advanced-algebra

[broken turtle]

1728

[hard kite]

Yea it is true

[hard kite]

Guys I found such an awesome proof of this I really want to share it

[ornate atlas]

But the character limit is too small to contain it?

[hard kite]

First for context, after I thought it was not possible i caved to weakness and asked to chatgpt (I want ot use such ''''tools'' at little as posble cause im opposed to it idealogically and i think its lame and anti human ) ot be sure fo it and it said it was actually true but not canonically and it gave me like a huge wall of text that i just was not gonna read but it said something that a fg torsion module over dedekind domain is completely determined by the isomorphism types of its localisations at primes (in a more elaborate and text wall way ofc). I asked why it gave another huge wall of text but it jad soemthing avoutn the classification of modules over a Dedekind domain in it (which I was not aware of at the time) and so i quit the the prompt cause i wa sfeeling really bad abt using gpt for mah thesis but i looked in curtis-reiner and it fr had a section on classification of fg moduels over Dedekind domain and it had some classification result in it and i used that and thought about and this proves what i wanted it. But i felt bad about ti all day

So at the end of the day I was writing all my results down and I typed a proof that was liek follows from this theorem in CR and this gives a non - canonical isomroopshism. Then sometime I was standing and listening to music and I was suddenly hit by a flash of insight I really don't know how and I instantly knew a direct proof that is really insightfull in my opinion and it is also kind of funny. Plus it gives a canonical isomorphism. This made me really excited and feel less guilty. This is more fun to me so from now on I vow not to use any ai '''tools''' for anyhtign related to my thesis. If you think this corny or stupid or whaytevr. I dont care and IDGAF 😝 this is mine so i can do whhat i want X) peace

ok now i will type it out

It wil follow from the following lemma:

Let $R$ be Dedekind and $M$ a f.g. torsion $R$-module. There is a (canonical) isomorphism [ M \cong \bigoplus_{\mathfrak{p}} M_\mathfrak{p}.]

Proof: Since $M$ is finitely generated and torsion, $\mathrm{Ann}R ~ M \neq 0$ (think about that -_0 ). Let $A = R / \mathrm{Ann}R M$. Then $A$ is Artin and $M$ is the restriction of a scalars of a finitely generated $A$-module $M_A$ along the quotient map $R \to B$. Writing $\iota:\mathrm{Spec}~A \to \mathrm{Spec}~R$ for the inclusion, we may rephrase this as $\tilde{M}=\iota\ast \tilde{M_A}$. Since $A$ is Artin, $\mathrm{Spec}~ A$ is finite and discrete. Hence,
\begin{align*}
\Gamma \tilde{M} &= \Gamma \tilde{M_A} = \prod{x \in \mathrm{Spec}~ A} \Gamma({x},\tilde{M_A}) = \bigoplus_{x \in \mathrm{Spec}~A} (\tilde{M_A})x = \bigoplus{x \in \mathrm{Spec}~R} (\iota_\ast\tilde{M_A})x = \bigoplus{\mathfrak{p} \text{ prime of } R} M_\mathfrak{p}.
\end{align*}
q.e.d.

By the lemma, it suffices to check that, writing $A=R/I$,
[
(I^{-1}/R)\mathfrak{p} \cong A\mathfrak{p},
]
for all primes $\mathfrak{p}$. As $R$ is Dedekind, $I$ is projective, hence flat. Thus,
[
I^{-1}/R = I^{-1} \otimes_R R/I = I^{-1} \otimes_R A.
]
Since $I^{-1}$ is invertible, $I_\mathfrak{p}^{-1} \cong R_\mathfrak{p}$ at all primes $\mathfrak{p}$. Hence,
[
(I^{-1}/R)\mathfrak{p} = I^{-1}\pp \otimes_{R_\mathfrak{p}} A_\mathfrak{p} \cong A_\mathfrak{p}.
]

Ok so maybe it is not tecthnically canonical since for the last step you need to choose a generator of $I$ but it is almost canonical lol ahah. Whatever if you read this, thank you for reading.

Maybe don't bother reading this unless you are really crazy about commutative algebra. The essential idea is to rephrase it in terms of sheaves on the spectrum of some artin ring, and then using that the global sections of a sheaf on a discrete space is just the product of the stalks at all the points. Which is kind of an unexpected direction for this proof to take since when I was writing these things I was working on some problem in representation theory.

sorry for the deranged walls of text people have a good day algebra chat

[broken turtle]

1728
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hushed bone]

Cute proof

[void plank]

Suppose f: C --> C' and g: D --> D' are chain maps

Then f (x) g is a chain map with components (f (x) g)_n = \osum_(i+j =n) f_i (x) g_j

Consider MC(f (x) g). Does it have any relationship with MC(f) and MC(g)?

I was thinking something like MC(f (x) g) = MC(f) (x) MC(g), but that's probably wrong

[hexed tangle]

Is the nilradical defined for non-commutative rings?

[ornate atlas]

Yes but you still have to define it for two sided ideals afaik, and for Noetherian rings it just coincides with the usual prime radical

[wide topaz]

Good point, but a small clarification: in noncommutative rings, the upper nilradical and prime radical are different in general, but coincide for Noetherian rings (and are equal in the commutative case). Both are two-sided ideals.

[ornate atlas]

Please don’t respond to me, or anyone else, with chatGPT output. Especially when you’re saying exactly what I have already said.

[hexed tangle]

Thanks!

[wide topaz]

I didn’t say ur wrong.. I was just clarifying your point, and don’t assume, I tried to give him a clarified response of his ques.

[lone jacinth]

I think the definition I've seen would be that the nilradical is the sum of all (left) nil ideals, which you can then show equals the sum of all right nil ideals and hence is two sided.

So you don't have to define it for two sided ideals, but you need to use ideals and not elements.

[hexed tangle]

I am to show that that the functor RIng -> Set: R |-> Nil(R) is not representable, so is this definition even functorial?

[lone jacinth]

It's not, no.

[hexed tangle]

🙁 F...k

[lone jacinth]

But maybe it's a more interesting exercise if Ring is the category of commutative rings

Or I guess Nil(R) could just mean the set of nilpotent elements.

Neither should be representable

[hexed tangle]

I asked my TA if the rings were assumed to be commutative and he said, no, he thinks his proofs work

[lone jacinth]

I mean, if something isn't a functor then it also isn't representable, so his proof might work

[hexed tangle]

yes, indeed

[lone jacinth]

But yeah, then I guess the follow up should be how he's defining Nil(R)

But whatever it is, surely if you restrict to commutative rings it should be the same. And just considering commutative rings should be enough to prove it not representable

So that would be an approach for you to solve it no matter what he answers

[hexed tangle]

true

[fierce steeple]

Is this meant to be the mapping cone?

[lone jacinth]

It's not as simple as MC(f) (x) MC(g) at least.

Just consider C=D=C'=D' = Z and f multiplication by 2, g multiplication by 3 for example

[nimble orchid]

Hi, is there anybody who knows what is pfaffian in linear algebra? What are the fundamental properties?

[foggy galleon]

it's like a square root of the determinant

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

it is used in the chern-gauss-bonet theorem

[nimble orchid]

Thank you, but how can I associate a polynomial to a Matrix, the determinant is not a Number?

[lone jacinth]

The determinant is a polynomial

And the pfaffian is a polynomial with
Pf(A)^2 = det(A)
for A skew symmetric

[nimble orchid]

Is It?Take R^2,2 Matrix ((a b),(c,d)) which is the polynomial?

[lone jacinth]

ad - bc is the polynomial

A polynomial is a function involving multiplication and linear combinations of the unknowns

[nimble orchid]

Ah ok, so It's not a polynomial in just One variable with fixed coefficients, but four variables.

[nimble orchid]

Ah ok, thank you!

[fierce steeple]

I feel this is potentially misleading saying it's a function

Maybe it's ok here in context lol

[lone jacinth]

Meh, always makes sense to think of them as functions as long as you're not married to the domain being the ground field

[fierce steeple]

sure ye you can get functions from them

ig here it's fine though lol

[eager hound]

A polynomial is stored in the map P^*: Spec(k[x]) \to Spec(k[x]) which is a function.

[fierce steeple]

Nice

[eager hound]

and if k is a field then you can just look at maxspec

[fierce steeple]

Don't you also need the data of the sheaf of rings here (i.e. the morphism of schemes)

to remember the polynomial

Or am I cooked...

[eager hound]

true

[fierce steeple]

No u

[eager hound]

A polynomial is stored in the map Y(P^*): Y(Spec(k[x])) \to Y(Spec(k[x])) then

[fierce steeple]

So much in that excellent formula.

[eager hound]

E = Yoneda + AI

[fierce steeple]

A polynomial is just a function * -> k[x]

[plucky arch]

A polynomial with variables in S is an element of the end $\int_R (R^S \to R)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[cloud marlin]

I haven't learned ends and coends, is this equivalent to saying "finitely supported function N^S --> R"

[plucky arch]

It’s essentially a set of natural transformations

Given any set $S$, we can define the $|S|$-times power of a ring $R$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Just the ring of functions S -> R under pointwise operations

This defines a functor CRing -> CRing

You can then look at the set of natural transformations between this and the identity functor

Hm you may want to compose with the forgetful functor to Set first

[plucky arch]

Yes you do this

So the set of natural transformations between this and the forgetful functor

[rose mirage]

it's actually quite impressive how quickly you can manipulate these categorical concepts. My dumbahh is still struggling to compute this as an equaliser

[plucky arch]

this is just the kan extension formula for a left adjoint

this end is $\text{Nat}(\mathbf{Set}(S, U(-)), U(-))$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[rose mirage]

right yeah I get that R^S is the S power of R or whatever but I want to see it

[plucky arch]

which is iso to $\text{Nat}(\mathbf{Ring}(\mathbb{Z}[S], -), U(-))$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

and by Yoneda that's $U(\mathbb{Z}[S])$

[rose mirage]

oh my science le heckin free forgetful adjunction

[broken turtle]

Pseudo (Cat theory #1 Fan)

[rose mirage]

this is the part I want to compute

[plucky arch]

ah sure

you can think of intuitively as "if I have a function from S -> U(R), what are all the possible ways to 'evaluate' it to produce an element of R?"

the polynomial ring Z[S] answers this question

it's the ring of all possible evaluators

[rose mirage]

I know more generally that the end of Hom(F(-), G(-)) is Nat(F, G) but I want to remind myself why

[rose mirage]

hmmm

this is a very "kan extension"-y view point

[plucky arch]

yes exactly

the kan extension formula for a left adjoint for a functor $G : D \to C$ is $F(c) = \int_d d^{\text{Hom}(c, Gd)}$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

i.e. "given a morphism c -> Gd, what are all the possible ways to 'evaluate' it to produce an element of d?"

is the rough idea

[rose mirage]

ok I've finally remembered what the maps in the equaliser are and yeah I can see why they're natural transformations

[rose mirage]

yeah I've definitely seen this before as well it's just oh so very fuzzy oh so very blurry

I shall compute this one as well

she twists my arrow category til my end co

[neat nexus]

If I replace Ring with R-Alg this ends up as R[S] instead of Z[S] right?

