advanced-algebra

[fierce steeple]

Like one interpretation of Ext is as homotopy classes of maps A -> B[n], at least after replacing with injectives

But you won't get something liek that for Tor

[fathom fox]

i see

[drowsy compass]

given a bunch of algebraic numbers, how can one determine if they're linearly independent over Q

[limpid horizon]

For a left exact functor F there is a recipe to construct its right derived functors RiF. If we have some other collection of functors {Gi}, how can we tell that {Gi} are also right derived functors of F?

I mean i guess i have a few questions then: what notion of "equivalence" in category theory even is that

[rose mirage]

derived functors are kan extenstions and are therefore unique up to iso

[limpid horizon]

I thought my question had something to do with “universal delta functors”

[rose mirage]

if you want it at that level of generality, then universal delta functors have a universal property (the one where any transformation from it to another delta functor is entirely determined by data from the zeroth degree - the precise details escape me), and are thus unique up to iso

although hmmm

I've kind of buried the lead there by asserting it has a universal property it's unique up to iso - but to show the property is universal you have to show it's unique up to iso anyway, although it shouldn't be too hard to use their defining property to construct an inverse to any two maps between universal delta functors

I'll have a think

[limpid horizon]

Btw, the context is I’m trying to understand why A.4 is true

[limpid horizon]

To understand A4 , i do need to learn something about delta functors or kan extensions?

[fierce steeple]

Im sure you dont. This will be a specific resolution computing Ext

[fierce steeple]

But yeah this is the point here, like the derived functors are the universal delta functors here

This should be the Koszul complex associated to x1,...,xn

[limpid horizon]

Really? Because I looked at bruns and herzog and the way they proved it was something like showing some long exact sequence holds, showing H^0 is the functor evaluated at the object, and showing H^i is zero on any injective resolution

And so my question then was why does that prove its right derived functor

Oh

[fierce steeple]

What is "it" here

Idk bit unclear to me what ur refeering to

[limpid horizon]

H_m^i(M) = H^i(M (x) Cech complex)

Smth like that

[fierce steeple]

I thought you're meant to take some limit here tbh but itll be that the Cech complex is what helps you compute "derived" quotients essentially

I can look later ig for an easier proof hm

[limpid horizon]

In a sec ill show u what im actually referring to in Bruns and Herzog it would probably be more clear

[fierce steeple]

I imagine they are basically showing you can compute derived functors via acyclic resolutions or smth

[lone jacinth]

Say you have some object M, then you can find a sequence
0 -> M -> I(M) -> C -> 0
with I(M) injective.

Then your have a long exact sequence
0 -> H0(M) -> H0(I(M)) -> H0(C) -> H1(M) -> 0

you can compare this with the right derived functor to see that R1F(M) = H1(M).

You also get H2(M) = H1(C) and R2F(M) = R1F(C), and so on. So it holds by induction

[summer quest]

Either view these as universal delta functors or better just work with derived categories or stable infinity categories

[rose mirage]

stable infinity categories is pushing it a bit boss 💔 but derived cats are a good framework for this level

[fierce steeple]

based lol

[limpid horizon]

I'm not sure if I understand this

[lone jacinth]

Okay, well you know how a short exact sequence gives you a long exact sequence in derived functors?

[limpid horizon]

yea

This was the proof in bruns herzog showing H_m(M) = H(M (x) C)

[lone jacinth]

And it vanishes on injectives, so when you apply it to this sequence you get 4 terms then 0 like I wrote above

So you get
0 -> Gm(M) -> Gm(I(M)) -> Gm(C) -> H_m^1(M) -> 0
and also
0 -> Gm(M) -> Gm(I(M)) -> Gm(C) -> H^1(M(x)C) -> 0

So the two last terms are equal

Anything specific you don't understand?

[limpid horizon]

Ok, im trying to understand what ur writing here vs what bruns and herzog are doing

firstly, last two terms are equal bc Gm(C)/im(Gm(I(M)) = H_m^1(M) and Gm(C)/im(Gm(I(M)) = H^1(M(x)C)?

