advanced-algebra

[fierce steeple]

Sure yee

[hushed bone]

Two sided tuff

[fierce steeple]

Lol

[lone jacinth]

Tough, I'm down with every property of a ring just being a property of their representation theory and then throwing out all other words

[steep rivet]

a simple ring elpmis is a ring for which...

[lone jacinth]

The matrix ring might as well be a field for all I care

[hushed bone]

I care a lot

[lone jacinth]

Then it is decided! We make it a field

[ornate atlas]

Commutative rings are basically fields

What do you mean you can just add inverses

[lone jacinth]

I like that Japan is separate from worldwide

[hushed bone]

It’s thanks to our thousand times folded theorems

[ornate atlas]

I do mean this somewhat seriously, division makes me genuinely uncomfortable

[steep rivet]

wait, $k$ and $k^{n\times n}$ are morita equivalent, right?

[broken turtle]

PKThoron

[steep rivet]

does that mean they have the same modules or only that their module categories are equivalent?

[fierce steeple]

What does same modules even mean

[steep rivet]

that every k-vector space also happens to be a module over its matrices and vice-versa

[hushed bone]

What I’m reading was just

Do you pronounce it potato or potato

[fierce steeple]

Ig like you can say for each R-module does the underlying abelian group admit the structure of an M_n(R)-module and vice versa

[steep rivet]

yeah

[fierce steeple]

But ye they are equivalent cats

[hushed bone]

There’s no canonicity in that

[fierce steeple]

Not this

[steep rivet]

i assume that's not the case though

[hushed bone]

Horrible definition

[steep rivet]

like how would 5x5 matrices act on k^2

[drowsy niche]

https://www.tandfonline.com/doi/abs/10.1080/00927879108824246

so the first DaRT example would suffice

[hushed bone]

Woah

[drowsy niche]

this is a crazy result ngl

[steep rivet]

so what IS the category equivalence between $k$-Vect and $k^{n \times n}$-Mod?

[broken turtle]

PKThoron

[drowsy niche]

haven't read the proof but will later tonight

[lone jacinth]

S = k^n is the unique simple Mn(k)-module.

Then you take the k-module k^m to S^m

[fierce steeple]

Given an R-module V, let M_n(R) act on V (+) ... (+) V in the obvious way

[steep rivet]

that's one direction

and EVERY module over the matrix ring is like that?

[fierce steeple]

Yes

[steep rivet]

how come?

[hushed bone]

You only need one direction for an equivalence of categories, who has ever in their life constructed the inverse

[fierce steeple]

You can just like apply e11 basically (matrix w top left entry)

[hushed bone]

😹

[steep rivet]

what's e11?

hm

[fierce steeple]

And construct an explicit inverse

But yeah also what chmonkey said

[hushed bone]

e621?

[fierce steeple]

e67

[steep rivet]

~_~

[hushed bone]

SIX SEVENNNNN

[fierce steeple]

There's also the monadic proof of this stuff

[lone jacinth]

Notice that Mn(k) is isomorphic to S^n as an Mn(k) module.

As every module is a quotient of direct sum of this, and this is semisimple every module is a direct sum of sums of S

[steep rivet]

crazy

[lone jacinth]

Another way to write down the equivalence would be
S(x)- : Mod k -> Mod Mn(k)
with inverse
Hom(S, -): Mod Mn(k) -> Mod k

Using the fact that End(S) = k for the latter map

[fierce steeple]

Ig this is the story of S being a compact generator

Fun times

Thank u for stating it as such jagr lol I forgot

[lone jacinth]

Moritas theorem

Every equivalence of module categories is on this form

[spice idol]

I second this
it's too nice

[unborn rampart]

I found this: https://www.sciencedirect.com/science/article/pii/S0747717183710515
which says that degree-reverse-lex has a complexity of d^(n^2), while pure lexicographic is d^O(n^3). But this paper is from 1989, so maybe there's a more modern reference somewhere

[tropic bobcat]

what's the site? all I can find is homepage for a programming language called Dart and not DaRT?

Ohh ok

thanks

[woven slate]

So, I'm thinking about enrolling in a non-degree course or two at a school two hours away from me to transfer in, as my current grad program covers no algebra and no topology. In particular, maybe Abstract Algebra II (60x)? I took Abstract Algebra I (4yy) in undergrad.

At the school two hours away:

MATH 40x. Introduction to Abstract Algebra. 3 Hours.
Semester course; 3 lecture hours. 3 credits. Prerequisites: MATH 3xx and MATH 3yy, each with a minimum grade of C. An introduction to groups, rings and fields from an axiomatic point of view. Coset decomposition and basic morphisms.

MATH 50x. Abstract Algebra I. 3 Hours.
Semester course; 3 lecture hours. 3 credits. Prerequisites: MATH 4xx with a minimum grade of a C, or permission of instructor. A study of groups, subgroups, quotient groups and homomorphisms, group actions, sylow theorems, direct and semi-direct products, rings, integrals domains, and polynomial rings.

MATH 60x. Abstract Algebra II. 3 Hours.
Semester course; 3 lecture hours. 3 credits. Prerequisite: MATH 50x. A study of modules, vector spaces, field extensions and Galois theory.

And the course I covered at my undergrad program:

MATH 4yy - Abstract Algebra I 3 credits
Introduction to the theory and applications of algebraic structures including groups, rings, and fields. Prerequisite(s): MATH 3qq or MATH 3zz

In 4yy I covered normal subgroups, quotient groups, Lagrange's theorem, Chinese remainder theorem on Z_n, ring ideals, the basics of unital and commutative rings, the basics of integral domains, basics of fields and finite fields, homomorphisms and isomorphisms, up to First Isomorphism Theorem.

I think we may have lightly covered centers, commutators, etc. I'll need to take a second look back through; I took it in 2024.

Should I take 50x, or can I just dive into 60x? I have Dummit & Foote in case I need to brush up on a few topics I didn't cover (Sylow theorems, direct and semi-direct products, anything additional with polynomial rings, etc.), but I can't tell from where I stand if the standard sequence includes extra group and ring theory in a class like 50x. 4yy was taught by a very sharp commutative algebraist, though the uni I took it at doesn't have a graduate math program.

Thanks so much :)

[worldly zealot]

so i can't tell if 50x has 40x as a prerequisite or if that's actually different. my assumption is that 50x and 40x are similar to a system we have at my school, where there's a "grad" and "undergrad" version (although both are basically only taken by undergrads) of abstract algebra, with the grad version probably being more rigorous. it probably includes more content which would depend on the course instructor.

but any case which ever you took in undergrad likely would function as a prerequisite for a standard algebra 2 course like modules/galois theory. you really don't need too much fancy stuff to get started

brushing up on dummit & foote seems wise, and there's certainly nothing worth taking an entire course to review

[woven slate]

When you say "take it online" do you just mean MIT OCW or a textbook and paper or something? I find I learn better with the structure of a course.

[woven slate]

Thanks so much!

Gotcha

Due to work and life constraints for where I'm living right now, I'm in an applied math master's right now (no pure math master's or PhD nearby)

But there were courses like abstract algebra II, algebraic topology, homological algebra, category theory, Lie theory, etc. that I wanted to take before I worked on a PhD thesis

And I thought it would be easier if I could take AA II, or abstract linalg, etc. as a real course at a uni rather than getting lost in the textbook/paper. I find I can spin my wheels with that stuff for way too long and it's hard to be as disciplined about it, esp. when I have many other life obligations

[worldly zealot]

hopefully you can continue to find ways to engage with the math you enjoy

[woven slate]

Thanks hk

Gang gang 🤝

I appreciate it man ❤️

[edgy jay]

i have no clue where to even ask this question, trying here

does anyone know if operator algebra is just a notational thing, a convenience of syntax, or does the perspective offer any new insights that are nearly impossible without it?

[ornate atlas]

You certainly could just jump in with 60 if you feel reasonably comfortable with the stuff you did in UG, I don’t think a lot of the content from 50 should be directly needed in 60. But I think they both seem worth doing if you’re able, they’re good courses and will just continue to build your algebra strength

In any case I wouldn’t take 40

[civic stone]

Can someone help me understand the conceptual difference with the following: if I'm looking at the k[x,y]-modules k[x,y]/(x) and k[x,y]/(y), then apparently Hom(k[x,y]/(x) , k[x,y]/(y)) = 0, but I thought that if I say k[x,y]/(x) = k[y] and k[x,y]/(y) = k[x], then Hom(k[x,y]/(x) , k[x,y]/(y)) = Hom(k[y] , k[x]) and a map k[y] -> k[x] would be given by sending y to some polynomial f(x) so Hom(k[y] , k[x]) = k[x]. I think I might be confusing maps of rings and maps of k-algebras or something?

[lone jacinth]

Your confusing maps of algebras and maps of modules.

A map of k[x, y] modules would need to be k[x, y]-linear.

So f(y) = y * f(1) for example

[near lantern]

I think Bourbaki Lie Algebras Ch IV-VI covers it pretty well (although Bourbaki's style of writing is not for everyone).

[last talon]

Yeah that’s the reference I’m given
Is there an english version?

[near lantern]

Should be

[last talon]

It does exist…
But I don’t have easy access to it 😢

[limpid horizon]

This is what theyre using for the lower bound, and that is A3 that they cited. Tbh, im pretty confused on how this is working

if you can explain some things the dumber the better 😛

[hushed bone]

These also hold for finite modules and (d)’s definition says it can only go down by as many elements you mod out by

[limpid horizon]

why do they also hold for finite modules?

[hushed bone]

Isn’t the left one trying to argue that in essence

Just ignore what I said and just read the left image lol

But it holds because you develop a theory using the Hilbert function and it works equally for modules

[limpid horizon]

I think I'm a bit unsure on why its harmless to replace M with R on the LHS

[lone jacinth]

dim(M/IM) = dim R/Ann(M/IM)
and
Ann(M/IM) contains Ann(M) + I

(I = (x1, ...))

So assuming Ann(M) = 0 (by replacing R) you have

dim(M/IM) >= dim R/I >= dim R - r = dim M - r

And you just need to prove the dim R/I >= dim R - r part

[hushed bone]

Don’t you get dim R/I >= dim M/IM

[limpid horizon]

Ann(M/IM) contains I

[hushed bone]

The issue is showing that Ann(M/IM) cant be so much larger than Ann(M) + I so that it dimM/IM can’t drop more than you want

[lone jacinth]

I guess you need Ann(M/IM) = I for Ann(M) = 0

[hushed bone]

So I think I can get it

But I think the ideals only agree up to radical

What you want is to be looking at the support of the modules because V(Ann M) = Supp M

[lone jacinth]

Well that's all you need anyway

[hushed bone]

When you mod out by I clearly Supp M/IM < V(I)

And so take a prime in V(I) p

You want to show that V(I) = Supp M/IM

(I’ve assumed Ann M = 0 by moving to R/Ann M)

But if (M/IM)_p = 0 then M_p = IM_p and I < pA_p so Nakayama says M_p = 0

[lone jacinth]

Can also use Cayley Hamilton I guess

rM < IM implies r^n + Ann(M) < I

[limpid horizon]

so this part wasnt true because of the first inequality right

[lone jacinth]

I mean it's true. Just the first inequality should be an equality

(and uses that M is finite)

[limpid horizon]

Ok I think it makes more sense. thank you

dim(R/(x1,...xn)) >= dim R - n even if R isn't local right. This is just Krulls ideal theorem?