[plucky arch]

i believe so, R[S] should be the free R-algebra

[neat nexus]

I see, thanks

[void plank]

Yes it is, my apologies for not saying!

[void plank]

Ah ok! I'll see if I can find the correct formula somewhere (if it exists)

Also, is it known how the mapping cone plays with quasi-isos? If you have two chain maps f ~ g, is MC(f) related to MG(g) in some way?

[lone jacinth]

If f and g are homotopic then the mapping cones should be homotopy equivalent.

[void plank]

Thanks!

[fierce steeple]

Also like the mapping cone is a thing that makes sense in the derived category (and oo-categorically it has the same universal property as cokernels do usually) so this will be true and you should expect it to work very well accordingly

[void plank]

Thanks!!

[nimble shell]

Why do we describe vectors as arrows initially but it’s not reflective of mathematical reality (i.e. they’re not arrows when dimension n>3)

[spice idol]

guy discovers intuition

[median halo]

i don't really see the issue with describing vectors in R^n with n > 3 as arrows

[nimble shell]

What would an arrow in R^4 look like

[summer quest]

lol

[fierce steeple]

Also vectors needn't be in R^n either

[lone jacinth]

It would look like a line segment with a little pointy thing at the end

[neat nexus]

ChatGPT said something (which I verified): h_1= f times 1_D, h_2=1_C’ times g, h=h_2 • h_1=f times g. T: MC(h_1)-a->MC(h)-b->MC(h_2)-c->ΣMC(h_1) is a distinguished triangle , which is isomorphic to T’: MC(h_1)->MC(h)->MC(a)->ΣMC(h_1), where a=(h_2, 0; 0, 1), b=(1, 0; 0, h_1), c=(0, 1; 0, 0). The isomorphism is given by (1, 1, w) where w=(1, 0; 0, 0; 0, 1; 0, 0), whose inverse is (1, 1, w’) where w’= (1, 0, 0, 0; 0, h_1, 1, 0).

[nimble shell]

True lol

[nimble shell]

Can it be represented visually

[lone jacinth]

Sure

[nimble shell]

I feel like for higher dims algebraic representation makes more sense

[lone jacinth]

I'm not sure 3d and 4d are different in that regard, but sure

[cerulean cove]

Can someone help me prove this

[void plank]

Let me parse this for a bit

[lone jacinth]

Maybe start by identifying the cyclic subgroups of U(12)

[cerulean cove]

(1,5) (1,7) (1,11)

[lone jacinth]

I'm not sure what your notation means here

[cerulean cove]

H= {1,5} a cyclic subgroup

[lone jacinth]

Okay, I see what you mean.

Then try to pick two of those subgroups and compute their product

[cerulean cove]

$${(1,1),(1,7),(5,1),(5,7)}$$

[broken turtle]

Haruki

[cerulean cove]

Got this from {1,5} X {1,7}

[lone jacinth]

Right, so what I actually meant, was computing their product in U(12).

But okay, can you imagine a homomorphisms from this direct product to U(12)

[cerulean cove]

I am stuck there

I can't

[lone jacinth]

So do you see how {1, 5} naturally sits as a subgroup inside the direct product?

[cerulean cove]

Idk what you mean by this

[lone jacinth]

Given two groups G and H, is there a subgroup of GxH isomorphic to G?

[cerulean cove]

I don't know

[lone jacinth]

Hmmm, do you know how elements in GxH multiply?

[cerulean cove]

Yes

(1,1) . (1,7) = (1,7)

(1,7). (5,1) = (5,7)

[lone jacinth]

So given g in G what could be an element in GxH that has anything to do with it?

[cerulean cove]

(5,1)

?

[lone jacinth]

Yeah, so do you see how
g to (g, 1)
is a group homomorphism of G into GxH?

[cerulean cove]

I don't see it

Ohh wait

But yes but what will we get from showing this

[lone jacinth]

Allright, so what do we need to show?

If g and g' are two elements, then the homomorphism should take
g g'
to
the product of
(g, 1) And (g', 1)

Does it do that?

[cerulean cove]

Yes it does

[lone jacinth]

Allright, so then the direct product actually behaves as if you have a copy of G (things like (g, 1))
and a copy of H (things like (1, h)) and you take their product
(g, h) = (g, 1)(1,h)

[cerulean cove]

Yup

[lone jacinth]

So now back to U(12)

The most natural choice for a homomorphisms might take (5,1) to itself (i.e. 5) and (1, 7) to itself (i.e. 7)

Does this give you enough to determine a homomorphism?

[cerulean cove]

Hmm i think I see it but we get two different?

One takes the first element from the pair

Another takes the second

I don't understand why

Also what about (5,7)

[lone jacinth]

So let's give names to things:

f: {1,5} x {1,7} -> U(12)

is our homomorphisms.

Now I'm saying we would like
f(5, 1) = 5 and f(1,7) = 7

Are you able to figure out what f(1,1) and f(5,7) should be from this?

[cerulean cove]

f(1,1)=1

And f(5,7) =11

[lone jacinth]

And is that a homomorphisms?

[cerulean cove]

It should be but how do I give it a generalised form so that I can prove

[lone jacinth]

Well what is the definition of a homomorphisms?

[cerulean cove]

That it is a homomorphism

[cerulean cove]

f(a*b)= f(a)*f(b)

[lone jacinth]

And does f satisfy this?

[cerulean cove]

Lemme check

[cerulean cove]

Yup

But I checked just one

f{(1,1)x(5,1)}

Do I need to manually check them all?

[lone jacinth]

Well it has to hold for all pairs a and b

If you don't see any pattern to it then that's the only way to find a pattern

[bold panther]

Idk why but that hits hard

[cerulean cove]

This is the correct way?

[bold panther]

Cool phrase

[lone jacinth]

If you really don't want to compute anything I guess maybe think about why you picked f(1,1) = 1 and why you picked f(5,7) = 11

[cerulean cove]

I see a pattern where i take the direct product to u12 after multiplying a and b

[cerulean cove]

I picked f(1,1)= 1

Cause every isomorphism does that

And took f(5,7)=11 cause 5*7= 35= 11 in Un

U_12

I mean

[lone jacinth]

Right so f(5,7) = 5*7 here

Is that a coincidence?

[cerulean cove]

I don't think so

[lone jacinth]

What would be a pattern that makes sense there?

[cerulean cove]

There?

Where?

[lone jacinth]

f(g, h) = ?

Matching g=5, h=7

[cerulean cove]

I don't know how to put it in mathematical phrase

[lone jacinth]

Well I wrote a formula here? Could we generalize it beyond just 5 and 7?

[cerulean cove]

Yes

g*h

[lone jacinth]

And is
f(g, h) in fact equal to g*h for all g and h?

[cerulean cove]

Yup

[lone jacinth]

So now you can use general variables for a and b instead of computing ever possible pair if you like

[cerulean cove]

Wait lemme do it and show u

[cerulean cove]

Got it

We have to use that u12 is commutative

[lone jacinth]

Excellent!

Btw for the future this question would be more suited for #groups-rings-fields

[cerulean cove]

Okay

[cerulean cove]

Can we just say that u12 is commutative?

Or we need to show it too

Someone said

That u12 is similar to Kleins 4 group

Also thank you

[lone jacinth]

I guess that depends on the assumed knowledge of your course / problem

[lone jacinth]

Yes Klein 4 is exactly the product of two cyclic groups of order 2

[limpid horizon]

Jagr is a man of many wise words …

[void plank]

Jagr is a wise man

[lone jacinth]

With wisdom comes age, soon I'll be a proper adult

[fierce steeple]

Jagr is 75

[lone jacinth]

Yearning for the day I'm no longer allowed to play those board games marked ages 9-99

[worldly zealot]

From your username I can conclude that you are -782 years old

[ornate atlas]

That’s how he knows so much maths, he came back from a time where homological algebra is taught in nursery

[lone jacinth]

Maybe I need to put BCE at the end

[eager hound]

Jagr is a man of many wise words …

[fierce steeple]

Ed Sheeran and Lego when they meet a 100 year old

[foggy galleon]

can this sequence be continued?

[subtle smelt]

Isn't this group cohomology?

You can identify H^i(G,M) with Ext^i_ZG and borrow its long exact sequence. Computing the individual terms comes down to putting your hands on the bar complex resolving Z or occasionally whatever description allows you to realize some H^i(G,M) vanishes.

[lone jacinth]

I'm assuming A, B and C are non-abelian groups

[hushed bone]

I know for non-abelian sheaf cohomology there is an extremely painful way to get to H^2 in terms of gerbes, this is in Giraud for example

I can only imagine that one could write non-abelian group cohomology as non-abelian sheaf cohomology over some space

¯_(ツ)_/¯

[fierce steeple]

Iirc for noncommutative groups you like maybe provably cannot extend much more

Hmmm

[ornate atlas]

I think my supervisor said something about being able to get H^2 and H^3 in like specific cases but anything more than that is hopeless (but I dont understand this at all so uhh no questions please)

[fierce steeple]

Tbh my opinion was it's a miracle this stuff makes any sense at all aha

[ornate atlas]

Something about like how things associate, he drew something which very much looked like the pentagon axiom and apparently that breaks shit if you try to extend things

[foggy galleon]

why

But like do the H2 tell you something about that sequence, or is there no connecting map

I never understood why nonabelian cohomology seems much more intractable and poorly behaved. Tho H1 is not bad

[scarlet prairie]

could anyone help me understand what the superscript zero means?

does $E_{p*}^0$ just mean the chain complex formed by the 0th column?