[lone jacinth]

Yup

[lone jacinth]

So I'm just answering why it's enough to show that you have LES and vanishing on injectives, which is the two things b&h shows

[limpid horizon]

yeah so if we show H^i(-(x)C) vanishes on injectives, then we can write a LES like what you did in the second one right

[lone jacinth]

That's right

[limpid horizon]

This continues like:

0 -> Gm(M) -> Gm(I(M)) -> Gm(C) -> H_m^1(M) -> 0 -> H_m^1(C) -> H_m^2(M) -> 0 -> H_m^2(C) -> ...

0 -> Gm(M) -> Gm(I(M)) -> Gm(C) -> H^1(M(x)C) -> 0 -> H^1(C(x)C) -> H^2(M(x)C)-> 0 -> H^2(C(x)C) -> ...

Sorry I guess im just trying to understand the induction. Induction stuff usually gets me confused

So we know that H_m^1(C) = H^1(C(x)C) so from 0 -> H_m^1(C) -> H_m^2(M) -> 0 and 0 -> H^1(C(x)C) -> H^2(M(x)C)-> 0 we have H_m^1(C) = H_m^2(M) and H^1(C(x)C) = H^2(M(x)C) so then H^2(M(x)C) = H_m^2(M) lol

Sorry looks ridiculous all spelled out like that lol just wanted to be sure i get the idea

And this is what b&h had in mind ?

I mean i appreciate showing it this way so i can avoid any more abstract category theory

[lone jacinth]

Yeah, I think it's usually called dimension shift.

H^i+1(M) = H^i(cosygy M) for i>0

[limpid horizon]

Hmm ok

[outer lion]

Is there a map between two associative algebras that preserves the induced Lie structure, but not the algebra structure?
As in, are there two algebras A, B, and an f : A -> B such that:
f(ab - ba) = f(a)f(b) - f(b)f(a)
but not
f(ab) = f(a)f(b)

[lone jacinth]

Take for example A and B commutative

[outer lion]

oh, yeah

that's a good point

[vapid axle]

Bro wtf is ur class making u read

[limpid horizon]

This is for my masters thesis 😂

[fierce steeple]

Local homology is homology when u zoom in

Real

[mental escarp]

As in, local homology is a specific homology group associated to a local ring?

[hushed bone]

This is just local cohomology, and the technique is pretty standard for lots of functors

[vapid axle]

Ah but of course

[limpid horizon]

Lol

[limpid horizon]

The technique in the proof there, or what exactly are you referring to is "standard"?

[hushed bone]

I guess the thing about the lim is what I was referring to, but the rest of the proof is just using some universal delta functor nonsense

And effaceability

[cobalt sonnet]

does anyone know why we require the evaluation and coevaluation map to have different types like V tensor V star for one and the opposite for the other. Why not both the same. We can still formulate rigidity equations without it no?

[fiery seal]

In KPCA please try to explain the formula to center the gram matrix

[rare walrus]

We can, but it's nice that the 'associativity' aligns nicely when we write it the other way around. This is precisely the composition below

[rare walrus]

For everyone else: this question has already been answered in #linear-algebra

[lone jacinth]

C is not necessarily symmetric monoidal right. So V(x)V* and V*(x)V aren't necessarily isomorphic

And I guess it makes sense that coevaluation is the dual of evaluation

[cobalt sonnet]

yeah that makes sense

[hexed tulip]

I want to proof the abels summation formula for \sum_{n=M+1}^{N} a_{n}b_{n} , but I get a little bit a different result (It should be N+1 apparently, but I don't understand, where I did the mistkae)

[golden osprey]

Why do you think it should be $N + 1$? 
The values $a_{N + 1}$ and $b_{N + 1}$ don't even appear in $\sum_{n = m + 1}^N a_n b_n$.
Anyways the last summation I think should be $\sum_{n = M + 1}^{N - 1} A_n (\mathbf{b_n - b_{n + 1}})$.