You needed R local and M finite for the one with modules to work?

[limpid horizon]

Maybe not R local

[hushed bone]

You need it to be local. If R is a product of say, a 10 dim ring S and a 1 dim one T then when you mod out by (1,0) you go from R, a high dimensional ring? To T a 1 dim one

[astral ginkgo]

Well first if R isn't local noetherian dimR may be infinite. Even if we assume that dim R was finite I don't see how krull implies this.

[limpid horizon]

dim R can't be infinite if R is noetherian right

[astral ginkgo]

It certainly can. It can't if it's local noetherian

Nagata has a famous example

[normal narwhal]

I think this should be true if R is a finitely generated k-algebra for a field k since ht p + dim R/p = dim R, but not sure what the minimal assumptions on R need to be

[limpid horizon]

why is that equation true in that case

[limpid horizon]

Of dim R being infinite even if R noetherian?

[limpid horizon]

How do you know S x T is not local?

[hushed bone]

1: it’s a product of rings

2: that theorem fails

[limpid horizon]

because m x T and S x m' are maximal ideals?

[hushed bone]

Yeah

[normal narwhal]

https://en.wikipedia.org/wiki/Noether_normalization_lemma#Refinement

[verbal panther]

Are there any other notable relations between two binary operations like distributivity?

[torn harbor]

idk if there's a name but there's this relation

which comes up in the Eckmann-Hilton argument

[verbal panther]

Eckmann Hilton argument?

[torn harbor]

if you have two operations satisfying this condition + both are unital

then the two operations are the same and are commutative and associative

[verbal panther]

Neat

I can see it

[dark widget]

you need some kind of associativity in bimodules:
(rm)s = r(ms)

[verbal panther]

It’s funny how powerful being unital is with some properties

[spice idol]

yeah

a unital self-distributive binary operation is necessarily associative for example

[astral ginkgo]

Yes

The intuition is just that even if you can't have infinite chains you can still have arbitrarily long finite chains of primes

[steep rivet]

Finite polynomials on infinitely many variables mayhaps?

[sacred sentinel]

That's not noetherian

[lone jacinth]

Nagatas example is sort of like that though. You just localize enough to make it Noetherian

[hushed bone]

Yeah

Making an example is a PITA

[lone jacinth]

The construction of the example isn't really that bad. I feel like the showing it's Noetherian part is the tricky thing

[hushed bone]

Yeah

That’s what I mean, constructing means showing it has the properties

I think it uses an infinite prime avoidance you need to prove in that case

[pastel shoal]

Can you give an example of such a group?

There is some discussion here: https://mathoverflow.net/questions/110208/understanding-groups-that-are-not-linear

But they only consider linearity over fields and finitely generated groups

[last talon]

If I can prove that “over integral domains” suffices, it isn’t that hard (take field of fractions)
But I’m not sure if there’s gonna be a problem with zero divisors
I’d be surprised if such a group doesn’t exist

[pastel shoal]

Also why is it enough to check if the group is linear? I don't see where monos come into the picture anywhere in condition 2

[pastel shoal]

Yeah I believe that there is no left adjoint but I'm trying to find an explicit counterexample

[plucky arch]

monos come into the picture for the special adjoint functor theorem

[weak lodge]

side note on a name for this relation: if we consider $(X,\otimes)$ as a magma, then this identity is precisely the statement that $\circ\colon X\times X\to X$ is a magma homomorphism
If $\otimes = \circ$ (which, as you said, happens in particular when both operations are unital), then there actually is a name for this identity; we say that the magma $(X, \otimes)$ is medial or entropic. For example, for an elliptic curve, the chord-and-tangent process $x\otimes y\coloneq -(x+y)$ (so that $x+y=0\otimes(x\otimes y)$) makes the curve into a nonassociative medial quasigroup

[broken turtle]

harmacist

[lone jacinth]

Anticommutativity and the Jacobi identity hold in lie algebras, so that's noteworthy I guess

[limpid horizon]

For "p consists of zero-divisors of M/xM" part, is this because, p kills z, and since z cant be in xM, its image in M/xM is nonzero so we know p is contained in ann(z) for z in M/xM?

Also, I know if p is in AssM then p is the annihilator of some cyclic submodule Rm. I suppose there could be other cyclic submodules that p also annihilates, but I'm unsure on why we are able to pick a maximal one. I guess why do we know a maximal one exists?

[spice idol]

commutative theories!!

[past cove]

Zorn's lemma :)

[past cove]

Also yes this is the correct idea

[past cove]

What are the assumptions here?
Noetherian R plus f.g. M?

Cuz I think if you wanna use Zorn you need assumptions like that

Cuz my idea is basically

Take the poset of all cyclic modules annihilated by p

Since p is in Ass(M) this is nonempty

Now take a chain

Since R is noetherian and M is f.g. M is noetherian and so the chain stabilizes

Hence it has an upper bound

And so by Zorn's lemma you have a maximal such cyclic submodule

If you didn't have those assumptions it be harder because you have a chain and finding the upper bound would be nontrivial

You could take the union of all the cyclic submodules in the chain and clearly this is a submodule annihilated by p, but idk how to show it's cyclic and I doubt you can even show it's cyclic without some extra assumptions

[limpid horizon]

Oh ok yeah I was thinking maybe M is noetherian. Then its immediate from that right? Nonempty set of submodules has a maximal element

[past cove]

yeah if R is noetherian and M is f.g. then M is noetherian

not a difficult proof

[limpid horizon]

I will try it

[real sinew]

Why are (group) representations not required to be injective?

It seems to me that a representation where the some group elements are represented by the same thing is not really a "representation"

[last talon]

Because that ends up being too restrictive
Like, I’m like 99% sure things like Maschke’s theorem fail if you require this

It’s kinda for the same reason as why we don’t require group actions to be injective, or homomorphisms

Because it turns out that studying quotients is kinda useful

[wanton spoke]

representation theory wants to study all ways a group can act

injective is a stronger property

if people required injectivity by default you’d lose tons of natural examples especially ones coming from quotient groups

etc ....

[last talon]

It’s also not a priori obvious every group should have a representation (to someone just starting learning rep theory) with that requirement

[lone jacinth]

Even if you only care about faithful representations you might need to use other representations (for example a faithful rep might be a direct sum of non-faithful ones).

At that point is just becomes a question of convenience. Should "representation" mean "faithful representation" and then make up some adjective for "not necessarily faithful" or the other way around. Choice is simple

[worldly zealot]

For an explicit example, it’s interesting information that a triangle exists inside a hexagon, and a cyclic group of rotations exists inside a dihedral group . These all arise as non injective representations D6

[real sinew]

Great answers, thanks everyone

[real sinew]

Is it true that every group has a faithful rep?

[last talon]

Yes (and thanks for proving my point because I was wondering if someone would challenge me on that)

Consider the vector space with basis G, and act by right multiplication sending the basis element h to hg

[real sinew]

That makes sense

Thanks!!

[wanton spoke]

yes if infinite dimension is allowed

[lone jacinth]

If infinite groups are allowed then why not

[wanton spoke]

the distinction is not whether the group is infinite but whether the representation is allowed to be infinite-dimensional

[lone jacinth]

Well,
findim reps of finite groups,
reps of groups,
reps of finite groups,
findim reps of groups

Would be the four distinct options

[tough helm]

does anyone know what a smooth algebra is?

wikipedia only gives the definition over a field

[lone jacinth]

Should be such that
R -> A
is a smooth map
https://stacks.math.columbia.edu/tag/00T1

[smoky wasp]

quick question, im trying to show that if R is integrally closed

R[x] must be integrally closed

im hung up on basically one small lemma I cant seem to sort out

Ive reduced it to the case where R[x] is integrally closed in S[x], where S = Frac(R)

I just want to show that if xf in S[x] is integral over R[x]

then f is integral over R[x]

[smoky wasp]

nvm I think i got it

[fierce steeple]

Pog

[lone jacinth]

How'd you do it?

I think my approach would be for f in S[x] integral over R[x] we show the coefficients are integral over R.

First take the polynomial with coefficients in R[x] witnessing f being integral, then take the subring R' of R generated by the coefficients of those coefficients.

Then R' is a finitely generated ring, so Noetherian and f is integral over R'[x].

This means R'[x][f] is an fg R'[x]-module. Now the set of leading coefficient of polynomials in R'[x][f] forms a subring of S, and if we take a generating set for the R'[x]-module R'[x][f], their leading coefficient generate this subring as a R' module.

Now if a is the leading coefficient of f, then R'[a] is a submodule and because R' is Noetherian it is also finitely generated, so a is integral. That means a must be in R.

Say f has degree n, then ax^n - f is integral and of degree less than n. By induction ax^n - f is in R[x], hence so is f.

[smoky wasp]

I proved a different thing to get to the same end result lol

[smoky wasp]

I'll post it later once I get home

I just hopped on the train lol

Lowkey procrastinated on this hw way too long but c'est la vie

[edgy pond]

Consider the action of the quaternion group Q8 inside the quaternions H on H. This action is irreducible over R. The endomorphism ring over this action is isomorphic to Q8. To see this one can calculate all the commutation relations that needs to be satisfies by a 4x4 matrix. But since Q8 acts by multiplication inside H, is there a quick way to see that the endomorphism ring of this action must be isomorphic to Q8?

I think endomorphism ring should be the set of elements in H that commute with Q8

ig this is not the right way to see this

[lone jacinth]

I guess you can notice that H is isomorphic to R[Q8]/(1 + (-1)) (where (-1) is the element of Q8).

In general when you have a two-sided ideal I, then the endomorphism ring of the R-module R/I is just R/I

Or it might be (R/I)^op depending on your conventions

But H is isomorphic to H^op

[lone jacinth]

Alternatively: you know the endomorphism ring is a finite dimensional division algebra by Schurs lemma.

As H is associative multiplication on the right is Q8-linear. So H is a subring of the endomorphism ring, making it at least 4-dimensional. As H is the only at least 4d division algebra it must be H

[edgy pond]

I see

[lone jacinth]

Another alternative: the endomorphism ring must be either R, C or H.

So tensoring with C, the endomorphism ring would become C, C^2 or M2x2(C).

These would correspond to the endomorphism ring of respectively: one irrep, sum of two distinct irreps or square of one irrep.

Computing the character table of Q8 we see that it has one 2d irrep over C and all the others are 1d. Since there's no 4d irrep the first case is out. For the second case you at most get something 2+1 = 3 dimensional, so it must be that we're in the third case.

