

[rapid cave]

so whats wrong with this?

[tardy hedge]

Young unc. lol

I am 25 soon 😶

[thorn jay]

Hom(-, C) commutes with coproducts in the sense that Hom(X ∐ Y, Z) ≈ Hom(X, Z) × Hom(Y, Z)

not with products, thats the job of the second variable

[rapid cave]

coproducts are tensor products?

[thorn jay]

for commutative algebras yes

[rapid cave]

yes thats the context

[thorn jay]

then ye Hom(A ⊗_k B, C) ≈ Hom(A, C) × Hom(B, C)

[tardy hedge]

OMG he said the name

[rapid cave]

but I am working with \times

[thorn jay]

ah classical AG huh

[rapid cave]

yes

[thorn jay]

how do products of algebras pop up there?

[karmic moat]

the coordinate ring of a product of affine varieties is the tensor product of each coordinate ring

[rapid cave]

I need to prove the tangent space of VxW for two varieties is the direct sum of tangent spaces of V,W

[thorn jay]

i know im talking about products of algebras

tangent spaces form an algebra?

[rapid cave]

no but how they are defined is based on the coordinate ring

[thorn jay]

how're they defined then?

[rapid cave]

V(k[eps]) is Hom(k[V], k[eps])

[thorn jay]

yes yes okay

well Hom(k[V × W], k[e]) = Hom(k[V], k[e]) × Hom(k[W], k[e]) so its a good place to start there

[karmic moat]

should still be the same for affine varieties

[thorn jay]

^

[rocky cloak]

k[VxW] = k[V] (x) k[W] though

[karmic moat]

taking tensor product over k

[rapid cave]

oh really?

why is that?

[thorn jay]

duality

[rapid cave]

I am thinking of SpmA \times SpmB

[thorn jay]

affine sets are dual to finitely generated algebras

and dualities turn products into coproducts, and vice versa

[rapid cave]

by duel you mean what?

[karmic moat]

milne didn't cover that?

[rapid cave]

probably did

[karmic moat]

there's an equivalence of categories between affine varieties and the opposite category of fg reduced k-algebras

[thorn jay]

how have you been doing AG without the duality between affine sets and coordinate rings, like its basically the only reason why things work out nicely :P

[rapid cave]

I just have the memory of a peanut

[karmic moat]

anyways yes you have an equivalence of categories that sends an affine variety X to k[X] and a morphism X -> Y to k[Y] -> k[X]

[rapid cave]

yes

[karmic moat]

that's what enpeace means by duality

[rapid cave]

shouldn't it be reduced instead of integral?

[thorn jay]

yes

[karmic moat]

yeah

if k is algebraically closed and your varieties are irreducible then you get integral

[thorn jay]

i wonder- is there a geometric characterisation of V(k[e])?

[rocky cloak]

What kind of characterization are you thinking of?

It's a point that's a little fat in one direction

[thorn jay]

my ever present hunger for generalisation to UA

or as best as i can

[rocky cloak]

Does UA deal with geometric characterizations?

[thorn jay]

i am working in universal algebraic geometry

[rocky cloak]

I see, do you have an example of what a geometric characterization is? I'm still not clear on that

[rapid cave]

I still don't understand why how is SpmA \times SpmB is the same as Spm(A \otimes B)

or the whole product turn into coproducts things

[fading acorn]

fiber product

[thorn jay]

maybe geometric isnt the right word, but like any characterisation that isnt "the quotient by the ideal (x^2 - x)"

[thorn jay]

well this is just category theory

[rocky cloak]

You might think about prime ideals as kernels of maps to fields.

So then Hom(A(x)B, k) = Hom(A, k)xHom(B, k)

This is not a full proof, but is like the rough idea

[rapid cave]

I would like to keep this a bit more concrete xD

[vapid vale]

does "a fattened point in an arbitrary direction" not count

[thorn jay]

that seems less a characterisation but more of an intuition

[rocky cloak]

2d connected algebra (assuming k alg closed)

Or 2d local alg with residue field k

[crystal vale]

you won't believe this but today someone ask me same question which i did exactly one year ago😀

[karmic moat]

#algebraic-geometry message and #algebraic-geometry message relevant maybe

[rocky cloak]

Do you see why an algebra homomorphism from A(x)B is the same data as one from A and one from B?

[rapid cave]

yes that is just the universal property

[delicate orchid]

the result follows

[thorn jay]

hmm

[vapid vale]

also vakil has a good section on how nilpotents feel geometrically

[karmic moat]

took the words out of my mouth

[rapid cave]

this is what I would like to understand

[karmic moat]

i'm trying to find the relevant section in vakil lol

4.2

[vapid vale]

4.2

lol

[thorn jay]

lmao

[rocky cloak]

And you understand that prime ideals are kernels of maps to fields?

[thorn jay]

thank you two

[rapid cave]

no

like Hom(A, k) is related to Spec(A)?

[rocky cloak]

Okay, then exercise for you: prove that the quotient of a ring by a prime ideal is an integral domain

[rapid cave]

of course yeah

basic fact from ring theory

[thorn jay]

i would surprise me if i could carry of this into UA (tangent vectors and, hell, the concept of a direction is inherently something about abelian groups), but one can try :P

[rocky cloak]

So this at least gives a natural way to take a prime of A(x)B and make a pair of primes from A and B

And I guess you really only want it for closed points anyway.