[broken turtle]

Spamakin🎷

[limpid horizon]

next up i think i have to learn about exterior algebras

have 0 clue what thats about

[golden osprey]

What context are those coming up in for you?

[limpid horizon]

Ill show u one sec

im trying to understand this theorem

i think that underlined guy is the "exterior face ring" of a simplicial complex

there is so much there that i have no clue abt atm lol. I need to review the combinatorial stuff from the ground up first i think

it took a while for me to get some idea of how u can compute local cohomology using cech complex

[hexed tulip]

\sum_{n=M+1}^{N} a_n b_n = \sum_{n=M+1}^{N} A_n (b_{n+1} - b_n) + A_N b_N - A_M b_{M+1}, my mathbook gives me this solution

[golden osprey]

Can you send a screenshot of your book

[hexed tulip]

1.9

[cerulean cove]

Anyone use Dummitt and Foote? How’s the quality/difficulty/pacing? my grad alg 1 course is using it next semester

[limpid horizon]

I liked dummit and foote when i was using it

Its very wordy though

Its not a difficult book tho its pretty friendly

Has lots of examples etc

[cerulean cove]

oh nice

I’m coming from a book considered bad (my DE class was gallian) so that’s nice to hear

[limpid horizon]

DE?

[cerulean cove]

dual enrollment just like undergrad algebra 1

[molten vale]

Let R be a subring of S. e:S to S be an embedding of rings, such that e(R)\subseteq R. under e, we can create a chain S\subseteq S\subseteq... and R\subseteq R\subseteq... Taking union we get two rings S' and R'\subseteq S'. If S is a flat R module. Is S' a flat R' module?

[fierce steeple]

Im confused. Do you mean the chain e(R), e^2(R),...? And do you mean intersection rather than union?

[molten vale]

no its S\subseteq S (the embedding sends the first S into a subset of the second S). So I am taking union

[fierce steeple]

Okay i mean not the best notation then lol

[molten vale]

mb 😂

[fierce steeple]

And so i guess this is more of a direct limit hm

[molten vale]

I believe so (although i haven't learnt about direct limit)

[fierce steeple]

But yeah i believe this true. I believe this is saying a direct limit of flat ring maps is flat

[molten vale]

thanks, ill search up what those means

[hexed tulip]

can someone pleas explain me, how the book got the result from (1.9)

This are my calculations, but I get a different solution

[earnest latch]

-# sorry for the basic algebra question!

is it true that, working in Z/p^2Z (p a prime), the multiplicative inverses of the numbers 1 through p-1 sum to 0?

[sly rune]

yes, as long as p > 3. this is wolstenholme's theorem, see wikipedia for a good proof

[nocturne bramble]

Man, @frank timber , you've been so helpful. You might be the greatest mathematician of all time.

[hushed bone]

Facts bro.

[agile kraken]

Are there fields K, L such that L = K(a) and |Gal(L/K)| = infty ?

[hushed bone]

Yeah, the linear fractional transformations if a is purely transcendental

[agile kraken]

oh wait

you're right

xd

i thought about that earlier but in my head i had the algebraic word stuck somewhere probably haha

ty

[hushed bone]

Swag

[rare walrus]

Let $\mathbb E : \mathbb F$ be a finite Galois field extension with Galois group $\mathcal G$.

\bigskip
I have heard that, when $\mathcal G$ is cyclic, the cohomology group $H^2(\mathcal G, \mathbb E^\times)$ is isomorphic to the 'norm classes' $\mathbb F^\times/N^{\mathbb E}_{\mathbb F}(\mathbb E^\times)$.

\bigskip
Is this also true for non-cyclic extensions? Is it true for all Abelian, or solvable extensions?

[broken turtle]

Pingutjie

[glass beacon]

Young people need to know the trajectory of Free Basic and QBasic and their achievements in politics ?