[broken turtle]

Former Rank 7 LLORT AJNIN

[last talon]

You’re doing spectral sequence stuff
You’re about to make infinitely many pages which look like this, and will be denoted E^n_{pq}

No

By the pth column

[scarlet prairie]

hmm wwhere would the pages come from

isn't a double complex just 1 flat sheet

[last talon]

Look at the bottom paragraph

You’re going to take the homology to get the 1st page
Then there is a way to get differentials on the 1st page, and you take homology of that to get the second page
Then there is a way to get differentials on the 2nd page, and you take homology of that to get the 3rd page
And so on

[hushed bone]

E^n+1_p,q is the cohomology at E^n_p,q

And the maps keep changing direction by a factor of +1,+1 so they get more and more down and to the right

[last talon]

Yes
But we’re going to consider a sequence of double complexes, which will be indexed by the superscript

[hushed bone]

Why these maps exist is a mystery that no one has ever been able to figure out, it was given to Leray during WW2 in a dream

[scarlet prairie]

so from my starting (first quadrant) double complex, i form another double complex by taking vertical homology at every index (p,q)

[hushed bone]

Yeah but now your maps go to the right by 1, and then you take homology along those maps

Then your maps go 2 right 1 down then you take homology among those

Then they go 3 right 2 down and you…

[scarlet prairie]

can i check what these maps are?

induced by the original d^h?

[hushed bone]

In special cases yes, in most cases no

If you have enough information sure, and often honestly you know what’s on E^1 or E^2 and it’s usually a lot of 0s

So you don’t need to know what these are

[scarlet prairie]

i mean im not very sure how we are getting the horizontal maps*

[hushed bone]

But in topology you use some spectral sequences which aren’t this nice and figuring out what the maps actually are is like, totally intractable

[hushed bone]

See this comment above

[scarlet prairie]

ah okay let me continue reading abit more, thanks for the help

[fierce steeple]

My advisor once joked like

Homotopy theory is the part of maths where you study spectral sequences which don't immediately degenerate

[hushed bone]

Yeah

[scarlet prairie]

ok i have understood up to page 3 now, this is really ugly

[silver goblet]

I thought all spectral sequences degenerate at page 2

[last talon]

They degenerate at page 2, or you run crying to your nearest homotopy theorist

[silver goblet]

shoot, I am my nearest homotopy theorist

[last talon]

I’m sorry :3

[past cove]

the technicalities are horrible, but the ideas are quite clever

[dark ibex]

In pow camp of Austria*

[eager hound]

If A is abelian you can continue this sequence and the vanishing of H^2(A) will tell you if the last map is surjective I think

[eager hound]

There’s kind of the standard doctrinaire reason from homotopy theory: degree n group cohomology is represented in the homotopy category of spaces by maps from X to B^nG where B^nG is a space whose nth loop group is G. Since BG always makes sense there is always some notion of H^1. On the other hand since \pi_2 is always abelian there is no space B^2G for G nonabelian

[forest turtle]

ask in #prealg-and-algebra

[limpid horizon]

from advanced algebra to prealgebra lol

thats pretty funny

[scarlet prairie]

slightly confused about the category of homology spectral sequence. Lets say we start at page 0, Is it right that it suffices to define a morphism as just a family of chain maps on page 0?

[scarlet prairie]

i understand how this can induce maps on between page 1, but its not clear to me that this will still be chain maps. or is it that we specifically also enforce this as part of the definition?

[lone jacinth]

It needs to satisfy both parts of the definition yeah

[crisp bone]

What's a good introductory book on lie algebra for QCD? For reference I've don't MV Calc (calc 3), diff eq, and linear algebra (with some proofs)

[dark ibex]

You should do them first

Especially linear algebra is most used

[scarlet ermine]

anyone have a quick reference for the bijection between extensions and Ext? I need the 2 case in particular, going from injective resolution definition to an extension

Or maybe I can ask about it here too. Given eps in Ext^2(A,B), I have a cocycle representative in Hom(A,I^•), where I is an injective resolution of B

then I try to build my extension… in the Ext1 case, I know that I would do something like this. If I want a n exact sequence

0->B->C1->C2->A->0,

I could try setting C2 to be the pullback of the cospan I1 -> im(I1->I2) <- A

do I keep just putting pullbacks C1? or in general is that the strategy?

[lone jacinth]

So you have a cocycle
A -> I2
then you take the pullback of
A
v
B -> I0 -> I1 -> I2

To get a sequence
B -> I0 -> pb -> A

[scarlet ermine]

hmm i see

oh yeah my indexing was off too

and the map from I0 to the pullback is like inclusion of a direct sum and before quotienting?

if we’ve constructed the pb that way

or I guess we’re pairing the map I0->I1 and the 0 map to A

and it works because I^• is a complex

[lone jacinth]

The kernel of pb -> A is naturally isomorphic to the kernel of I1 -> I2.

This is true in general for pullbacks.

Then you already have a map from I0 to the kernel

[lone jacinth]

Yes exactly

[scarlet ermine]

i see, thanks!

[crisp bone]

Typo I meant “done” Grammarly is being strange on my phone that le moment

[solar linden]

what would "order" be here?

like, what's the order being preserved

[digital parcel]

Probably containment?

[rose mirage]

the idea is that the ideals look like \prod_i \in J R_i with J a subset of I, fliters are ordered by inclusion and hopefully it is clear why this induces an isomorphism of posets to the set of ideals under containment

[spice idol]

is very complicated *if I is infinite

[solar linden]

ah

[solar linden]

ah

[solar linden]

yea that was an answer to the question of what was the structure like when the product is not finite

thx y'all!!

[spice idol]

oh im sorry lol didnt know the context

[solar linden]

it's ok

[spice idol]

what book is this?

[solar linden]

stack exchange

[spice idol]

thats a big book

[solar linden]

yea it has a bunch of different subjects it's kinda crazy

I went researching about prime ideals on google because I wanted to find a really forced counter-example to a thing I made up in my mind

so I thought "can I make up a ring with an arbitrarily big cardinality of prime ideals?"

and I can, apparently

[spice idol]

yes

[solar linden]

I actually think I can use that thing to make up rings with any (infinite) amount of prime ideals which I think is interesting
it's also a thing I started thinking about

I think the set of filters is <= RxP (where P is the poset) and producting any P greater than or equal to R leaves it at P so this will end up having any cardinality I want

which also means conceptually that I can make up a ring with any set of prime ideals I want since yk bijections

although I think constructing it like that would be ugly asf

maybe

haven't thought too much about it

[drowsy compass]

containment

[solar linden]

wait I think I'm stupid

thought filters were sequence related for some reason

[spice idol]

the filter thing is related to ultraproducts

[solar linden]

mhgmhmg

[spice idol]

so true

[wanton heron]

Why is the intersection being K enough here for them to be linearly disjoint, we are not given that the other extension is separable

[lone jacinth]

As L is Galois
LM/M is galois and you have the natural map
Gal(LM/M) -> Gal(L/K).

This map is clearly injective and the image is Gal(L/E) where E is the things in L fixed by the Galois group of LM, i.e. LcapM, i.e. K.

So this is an isomorphism, hence [LM:M] = [L:K], so [LM:K] = [L:K][M:K]

Which is linear disjointness

[wanton heron]

So you're saying when we have something like

L      M
\.      /
  \.  /
    K

And L/K is galois then for linear disjointness it's enough to show L \cap M=K?

[lone jacinth]

That's what I'm saying

[wanton heron]

What is the natural map btw

[lone jacinth]

Restricting the automorphism to L

[wanton heron]

Oh ok

Shit I'm rusty on galois theory

I think I get it now

[lone jacinth]

You can generalize the result a little: as long as one of L/K and M/K is seperable then they're linearly disjoint of the intersection of M with every conjugate of L is just K.

Then if one of them is normal all the conjugates are the same, so the condition just becomes for the intersection to be K

So yeah one being galois is good or you can have one of them be seperable and the other normal

[wanton heron]

Right

Thanks

[edgy pond]

If V is a representation of G and H is a subgroup of G, then is V a subset of Ind_H^G (Res_H^G(V))?

[wanton spoke]

nope

[edgy pond]

hmm

i am trying to prove that if V is an irreducible complex representation of G and H is a subgroup of G then V is contained in the induced representation of an irrep of H. And I found a surjection from some Ind_H^G U into V using the fact that Ind is left-adjoint to Res so now I was hoping V is a subset of Res(Ind(V)) to conclude V is actually a subset of Ind_H^G U

[lone jacinth]

As long as the index of H is relatively prime to the characteristic of your field it should be a direct summand

Let g run over a set of representatives for G/H. Then a splitting of
kG (x)_H V -> V
is given by
v |-> 1/[G:H] Sum[g] g(x)g^-1 v

$v \mapsto \frac1{[G:H]} \sum_g g\otimes g^{-1}v$

[broken turtle]

jagr2808

[lone jacinth]

The other order is easier.

If W is a kH-module, then Res(Ind(W)) always has W as a summand independent of index.

[lone jacinth]

I would actually be curious if someone has a reference for this result.

I had to use it in a paper once and wasn't able to find it in the text books I looked through. Ended up just inlining this formula.

[edgy pond]

i see thank you for the explanation

[edgy pond]

in this question, is finding a surjection onto V enough to say V is contained in some induced representation?

[fierce steeple]

Damn was gonna joke about how you need good characteristic but you already covered that

Man after my own heart

[edgy pond]

like V is contained in the sense that there is a surjection onto it

[lone jacinth]

If G has order relatively prime to the characteristic of the field then kG is semisimple, so every surjections splits.

[edgy pond]

oh

i see

thank you so much!

[drifting niche]

Anyone around that can help me understand solving radical/quadratic equations?

Kinda need a tutor.

[spice idol]

im sure #prealg-and-algebra is better for you lol

[dark ibex]

Why you have undergrad tag then?

Tags are not meant to be a flex

[urban granite]

because bro thought what they're doing is too advanced

[queen tide]

https://grossack.site/2021/08/06/cyclic-extensions.html

[spice idol]

this is a great blog

[cloud karma]

They clearly are asking in the context of Galois theory

[urban granite]

Lol

[fierce steeple]

I just find it funny how like the way it's written makes me feel the person doesn't understand Galois theory anyway

[urban granite]

yeah lmao

[wise sedge]

Okay, I don’t think this is actually true

[potent frigate]

Does anyone have any recommendations on textbooks/resources for Homological Algebra? Currently using Sophie Morels book however at times it can feel too abstract. I’m just looking for something to use kind of along side what I’m already doing for some more examples/results/stuff lol

[ornate atlas]

Rotman is pretty gentle, its nice

I think most books on homalg will feel pretty abstract though, it just kinda is pretty abstract (though there are extremes, and Morels does seem to be on that other extreme lol)

Weirdly, even Weibel may be more approachable

[potent frigate]

Great, thanks guys!

[steady lance]

The Bob Ross of math I stg

[dark ibex]

Borcherd's vids are nice to start into, but quite incomplete

I believe

[torn harbor]

They aren't meant to be a full source more of a complement to a text

[dark ibex]

Indeed

[lone jacinth]

I don't know if I really feel like they're usually a good intro either.

The main take away is intuition and examples, so good for when you've started but haven't finished

[near lantern]

Consider the derived category D(R-Mod) for R a ring (i.e. chain complexes of R-modules with quasi-isomorphisms inverted). What limits and colimits does this category have? How can I compute them?