In which case it's just maps to k

[rapid cave]

sorry I am not following

like are you saying Hom(A (x) B, k) has kernel some prime ideal

[rocky cloak]

I'm just guessing what definitions you're working with

[next obsidian]

A -> A (x) B -> D where D is the quotient of A (x) B gives you a prime ideal of A

By taking kernel

Similar for B

[rapid cave]

oh ok

[rocky cloak]

Like are your spectrums subsets of k^n?

[rapid cave]

no

A is a reduced fg k-algebra

unrelated to k^n

[rocky cloak]

Okay, well if you're also doing generic points then VxW shouldn't be interpreted as the product of the topological spaces

But if that's the case I'm not sure why you would only be doing fg k-algebras

[rapid cave]

then what it would be

[rocky cloak]

Let me ask you this, what are the elements of Spec(k[x])

[rapid cave]

these are the varieties I am working with

[rocky cloak]

So as I expected you do define it as subset of k^n

In which case points are exactly homomorphisms k[V] -> k

So there's no subtleties to deal with

[next obsidian]

k is alg closed right?

[rapid cave]

yes

[rocky cloak]

Are there other types of fields

[next obsidian]

Okay

[next obsidian]

Well over non-perfect fields you can have issues with tensor product

Non-reduced 🙁

And I think F_p(t) is a field

Some people care about

[rocky cloak]

Non-perfect field. Blasphemy

[karmic moat]

can't believe milne is so gracious to work over algebraically closed fields in those notes

in his algebraic group notes he works in full generality

[rocky cloak]

At that point best to just go to affine schemes I feel

[next obsidian]

Yeah

But I had a prof who didn’t

And his hw was wrong because of this

Lol

[rapid cave]

wait so how should I interpret VxW

[rocky cloak]

In your case, as a set it's just the product. However, The topology is not the product topology, but the Zariski topology.

In a more general setting where you also have generic points you'll get new generic points as well

[vapid vale]

do you understand generally how points "correspond" to fields

[rapid cave]

yeah, by a quotient

points would be A/m where m is maximal

[vapid vale]

ok so you are not including nonclosed points

[rapid cave]

wdym?

[vapid vale]

every prime ideal is a point

[rocky cloak]

This would be affine schemes as opposed to affine algebraic sets I guess

[vapid vale]

oh

[rocky cloak]

Maybe definitions vary for the latter, idk

[rapid cave]

why do we consider that

[rocky cloak]

Kinda necessary if you want non fg stuff, or want k not alg closed

[vapid vale]

but also its nice to have generic points. if you have the closed set V(y-x^2), then you have your closed points (x-a^2, y-a) sitting along the parabola but you also have the parabola itself, with the property that its closure is the entire set. algebraic geometers have talked about things "generically" long before the modern AG language existed

[rapid cave]

but why would we call the parabola itself a point

[vapid vale]

ig the dumb answer is like, you are free to call anything anything – Spec A is at its core an abstract topological space where the points are prime ideals.

another answer is, if we really want to see this as a single point, we want a way to extract a field from the prime ideal, since Spec k is a point. well, there are two ways to do this. you can localize, and take the quotient by the maximal ideal of the local ring. you can also take the quotient and then take the field of fractions of the integral domain.

these processes should feel opposite to one another – one turns the prime ideal into a thing with a unique maximal ideal, and then a thing with a unique prime ideal; and the other turns the prime ideal into a thing with a unique minimal ideal, and then a thing with a unique prime ideal.

in any case, if youre a fg k-algebra over an algebraically closed field, you just get k back. and in this case Spec(R) is really isomorphic Hom(R,k)

[rapid cave]

nice answer

thanks I will let this sink in a bit

[vapid vale]

for the geometric picture, you can imagine the parabola as a line of glue, and the closed points on the parabola as glitter which gets stuck to the glue (i.e. in the closure). maybe to expand the picture, the parabola is sitting on a sheet of glass (some surface), the surface is sitting atop globules (higher dimensional things)...

if your bridge between geometry and algebra is good enough, its a good exercise to convince yourself what happens sequentially in those two processes of getting the residue field at the parabola – interpret the glitter above and glass and globules below as some containment/chain of prime ideals, and the primes which pass through quotienting and localizing are easily described in this way

[karmic moat]

relevant

[thorn jay]

are there situations where Spec A → Spec B is a covering? lol

[karmic moat]

no idea, haven't gotten that far

but i'm sure there is

wait im being self contradictory let me rephrase

i'm sure there are, but i don't know what situations

[thorn jay]

ah it appears the correct notion of this for schemes is etale morphisms (with the accent placed on one kf the e's)

i shouldve known

[south patrol]

Yea there is the notion of étale covers which in many ways are the "correct" analogue of covering spaces

[thorn jay]

i can imagine the Zariski topology being way too "ungeometric" for a topological covering map to behave well

[rotund pine]

How do we come to the conclusion that Z[i]/<2+2i> has 8 elements? I get the norm thing, but when trying to prove it through elementary proofs, I'm getting stuck. I can move beyond 2 is congruent modulo (2+2i) to -2i, I don't get the reasoning behind using division algorithms

Can't move beyond*