[foggy galleon]

Yes, this follows simply because G is cyclic

Like H^n=H^(n+2), for n=0 one has to take the Tate group

[rare walrus]

Yippee thank you for replying Croq

Just to be clear, you're saying that this isn't true for noncyclic extensions?

(at least typically)

[foggy galleon]

Yeah, typically such result shouldn't hold I think

[rare walrus]

Great OK thank you

[foggy galleon]

Here's an example from local class field theory. If your extension is of local fields then H^2 is always cyclic

[vapid axle]

I know this probably belongs in #groups-rings-fields but it looks like it's taken. Does part c) necessarily use a) and b)? I was able to show that B is free abelian without using a) and b).

[rare walrus]

Sorry keyo I'm lookin now

Well I'm impressed that you managed that because I think (a) and (b) clearly lead to (c)

I don't know if it is meaningful to ask if you can prove it without proving (a) and (b), since all are true facts so they are logically equivalent lol

But in any case, maybe you should try proving it using (a) and (b) as intended just to see if you can figure out why they're helpful

It's pretty much just (b) frankly...

[vapid axle]

Oops sorry

I meant I used a and b lol

In my earlier message

#groups-rings-fields message

[rare walrus]

OK so you did use a and b

Let me just typeset it here

Let ${\mathbf{x_1}, \dots, \mathbf{x}_n}$ be the set of nonzero generators for $B_n = B$. Then ${\mathbf{x}_1, \dots, \mathbf{x}_n}$ is a basis for $B$. Moreover, the subgroup generated by each $\mathbf{x}_i$ is infinite cyclic, for otherwise the subgroup $\pi_i (\mathbf{x}_i)$ of $\mathbb{Z}$ is not infinite cyclic.

Wait give me five mins sorrry

OK back

[vapid axle]

Yeah I was just curious why we proved it for all m <= n, seems like we just used the case m = n

Probably need to use it in the second part of the question

[rare walrus]

We essentially did proof by induction

You needed some reference to m < n for part (b) I reckon

[vapid axle]

Oh I just fixed some m and proved it for that lol, I didn't do any inductive argument

[rare walrus]

Hm OK

[vapid axle]

There's probably some flaw in my argument then

[rare walrus]

Can you show me your proofs?

[vapid axle]

Sure

[rare walrus]

Why may we assume that the \mathbf x_i are of the form (0, ... x_i, ..., 0)? You have simply stated that you may assume it but you haven't really justified it

I think this is false

For example, consider the subgroup of Z^2 generated by (1, 1)

Indeed if you assume that all subgroups of Z^n are 'nice' in the sense you have implicitly done above, then the proof is trivial.

Glad we caught that!

[vapid axle]

So (1, 1) is contained in B_2, and \pi_2(1, 1) = 1 is a generator of \pi_2(B_2), but (1, 1) is not of the form (x, 0) or (0, x)? Is that what you mean?

[rare walrus]

Yes, and you assume in your proof for 6a that (in particular) we may choose the generator of B_2 to be of the form (0, x)

Which is false

The only element of that form in this subgroup is (0, 0)

[vapid axle]

Ohhh I see we took B = <(1, 1)>

Makes sense now, thanks

Back to the drawing board ig lol

[rare walrus]

You really don't have to assume that the \mathbf x_i are of any particular form

Use induction!!

[vapid axle]

Okay I'll try now haha

[vapid axle]

Hopefully this is better?

[rare walrus]

OK so you have some (b_1, ..., b_j+1, 0 ...)

and you've written it as a sum of things in Z^n

namely (b_1, ..., b_j, 0, 0 ...) + (0 ..., 0, b_j+1, 0 ...)

but then you assume without mentioning it nor proving it that this first thing is in B_j when you haven't even proved it's in B

You're making the same assumption as you did in your first proof, just now using inductive language

Again, you cannot assume that \mathbf x_{j+1} is of any particular form except that it has those trailing 0s . Start by fixing the \mathbf x_i once and for all, and don't assume it's of a nice form!