As an additive category, obviously it has finite (co)products (which are the same as those of complexes, since the localisation functor is additive). From https://stacks.math.columbia.edu/tag/05QU I believe lots of morphisms don't have kernels and cokernels (unless R-Mod is semisimple, in which case I think they all exist?). I'm guessing that direct sums, direct products and filtered colimits all exist and can be computed, but I don't see how to check this easily because homsets/groups in D(R-Mod) are complicated.

[fierce steeple]

I can give a slightly ideological answer lol

Well okay not ideological

[near lantern]

OK IG for these questions there's always a comparison map, e.g. for a colimit the colimit cocone in Ch(R-Mod) becomes a cocone in D(R-Mod). If I know the colimit exists in D(R-Mod) and is exact in Ch(R-Mod) then it follows that the comparison map colim_i [C_i] to [colim_i C_i] is an isomorphism on homology, so an isomorphism(?).

[near lantern]

Sure.

[fierce steeple]

But the correct things to consider in most cases are undoubtedly homotopy limits/colimits

Then it has all limits and colimits

Idk a situation where you'd want to do something else

[near lantern]

Presumably a situation where the un-homotopy thing is easier to compute, in which case one hopes for a result telling one when it's close enough to the good homotopy thing for the difference to not matter?

[fierce steeple]

Sure but then you would probably do it within the category of chain complexes, where things are more computable

Not in the derived category

[lone jacinth]

Something not particularly hard to prove from the axioms is that in a triangulated category all epimorphism/monomorphisms split.

So using his (co) limits are (co) kernels of (co) products you get a strong condition on which (co) limits exist

[near lantern]

Is this some infinity-category thing where Ch has an infinity-category sructure with homotopy category D? So to do infinity-category constructions you should stay in Ch?

[fierce steeple]

It's mor that like D has an enhancement to an infinity category and it's what you get from inverting quasiisos in Ch

But you can also do stuff in Ch and pass to the oo cat D which is common

But what you're saying is sort of inbetween

[near lantern]

I think if the (co)product exists, the (co)limit exists iff the (co)kernel exists. So if I know the (co)product always exists, I can check which (co)limits exist (at least if I'm good at finding splittings).

[fierce steeple]

It's like the difference between topological spaces, the oo-category of spaces, and spaces modulo homotopy. The last one is pretty badly behaved categorically but appears

[fierce steeple]

Though I find it funnier to say that the oo-category has an impoverishment to a triangulated category

Joking. Just bants

[worldly zealot]

lol

[round seal]

is it a consequence of this that R is regular iff the resudie field has finite projective dimension. I think this is true but i guess im not sure how to piece the last two statements together to get this.

[hushed bone]

Yes

I mean I’m really confused what you’re asking

1. is the first condition and 4) is the definition of a regular local ring I’m guessing

@round seal

[round seal]

oh im being slow I think

[small pond]

does anyone know a good linear/abstract algebra prof on yt

[wanton spoke]

a good linear/abstract algebra prof on yt? lol if you’re not sitting with an actual textbook proving everything yourself getting stuck for 3 hours on a trivial lemma and rewriting the proof four times until it clicks you’re not learning math

[urban granite]

Daily reminder to use noggin

go watch youtube shorts or instagram reels

[toxic trout]

thats why tiktok exists

[analog schooner]

Is Michael Penn a good AA yter

I watch some of Borcherds videos

[spice quiver]

I didn’t even know he was an alcoholic

[limpid horizon]

Lol

Good one

[spice idol]

i miss when he did backflips to clear the board

[analog schooner]

what are elementary abelian p groups isomorphic to

should be like (Z/pZ)^n where the group is generated by n elements rihgt

how does that work and how could you even do that like idk Aut((Z/pZ)^n)

[rose mirage]

GL(n, p)

Dunno what you mean by “that work” or “could you even do that” of course you can take the automorphism group

[analog schooner]

for a group representation from (Z/pZ)^n to Aut((Z/pZ)^n) how would you represent the elements on p (the elements are linear transformations but how can you find them)

for representations of elementary abelian p groups

[rose mirage]

I still have no idea what you’re asking but I’ll try anyway. The indecomposable representations of Z/pZ over F_p are given by sending the generator to a n by n pascal matrix with n < p-1. So you can easily map (Z/pZ)^n inside of GL(n, p) for any n by just choosing indecomposable representations of each Z/pZ and adding them. Unfortunately, if n > 2 this p-group has wild representation type and it’s essentially impossible to classify all of the indecomposable representations

[dark widget]

what's GL(n,p)

[wanton spoke]

general linear group of degree n over the finite field F_p where p is prime

[dark widget]

nice

[wanton heron]

https://math.stackexchange.com/questions/4394652/completion-of-a-number-field-is-a-finite-extension-of-the-same-completion-of-rat.

Does this hold incase of a finite extension L/K which is not necessarily separable

I was trying to complete the argument given in the comments. By showing that a K-basis of L would still be a K_v generating set of L_w

But i don't immediately see why it has to be the case

[wanton heron]

🥀💔.

[near lantern]

This might not be the most elementary way to do it, but any finite-dimensional vector subspace of L_w (as a K_v-vector space) is complete, hence closed in L_w. So if you take the K_v-span of a K-basis of L, it it is both closed and dense (the latter because it contains L).

This is assuming a discrete (or maybe rank-1) valuation; general valuations behave more weirdly and i'm not sure if this in particular breaks.

[wanton heron]

Thank you.

[fierce steeple]

Isn't this essentially the definition

[buoyant fox]

If you have an infinite indexing set, are direct sum and direct product ALWAYS not isomorphic?

ie, does a infinte direct sum being of vs being isomorphic to an infinite product of vs imply that the indexing set of the direct sum and product must be finite?

[torn harbor]

yes unless they aren't for dumb reasons (like if you are taking a direct sum and all but finitely many factors are zero)

actually wait you said isomorphic not equal

hmm that's maybe trickier actually

yeah so for example if you take the infinite direct sum of Q vs the infinite direct product of Q

over a large enough index set they should be isomorphic

just because eventually they'll have the same dimension as Q vector spaces

actually nvm that isn't true the infinite direct product always has strictly larger cardinality basis

[lone jacinth]

If all the summands are non-zero, then the canonical inclusion of the direct sum into the direct product will not be an isomorphism.

They might still be isomorphic in some other way though.

Like the countable direct sum and direct product of copies of Q^N should both just be isomorphic to Q^N

[torn harbor]

yeah I was going to say I think if you let A = some big product of Q over some huge index set, then the countable direct sum and direct product of A should be isomorphic

idk if countable is enough but eventually I think it works

[buoyant fox]

I guess I amt rying to make the claim that they alwyas have differen tdimensioon

so what you are saying seems to imply thats not necessarily true?

[lone jacinth]

Something to note is that an infinite dimensional vector space over Q has dimension equal to its cardinality.

So if the dimension of what you're summing is much bigger than your indexing set it should win out, be it product or sum

[unkempt idol]

Is this the most appropriate text channel to discuss C*-algebras?

[little cradle]

Consider the conjugation action of GL_3 on K^{3x3}, it induces an action on the set of its associative subalgebras. I want a complete list of GL_3-orbits of the associative subalgebras of the lower triangular 3x3 matrices. Any ideas/if it’s known can you give me some references? Many thanks!

[urban granite]

#advanced-analysis

[lone jacinth]

It shouldn't be too hard to see that upper triangular matrices don't fix the lower triangular matrices as a subalgebra.

So then from LU decomposition you get must of the orbits uniquely presented by the upper triangular matrices.

Then you just have to deal with the case of matrices without LU decomposition.

[silver goblet]

are there any natural transformations between the identity functor on commutative F_p-algebras and itself that aren't powers of the frobenius?

[unkempt idol]

Okay thanks!

[lone jacinth]

No, consider what the natural transformation does to Fp[x]. It must send x to some polynomial f(x). Assume f(x) is not equal to x^n. Then f has a non-zero root (in some field extension E of Fp).

Consider the map Fp[x] -> E sending x to said root. Then the natural transformation would need to take that root to 0 which wouldn't be a homomorphism from E.

So such a natural transformation is given by sending anything to x^n and this is only a homomorphism (on for example the alg closure of Fp) for n a power of p

[silver goblet]

ah very nice, thanks!

yeah i was staring at F_p[x] but it didnt occur to me to think about roots of a thing in it. very slick

[near lantern]

To add a comment to this, observe that a natural transformation of id induces a natural transformation of the forgetful functor U by applying U to its components. But U is represented by Fp[x], so Nat(U, U) = Nat(Hom(Fp[x], -), U) = U(Fp[x]) = Hom(Fp[x], Fp[x]) by Yoneda's lemma. In other words, a collection of set maps mu_A: A → A for all Fp-algebras A, natural in A, is always given by a polynomial f in Fp[x] by substitution.

Moreover, U is faithful, so the map Nat(id, id) → Nat(U, U) we applied earlier is injective, so we can view Nat(id, id) as a subset of Nat(U, U). (Concretely, a natural algebra endomorphism is a natural set map which is always an algebra homomorphism, so the set of such is a subset of all natural set maps.)
In fact, it is the set of Nat(U, U) satisfying some equational identities (those asserting that the components are algebra homomorphisms, e.g., for addition we get that the two natural transformations a, b from U ⨯ U which are componentwise a_A(x, y) = mu_A(x) + mu_A(y) and b_A(x, y) = mu_A(x + y) must be equal).

But all of these identities will be between U^k and U for some k, which means by Yoneda's lemma they can be identified with elements of U(Fp[x1, ..., xk]) (Fp[x1, ..., xk] represents U^k).

What this means in practice is that you only need to check the equational identities in a single, suitable "universal" case. For example, to check if mu given by f in Fp[x] is additive, you can just check additivity for the universal case of the algebra A = Fp[x, y] and elements x, y. In other words, f(x+y) = f(x) + f(y) as elements of Fp[x, y].

This is why it's a natural idea to look at polynomial rings.

Actually you can use this approach to classify natural additive maps, multiplicative maps, algebra homomorphisms on the category of (unital associative) commutative k-algebras for any commutative ring, though I think the answer only differs non-trivially from what we've seen for additive maps (and even then you can kind of see it's related to "Frobenius in char p, nothing but id in char 0").