[trim basin]

Why do we like exact sequences

[silver goblet]

because they let us calculate things about unknown terms in the sequence

[trim basin]

Thankss

Ill check this out

[queen tide]

de rham complex of R^n is exact

and R^n is usually nice in general

so its the prototype

[steep rivet]

Long exact sequences often have a threefold repeating pattern

Like you go through objects X, Y and Z in one degree and it loops around to the next

$\cdots \to C^{n-1}(X) \to C^{n-1}(Y) \to C^{n-1}(Z) \to C^{n}(X) \to C^{n}(Y) \to C^{n}(Z) \to C^{n+1}(X) \to \cdots$

[broken turtle]

PKThoron

[steep rivet]

Now if you find that one of the objects repeatedly yields zero terms (say Z), then you find that the other are always isomorphic (here X and Y)

Even more excitingly, if it's the Y terms that yield zeros, you can establish an iso between the X term and the previous degree's Z term!

That often carries a character of "what happens in X in dim n already happens on Z in a lower dimension"

It sounds kinda dumb, but in topology, that proves "the 2-sphere is 2-dimensional because the 1-sphere is 1-dimensional"

[smoky wasp]

Im trying to prove this

I think I have the right idea

find the discriminant

argue its square free (should be ab probablY)

bada bing bada boom we are done

but im running into issues

there are clearly the embeddings

so this should be the discriminant matrix

but the determinant isnts quare free

am I messing up here or sth?

[lone jacinth]

Isn't the determinant just ab?

Or what are you getting?

The discriminant of Q(sqrt(a), sqrt(b)) would be a^2 b^2 though.

I don't if/why you wanted this to be square free(?)

[floral bluff]

I think it can be proved easier. Firstly it is easy to prove they are linear independent. Secondly they can be easily linearly represented by a basis 1，sqrt a，sqrt b and sqrt （ab）， so those are basis QED

[lone jacinth]

It's clear that it's a basis, the tricky thing is showing it's an integral basis

[leaden moat]

Marcus' Number Fields does this exercise with hints. I'm fairly sure there's a silly way of doing this using the conductor-discriminant formula and just comparing discriminants but that's ott

[smoky wasp]

Since if it is square free it is an integral basis

[leaden moat]

The converse isn't true though

[smoky wasp]

yeah, i just figured the assignment was trying to lead us in that direction

oh well

[smoky wasp]

ill check it out

[leaden moat]

Your exercise is like the final section of the one in that book. The exercise in there deals with all cases for a, b, not just both \pmod 1 mod 4

soz I mean \equiv

[lone jacinth]

I guess you can also use something like the ring of integers of a composite of linearly disjoint number fields is the composite of their number rings applied to Q(sqrt(a)) and Q(sqrt(b)), but maybe that's not accessible.

I guess it's also possible to just brute force exactly which elements are integral

[smoky wasp]

could one maybe argue that each of these elements are integral

and they are linearly independent

therefore they must form an integral basis?

[lone jacinth]

Well by that logic 1, sqrt(a), sqrt (b), sqrt(ab) would also be an integral basis

[smoky wasp]

ah yeah thats an issue

hm

could I do something knowing integral basis for Q(root(a)), Q(root(b)), and Q(root(ab))?

since I can find that

[leaden moat]

in theory if you know that disc(K) = \prod disc(quadratic subfields) for K biquadratic then you can do what I said above because for an order O in K, disc(O) = [O_K : O]² disc(K) (so since you know that the discriminant of your set of elements is (ab)^2 and the discriminant of your field is (ab)^2 you can conclude that [O_K : O] = 1), the problem is that this is circular if you don't use some junk from class field theory or smth. Marcus proves that product formula by working out that your elements form an integral basis in the first place

correct me if I'm wrong, it's been a while

[lone jacinth]

Yeah, an integral basis for Q(sqa, sqb) will be the product of an integral basis for Q(sqa) and Q(sqb) because they are linearly disjoint

[smoky wasp]

oh wait

real then I think im done

because I can pretty easily show (1, sqrt(a) + 1/2) is integral basis for Q(sqrt(a))

[smoky wasp]

Which gives exactly this

[steady lance]

So I just realized whenever I work with depth I don't think of it geometrically. Quick question: If I have a non-cohen-macaulay ideal I what is the geometric interpretation of depth(R/I)?

[smoky wasp]

Wait how do you show this

[lone jacinth]

It's in lemma 1.2 here at least
https://www.jstor.org/stable/pdf/1970526.pdf

[worldly zealot]

so if you don't know how a quasicoherent sheaf arises from a module, you can crudely think of R-modules as a vector bundle over the spectrum of your ring (not exactly what it is due to torsion but sufficient). a regular sequence (locally) then cuts out successive hypersurfaces in a way that reduces the dimension each time, the depth is the first time this process introduces nonzero local cohomology, i.e. creating a hole

for R/I this is just the structure sheaf of V(I)

[limpid horizon]

why does nonzero local cohomology detect a hole?

[worldly zealot]

for Z = V(I) and the sheaf M we have the long exact sequence ... -> H^i_Z(X,M) -> H^i(X,M) -> H^i(X / Z, M) -> H^{i+1}_Z(X,M) -> ..., so if H^i_Z(X, M) =/= 0 then we cant extend classes on X/Z to X

for an explicit picture, let R = k^n, Z = V(x_1, x_2, ..., x_n). local cohomology of this ideal is nonzero for the nth term, since locally around the hole at the origin we have a sphere

[steady lance]

Huh very interesting! Thanks!

[spice idol]

do you perhaps suffer from delusions of grandeur

[hushed bone]

<@&268886789983436800>

Nice AI

[proud veldt]

He also posted something similar in #point-set-topology

[hushed bone]

We have a point-set channel????

[proud veldt]

Yep

He even admitted to using AI

[hushed bone]

Epic.

[smoky wasp]

Consider this polynomial

assume p divides ai for all i

but p^2 does not divide a0

I am trying to show that the discriminant is not divisible by p

I have that each root individually is not divisible by p

thats easy to see

but im super stumped extending it to differences of roots

how can I do that?

like in terms of hints

[near lantern]

Do you mean divisible?

[smoky wasp]

wait what am I on about

yeah

[smoky wasp]

how would divisibility of roots even work

[near lantern]

Consider x^2 - 2x - 2. Its discriminant is 12.

[smoky wasp]

ah

[near lantern]

There are ways to do it.

[smoky wasp]

how can we?

[near lantern]

Actually I think this is the best way to do it.

[smoky wasp]

the initial question im trying to solve is im trying to prove the discriminant of the finite extension generated by such a minimal polynomial isnt divisible by p

if I turn my brain off and treat x as an integer I can by considering divisibility

but idk how that works

[near lantern]

Consider a ring B extending your initial ring A such that f splits completely over B.

Be careful because p may not be prime in B.

Prove that the discriminant D, as an element of B, lies in the radical of (p) (i.e. some power of D is divisible by p in B).

But D^{large}/p is in the field of fractions of A. We can choose B such that it doesn't contain any such element outside A (for example, if B is an integral extension of A). Then some power of D is divisible by p in A.

[narrow kraken]

For an arbitrary (i.e noncommutative) ring R, is there a more general condition than R being Artinian for J(R) to be nilpotent?

[lone jacinth]

R is called semi-primary if J(R) is nilpotent and R/J(R) is semisimple.

You also have perfect where R/J(R) is semisimple and J(R) is T-nilpotent (meaning any infinite sequence of elements in J(R) has an initial segment that multiplies to 0).

I don't think there is a nice condition without R/J(R) semisimple, as then you would also include Jacobson rings like Z and the polynomial rings, which are not really generalizations of artinian rings in any way

[narrow kraken]

for some reason, I don't see why Artinian and Noetherian are so fundamentally different. for example, so many properties hold for Artinian that are nowhere close to holding for Noetherian. Initially, I thought if R is Artinian, then R^op is Noetherian, but that's obviously not true. so, is there any way to turn a Noetherian ring into an Artinian ring?

[spice idol]

Ideals are in a way fundamentally upwards directed

because they are the closed set lattice of a closure operator

this means there's a huge difference between looking down and looking up

[lone jacinth]

I think it's worth noting that being Noetherian/artinian is defined in terms of properties of the R-module R.

Now in a general abelian category, then M in C is Noetherian/artinian if and only if M in C^op is artinian/Noetherian.

But (R-Mod)^op is not the same as R^op-Mod. The category R-Mod and the R-module R have several nice properties that are not preserved by duality. The fact that R is compact means that R artinian implies R being Noetherian. But there is no converse implication because R doesn't satisfy the dual condition to being compact (or finitely generated)

[spice idol]

yes, for rings the top ideal is finitely generated, so it is a compact element in the ideal poset, but the bottom ideal is not compact in the opposite poset

[near lantern]

Wait so compact Artinian implies Noetherian?

[drowsy niche]

so tbc compact = finitely presented here right?

if so then it's pretty immediate that compact artinian implies noetherian bc fg artinian implies noetherian

just by induction on generators + Hopkins-Levitski

[near lantern]

Well, that's for modules. I'm asking about an arbitrary abelian category (possibly where subobject lattices are complete and directed suprema respect intersections or some such conditions).

[lone jacinth]

So say the class of subobjects forms a set, and that we have appropriate cocompleteness.

Artinian implies the socle is nonzero. Mod out the socle, take the socle again, repeat, take colimit to get an ordinal indexed filtration.

Since we only have a set of subjects, this covers everything. Since we're compact the union is finite.

Then we are finitely filtered by semisimple artinian objects, which have finite length.

Hmmm, okay this argument doesn't quite work.

Maybe you actually need a little more

[lone jacinth]

Alright, it's not true. You need more.

Consider the poset P = N u {infinity}, and the category of P^op modules (over a field k let's say). Let M be the constant functor taking value k. Then M is compact and its subobjects are well ordered, but M is not Noetherian.

[spice quiver]

First time I’ve ever seen jagr not know something

[near lantern]

Isn't this R-Mod for R the path algebra, so that Hopkins-Levitsky applies?

In particular, M should be generated by the k at infinity and at n for arbitrarily large n, but not for finitely many n.

[lone jacinth]

The path algebra here isn't unital, but I think you can make this example work in R-Mod by adjoining a unit.

Not sure how you're applying Hopkins-Levitsky

[near lantern]

I'm suggesting that it's impossible to find a counterexample in this category.

[lone jacinth]

You're suggesting it, but I'm not sure why/how

Like if R is commutative, then cyclic modules are also rings, so fg+artinian = finite length.

But for R not commutative I'm not sure what an argument with Hopkins--Levitzky would be.

[near lantern]

I think I was misremembering fg Artinian ⇒ Noetherian.

[lone jacinth]

It should work in Mod-R if the simples are finitely presented:

Call a module locally finite if every fg submodule has finite length. For a module M write lf(M) for the sum of all finite length submodules.