[vapid axle]

Oh yeah you’re right

[rare walrus]

@vapid axle Hint: everything in B_{j+1} differs from something in B_j by some multiple of \mathbb x_{j+1}

[spice quiver]

Heh B_j

[arctic raft]

@arctic raft

[hushed bone]

#prealg-and-algebra

[urban granite]

💀

this channel is not for solving preuni algebra problems, go #prealg-and-algebra

as you can see the description of this channel, here is for things like homological algebra, comalg, etc

[cerulean cove]

ok😭

[granite cradle]

How are you 15 and have a pending pg

[arctic raft]

Huh

[granite cradle]

Yer bio

[arctic raft]

What about my bio?

[granite cradle]

I thought 15 was your age

[arctic raft]

It is

[granite cradle]

And you are a pending postgrad at 15?

[arctic raft]

So

[granite cradle]

I was just wondering if it is possible

[arctic raft]

Oh

[granite cradle]

If so then you are crazy smart imo

[arctic raft]

I mean, I have a 3.2

[spice idol]

I also have 3.2 euros to my name

[limpid horizon]

Congrats bro 👏👏👏

[spice idol]

Ikik I'm saving up 🔥 🔥 🔥

[distant harness]

The "pending postgraduate" role means that the user has pushed the button on the welcome screen saying they intend to apply for the "Postgraduate Math" role. It doesn't imply that the person behind is about to start postgraduate studies.

[granite cradle]

But it does give the impression that they are

[distant harness]

Many have it because they pushed the button and then thought better of it when the bot sent them a questionnaire. It's not terribly well communicated (that is, it's practically unknown) that they're then expected to go back to the welcome screen and downgrade to "Undergraduate" manually,

[distant harness]

Bad role naming, but not the user's choice.

Feel free to go to the meta channels and heckle the powers that be over the role naming and confusing joining experience -- but there's no good reason to give new joiners grief over it.

[granite cradle]

Well it is their fault for trying to pick roles they shouldn’t be
Also I’m pretty surprised people are just allowed to choose these roles(except for pg one ofc)

[distant harness]

It is pretty opaque to new users what will happen when they push the buttons on the welcome screen. (It looks like it has improved slightly recently, but the names of the roles you end up with is still not explained well (due to length constraints on the button texts), so many people with the Pending Postgrad roles genuinely have no idea how they got it or what they could do to avoid having it.

[granite cradle]

[ornate atlas]

It’s also just really not that deep

[limpid horizon]

Lol

[spice idol]

"put the fries in the bag bro"

[white bolt]

Let $Z$ be a finite additive group, and $A$ a subset of $Z$. I can't manage to show that there exists a subset ${v_1, ..., v_d}$ of $Z$ such that $d = O(\log{\frac{|Z|}{|A|}})$ and $$|{a + e_1 v_1 + ... + e_dv_d, a\in A, e_1,...,e_d\in {0, 1}}|\geq \frac{|Z|}{2}$$

[broken turtle]

TimourX

[white bolt]

I tried taking $V={v_1,...,v_d}$ a random subset of $Z$ with uniform distribution, and compute $$\mathbb{E}(|{a + e_1 v_1 + ... + e_dv_d, a\in A, e_1,...,e_d\in {0, 1}}|)$$

[broken turtle]

TimourX

[charred magnet]

Is there a good, standardized notation for a map $A \to A \times A$, sending $a \to (a,a)$? I'm tempted to call this the product inclusion map, and google shows $\pi$ used a bunch, but I'm not sure if it's standard. thx <3

[broken turtle]

haseeb

[south orbit]

diagonal map or diagonal embedding if i remember

[weak lodge]

Usually notated as $\Delta:A\to A\times A$

[broken turtle]

harmacist

[vapid axle]

Can somebody give me an example of problems involving tensor products where the tensor product is not explicitly mentioned in the problem statement

I'm trying to see how it's useful

[fierce steeple]

I wouldn't really view it as a thing that "solves problems" but more as a super ubiquitous/useful construction more generally.