[onyx imp]

this is the construction of a spectral sequence from a filtration of a chain complex

i dont understand how they got the very last isomorphism at the bottom

[zenith lichen]

I've been looking at loop algebras (the nonassociative analog of group rings, not the typical meaning) and i think i found a moufang loop algebra that contains a 54-dimensional ideal, but i'm not sure how to go about constructing this

[lone jacinth]

This is the second isomorphism theorem coupled with the intersection of A^r_p and Fp-1 being A^r-1_p-1

[onyx imp]

ah i gave up eventually but i see what you mean by that now

between the 3 hours and now, i just started my own chasing and actually somehow manage to chase the result from that the spectral sequence constructed from filtration of chain complex is indeed a spectral complex

[wise sedge]

Oh hey you’re here too
There was a discussion about nonassociative stuff in #advanced-lounge a few hours ago

[zenith lichen]

Really? Where

[wise sedge]

#advanced-lounge message

[spice idol]

what are even the kind of algebras arising from this

I can't imagine they'd be nice lol

[zenith lichen]

the moufang loop i found was commutative, so i think the algebra was a jordan algebra

[spice idol]

interesting

[zenith lichen]

if anyone knows of any other 54D commutative algebras of importance let me know

[analog schooner]

what are jordan algebras

the only thing i remember is that associative algebras are jordan algebras iff they are commutative

[zenith lichen]

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

[zenith lichen]

That's not right

[analog schooner]

oh wait no nvm

thats for an associative algebra is a jordan algebra

i mixed the two up (i said jordan algebras are associative iff they are commutative)

nonassociative algebras

[modest prawn]

Is there anyone who is studying sigma injectivity?

[thick creek]

in general, is the socle of a module M the set annihilated by the radical rad(R)? what about if we let M be a ring R as an R-module over itself?

[lone jacinth]

Not in general no. Take R = Z, then rad(R) = 0 which annihilates everything.

The condition you need is for R/rad(R) to be semisimple, for example for R artinian this would be true.

(and I guess you meant to write rad(R) not rad(M))

[thick creek]

i see, thank you

[thick creek]

where could i find some examples of socle and cosocle filtrations?

[thick creek]

also does anyone know any good references on noncommutative algebra? i am self-studying from 18.706 notes on mit ocw, but those get really hard to understand sometimes..https://ocw.mit.edu/courses/18-706-noncommutative-algebra-spring-2023/resources/mit18_706_s23_full_lec_pdf/

[neat nexus]

Kasch modules and rings

[worldly zealot]

Lam’s noncommutative rings

[thick creek]

Is it true that if M is a finitely generated module over a ring, then rad(M)=J(R) M? If not are there similar statements

[drowsy niche]

no. let R = k[x] and M = R/(x^2)

(for k a field)

[round seal]

Can I get some feedback on this? This was a hard problem.

[lone jacinth]

This is another thing that becomes true for R/J(R) semisimple.

In general J(R)M <= rad(M)

[storm basin]

sanity check: the highlighted statement as written is not strictly true, n is not just the number of elements in a generating set but the MINIMAL number of elements in any generating set

Context is this problem. Clearly (xy, xz, yz) = I(X) = (x, y) cap (x, z) cap (y, z) is a decomposition of I(X) into its minimal primes. I would want to conclude here: but these each have height 2 in k[x, y, z] not height 3...

there is probably some not dimension theoretic approach, I feel the best I can get with dimension theory is that it has to have at least 2 generators

[storm basin]

actually, I am tripping now; (x, y, z) has height 3 in k[x, y, z] and is prime but is clearly not a minimal prime over I(X) [because we explicitly know what all of these are]. On the other hand, the exercise is claiming that I(X) has a minimal set of 3 generators. Therefore the converse as written to the hieght theorem in the wikipedia article just doesn't make sense. Can someone point out where my (probably obvious) mistake is?

ah ok. The phrasing of the converse is just really bad on the wikipedia article. It's not true that a prime of height n is minimal over EVERY ideal generated by n elements (then this would give a counterexample). Rather, it is true that for every prime of height n, there exists an ideal generated by n elements (depending on the prime) over which that prime is minimal. IMO the wikipedia phrasing is very confusing

[thick creek]

can a socle filtration of an R-module M just stop without reaching M? in what cases does it have to reach M?

[frosty wagon]

yeah it stalls if the quotients have zero socle, usually needs to be semi-artinian

[thick creek]

right

so I'm trying to find the socle filtration of R = k[G] as a module over itself, where G is a finite p-group and char k = p

I found M1 = soc(k[G]) = k (sum_{g in G} g)

Then to find M2, I want to find soc(R/M1). But using the general property that Soc(M) = Ann_M (J(R)), where M is an R-module and R is Artinian, I got that Soc(R/M1)=0

but then the filtration just stops at M1 right, since M2 = M1

but I thought since R is Artinian it should reach R

[lone jacinth]

Not sure how you get that. But it shouldn't be 0

For example let G = Cp generated by g.

Then the socle is as you say generated by
1 + g + ... = (1 - g)^p-1
Then M2 would be generated by (1 - g)^p-2 etc

as J(R) is generated by 1-g

[thick creek]

why is M2 generated by (1-g)^{p-2}?

sorry im very new to this so i know almost nothing

[thick creek]

ah nvm i see

you can look at R=k[C_p] as k[x]/(x^p) basically

[lone jacinth]

Well certainty if you multiply it by (1-g) you end up in M1

[ornate atlas]

Im trying to get comfortable with the universal property of the tensor product, so im trying to show that $R[x_1,\ldots,x_n]/I \otimes R[x_{n+1},\ldots,x_{n+m}]/J \cong R[x_1,\ldots,x_{n+m}]/(I+J)$ (I think this should be the correct statement, this isnt an actual exercise so I could be wrong, but I think this is how it works in AG)

So to ease notation let me call the first factor $S_n$, the second $S_m$ and the RHS $S_{n+m}$. I think what I want to say is something like the universal property of the tensor product tells us that $$\mathrm{Hom}(S_n\otimes S_m,S_{n+m})\cong \mathrm{Hom}(S_n,S_{n+m})\times\mathrm{Hom}(S_m,S_{n+m})$$ and then Im a bit confused. I thought maybe I want to say something about the universal property of quotient rings, and apply Yoneda, but maybe this is way over complicating things (and im not quite sure how that goes either)

[broken turtle]

Nope

[plucky arch]

I would recommend a different approach

Show that the corresponding hom-functors are naturally isomorphic, and identify what the isomorphism is

[ornate atlas]

Which hom functors do you mean?

[plucky arch]

$\text{Hom}(S_n \otimes S_m, -)$ and $\text{Hom}(S_{n + m}, -)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[thick creek]

How is the associated graded module of a composition series (of R-modules) defined exactly? Since we did not give R a grading, I don't see how the grading of the associated graded module contains any nontrivial information

[ornate atlas]

Its the direct sum of each of the consecutive quotients in the filtration, gr R = \oplus R_i/R_{i-1}. You then equip this with the "obvious" multiplication. One of the nice things is that if your ring is positivley filtered, then you know its Noetherian if the associated graded ring is, and sometimes this is much easier to work with

This is how you (or at least how ive seen it done) prove that like the Weyl algebra is Noetherian

[plucky arch]

Do you have an idea for how to get started with this

[ornate atlas]

Sorry I havent started it yet, ill go try now lol

[fierce steeple]

Lol I was so confused by Pseudo's message and then looked at nope's message and was just as confused until I realised S_n was notation for a ring and not the symmetric group

i think here it is a bit nicer to think of this more informally. like "a map out of S_n (x) S_m is the same as ..." and when you work through the things it should hopefully be clear

[plucky arch]

Yeah that’s the Hom functor approach

[ornate atlas]

Like I think I just want to say that if a map sends f(x)g to c, then the image should send fg to c, but idk im a little confused

[plucky arch]

To start off with, you want to use the universal property of the tensor product

$\text{Hom}(R_1 \otimes R_2, P) \cong \text{Hom}(R_1, P) \otimes \text{Hom}(R_2, P)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

Naturally in P (in fact in all three variables, but you don’t need that here)

What does that get you here

After that you can use the universal property of the quotient

Did that help @ornate atlas

[ornate atlas]

I think so yeah im just running through the details now

[ornate atlas]

Yeah ok sure expanding it out it just kinda has to be true lol

[bronze timber]

rude!!!

[digital parcel]

excellent ring

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

[bronze timber]

a quasi-excellent ring "behaves well," and does not get time out.

[lime bloom]

We're running out of context-neutral adjectives!!!

[near lantern]

To be clear (I + J) should mean (and it does if we're on the same page wrt the notation) I[x_{n+1}, ..., x_{n+m}] + J[x_1, ..., x_n] or in a way that's valid more generally, I (⨯) R[x_{n+1}, ..., x_{n+m}] + R[x_1, ..., x_n] (⨯) J.

[thick creek]

is cosocle always just the socle filtration "reversed"? i did a few examples and it looks like it

[scarlet ermine]

with the stacks project's notion of classical generator (https://stacks.math.columbia.edu/tag/09SI), I can see that for R commutative Noetherian, R regular implies R itself is a classical generator of Db(R-mod) by replacing every complex with a bounded free resolution. I'm wondering if R being a classical generator implies R is regular (like there aren't weird ways to generate somehow)?

[scarlet ermine]

oh yeah okay I'm seeing other references prove that the compact objects of D+(R-mod) are perfect complexes, and being a compact generator of D+(R-mod) implies you are a classical generator of Perf(R), so this kinda matches up with what I would want

[dark ibex]

Which paper is this

[scarlet ermine]

actually I think the stupid filtration of a complex is semi-standard terminology haha, it actually is a legit adjective. At least I've also seen Gelfand and Manin's methods of homological algebra (III.7.5)

[silver goblet]

Have also seen this called the bad filtration so honestly the stupid filtration is more illuminating lol

[lone jacinth]

No, for a simple example take some M with a nontrivial filtration and consider M(+)S for a simple module S

[lone jacinth]

I've seen stupid and brutal as the adjectives used for this.

In contrast with the good truncation

[scarlet ermine]

that's the standard t-structure one?

[fierce steeple]

Lol it was just that Nope called these rings S_n

[fierce steeple]

I've seen stupid, brutal, foolish, Hodge lol

Though Hodge I insert here mostly cause it is funny

[lone jacinth]

Idk what that says about Hodge xD

[fierce steeple]

Yeah it's not even bad lol

Lol @silver goblet do you know this fun thing of like uhh

Stupid filtration and the Beilinson t-structure lol

I would guess maybe from A. lol

[ornate atlas]

I was at a talk about this the other week lol, amusing naming

[fierce steeple]

Nice

[dark ibex]

Not more amazing than StRing

Physicists would be hella confused in gestalten

[silver goblet]

yeah it's really fun

and yes if there is something to say about the beilinson t structure then A has said it

[fierce steeple]

lmao

[near lantern]

I feel like naive is less unnecesarily harsh and IIRC used in Weibel.