Take M fg and artinian. If lf(M) were done. Otherwise M/lf(M) has a simple submodule S. The preimage is then an extension

0 -> lf(M) -> E -> S -> 0

as S is finitely presented, an fg submodule of E is an extension of S with an fg submodule of lf(M) hence has finite length.

This contradicts lf(M) being the sum of all the finite length submodules.

[near lantern]

Do you use fp to get that the kernel of (the fg submodule of E) → S is fg?

[lone jacinth]

Yes

[near lantern]

I think you might need to assume S is coherent.

Although maybe I'm just being a bit slow finding the argument with fp.

OK no quotient of fg kernel of relations should be fg

[near lantern]

Commutativity of the ring try not to magically substitute for completely random properties of the modules challenge (impossible difficulty level)

[lone jacinth]

I guess S is automatically coherent as it doesn't have many submodules. But yeah you shouldn't need it

I'm wondering what's the right way to translate this to a general abelian category. Like I don't think it's enough to just say that the simples are compact...

[near lantern]

Why not?

Assuming subobject lattices behave like for modules.

Oh IG you really do need the finite presentation...

[lone jacinth]

Well just in the proof I'm sort of using more properties of ModR

Maybe you can frame it as Ext(S, -) preserving direct limits, though that should be slightly stronger than being compact

[hushed bone]

This is backwards

[near lantern]

It does suffice that S is a cokernel of a map between fg objects, right?

[hushed bone]

Uncommutativity of the ring makes random properties of modules become UNTRUE

[near lantern]

Commonkey

[lone jacinth]

You might also need that every object is a sum of fg objects

[hushed bone]

Deadass anytime Jagr types it’s scary to comment

[fierce steeple]

I am compact

[hushed bone]

Cuz I’ll say “oh this is true because fact”

And he hits me with a “these aren’t commutative so sentence with 4 adjectives that don’t exist for commutative algebra”

[lone jacinth]

Or something like that so that E actually has an fg subobject mapping onto S

[fierce steeple]

Yeah lol

[hushed bone]

This is such an interesting sentence I love stuff like this

[near lantern]

Is this not just true

[hushed bone]

Because your use of an tells me you think in your head fg = eff gee

Not finitely generated

[lone jacinth]

In ModR, sure

[hushed bone]

But I read fg as being “finitely generated” so I’d have used a

[near lantern]

OK it says something about the subobject lattice

That it's continuous I believe

[hushed bone]

I have a legit question, where do modules over noncomm rings show up naturally

Is this a rep theory thing primarily?

[near lantern]

I think any condition on subobject lattices is fine

[near lantern]

I would say any symmetry or action is liable to produce one somewhere

Like if you do something with a vector space V then End(V) = Mn(k) probably acts on the thing you made.

[hushed bone]

Hmmm

[near lantern]

So maybe?

[lone jacinth]

I mean, if there is a module involved then it is rep theory I guess

[near lantern]

Definitely rep theory is full of them

[hushed bone]

that’s weird because somehow in comm alg I don’t think we wouldn’t think about matrices acting on a vector space

[near lantern]

Not sure about the converse

[hushed bone]

But we avoid ever trying to say stuff about noncomm stuff

Like, for this whole wide world I live in everything is always commutative and I have never felt held back by that

[lone jacinth]

I mean the endomorphism ring of a module should have significance in commalg as well

[near lantern]

Yeah I think that's weird. The point of studying modules over comm rings is that you can see extra data as modules over some new ring. This is just an extension of that.

E.g. quite common to view k-module + endomorphism as k[t]-module

[hushed bone]

I think we just can say enough about them as is

Idk

Or the data you capture looking at things over noncomm rings just aren’t what we care about ¯_(ツ)_/¯

[near lantern]

The philosophy of doing this is that module theory for classes of rings has enough results that you can use things you would have proved for "just" modules to now prove things about modules with an endomorphism or whatever else

It's very natural to not stop at "extra data such that the ring whose modules it turns out to be is commutative"

Ofc you can say less but you ca always say the fewer things in general and also the more things in particular, just like you're happy to say a few things about all modules over comm rings and a lot more about vector spaces.

[near lantern]

Who is the "we" here, actually?

Like what are the questions that define the scope of your interests for the sake of this discussion?

[lone jacinth]

Isn't the goal of commalg sort of to study commutative rings? So then looking at them would make sense.

Whereas the goal of representation theory/ homological algebra isn't so much to study rings, but to study their modules. So then you add module-theoretic restriction (like Noetherian, artinian, semisimple) instead of something ring theoretic like commutative

[hushed bone]

Well you end up studying modules over commutative rings too

They end up being really really important, and there’s statements relating structure on modules to properties of the ring

Maybe it’s not as emphasized tho idk

[lone jacinth]

Sure, but you're studying the modules to learn about the ring(?)

[hushed bone]

Hmmmm

The two are pretty related

But I would say I imagine it is a little more focused on the ring itself

But also Noetherian hypotheses are almost omnipresent in the field

And that ends up mattering most often because it makes a lot of things work with modules

[lone jacinth]

I guess since two commutative rings are morita equivalent iff they are isomorphic, a commutative ring is a little more determined by its representation theory

[hushed bone]

You need finite generation to keep persisting so you need the fg modules to be coherent

[steady lance]

I've actually been thinking about non noetherian rings a lot recently and having to systematically go back and check if every little condition still holds is so tedious

it's kinda suprising how many Noetherian conditions I take for granted lol

[little cradle]

Do we have a complete classification of subalgebras of GL(3, C) up to isomorphism? If yes, can you send me some source?

[golden osprey]

There's this massive constructive commutative algebra text

which does alot of work to get around Noetherian hypotheses

https://arxiv.org/abs/1605.04832

I make no claim I have looked at much any of the 1000 pages though

But Lombardi is a good writer IMO (I have read other stuff by him)

[worldly zealot]

to add on to people mentioing endomorphism rings, i guess some motivation is that people have started to consider derived categories of coherent sheaves as a useful setting for understanding varieties. mukai showed some geometrically significant connections between derived equivalent varieties, and from what i understand homological mirror symmetry is fundamentally about some derived equivalence.

so it turns out that tilting theory is a very powerful tool for considering derived equivalences, and it boils down to the endomorphism algebra of some sheaf containing all of the variety's derived information in its endomorphism algebra (e.g. for P^n this is the sheaf O+O(1)+...+O(n)). so on some level, its very nice when you can mediate these derived statements with modules over the (noncommutative) endomorphism algebra

i guess in "nature" some things in quantum physics aren't supposed to commute but theoretical physics could be one big prank and i'd be none the wiser so i can't really claim that as motivation

[wary elbow]

It depends on what you mean by "classification." Classifying them up to inner automorphism makes more sense and has a long history. Here's a recent paper on the topic; you can probably find explicit calculations in the references.
https://www.sciencedirect.com/science/article/pii/S002186931000534X/

[little cradle]

Classifying up to conjugation

[wary elbow]

Ok, so that's exactly what that paper (& the references therein) addresses.

[hushed bone]

Yeah it’s not that they always exist per se, and I thought I wanted to think about non-Noetherian stuff, but then I just realized how much it just kind of sucks ass and idk what’s going on haha. And, for a lot of stuff comm alg is used for, they’re just there anyway

[near lantern]

OMG this book

I love it

[golden osprey]

this should be easy but what is the contradiction here 💀

[golden osprey]

oh you've looked at it? I wish I had the time lol

[near lantern]

Yeah just skimmed parts that seemed like they would have the most aha moments

[golden osprey]

nice

[near lantern]

The first 1-2 chapters on local-global principle were nice

And somewhere in the meat of the book they describe how to translate classical non-constructive arguments to be constructive (the basic idea that I took home from it is that whenever you do cases on a = 0 or a ≠ 0 to prove something for fields A, you should instead base-change to A/(a) and A[a^{-1}] for a general commutative ring A (which is algebro-geometrically the zero set of a and its complement). This eventually results in a proof over a bunch of quotients and localisations of A (which in AG form a partition of Spec A). Then you have to see whether this can glue to the result for A (which depends on the details of what you are trying to prove).

[near lantern]

By hypothesis F is generated by the yi's. But you found h^{-1} in F which can't be generated by y's.

[golden osprey]

why not?

[lone jacinth]

Each y is off the form fi/gi so any polynomial expression can be written with demoninator some power of products of gi. But this is always relatively prime to h

[golden osprey]

ah

thanks

[storm basin]

notation question: is R[1/s] for some element s in R just notation for localizing at the multiplicative set powers of s

[sacred sentinel]

They're the same ring, yes. The notation is the same as R[x] for an indeterminate: elements of R[1/s] are polynomials in 1/s which is the same as the localization R_s

[storm basin]

Yeah, makes sense, thanks.

[foggy galleon]

what are examples of "infinite group theory" (by which I mean using groups that are infinite) proving theorems about finite groups?

[wanton spoke]

finite groups such as PSL_n(F_q) are understood by first studying the corresponding infinite algebraic group over an algebraic closure or over C

[foggy galleon]

do you know a reference that explains this theory?

[wanton spoke]

not really but I think Serre's book (Linear Representations of Finite Groups) covers examples of studying finite groups by moving to linear or representation theoretic objects over C

[wary elbow]

The key phrase here is "finite groups of Lie type." There are many books on the topic. Carter's book is well regarded, so you could take a look at that one. There's also a more recent book by Malle and Testerman.

[foggy galleon]

are there any other examples of this btw

[wanton spoke]

for a finite group G you study the infinite space BG instead of G directly

[foggy galleon]

but is it possible to prove something interesting about G in this manner?

[wary elbow]

I'm not entirely sure that this is a well formed question, because you're making a distinction between finite and infinite groups that doesn't exist in practice. That is, finite and infinite groups always mix when you're doing group theory

For example, you can't understand cyclic groups (or even define them) without appealing to an infinite group -- namely the integers -- so you could say that any theorem about cyclic groups (and, by extension, any finite abelian group) is essentially proved via the integers (by taking quotient groups).

Put another way, you're asking for theorems where both finite and infinite groups appear in the statement of the theorem, to which the only answer is.... there's a lot of them

[foggy galleon]

there's actually a distinction and it is even possible to write a precise version of my question, but I believe it wouldn't exactly capture what I'm asking anyway. It is not well formed as is, but that's on purpose

You can classify finite abelian groups without "using" Z

[foggy galleon]

no, that's not it

[near lantern]

It's not a reference but I think Deligne-Lusztig theory is mainly about constructing the characters of these G(Fq)'s by treating them as subgroups of G(\overline{Fq}) (as an algebraic group actually).

[wanton spoke]

cohomology of G is literally the cohomology of BG

[foggy galleon]

but that's not proving anything in itself about G. My question is more like, there's a property of G that translates to a property of BG, you can write some argument about BG that doesn't automatically translate to an argument of G, and prove this property

[near lantern]

Does it count if the solution can technically be written without infinite groups but it's impossible to come up with without them?