But one type of example is this: often it is desirable to extend scalars to simplify things, which uses a tensor product. For example, often it's enough to prove things which depend on a field by "tensoring up" to the algebraic closure and using facts about algebraically closed fields

[digital parcel]

if X and Y are affine varieties, then the coordinate ring of X \times Y is the tensor product of their coordinate rings

[vapid axle]

I see, I guess I need more exposure then

Is (M \otimes_A N)_k isomorphic to M_k \otimes_k N_k?

[lone jacinth]

Yes, it's something that can be seen from
k = k (x)_k k = k (x)_A k
and swapping around the order of tensor factors.

[broken turtle]

bostock

[scarlet ermine]

@willow seal
This argument isn’t quite right because V(I) irreducible does not imply I is prime.

For example, even for C the complex numbers, in C[x]/(x^2), V(x^2) is just the singleton of the prime ideal {(x)} (ie the point 0 on the affine line), which is an irreducible topological space, but the ideal (x^2) is not prime (eg x*x is in it, but x is not). This example is really taking advantage of V(I) = V(rad I). On the other hand, if you somehow knew I is radical (which is usually tricky), it is true that V(I) irreducible implies I prime.

So to argue that I is maximal/prime, I would instead try to show that Z[x]/I is a field/integral domain. Now, to compute Z[x]/I, we can do something like the third (?) isomorphism theorem, ie quotient one relation at a time, since
I = (5) + (x^3+2x-3),
we can say
Z[x]/I = (Z[x]/(5)) / (x^3+2x-3)
(where now the polynomial is really the image of the polynomial in the quotient space). A more familiar description of this ring is
(Z/5Z)[x]/(x^3 +2x-3).

Now, you need to see if this ring is a field/integral domain/neither.

also, you have to be careful about whether an ideal is proper or not if you want to show something is not maximal by exhibiting a larger ideal. Maximal means no larger proper ideals, so you’d need to be very careful that, eg, (5, x, x^3 + 2x -3) isn’t accidentally the entire ring!

[willow seal]

thanks for the reply, re the second para this seems to contradict you

[spice idol]

It means that rad(I) is prime

[willow seal]

^ this is for algebraically closed fields like C

[golden osprey]

V(x^2) = V(x) but (x) is prime and (x^2) isn't

[scarlet ermine]

The tricky bit is that I(V(J)) is not necessarily J, it’s the radical of J. Whenever you get ideals by doing I(Y), you always get radical ideals, but general ideals need not be radical

[willow seal]

so if i understand irreducible locus => the radical of the ideal is prime

[scarlet ermine]

yup

[fierce steeple]

If and only if

[willow seal]

good point thanks

[willow seal]

re third para, trying to understand the use of the 3rd iso thm here. basically you did $\mathbb{Z}[x]/I\cong [\mathbb{Z}[x]/(5)]/[I/(5)]$, then for the top quotient you're basically modding all the coefficients of integer polynomials mod 5, and for the bottom quotient the ideal $(5)$ gets quotiented out, and modding by 5 does nothing to the coefficients of $x^3+2x-3$? something like that?

[broken turtle]

bostock

[scarlet ermine]

So in the top quotient I’m using
Z[x]/(5) is isomorphic to (Z/5Z)[x], where the isomorphism is sending a coset
f + (5)
for f in Z[x] to a polynomial in (Z/5Z)[x], where I’m replacing each coefficient of f with an element of Z/5Z (using the projection Z->Z/5Z). (you can check that this is a well-defined isomorphism)

Now, under this isomorphism, the ideal I/(5) inside Z[x]/(5) becomes an ideal inside (Z/5Z)[x]. To see which ideal it becomes, you track where generators of I/(5) go under the isomorphism. The ideal I/(5) is generated by the coset
x^2 + 2x - 3 + (5),
and the isomorphism
Z[x]/(5) ≈ (Z/5Z)[x]
sends this coset to the polynomial
1x^2 + 2x -3
but now I’m considering 1, 2, and -3 as elements of Z/5Z (if you want, you could but bars on top to indicate that they are equivalence classes/cosets for 5Z).