[hushed bone]

Stupid is on stacks project

Which means that’s the real name

[lone jacinth]

Naive is good, but I think brutal is pretty descriptive.

And stupid is just 100% funnier

[scarlet ermine]

I like stupid because my perspective is that the stupid filtration has associated graded pieces which are just the terms of the complex itself, so you didn't get anything interesting (like homology), so it's kinda stupid. and I like studying stupid things, it's not like that's a down side haha

[hushed bone]

You know what your pfp reminds me of

Something like

“You have no fucking idea what you’re up against”

[scarlet ermine]

haha

I'm just a big fan of these random silly dogs on instagram

[dark ibex]

Yes stacks are stupid /s

[hushed bone]

He kinda has founding father hair

[hushed bone]

https://tenor.com/view/funny-or-die-will-ferrell-watch-your-mouth-filthy-mouth-mouth-gif-4427315

[vocal saffron]

Snacks project

[hushed bone]

That sounds yummy

[vocal saffron]

Pursuing snacks

[floral bluff]

how can I learned advanced algebra now. I have only math foundation of physics institution. And I spend too many years struggling for surviving. Now I want to learn pure math agian. May I get some advice?

[hushed bone]

Call me Grothendieck the way I be pursuing these stacks

Photo of a gangster SpongeBob dripped to the grills

[hushed bone]

Pick up like, Pinter’s book on algebra and try to read it

[dark ibex]

Most of advanced etale cohomology haven't been stacked yet

[dark ibex]

Start with artins book

[floral bluff]

thank you

[floral bluff]

how can I understand the homology algrbra level

[dark ibex]

Start with pre algebra ||If you know abstract algebra, then you can start doing it with some basic cat theory knowledge, if not start dummit and foote||

[floral bluff]

category theory?

I have read it. But I find it too abstract and I don't know how to use it

[dark ibex]

Technically you can start homo algebra

Tor and ext stuffs

[floral bluff]

thank you

I feel tor and ext are a bit difficult

[worldly zealot]

#1203471755449073774 message great minds

[worldly zealot]

you should learn from a textbook that's a first course in abstract algebra, the content in this channel is generally stuff only accessible after a good foundation in groups, rings, fields

[floral bluff]

I have learned abstract algebra and I am stop at here in my undergraduate

[astral ginkgo]

This but unironically

[past cove]

only in biden's america can someone think stacks are bad

[steep rivet]

inchrasting

[urban granite]

yeah these are good things, for example BG classifies principal G-bundles, i.e., X\to BG correspond to a principal G-bundle
in particular if G=G_m then this classifies line bundles, and if G=GL_n then this classifies vector bundles of rank n

there are even more features, for example:

vector bundles on BG_m are Z-graded vector spaces

vector bundles on the stack [A^1/G_m] are Z-filtered vector spaces

there are more examples where you can transfer linear algebra terms in terms of quasi-coherent sheaves on some stacks (idk that so much about other examples, though)

[novel nimbus]

what's this ?

[urban granite]

LLMs ig

[novel nimbus]

well the llm was nit too great at defining it

@urban granite its like llms use stacks fr covers and then the buncles are a stack umm noope they are kinda but not exactly also what soace whats the top ??

btw here's an honest q - - i am teaching at columbia in the fall complex analysis alfohrs -- this is my first class where i make all the notes

any advice

?

[urban granite]

I'm not even doing TAs lol

not even postgrad either i mean im undergrad student now

[novel nimbus]

@urban granite lel i saw yr bio and the def in yr bio is exactly what a sheaf is lel

and a stack

it's super hard to teach first yrs honestly @urban granite cuz they're always wngineers who'll tryn tell me i dont know basic anlaysis like

\zeta(-1

)

and then i have to be super nice cuz else i get shitty reviews

[round seal]

If i have a regular R sequence and i localize, is it still regular in the localization, up to discarding the elements of the sequence that become invertible

i feel like that should be true but i’m not sure how to justify this

[near lantern]

I think yes because localisation is exact.

So if r1 is cancellative on M that remains true after localisation and S^{-1}M/r1 S^{-1}M = S^{-1}(M/r1M). Now induct.

[hushed bone]

If it’s Noetherian it follows by looking at the Koszul complex and taking homology as well, again using that localization is exact

[lone jacinth]

Do you need Noetherian for that?

[hushed bone]

I was being safe

I’m pretty sure you need it for something

[lone jacinth]

Never wrong to rely on our old friend Emmy aye

[hushed bone]

Hmm, it also needs local

Which kind of defeats the purpose if you’re asking about localizations

But thinking about it

I feel like if all Koszul homology is 0 it must be a regular sequence kind of by like

Definitions

Nah

https://stacks.math.columbia.edu/tag/062D

Stacks project says none of the arrows can be reversed in general

And this is asking if Koszul-regular => regular

[limpid horizon]

Chmonkey finger

[pastel shoal]

Does GL_n : CRing \to Grp have a left adjoint?

[plucky arch]

Group ring Z[G] I would assume

Wait

That’s not commutative

[last talon]

Nope

[plucky arch]

I realised

[last talon]

Nope, doesn’t preserve limits or colimits

(Can’t remember which way around it is, but it preserves neither so)

[pastel shoal]

It preserves products right? It doesn't preserve equalizers?

[last talon]

It doesn’t preserve products

[random oxide]

why universal cover Sl_2(Z) has no faithful representation

[pastel shoal]

I think I'm slow what's a counter example for this

[last talon]

Actually I think it does now I think about it

[spice idol]

doesnt it?

yeah

[last talon]

Check the conditions of like SAFT or something

Yeah and it preserves equalisers so it preserves limits

[pastel shoal]

idk SAFT but I remember reading some mathoverflow post saying Grp doesn't satisfy the hypotheses or something like that

[last talon]

SAFT fails
We don’t have a coseperating set

GAFT also fails
Take a group which isn’t linear over any commutative ring
Then we have no solution set

So t here’s no left adjoint

[astral ginkgo]

Interesting thought I had: In the definition of the universal property of the tensor product, the map from M x N to the tensor product is required to be bilinear. What happens if you drop this requirement?

I haven't found any information on this and haven't figured it out myself yet

You can prove the map has to be bilinear by explicitly constructing the tensor product, but can you do it without this explicit construction?

[torn harbor]

bilinearity is central to the tensor product, if you drop it there's basically nothing left

Oh I guess you mean if you have the same universal property but drop the assumption that the map into the tensor product is bilinear ok

[spice idol]

then it wouldn't be a universal property anymore

[torn harbor]

well it still would

oh yeah it isn't equivalent to the usual tensor product anymore if you do that I believe

since the free vector space on V\times W would satisfy that universal property

its not even a true universal property since it doesn't determine the object up to isomorphism I guess is your point enpeace

[spice idol]

yes that's what I meant lol

any quotient between the free vector space on V \times W, and V \tensor W would work in fact

[torn harbor]

yeah

[hushed bone]

Can you specify what exactly you mean?

Are you taking the manual construction of tensor product and the same map?

Are you asserting that the tensor product factors all bilinear maps via this map, but just dropping the requirement that the map to the tensor product be bilinear?

something else?

[thick creek]

is it correct that cosoc(M)=M/rad(M), and the cosocle filtration of M is given by M =M^0 \supset M^1 \supset M^2 ..., where M^{i+1} = rad(M^i)?

[astral ginkgo]

I'm saying if you take the manual construction and the regular map there sending (m,n) to m(x)n, then if you have another tensor product (T,f), f will factor through this map and thus also be bilinear.

So it seems to me that requiring the map from MxN to the tensor product in the universal property to be bilinear is unnecessary since it's implied. In this case, can we do it without using the explicit construction as a quotient of a free module

[near lantern]

What is it universal for? Just any map? Then you just get the free module on M ⨯ N.

[astral ginkgo]

No, still universal for bilinear maps

[near lantern]

I see.

[near lantern]

As pointed out here and earlier, it would no longer be a universal property.

A good universal property should include being one of the things that uniquely factor through it; otherwise it usually becomes too easy to satisfy vacuously.

[astral ginkgo]

I convinced myself it doesn't satisfy it but yea of course it does

Fair enough

[astral ginkgo]

In retrospect this also assumes implicitly that the maps into the tensor product are bilinear

[near lantern]

For another example, consider abelianisations. A map G → H uniquely factoring any map from G to an abelian group could be any quotient between G and G^ab, i.e., G/K for any normal K ⊆ [G, G]. It's only when you force H to itself be abelian that you pin down G/[G, G].

[astral ginkgo]

Just had a brainfart

[near lantern]

The unique factoring property is like an upper bound and itself having the property is like a lower bound (or they're constraints on opposite sides, whether upper/lower or lower/upper or source/target or big/small or whatever), which together intersect in just one object (up to unique isomorphism).

[astral ginkgo]

The equivalent thing here would be not requiring the map G->H to be a group homomorphism

[near lantern]

Well, one possible analogous thing.

I was thinking in terms of the adjunction between Grp and AbGrp.

[astral ginkgo]

Sure, just seems like the more natural analogy to me

[astral ginkgo]

Sure

[near lantern]

I think this is just pedagogically more illustrative for the point I wanted to make, nothing more.

[astral ginkgo]

Here the free abelian on G would work in that case

I mean it becomes the same univ property

[near lantern]

You are correct but I don't like it.

[astral ginkgo]

Lmao

[grave onyx]

Can someone help
What is 5x - 7x + y = 0

I am learning agelbra

how to make y the subject

[honest gulch]

y = 2x

[random oxide]

let G be a simple lie group and g be its lie algebra , say g=g^{+} + g^{-v} + g^{0} where g^{0} is space of eigen value with abs value 1 (of Ad_g) then want to know if G has a decomposition of the form exp(a) exp(b) exp(c) where a , b, c are in respective subspaces or not?

[wary elbow]

It's not clear what you mean here, but it looks like you're describing the decomposition of g into generalized eigenspaces for a fixed element in the Lie algebra. If that's what you mean then no, elements of G cannot be written like this in general. I'm going to assume the element is nonzero because it is trivial to see this is false if the element is 0 (by nonsurjectivity of the exponential map).

Consider Sl_2(C). For a nonzero element H of the Cartan subalgebra, the associated decomposition of g is the usual weight space decomposition, and the elements of G in the form exp(a) exp(b) exp(c) are the big cell U^- B (or some conjugate of the big cell, depending on the order of a, b, c). The big cell is dense in G but is not equal to G.

[random oxide]

i was thinking of something similar KAK decomposition

[wary elbow]

The generalized eigenspaces in your picture are not the KAN decomposition though. They appear to correspond to generalized eigenspaces of a single operator

What is the element g in your picture? That is, what is Ad_g?