I don't have an example I'm sure of ready either way, but it might help others who answer.

[foggy galleon]

if calculating the cohomology on BG could be simplified via topological things that don't have a (obvious?) reflection in the finite group world then that'd count I guess. But what is an example?

[foggy galleon]

yeah

[wary elbow]

Ok, I understand what you're looking for, but it still feels awkward to me. In my experience, it always goes the opposite way: finite groups are easier to deal with than infinite ones, so you leverage facts about finite groups to get similar facts about infinite groups (eg, Frobenius kernels of algebraic groups). Going the other way feels artificial.

I'm not aware of any properties of infinite groups that are somehow easier to prove than the same properties for finite groups

[torn harbor]

I think the main advantage for groups of lie type at least is that your field is algebraically closed

when you work with infinite groups

[wary elbow]

oops, deleted some wrong statements, sorry

[torn harbor]

all good, I'm not really an expert on this stuff tbh

I do think in general though any time when using infinite groups helps with finite group theory

its moreso that there's some important other powerful tool you are using, but as a side effect you have to deal with infinite groups

all else equal you would basically always prefer to work with finite groups

[wary elbow]

I agree with this. However, this isn't what croqueta is asking for: they want something more precise

[torn harbor]

you might want to look at ultraproducts

I don't know if there are applications to finite group theory

but I do know that there are examples of results about finite graphs that can be proven by taking some sort of ultraproduct of finite graphs and using infinite theory

[wanton spoke]

take G = C_2 then BG is just RP^infnty and at that point the computation becomes a pretty concrete topological one at least to me it feels much easier to work with the topology of BG than with the abstract finite group itself

[torn harbor]

though most often ultraproduct techniques are useful for proving asymptotic-type results, which may not be what you are looking for

[summer quest]

tons of modular representation theory of finite groups (among other things) these days is being done using insane methods from p-adic geometry lol

[rose mirage]

she reflects on my levi datum til my reflection group spetsial

[random oxide]

She cut you man

[foggy galleon]

this seems interesting

but I could not find the paper lmao

the other reference is Hsu Quilts, but I'm not sure why it's even related lol

[ornate atlas]

Unsure if this is better for here or #⁠category-theory , but I suspect Jagr or Psuedo knows and they prowl both anyway lol

I suspect that im being a bit silly and falling at the last hurdle here, but im trying to verify that Ch(Mod-R) is a model category, and im struggling to see why the cofibrations are closed under retracts. Injective is easy, but im struggling to see why the cokernel of the retract needs to be projective

I was thinking something like, we have
$$\begin{tikzcd}
A \arrow[r] \arrow[d, "f"] \arrow[rr, "=", bend left] & X \arrow[r] \arrow[d, "g"] & A \arrow[d, "f"] \
B \arrow[r] \arrow[rr, "=", bend right] & Y \arrow[r] & B
\end{tikzcd}$$

and we get a the ladders of SES's
$$\begin{tikzcd}
0 \arrow[r] & X \arrow[r] & Y \arrow[r] & \mathrm{coker}(g) \arrow[r] & 0 \
0 \arrow[r] & A \arrow[r] \arrow[u] & B \arrow[r] \arrow[u] & \mathrm{coker}(f) \arrow[r] \arrow[u, "\alpha"] & 0
\end{tikzcd}$$
and the same with the "rungs" flipped. From which we can write
$$\begin{tikzcd}
\mathrm{coker}(g) \arrow[r, hook] & Y \arrow[d, "\pi"] \
\mathrm{coker}(f) \arrow[r, "\alpha"] \arrow[u, "\alpha"] & \mathrm{coker}(g)
\end{tikzcd}$$
which gives a lift.

But like im not really conviced by this, in particular im not using that coker(g) is projective which feels very much wrong

[broken turtle]

Nope
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ornate atlas]

In fact im atually completely unconviced by this, it should work for any map and any surjection, so this is way off

but yeah im a little lost here if anyone could point me in the right direction

[hushed bone]

From my memory looking at Quillen’s book homotopical algebra, verifying this part is absolutely hell

I think it’s done a lot later in the book because it’s not obvious at all

[ornate atlas]

It isn’t even proven in the paper/notes I’m reading because it’s “straightforward” lol

I’m pretty sure it should be too, I think this is just like actually knowing what the maps are I don’t think there’s really anything going on

[hushed bone]

My memory is that it’s not obvious lol

I remember trying to show it by hand cuz it’s just mentioned that this gives a model structure and it’s wonky

[lone jacinth]

So cok(f) will be a direct summand of cok(g), and summand of projective is projective

I'm not exactly sure what you're doing in your argument there, but it is indeed straight forward

[hushed bone]

Grrrrrr

Wait closed under retracts wasn’t the hard part

For when I was putzing around with this

I think it was something about having factorizations

Idk, whatever

[lone jacinth]

Factorization should be a little more annoying yeah

[ornate atlas]

I am not either, was a silly thought at the end of the day lol, just nonsense

Why exactly is it a summand though?

[lone jacinth]

When you have maps
C -> D -> C
where the composition is the identity, then C is a summand of D

[ornate atlas]

Is this just the splitting lemma (I’m home in bed now so I can also just work that out in the morning if that’s something that should be clear)

[lone jacinth]

I mean, sure that's the splitting lemma.

But the fact that summand of projective is projective, really only uses that you have this splitting C -> D -> C, so is even true in additive categories that are not idempotent complete.

[hushed bone]

tf is idempotent complete

[lone jacinth]

Every idempotent morphism has a kernel and image.

So basically every idempotent e: D -> D makes D = ker e (+) Im e

[hushed bone]

Jesus lmfao

Jagr, all the cats I am comfortable with always have kernels and images lol

I guess uh

Ring doesn’t and this holds there?

By the splitting (1,1) -> (1,0) or whatever

[lone jacinth]

Same, I just got to remember to add it to the assumptions in the notation and conventions

[lone jacinth]

Well, Ring isn't even additive

But a stupid example could be something like even dimensional vector spaces

[hushed bone]

Ring is a bad category

[lone jacinth]

Less stupid I guess, the category of free modules

[hushed bone]

Lmfao

Does that not have kernels?

Oh

Yeah lmfao you can have

Non-full rank maps

[lone jacinth]

Yeah, so a projection onto a 1d subspace would be an idempotent without kernel nor image

[hushed bone]

My reaction to that information

[lone jacinth]

And if infinity counts as even, then you can make idempotents with image, but no kernel and vice versa.

[ornate atlas]

That is a very nice very terrible example lol

[hallow bone]

are you comfortable with the category Set?

lol

[hushed bone]

I don’t think about that as a category lmao

[hallow bone]

its the prototype 1-category!

fine then

Top

[lone jacinth]

I think you misspelled Ab

We don't do Set around here

[hallow bone]

cowardsss

you cant even function without zero objects

[digital parcel]

u mean TopGrp?

[fierce steeple]

Idempotent complete is important dawg

[fierce steeple]

Thinking about Top as a category is a mistake

[hushed bone]

Exactly. 1-category aka sets

[hushed bone]

I don’t think about this

[fierce steeple]

I only think about Top to talk about open covers of a scheme.

Chmowned

[hushed bone]

Legiterally

I think about one topological space at a time

[fierce steeple]

And this is avoidable anyway

[fierce steeple]

Average modern day homotopy theorist

[hushed bone]

Keep your lgbtq away from my algebraic geometryyyyy

[fierce steeple]

Why

[hushed bone]

https://tenor.com/view/winking-yee-yeeyee-wink-spit-gif-18514536

[fierce steeple]

Homotopy theorists do all the best AG

[hushed bone]

Your homo-what’s its

Also I don’t lik that you have a “theory”, that’s jus matrix bullshit designed to keep us fed with lies

I build a mind palace and draw the lines inside my mind

I wasn’t able to drive this bit to the finish line

I think there was potential somewhere in there but it requires a better meter than I

[fierce steeple]

Lol

Lol

[silver goblet]

is this the newest 2000 page long jacob lurie book

[silver goblet]

wdym 💀

i guess "basically just Top" anima is a better behaved \infty-category

[hard kite]

Let $R$ be a Dedekind domain with fraction field $K$, $L/K$ a quadratic extension and $S$ the integral closure of $R$ in $L$. Then $S$ is Dedekind. Is it always true that $S=R[\alpha]$ for some $\alpha \in S$?

[broken turtle]

1728

[lone jacinth]

It is not true
https://kconrad.math.uconn.edu/blurbs/gradnumthy/notfree.pdf

Apparently K = Q(sqrt(-6)), L = K(sqrt(-3)) is a counterexample

[hard kite]

thank you that is great to know

[fierce steeple]

I mean anima ye lol

[hard kite]

Let $R$ be a noetherian domain of Krull dimension 1, $I$ an invertible $R$-ideal and $P$ a prime of $R$ dividing $I$. Is it true that $P$ is invertible? If this is the case, is it true that an integral $R$-ideal $I$ is invertible iff it is product of powers of invertible primes ?

[broken turtle]

1728

[lone jacinth]

If I = PJ, then
P(J I^-1) = (1)

[past cove]

does divide here always imply that there exists such a J?
Because I interpreted it as just meaning that I is a subset of P, but idk if this a priori implies there exists such a J

[lone jacinth]

That's how I interpreted divide 🤷‍♀️

[past cove]

well that's fair haha

[lone jacinth]

They talk about I being a product of prime ideals so

[ornate atlas]

Just to double check I understand this correctly, we have a diagram as above, and since A->X->A and B->Y->B compose to the identity, the induced maps on cokernels must too (this im slightly iffy on, I think it should be true but I could be wrong), but then this implies the sequence splits, so coker(g) = coker(f) (+) ker(induced retract), hence coker(f) is projective since coker(g) is

[lone jacinth]

Yes

The first part just follows from the cokernel being functorial

[ornate atlas]

Yeah that makes sense, I thought it couldnt really work any other way but checking seemed annoying lol

[hard kite]

oh oops

I meant dividide as in P contains I sorry wrong terminology

I am so used to Dedekind domains that I don't distinguish between the two

[lone jacinth]

And you don't really need the splitting of the sequence, it's just

     cok(f) <-> cok(g)
         v

M ->> N

then use that cok(g) has the lifting property

[ornate atlas]

How do we get that diagram?

[lone jacinth]

The diagram is the definition of being projective

Or for cok(f) to be projective

[hard kite]

Let me clear up what I am wondering about. I have an invertible integral R-ideal I and a prime ideal P of R that contains I. Does this imply that P is invertible? I am asking this because if every prime ideal containing I is invertible, then I itself is a product of prime powers and invertible. I am wondering wether the converse also holds

[lone jacinth]

Any principal ideal is invertible

So that would just imply all P are invertible

[ornate atlas]

Oh yeah sorry I was confused about why we had maps in both directions with the cokernels but that’s just from the retraction

[hard kite]

hmmm I agree but I am thinking about how that helps

[lone jacinth]

I know all primes invertible implies dedekind, so if R isn't dedekind this won't be true

[hard kite]

Why ?