[random oxide]

g is some element of group , Ad is Adjoint ,

[wary elbow]

Ok, so it's exactly what I described in my earlier comment. You can't do this in general. The SL_2 example I gave is a counterexample.

[random oxide]

can we make a local decomposition

[wary elbow]

What do you mean by local decomposition?

[random oxide]

a nbd of g as a product of open sets , OPEN SET coming from g+ g- g^0 through exp map

[wary elbow]

No, this isn't possible either. The exponential map will generally not take open sets to open sets and it won't be injective either.

[random oxide]

locally maybe

[wary elbow]

Ok yes, this is always possible, but it's not interesting. In fact, this is always possible for general differential geometry reasons: if you take any direct sum decomposition of the Lie algebra, then the induced map by applying exp on each component is a local diffeomorphism.

This is because the differential of this map is obviously surjective at the identity, and hence is an isomorphism of tangent spaces.

What makes the KAN decomposition interesting is that it's global.

[random oxide]

yes iwasawa is nice,

[random oxide]

can we expect it almost everywhere

[wary elbow]

What do you mean by almost everywhere?

[random oxide]

w.r.t haar measure on lie group

[wary elbow]

No, I don't think so. Here's a rough idea: the image of the exponential map is not always dense, which means the compliment of the image is not measure zero

[random oxide]

okay it might setlle

[sand falcon]

hello, i got an introduction on Hopf algebras and the teacher said that one motivation for the study of Hopf algebras is that the tensor product of 2 modules over a Hopf algebra always makes sense. I was wondering why we needed our algebra A to be a Hopf algebra for the tensor product to make sense, is it not possible to generalize the construction even if A is not commutative ?

[astral ginkgo]

If A is not commutative, then a tensor product over A is generally not going to have an A-module structure

The comultiplication allows you to define a module structure on the tensor product

[scarlet ermine]

https://math.stackexchange.com/questions/4581569/can-we-define-tensor-product-of-modules-over-an-algebra
this might be a good read!

the idea is that to tensor modules M,N over A, when A is not commutative, to get a good theory we really want M to be a right A-module and N to be a left A-module, so that M tensor_A N has scalars moving in the middle between symbols m and n

so instead, you ask that A be a k-algebra, where k is a commutative ring inside the center of A, so that when M, M' are both left A-modules (or you could pick the convention of right A-modules), and at first only form the tensor product M tensor_k M' (which makes sense because k is commutative, so gives left AND right scaling)

you then are sad because M tensor_k M' doesn't have a canonical A-module structure. However! Leftover from M, M' being left A-modules, you get a left (A tensor_k A)-module structure on M tensor_k M', and then you use comultiplication A -> A tensor_k A and restriction of scalars to make M tensor_k M' into a left A-module

so exactly the choice of comultiplication ended up giving left module structures on tensor products of left A-modules/k

also maybe read here https://mathoverflow.net/questions/107844/left-module-structure-on-the-tensor-product-oftwo-left-modules

it's also good to keep some examples in mind, like group algebras (where modules = group representations, and tensor product is the tensor product of representations)

[random oxide]

what are some applications of grothendick galois theory

[random oxide]

As i remember , i studied hopf algebra because of duality arising from algebraic group schemes and k- algebra homomorphism so , hopf map is dual to group multiplication via YONEDA lemma

thanks yj

will see thanks

rn i am using it only for algebraic group schemes, very new to it

[random oxide]

Need a simple proof of " Hopf alg over field of char 0 is reduced"

[near lantern]

Note: the mechanism of becoming a module is very different in these two cases. In both cases a priori you have a A (⨯) A module stucture. One applies extension of scalars, changing the module to turn this into an A (it base-changes by the algebra map A (⨯) A → A). The other applies restriction of scalars to change it into an A (by the comultiplication map).

OK that's the first time I realised this similarity between the two.

[near lantern]

Hmm, tensor product in the category of A-bimodules changes the base tensor product from a A (⨯) A^op (⨯) A (⨯) A^op module A (⨯) A^op by annhilating the action of the subgroup I (or equivalently right ideal) of the middle A^op (⨯) A generated by {a (⨯) 1 - 1 (⨯) a : a ∈ A} (the left ideal generated by I is the kernel of mul: A^op (⨯) A → A: b (⨯) a ↦ ab). So this cannot be realised as a base-change; you are really tensoring with a bimodule to shift categories.

Anyway

Let G be a finite group and k a field such that |G| ∈ k^⨯. If two representations of G over k are isomorphic over the algebraic closure of k (i.e. they have the same character), must they be isomorphic over k?

[spice idol]

both are cases of tensor products for Beck-modules
I saw some slides on the problem of deciding when something like that exists but i cant recall exactly their results

(A Beck-module over X in C being an abelian group object in C_/X)

[near lantern]

What.

[spice idol]

lol

[near lantern]

What's C and X here?

[spice idol]

C category and X ∈ Ob(C)

my bad

[near lantern]

No duh

In this setting which category and object are they?

[spice idol]

for groups its C = Grp and X = the group youre looking at the G-modules of (of course for reps over a field you take the k-vect objects rather than abelian group objects)

and for commutative rings, C = CRing and X = your commutative ring

[spice idol]

the relevant slides

[near lantern]

Wait but what does the slice category C/X have to do with anything?

Oh Mod(X)

[spice idol]

yes

its tied intimately with the idea of module of differentials? apparently, its what the nlab page said :P

[near lantern]

OK this is a generalised settings to defin derivations

acc to nlab

[spice idol]

and give general coefficient modules for cohomology of algebraic structures
and in the case where the variety is congruence-permutable this cohomology actually is also related to extensions of the algebras? for a sufficient notion of extensions

[thick creek]

is it true that if R is a local algebra (and we're viewing R as a left R-module over itself), Soc(R) = J(R)^{n-1}, where n is the minimal value for which J(R)^n = 0

[near lantern]

So the intuition is that you should take cohomology of a thing to be Ext(Kahler, -) for a suitable Kahler (often the "trivial rep").

[near lantern]

What's n?

[near lantern]

Also bumping this question because I yapped too much.

[thick creek]

sorry updated it

[lone jacinth]

Not necessarily no. Take
R = k[x, y]/(xy, x^2, y^3)
Then
J = (x, y)
J^2 = (y^2)
J^3 = 0
but the socle is (x, y^2)

[lone jacinth]

Yes. Note that the the splitting field K is a finite extension of k, so we can think about that instead of the alg closure.

Then for two representation U and V, then if U(x)K and V(x)K are isomorphic KG modules they are also isomorphic kG, modules. But as kG modules they are just U^n and V^n. Which by KRS means U=V.

Alternatively, if k is an infinite field (and U and V are finite dimensional) there's a fun argument.

Hom_kG(U, V) is an algebraic variety just equal to affine space. And Hom_KG(U(x)K, V(x)K) = Hom_kG(U, V)(x)K.

The determinant is a polynomial on this variety. If it takes a non-zero value on the latter it is a non-zero polynomial. For an infinite field a non-zero polynomial is never constantly 0.

[scarlet ermine]

what is the determinant here? are we identifying the left and right as k vector spaces using the isomorphism to define the determinant? this seems really cute, I like this

[lone jacinth]

Yeah, pick some basis such that
U = k^n = V
then the usual determinant

[scarlet ermine]

i see nice

[fierce steeple]

Ye

Quillen, homology of commutative rings

[hushed bone]

I think you mean

On the (co-)homology of commutative rings

🤓

[fierce steeple]

Yes

Well I'm pretty sure there is also a paper called homology of commutative rings by Quillen

But unpublished or smth

[hushed bone]

Grrrr

[fierce steeple]

Lol

[near lantern]

Fun stuff. (My question was if there's a Galois theory in the ℤ-graded setting and @ nHecke(G) got me started.)

For $L/K$ finite, the automorphism group has at most the expected cardinality $[L[n] : K[m]] = [L : K] (m/n)$ but equals it only when $L/K$ is Galois \emph{and} the coboundary (for $G = \operatorname{Gal}(L/K) \curvearrowright L^{\times}$) of $\phi$ is valued in the subgroup $(L^{\times})^{m/n}$.

[broken turtle]

Raghuram

[near lantern]

Oh, and L should have all (distinct) m/n^th roots of unity.

[thick creek]

hm I see. I was thinking about the socle filtration for group algebras k[G], where char k = p and G is a finite p-group. would this hold in that case?

[lone jacinth]

Yes, in this case the socle should be simple.

[thick creek]

so in this case we have R=k[G], since R is Artinian we have soc(R)=Ann_R (J). and also we can just directly observe that soc(R) is the 1 dimensional k-vector space spanned by (sum over g in G of g). but how do we know Ann_R (J) = J^{n-1}?

[lone jacinth]

Well clearly J^n-1 is a subset of Ann(J)

So if it's 1d they have to be equal

[thick creek]

ohh oops thank you!

[near lantern]

soc(M) is the sum of all simple subreps and the only simple rep is the trivial one. So soc(M) = M^G for any M. In particular for the regular rep that finishes it.

[near lantern]

OK I can't see how to make this constructive/computational.

[lone jacinth]

What do you mean exactly?

[near lantern]

Uh actually I think my problem is with making something else explicit.

My setting is the Chevalley-Shepard-Todd Theorem. You have (i) polynomial rings A ⊆ B both in n variables over a field K and a finite group G (order coprime to char(K), take char(K) = 0 if it's easier) acting on B such that A = B^G. You also know (ii) B is a free A-module (of finite rank because B is integral over A and fg) and that (iii) we can choose a basis such that the diagonal entries of the G-representation on B are in K rather than A.

Now if we base change to Frac(A), we get the finite Galois extension Frac(B) = B (⨯)_A Frac(A)/Frac(A) = Frac(B)^G. By the normal basis theorem, this is isomorphic to the regular representation over Frac(A). But this implies by going via the character that the same is true for B (⨯)_A K over K, where A → K is by setting the variables to 0. (Maybe this doesn't really use (iii).)

This is mainly out of curiosity, but I would like to explicitly construct a normal basis of the algebra B (⨯)_A K over K from a normal basis for the Galois extension Frac(B) over Frac(A).

[spice idol]

yippe! Beck modules are also nice because they let you do hom alg for general algebraic structures

[near lantern]

TBF you can do that as soon as you get an abelian category. To me the idea of picking out what the "trivial module" is seems the most interesting.

[spice idol]

right but beck modules give a way to nicely associate an abelian category to an algebra

in the sense that each morphism f : A → B functorially induces a functor f_* : Mod_V (B) → Mod_V (A), which has a left adjoint

[near lantern]

Hmm. IG it's true that you get abelian categories without specifying anything other than C. So it's somehow a "canonical" abelian category associated to C's.