I know why all primes inv => dedekind

but why would it be a problem if im just considering the primes dividing some invertible ideal

[lone jacinth]

Because every ideal contains a principal ideal

[hard kite]

oh duh

lol

lool

ok thank you

But just to be sure since im slow today, if every prime that contains I is invertible then I is invertible and a product of powers of these primes right

I think so dw abt it ac

[lone jacinth]

Idk, sounds reasonable I guess

[lone jacinth]

There are finitely many primes p1, ..., pn above I.

They are maximal, so their intersection is just their product, so I is contained in the product.

Multiplying by p1^-1 gives something in p2*...*pn so still an ideal in R. Repeat until some product of primes times I equals R

So yes, true

[hard kite]

thankz 😸

[narrow kraken]

Let's suppose we are working in the category of left R-modules (R can be noncommutative)? If I am correct, why do we have Hom(A, B \oplus C) = Hom(A,B) \oplus Hom(A,C), but not Hom(A \oplus B,C) = Hom(A,C) \oplus Hom(B,C)? Is there an example where the second is not true? I think direct sum in the first variable sends direct sums to products, while direct sum in the second variable preserves direct sums. I know this has something to do with abelian categories more generally and (co)products, but I have no intuitive understanding of co(products). Also, I don't really get why the difference between direct sum and product is so significant, given direct sum is just the product but only finitely many components can be nonzero.

also, does anyone have any good references for gaining more intuition behind these abstract categorical definitions?

[plucky arch]

https://pseudonium.github.io/2026/01/18/Products_Categorically.html

[plucky arch]

i think both of these identities should hold?

finite direct sums and finite direct products should coincide i think...

direct sum does a different thing to the direct product

lmk when you get back

[lone jacinth]

It's true that direct sums in the first argument turn into products and that products in the second argument also turn into products.

But as Pseudo says, finite direct sums equals finite direct products. So both of those identities hold.

The second argument does not in general preserve arbitrary direct sums, but does for A a finitely generated module.

(but it always preserves finite direct sums, as those are products)

[summer quest]

yeah for finitary direct sums both statements are true

[silver goblet]

as for an example where it is not true, as others have said Hom(-, C) sends coproducts to products, so take any category where these are different. For example, if I want to map out of a coproduct of sets, I just need to map out of each set. Hom(A \sqcup B, C) has |C|^(|A| + |B|) elements (which is |C|^|A| × |C|^|B|) but that is different from the disjoint union of Hom(A, C) and Hom(B, C), which has |C|^|A| + |C|^|B| elements

you're right that the difference between coproduct and product is not too bad in left R-modules but the difference is pretty drastic even in other important categories

[cloud karma]

Consider the identity cot(x) dx = log(sinx)+C. I'm trying to find a suitable vector space over R where this identity makes sense. So I take smooth functions from R\pi Z to R and quotient it by W=R^Z since R \piZ is disconnected and will be a disjoint union of connected sets which look like I_n = (n pi, (n+1) pi). My question is how to prove that the dimension of W is infinite countable?

[lone jacinth]

Did you mean W to be the quotient space or R^Z ? In either case it should not be countable dimensional.

Why do you need to prove that?

[cloud karma]

Not the quotient space. Just the R^Z portion

Oh it's uncountable?

[lone jacinth]

Yes

[cloud karma]

I think the core idea of this exercise is just to distinguish between direct sums and direct products in infinite case

[cloud karma]

Why so?

[lone jacinth]

Consider first Q^Z, this set is uncountable, but a countable dimensional space over Q must be countable, so Q^Z has uncountable dimension.

Now tensor with R, then R(x)Q^Z is a subspace of R^Z, so that must have dimension at least as big

[cloud karma]

Is there a way to do this without tensor products?

[lone jacinth]

Sure, not that R^Z = R^Q by picking a bijection between Z and Q.

Now consider for each real number r, the dedekind cut {q in Q : q < r}.

These define an uncountable chain of subsets of Q. Now let v_r be the vector that is 1 on this dedekind cut and 0 elsewhere. Then all the v_r are linearly independent

[cloud karma]

Okay yeah thank you

[narrow kraken]

is there a nice description of the set of cyclic elements of an R-module M, i.e, m in M for which Rm = M?

[fierce steeple]

I'm not sure what sort of better description you can want – typically the set is empty

[lone jacinth]

So let's say R is commutative and M = R/I, then an element is cyclic iff it is a unit in the ring R/I

[hallow bone]

is the commutator on normal subgroups an associative operation?

no its not; if it were then the definitions of nilpotent and solvable would coincide

interesting

for rings the commutator is associative; its just product of ideals

[hidden shore]

commutator is abb^-1a^-1?

[hallow bone]

If H and H' are subgroups on G, then [H, H'] is the subgroup generated by X = { [h, h'] | h ∈ H, h' ∈ H' }

[hidden shore]

oh are those lie brackets

[hallow bone]

if H and H' are normal, then [H, H'] is also normal and so you get a binary operation on the set of normal subgroups

[hallow bone]

? no

[g, h] is here ghg^-1h^-1

[hidden shore]

Never mind sorry

[hallow bone]

lol its cool

[hallow bone]

could be interesting to look into conditions exactly when the commutator is associative

[cloud karma]

Lmao still thinking about the tensor proof jagr gave

It's such a short and nice argument

[urban granite]

proof by asking a toddler on the street

[plucky arch]

I think there’s also a direct diagonal argument proof you can do

[unborn rampart]

Do ring homomorphisms commute with polynomial remainder? I mean if f, g is in R[x] and phi is an endomorphism of R[x], is phi(f) rem g equal to phi(f rem g) ?

[lone jacinth]

I guess you can just consider f=g

So no

[unborn rampart]

Hmm, what about phi(f) rem phi(g) versus phi(f rem g) ?

[silver goblet]

another example over any characteristic p ring. Take f is x and g=x^p, and phi to be the frobenius

[lone jacinth]

Well if
f = qg + r
then
phi(f) = phi(q)phi(g) + phi(r)

So I guess depending on how you're defining remainder you just want to show that the degree of phi(r) to be less than the degree of phi(g).

If R is an integral domain then the degree of phi(f) should be the degree of f times the degree of phi(x)

Otherwise it could get complicated I think

I guess remainder becomes less well defined then anyway

[unborn rampart]

That makes sense, thanks I want to generalize it to noncommutative multivariate division, but I wanted to make sure it made sense in the simpler case first

[ornate atlas]

Im trying to prove that the adjuction between restricton and extension of scalars is a Quillen functor ChR->ChS (projective model structure), but im a little confused as to why the extenstion of scalars needs to preserve cofibrations (but im guessing this is just an easy algebra issue)

So like specifically what im trying to show is that if M_n->N_n (n>=0) is an injection with projective cokernel, then so is S(x)M_n->S(x)N_n. Its clear that the cokernel is preserved, because left adjoints preserve colimits, but I dont see why the induced map must still be an injection. Does that follow because the cokernels are preserved? Because in general the tensor product of an injective map need not be injective right?

Like we don’t run into this issue with the weak equivalencies between cofibrant objects (quasi isomorphisms of projective chain complexes) because projective modules are flat right? But I’m not sure I see why it works before, I’m guessing specifically the cokernel being projective forces it but I’m not sure exactly

[scarlet ermine]

Sorry maybe I don’t understand the context, but in your setup don’t you have a split exact sequence, and any additive functor preserves split exact?

[ornate atlas]

This is quite possibly the case yeah, why do I have a split exact sequence?

The situation is that I have some chain complexes, and in positive degrees the chain map is injective with projective cokernel, is that equivalent to this being split exact? (My knowledge of that kinda stuff is pretty bad which is why I suspect this is just a basic algebra issue)

[scarlet ermine]

Ah yeah a SES with the right term projective is split (use the definition of projective to produce a splitting)

[ornate atlas]

Ahh I see

[scarlet ermine]

since N_n subjects onto the cokernel, you lift the identity map coker->coker to N_n

[ornate atlas]

So yeah take 0->M_n->N_n->coker->0, that splits, additive functor thing

[scarlet ermine]

yup

[ornate atlas]

I see, yeah this is kinda obvious now that it’s been pointed out but I just didn’t realise that

[scarlet ermine]

ah yeah no worries

[ornate atlas]

I for sure need to brush up on my basic homalg and abelian categories stuff lol

[daring carbon]

My apologies, my background in algebra is kinda weak

I only read like module theory

I am interested in compact abelian groups

Like T, or Z_p. While the dynamics of a rotation in T depends on the type of irrational number, on Zp I noticed all rotations by units i.e. x-> x+unit are equivalent to x->x+1. This is essentially because Zphat = Qp/Zp is acted upon naturally by Zp^x.

I am wondering if this scenario of having the field of fractions /Zp for the dual is something that's kinda "unique" to p-adic integers of if it's something I should expect from compact abelian rings.

This has me wondering for examples of compact abelian rings, I know that if I have a local ring I can complete it into a compact ring. The best example of these types of rings I know are localizations of number rings at primes.

How do completions of localization of number rings at prime ideals look like? What is their Pontryagin dual?
Do I also have essentially "trivial" dynamics?

I'm not sure if these are good questions to ask, or if I'm even in the right channel

[narrow kraken]

sorry for the wall of text, but my professor recently introduced the Yoneda lemma and (co)products. I am confused on some of the examples he gave though, as well as the formulation of the coproduct.

Yoneda lemma:

Given a category C, consider the functors R : C^op → Fun(C, Set) mapping X to R(X) in Fun(C,Set), where R(X) : T → Hom(X, T). Also, consider C : C → Fun(C^op , Set) mapping X to C(X), where C(X) : T → Hom(T, X). Both R and C are fully faithful (i.e, embeds a full subcategory). Thus, an object in C is uniquely defined (up to isomorphism) by the functor it (co)represents.

Example 1: Take a ring R and let C be the category of R-modules. Take the forgetful functor F: R-modules -> Set that sends M as an R-module to M as a set. Then, which object X represents this functor, i.e for which X do we have M = F(M) = Hom(X, M)? Clearly, we have X = R.

**Question: How do we know we can do this, since we do not know that the representation functor is essentially surjective? Not every functor F in Fun(C,Set) can be represented by an object in C, right? Do we not need to check any of the other axioms like compatibility with morphisms, i.e, why are we immediately done? **

Example 2: Again let C be the category of R-modules. Fix some r in R, and consider the functor F: M -> {m in M | rm = 0} =. Ann_M ((r)). Then, which object represents this functor, i.e, for which X do we have Ann_M (r) = F(M) = Hom(X,M)? Clearly, the object is R/(r).

Definition: The coproduct is the object that represents the product of Hom-sets.

**Question: I’m overall just kind of confused on this, a concrete example (especially where the coproduct is not just the product) would be really helpful. **

[hallow bone]

for the first question, you can explicitly write down the isomorphism F(M) -> Hom(R, M) and check it is natural

it is equivalent to saying that R is the free module on a single element

[hallow bone]

think about coproduct in sets

that will be the disjoint union

[narrow kraken]

yea i agree that M = Hom(R,M), but like don't we have to check the other axioms like compatibility with morphisms and stuff

maybe i'm just overthinking it actually

[hallow bone]

yes you can write those down explicitly too

[plucky arch]

Using “=“ here is a little sus right

Really you want a natural isomorphism

They might not be literally equal as functors

[worldly zealot]

Could anyone explain the Karoubi envelope of a category? I see the definition and some examples but I am kind of lost on what it’s doing and how to think about it

[hallow bone]

true

[plucky arch]

This is the idempotent completion right

[worldly zealot]

Yeah

[hallow bone]

however the naturality is so natural and naturally canonical that they may as well be equal

[plucky arch]

In the category of vector spaces, idempotents correspond to projection operators

Linear maps P with P^2 = P

You can generally think of idempotents in categories as a kind of projection

Now, in linalg, you often want to study the subspace corresponding to a projection, im(P)

This has an inclusion map iota : im(P) -> V, which has an inverse P : V -> im(P)

In fact, this turns out to be a splitting of the idempotent

So, by analogy, you can think of idempotent splitting as identifying the subobject the idempotent projects onto

In fact this is often a convenient alternative representation of subobjects, you use a projection operator

[undone idol]

for nontrivial example of coproducts, you can look at

• disjoint union in Set
• disjoint union in Top
• free product in Grp (this one might be a little harder to see why it satisfies the representability property)

[summer quest]

a good example to consider is the Karoubi completion of a preadditive category (categories enriched in Abelian groups), in which case the Karoubi completion is a pseudo-Abelian category (one where every idempotent morphism has a kernel, and hence also a cokernel)

[worldly zealot]

I see I see

[plucky arch]

This works cause im(P) = ker(I - P)

[worldly zealot]

Right I saw projective modules as the completion of free modules, this makes total sense now

[summer quest]

this pseudo-Abelian example shows up when constructing various categories of motives for example

[worldly zealot]

The example I am needing to work with is the singularity category

[plucky arch]

Right exactly, retracts correspond to subobjects with a projection operator

The issue is that a category might not have a splitting for its idempotents

[worldly zealot]

Which is some triangulated category and the completion is supposed to be akin to completing local rings

[summer quest]

right yeah singularity categories are kind of similar to this example around free versus projective

[plucky arch]

So the karoubi envelope formally adds in any missing idempotent splittings

You can think of it as adding all “virtual” subobjects for projection operators in your category

[worldly zealot]

Right

[plucky arch]

If your category already splits every idempotent then its equivalent to the karoubi envelope

[undone idol]

there is an alternate interpretation of Yoneda that I like a little bit more; Yoneda is essentially saying that if Hom() is some represented functor (let's call the represented element x) from C to Set, and F is some other functor from C to Set (of appropriate variance), then there is an extremely easy way of understanding how many natural transformations there can be from Hom to F (reverse the direction for the other variance).
And that correspondence is exactly given by the elements of F(x), to which you associate the NT given by sending id_x to that element. and this correspondence is natural in both C and Set (ie behaves well when you vary both x and F)
so the statement is saying that

• you can always do this to get a well-defined NT (faithful)
• every NT can be recovered this way by reversing the process described above (full)

[plucky arch]

Yeah there are lots of cool ways to view yoneda

[plucky arch]

It’s worth going through this explicitly

Let $G : \mathbf{R}-\mathbf{Mod} \to \mathbf{Set}$ be the functor $\text{Hom}(R, -)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

We want to show that there is a natural isomorphism $\alpha : G \Rightarrow F$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

So for every R-module M, we want a function $\alpha_M : \text{Hom}(R, M) \to F(M)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

I’ll define this through “evaluation at 1”, i.e. $\alpha_M(\phi) := \phi(1) \in M$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

This has an inverse $\alpha_M^{-1} : F(M) \to \text{Hom}(R, M)$ defined by $\alpha_M^{-1}(m)(r) = r*m$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Naturality of alpha says that for a homomorphism $\psi : M \to N$, you need to check $F(\psi) \circ \alpha_M = \alpha_N \circ G(\psi)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Following the definitions, this boils down to $\psi(\phi(1)) = (\psi \circ \phi)(1)$, which is true, so alpha is natural

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Note that since alpha is invertible, you can deduce $\alpha_N^{-1} \circ F(\psi) = G(\psi) \circ \alpha_M^{-1}$, meaning $\alpha^{-1}$ is natural too! So we have a natural isomorphism, as required

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

There are many ways to simplify the argument above, but i think it’s helpful to see the full details at least once

( @narrow kraken )

[narrow kraken]

thank you so much! (I also saw your answer to my old question about hom distributing over direct sums/products, i haven't been able to get to it yet oops)

[plucky arch]

Can i give my preferred way to think about yoneda? It might help you

[narrow kraken]

sure

[plucky arch]

Personally, i would say yoneda is about is-does duality

These refer to two types of perspectives you can take on something - what it is, and what it does

For example the “is” of a word corresponds to its definition, while the “does” corresponds to its usage

What money “is” might just be fancy paper, disk-shaped pieces of metal or 1s and 0s on a bank computer; but what it “does” is allow you to trade for goods and services

In that way it gains value beyond its intrinsic physical worth

This comes up a lot in mathematics as well

[narrow kraken]

ic

[plucky arch]

A matrix “is” a grid of numbers, but what it “does” is represent a linear transformation

In general the “is” side tends to be passive, about the internals or intrinsic properties of your object

While the “does” side is active, about how your object relates to and interacts with other things

A good toy example are numbers themselves

Using addition, you can view numbers actively as translations of the number line

So “3” means a translation of 3 units to the right

Things like negative numbers make a bit more sense in the does-world - you can’t have -2 apples, but you can definitely translate 2 units to the left

0 plays a special role too - in the does-world, it corresponds to leaving the number line alone

But more importantly than that, 0 allows you to convert back from does to is

If you have a translation but you’ve forgotten what number you’re translating by, you can recover it by following where 0 goes

Since 0 maps to 0 + n = n

does that example make sense?

(i promise this is crucial for understanding yoneda)

[narrow kraken]

i think so

[plucky arch]

Another way to view numbers actively is with multiplication

In that case, you identify a number k with the operation x -> kx, a stretching of the number line

[narrow kraken]

instead of treating an object (a number) as some intrinsic thing, you can identify it by seeing what is does to another object (0)

[plucky arch]

And again, the neutral element for multiplication, “1”, allows you to convert back from does to is

One cool thing is that “-1” corresponds to rotating the number line 180 degrees, in the does-world

So solving “x^2 = -1” in the does-world becomes “find a transformation that, when applied twice, gives you a rotation of 180 degrees”

[narrow kraken]

mm

[plucky arch]

But that’s easy! Just rotate by 90 degrees

If you want to convert back from does to is, you follow where “1” goes

And it lands right where “i” should be

So by working in the does-world you can discover complex numbers for free

[narrow kraken]

ic

[plucky arch]

Yoneda is the same idea, just applied to category theory

[plucky arch]

It comes down to this definition

The key observation is that you can view the element $1 \in R \textit{ actively}$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

What it does is take a homomorphism $\phi : R \to M$, and give you an element of $M$, namely $\phi(1)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[narrow kraken]

right

[plucky arch]

In fact, this works for any element $r \in R$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Instead of viewing it passively as just an element of $R$, you can view it actively as something that takes a $\phi : R \to M$ and produces an element $\phi(r) \in M$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

This should remind you of the double dual map from linalg

Where you can view elements of V “actively” as linear maps V^* -> K

[plucky arch]

Yoneda tells you that in fact you have an equivalence between the passive and active points of view

If you have a natural transformation $\alpha : \text{Hom}(R, -) \Rightarrow F$, then you can go to the passive pov by following the identity element!

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

It’s exactly like how we followed 0 and 1 previously

We get $\alpha(\text{id}_R) \in R$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

And more generally, $\alpha(\phi) = \phi(\alpha(\text{id}_R))$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

So, we have a natural correspondence between elements of R, and natural transformations Hom(R, -) => F

The former is the passive pov, the latter is the active pov

Does that make sense?

[narrow kraken]

ic, makes sense

[plucky arch]

This is why yoneda is so fundamental to category theory

Cat theory tends to focus a lot more on what things “do” over what they “are”

Yoneda is essentially what allows us to do this, since it tells us that there’s an equivalence between “is” and “does”

So you can switch between the passive and active perspectives depending on what’s more convenient

Like how you can switch between viewing a matrix as a grid of numbers, or as a linear transformation

[hidden shore]

is this the same type of idea behind a vector and its dual?

[plucky arch]

Yes actually

If you have an inner product on V, then you can view a vector v “actively” as the covector <v, ->

The analog of is-does duality here would be the Riesz representation theorem

It tells you there’s an equivalence between viewing a vector passively and viewing it actively

[hidden shore]

cool!

[plucky arch]

It’s not a direct consequence of yoneda since yoneda works when you have a “neutral element” that lets you convert back from does to is

But I would say they’re both examples of the same underlying idea, namely is-does duality

[plucky arch]

In fact, the composition in a category lets you go from is to does, while the identity lets you go from does to is

[grave pumice]

is-does duality is awesome

wait im reading up the messages this is an awesome explanation

thank you so much @plucky arch this really cleared up a lot for me

[plucky arch]

This is my preferred way to understand yoneda these days

[steep rivet]

it's the clearest way to me so thank you for putting it this way

[plucky arch]

[blissful field]

Not quite sure what is meant by taking the transpose of a function here. I thought I did but I don't.

f and g are matrices in SO(3)

[storm basin]

I think that the claim that this isomorphism is not canonical is wrong, and the isomorphism Z/mZ otimes Z/nZ -> Z/gcd(m, n)Z IS canonical (it's just induced by multiplication then projection). Instead the isomorphism that's not canonical is Hom(Z/mZ, Z/nZ) -> Z/gcd(m, n)Z. Can I get a sanity check that this is correct? Or maybe he's saying that the isomorphism with domain Tor is not canonical?

[lone jacinth]

g is a matrix, and f is a polynomial.

[lone jacinth]

The isomorphism Tor(Z/m, Z/n) ~= Z/m (x) Z/n is the one they're talking about yes.

(I mean it's the only isomorphism symbol in the paragraph)

[storm basin]

Ok good point

im being dumb

[golden osprey]

dumb question time!

Ok so
Q1: (i) => (ii) => (iii) are true for an A right? Not just Noeth maximal?
Q2: What breaks in (iv) => (i) if we just have A is Noetherian? Can we not just use any maximal ideal m to get what we want?