[golden osprey]

(mainly asking here to not interrupt #groups-rings-fields)
https://en.wikipedia.org/wiki/Monomial_order

There are alot of statements in this page about how certain monomial orders are harder to compute Grobner bases for than others

Is there a good reference for differing algorithms and complexity bounds for different monomial orders? The two CLO texts don't talk much about complexity

[fierce steeple]

Ye I mean a nice thing from my perspective is that it all generalises to ∞-categorical settings

So like the same statement is true for some flavours of homotopical rings and this is very important for deformation theory

[spice idol]

oh peak

[astral ginkgo]

Oops, I forgor that you initially need to take the tensor product over the base field

[tropic bobcat]

Hi. So there's a course they're offering and pretty much they're going to cover these things:

The book they're following is https://drive.google.com/file/d/1eICEspJQeTOi2xLzUFkQ3mM-gxcBjZ4I/view?usp=sharing (so basically chapters I-IV, and an optional ch. VI)

I have not learned commutative algebra, but I want to. Is this course worth enrolling in if that's my goal?

I understand that commutative algebra requires understanding of rings & modules and pretty much the topics mentioned, but I'm asking if the materials covered here is worth spending one semester on, keeping in mind that my goal is to learn some commutative algebra some time this year?

[near lantern]

Looks useful. I will say that depending on the subset of commutative algebra you want to know a course marketed as "commutative algebra" may get to topics closer to what you want.

This is very module-theoretic. There's also a very ideal-y/algebro-geometric set of basic theory to know (e.g. skim the table of contents of Atiyah-MacDonald).

[ornate atlas]

It probably misses a lot of what a first course in comalg covers, but hits a lot of what a second course does funnily enough lol

But also it just looks like a sick course

[limpid horizon]

hey can anyone let me know how I can understand this?

[steady breach]

i dont

x^2+5x-3=0

solve for x

[limpid horizon]

i would be careful mr. AgNO3

[subtle smelt]

They are lurking and stalking when you least expect it mr. AgNO3

[jaunty sparrow]

#prealg-and-algebra

[subtle smelt]

Depth as in regular sequences?

[limpid horizon]

yeah but right now im trying to understand dim M/xM = dim M - n

[thick creek]

for an R-module M and submodule N, is rad(M/N) = (rad(M) + N)/N

[steady breach]

i was just just JK

[rare walrus]

Shitposts go in #chill, and if by chance you do need help with this, please see #❓how-to-get-help. We try to keep the advanced channels, like this one, apart from this kind of stuff. Let this be the last time please.

[steady breach]

are you a mod?

[rare walrus]

Yes.

Another mod has already asked you not to do this

[steady breach]

are mods pink?

[rare walrus]

That's right.

[steady breach]

ok ill go torture the pre uni people thank you and goodbye

[subtle smelt]

Actually wdym by dim M

I thought that was the Krull dimension of R at first

[steady breach]

oh wait are we allowed to swear?

[rare walrus]

Yes. Please see #rules

[unborn rampart]

I'm also interested in this, do you mind posting here if you find anything? I remember reading that if you want a Gröbner basis in something other than grevlex order, it's usually faster to compute it using grevlex and then convert it afterwards, but I haven't seen any justification for this

[thick creek]

if M is a finitely generated R-module, then is rad(M) = J(R) M

[hushed bone]

That x is a sequence of elements of length n

In the proof they showed if x is M regular then dim M/xM = dim M - 1

It decreases by at least 1 because x is not in any minimal prime of M, and it goes down by at most 1 by like, the Hauptidealsatz

[hushed bone]

What is the definition of rad(M)?

[thick creek]

the radical of the R-module M, i.e the intersection of all maximal left R-submodules of M (note that R doesn't have to be commutative)

[hushed bone]

I think no

I believe Q/Z as an abelian group (Z-module) has no maximal subgroups

Meaning that the radical should be all of Q/Z

But J(Z) = 0 so you get a contradiction

With finite generation it might be possible, but then a module with no maxima submodules will give a contradiction

So it would necessarily be over a non-Noetherian ring

This is all assuming you have commutative rings, which is fine to make a counterexample

[thick creek]

hm ic

[lone jacinth]

This is another thing that is true when R/J(R) is semisimple.

[lone jacinth]

The left contains the right, but they're not equal in general
For example
rad(Z) = 0
rad(Z/4Z) = 2Z/4Z.

If R/J(R) is semisimple then you can use rad(M) = J(R)M to show they're equal.

[drowsy niche]

is there a name for rings with this property?

almost artinian?

[lone jacinth]

I mean you have perfect and semiperfect which are kind of close, but that's still more restrictive

I'm not sure Ive seen an example where R/J is semisimple but it's not semiperfect though

So maybe that would be fun to cook up

[drowsy niche]

https://en.wikipedia.org/wiki/Semi-local_ring

[lone jacinth]

So that's a name for it.

Interesting that they don't give examples that aren't semi-perfect.

Except maybe endomorphism ring of some artinian modules could work...

[drowsy niche]

aren't those already going to be semiperfect though?

[lone jacinth]

Idk

But all the other examples they listed would be

[ornate atlas]

DaRT doesnt reckon so

[drowsy niche]

oh wow

that's convenient

how have I never heard of this before wtf

[ornate atlas]

Heard of DaRT or of these examples?

[drowsy niche]

never heard of DaRT

[ornate atlas]

Ah yeah its a nice website

[drowsy niche]

the localization at two primes example is really clean here

[ornate atlas]

Pi base is the one I always forget about, topospaces is nice but doesnt always have what im after

[drowsy niche]

oh yes back to what I was saying: by "those" I meant endomorphism rings of artinian modules

bc they are in particular semiprimary

[steep rivet]

how do you remember all these ring characterizations

semiperfect, semisimple, artinian...

[drowsy niche]

I'm in a noncommutative alg class rn

I will forget all of this in a few months

[steep rivet]

fair

[ornate atlas]

I dont, I open my noncom rings notes minimum once a week tbh

But also like semisimple and artinian are big hitters, those are ones youll remember

Though in a PhD interview the other week I got the definition of a simple group wrong because I forgot groups arent actually just sucky modules and you need to specify that theyre normal...

[last talon]

What’s a good source on BN-pairs?
I’m reading Trees by Serre, and it assumes some knowledge now (chapter 2, section 1.7), but I don’t really have any

[lone jacinth]

Isn't the endomorphism ring of the p-Prüfer group the p-adic integers?

[drowsy niche]

ah right hm

[hushed bone]

Jesus Christ all these noncomm alg adjectives

Must be what ppl feel seeing alg geo people talk

Quasi-syntomic

[drowsy niche]

oh ok endomorphism rings of finite length modules are semiprimary

but not artinian modules in general

[hushed bone]

What is a (semi)primary… ring, module?

Primary means something for a submodule in commalg

[drowsy niche]

R is semiprimary if J(R) is nilpotent and R/J(R) is semisimple

[hushed bone]

Semisimple = finite direct sum of simple rings right?

Simple meaning no nontrivial proper ideal?

Hmm that can’t quite be right

I just remember finite product of matrix rings

From a rep theory course

[drowsy niche]

why not?

[hushed bone]

I guess not

I was thinking of the simplest ring I can think of that’s not a field

And decided it was a local Artinian ring which isn’t simple

But that’s cuz I was thinking in the wrong direction

And also still being comm alg brained

[lone jacinth]

Here semisimple means semisimple as R-module.

So it's not really about simple rings (though Artin-Wedderburn gives you that a semisimple ring is exactly a finite product of artinian simple rings)

[hushed bone]

What’s a semisimple R-module

[lone jacinth]

Direct sum of simple R modules

So yeah R is a semisimple ring if R is a semisimple R-module, in the same way as R is defined to be artinian/Noetherian if R is artinian/Noetherian module.

But contrary to R being a simple ring, which means it doesn't have any nontrivial ideals

[hushed bone]

And simple means no nontrivial submodules right?

So idk how simple ring is different from simple as an R-module

[lone jacinth]

Equivalent definitions include
Every submodule being a direct summand
or
Sum of its simple submodules

[hushed bone]

🗿

[lone jacinth]

Well the ring of 2x2 matrices is a simple ring for example

But as a module it is equal to the sum of two copies of the simple module

[hushed bone]

Over a field?

[lone jacinth]

Yes

[torn harbor]

as a left module

[hushed bone]

As a module over itself???

[lone jacinth]

Yes

[hushed bone]

But how can that be

Ideals are exactly submodules

[torn harbor]

As a two sided module the definitions coincide

Submodules are left ideals if you only consider it as a left module

[lone jacinth]

Left ideals are submodules

The matrix ring has infinitely many left ideals, but only two ideals

[hushed bone]

I’m so confused. It seems like for one you’re talking about left-stuff and the other two-sided

Why would you change that

[lone jacinth]

Change what?

Modules are not two-sided

[hushed bone]

Like it seems stupid to me that simple ring is about two sided ideals and simple module is about left submodules

[lone jacinth]

What else could simple module be about?

[fierce steeple]

Subbimodules

[lone jacinth]

Simple ring is a ring without quotient rings
A simple module is a module without quotient modules

[fierce steeple]

Jk then it isn't even modules

[hushed bone]

Hmmm ok

[fierce steeple]

Anyway chmonkey dw I am in the same boat lol this has always confused me as a comm alg person

[lone jacinth]

Now it is confusing that a "semisimple ring" is not a decomposition of simple rings, but instead a ring that is a semisimple module, but I don't know what a better terminology would be

[fierce steeple]

A funny thing I guess is how like

I guess simple => semisimple for rings isn't a complete triviality even though the naming suggests it lol

[lone jacinth]

It's also not true right

[fierce steeple]

Oh maybe I am thinking of e.g. among fd k algs

[lone jacinth]

semisimple implies artinian

[fierce steeple]

Maybe some artinian condition needed

Sure

[hushed bone]

I mean this makes sense. I didn’t get why you quantify over left ideals but these don’t get you rings as a quotient

[fierce steeple]

Okay well that just makes the terminology worse lol

[lone jacinth]

Artin-Wedderburn tells you that semisimple = finite product of artinian simple rings

[hushed bone]

I’m just glad my abs are my bas

[lone jacinth]

It's not amazing, but what's the alternative
"Module-semisimple"?
Kinda long

[fierce steeple]

It seems the terminology simple ring is the bad bit as it suddenly brings in two sided stuff but ye

[ornate atlas]

Do I need to find the hobgoblin meme

[lone jacinth]

Well it just follows the convention of simple algebraic structure being one without quotients