[astral ginkgo]

Q1: yes, free implies flat which implies 3 as a special case for any ring.
Q2: Nakayama's lemma requires that the ideal you're quotienting by be contained in the Jacobson radical

In the case of one max ideal that's just the ideal

[golden osprey]

ahhh

[astral ginkgo]

When you have 2 or more you can't guarantee this

[golden osprey]

right right

thank you

[astral ginkgo]

Np

[golden osprey]

actually (iii) => (iv) also holds for any ring right?

just basically definition of Tor

yea

[icy sluice]

What’s Tor(A,B) ?

Oh wait ok

[fierce steeple]

Seeing Tor undecorated makes me uneasy lol

[digital parcel]

which Tor is the undecorated one supposed to denote?

Tor_0?

[past cove]

Me when hypertor

[digital parcel]

is that just Tor of complexes

[past cove]

Yes

[digital parcel]

scary stuff...

[past cove]

Tho I prefer (x)^L

[digital parcel]

hopefully i dont have to think about "hypertor" anytime soon

[rose mirage]

The fuck is hypertor

Is it $\varprojlim_{n \in \mathbb{N}} Tor^n(-,-)$?

[broken turtle]

The Real Wew Lads Tbh

[foggy galleon]

what would the projections even be? They said above what hypertor is

[fierce steeple]

Tor

But I mean this is terminology sometimes used to emphasise like you are taking Tor of two complexes

Similarly for hyperext, hypercohomology

Though it is a bit redundant imo

[fierce steeple]

Usually denotes Tor_1

[rose mirage]

I was going off of the definition of hypercentre

[rose mirage]

oh like the spectral sequence nonsense

[digital parcel]

I see

[fierce steeple]

Idk what you mean

They can be computed via spectral sequences

[fierce steeple]

But kinda eh terminologynimo

Just outing yourself as someone scared of Tor_n for n > 1

[rose mirage]

am I remembering it correctly? it's the cohomology of the total complex in my head

[fierce steeple]

I mean there are a few things between this, hypertor and related spectral sequences

[quick whale]

Why does defining Ext through

• projective resolutions in the first argument, apply contravariant hom, and take the cohomology

give the same result as

• injective resolutions in the second argument, applying covariant hom, and taking cohomology

i haven't been able to find a good proof of this fact

[past cove]

Oh god

[fierce steeple]

There are proofs in Weibel and stuff but I think the usual proofs are a bit ad hoc, basically because it is a standard spectral sequence argument in disguise. The perspective here is to consider resolving both things at once and computing it in two different ways by truncating each of the complexes (individually) which recovers each of the two approaches

[past cove]

I think for this you need some like horseshoe lemma

[quick whale]

i understand that with a projective res of the first argument and an injective resolution of the seocnd argument you get a double chain complex whose cohomology groups across every row is 0 except the first chomology group, and same with the columns, but from there I do not know how to show this gives an isomorphism between the two...

i'll take a look at weibel though, thanks

[fierce steeple]

There are canonical map between the thing where you resolve both and the ones wheee you resolve one. You can run the spectral sequences for all and use functoriality of this spectral sequence

[wise sedge]

Is every HNN extension of a finite group linear?

[limpid horizon]

Do you mind explaining what was happening there again? I didn't really understand this the first time round and I want to understand it

Its hard because im not even really sure what I don't understand (a lot about this material is fuzzy to me)

I guess that discussion was about why to show dim(M/(x1, ... xr)M) >= dim M - r its enough to show dim(R/(x1, ... xr)) >= dim R - r?

[lone jacinth]

So rad(Ann(M/IM)) = rad(I + Ann(M)).

And R/I and R/rad(I) have the same dimension

So once you've reduced to Ann(M) = 0, it's just
dim M/IM = dim R/Ann(M/IM) = dim R/rad(Ann(M/IM)) = dim R/radI = dim R/I

[limpid horizon]

Do you mind explaining a bit more why we can assume Ann(M) = 0? By replacing R I guess you mean view M as an R/ann(M)-module so that now ann(M) = 0?

[lone jacinth]

We just replace R by R/Ann(M). Now Ann(M) = 0

[limpid horizon]

In A3, for (c) <-> (d), I originally thought we needed to be careful on showing that the infimum is the same, but any x1, ... xm satisfying (c)'s condition also satisfies (d)'s and vice versa so its just the same thing I guess right

[narrow kraken]

why are these two equivalent? i've been stuck on proving this for days. Let's say R is Artinian. M is a quotient of an indecomposable projective R-module <--> M has a unique maximal submodule

[hushed bone]

Holy shit I didn’t know these are equivalent. I am gonna think about this now

I mean maybe this could be clear gonna think now

[narrow kraken]

well they give the condition that M\C(M) is a submodule but that's equivalent to M having a unique maximal submodule I believe

[lone jacinth]

These are not equivalent. Take R=M=Z for example.

Maybe you have some extra assumptions on R you're missing....(?)

[narrow kraken]

ah R is Artinian

[lone jacinth]

Then it's true yes

[narrow kraken]

is there a proof anywhere for this?

in particular showing the 2nd direction M is a quotient of an indecomposable projective R-module -> M has a unique maximal submodule seems difficult

[hushed bone]

Oh, R is Artinian lol

[drowsy niche]

let M = P/N. decompose R into components Re_i for primitive idempotents e_i. Then P is projective iff direct summand of some free R-module R^n. mess about w/ the idempotent decomposition in R^n and show that since P is indecomposable it should be isomorphic to one of the Re_i

the P is local (unique max ideal J(R)e_i)

hence a quotient of it is local

I don't have a reference but these are all pretty standard tricks you might find in e.g. Lam's first course on ncrings

[narrow kraken]

hm let me think abt this

ah ok I was actually reading that, I guess I just have to read more of it

[drowsy niche]

I think this should all be within the first two sections

or well.... maybe 4 if you haven't seen the jacobson radical yet

[narrow kraken]

i have seen the jacobson radical elsewhere, but i've only gotten through around section 1 of the book iirc

[drowsy niche]

yeah section 1 is a bit of a mess lol

[hushed bone]

Does this work for noncomm R?

[drowsy niche]

don't let the examples discourage you

[hushed bone]

Like the word “local” seems hard for me to swallow for a noncomm ring

[drowsy niche]

it should

[narrow kraken]

it just means all the non-invertible elts form an ideal no?

[hushed bone]

As is localization, and in comm alg you usually see local rings by localizing at primes lol

I guess like it shouldn’t matter but I know the notion of primality is bad for noncomm rings

[drowsy niche]

yeah the word local is a weird choice on my part lol

I just meant has a unique maximal submodule

[narrow kraken]

oh oops I didn't realize you were talking about a module

ic

[hushed bone]

Like the way I’m familiar to do this is to note that J(R)^n stabilizes then show it stabilizes at 0 (I would say it’s the same as the nilradical cuz all primes maximal and…, but this is already hazy for noncomm rings)

Then note that for some n the ma into the product of all R/m^n over maximal m is injective

And note that R/m^n is local with maximal ideal m

[narrow kraken]

wait sorry what do you mean by "do this"? also what is "the ma"

[lone jacinth]

R is local iff R has a unique maximal left ideal iff R has a unique maximal right ideal iff R modulo Jacobson is a division ring

[hushed bone]

map

[fierce steeple]

mapo tofu

[lone jacinth]

So the main idea is that simple modules have projective covers.

If P is indec projective and N is a maximal submodule, then P/N is simple. Then the projective cover is a direct summand. If P has two maximal submodules then P maps onto P/N (+) P/M, so P has a direct summand of the direct sum of two projectives and hence not indecomposable.

Then the only thing missing is P having any max submodules at all, which you get from R/J being semisimple and R/J (x) P being nonzero because J is nilpotent

[narrow kraken]

can you explain this step "mess about w/ the idempotent decomposition in R^n and show that since P is indecomposable it should be isomorphic to one of the Re_i"?

i got that we need P = e_i P but idk how to continue

[drowsy niche]

if you decompose R into Re_i and have R^n \cong P \oplus Q for P indec projective, then decompose Q into indecomposable components. If R = (+)_i^t Re_i then R^n \cong (+)_j^n (+)_i^t Re_i is a decomposition into indec's. Indecomposable summands for R^n (of finite length) are unique up to iso and permutation here so we must have P \cong Re_i

for some i

[narrow kraken]

why does n have to be finite, like P can also be a submodule of R^I for some infinite index set I right

[drowsy niche]

damn

yeah

well

We can still express P as a direct sum of indecomposable fg projectives right?

even if it's a summand of R^(I)

[narrow kraken]

sure

but krull-schmidt doesn't apply anymore right

[drowsy niche]

well it would apply to the fg summands

and then you could use that P is indecomposable

[narrow kraken]

wait why is each one fg

[drowsy niche]

this is a characterization of perfect rings more generally

[narrow kraken]

oh ic

[narrow kraken]

wait am i dumb if P is indecomposable how could you express it as a direct sum of indecomposable modules

[drowsy niche]

well they're all going to be trivial except one Re

[narrow kraken]

wait so in the decomposition of P into a direct sum of indecomposable fg projectives, we know that there's only finitely many summands?

[drowsy niche]

not a priori

but yeah this is very janky lol

[narrow kraken]

hm

[drowsy niche]

I imagine this is probably a well known result abt projective modules over semiperfect rings

[narrow kraken]

that there's finitely many summands?

[drowsy niche]

(that they can be expressed as direct sums of Re)

[drowsy niche]

not necessarily (or at least I don't see why that would be the case)

[drowsy niche]

apparently this is a result for exchange rings which are rings R such that R as a left R-module satisfies the exchange property defined in the first screenshot.

semiperfect rings are exchange rings and theorem 1 is our desired result
https://link.springer.com/article/10.1007/BF01419573
@narrow kraken

no clue what the Crawley-Jónsson theorem is though lol

I will also note that the name "exchange rings" does have excellent meme potential

[narrow kraken]

interesting tysm for finding it

[narrow kraken]

so here, "then the projective cover is a direct summand," what projective cover is it referring to? of the module Q that's the quotient of an indecomposable projective R-module?

i also don't get this part: " If P has two maximal submodules then P maps onto P/N (+) P/M, so P has a direct summand of the direct sum of two projectives and hence not indecomposable. "

[blissful field]

what could the hat mean

v are vectors and lambda are covectors

[hallow bone]

means omit

[blissful field]

Thank!

[hallow bone]

it shows up here and there lol

whats the context here?

[hallow bone]

(to be clear: it means removing the element from the sequence)

so (v1, ..., \hat{v}_i, ..., vn) = (v1, ..., vi-1, vi+1, ..., vn)

[blissful field]

Yeah I said this because "Remove the ith vector and jth covector" makes sense here

[hallow bone]

coolcool

i usually see it in the context of homology stuffs

[blissful field]

Group homology I assume 😄

[hallow bone]

or singular homology

or rack and quandle homology

[past cove]

group homology isn't really a useful thing

or at least

it doesn't work that well lol

[blissful field]

Do you know of any resources for that last one?

[hallow bone]

homology of classifying space or wtv

[blissful field]

wut

[hallow bone]

group cohomology on the other hand