Essentially, at first when we do the quotient you put bars on entire polynomials, but under the isomorphism we only have to put bars on the coefficients

[willow seal]

ok i see what you mean, thanks for the explanation

[scarlet ermine]

yeah no problem!

[lyric berry]

It's used to construct Dehn invariants of polyhedra.

Also tensor products and direct sums of vector bundles can be used to define a pretty important topological invariant called k-theory.

[cobalt sonnet]

Does the center of a Frobenius algebra inherit the structure of a Frobenius algebra in some way?

I have been thinking about the example of the group ring but I can't figure it out

Multiplication works out nicely

i don't get how the comultiplication would work in the center though

the usual frobenius structure is just delta(g)=sigma_h gh^-1 tensor h

[lone jacinth]

Consider k a field of characteristic 2 and G = Q8 the Quaternion 8 group. Then the center of kG should be
k[x, y, z, w]/(x, y, z, w)^2, which is not Frobenius.

[tidal zealot]

Or does completeness forces the field to be of characteristic zero

[lone jacinth]

Being ordered implies characteristic 0. You can try to prove it yourself

[jaunty bison]

Is the following statement true? Let A be a commutative ring, M an A-module and p a prime ideal. If m_1,...,m_r in M generate M_p as an A_p-module, then there exists f in A \ p such that m_1,...,m_r generate M_f as an A_f-module?

I tried to come up with some example about this, but could not.

[lone jacinth]

Consider X = M/(m1, ..., mr). Then m1, ..., mr generating Mp is equivalent to Xp being 0.

Xp is 0 if for every x in X there is an f in A\p such that fx = 0.

If X is finitely generated you can pick a single such f (just take the product of each for a finite generating set).

If X is not finitely generated it may fail. For example take A = Z, M = Sum_p Z/p

[jaunty bison]

Is there some canonical example? I have difficulties understanding this

[tidal zealot]

Why does no book mentions this?

[lone jacinth]

I'm not sure I believe that no book mentions this.

It's here on Wikipedia for example
https://en.m.wikipedia.org/wiki/Ordered_field#:~:text=An ordered field has characteristic,order of the rationals themselves.

[lone jacinth]

The example I gave would be a good start

The sum of Z/p over all primes p. Localizing at a given prime just gives Z/p for that specific prime.

Localizing at an element f gives the sum over Z/p for every p that doesn't divide f, which is an infinite sum

[cobalt sonnet]

But over C this should be true right?

[cobalt sonnet]

For context I am studying TQFTs and only a commutative frogenius algebra defines a TQFT so I was wondering if I can always use the center as a frobenius algebra

[lone jacinth]

Specifically for group algebras?

Over C they are always semisimple, so the center is just C^n for some n

In fact n=the number of conjugacy classes of your group

[cobalt sonnet]

Yeah, the n is just the number of conjugacy classes

I can give it a Frobenius algebra structure but does it inherit one from the algebra itself?

I understand how the algebra restricts but the coalgebra part is not clear to me

[lone jacinth]

Okay, I see what you're asking.

Yes, for the group algebra in characteristic 0 you should be able to just directly restrict the Frobenius structure to the center

I wouldn't think this works for arbitrary Frobenius algebras though

For example consider the cyclic Nakayama algebra of rank 2 with radical square 0. I.e. algebra with basis e1, e2, a, b with
e1^2 = e1, e2^2 = e2
e1 a = a = a e2
e2 b = b = b e1
and all other multiplications 0.

Then the center is C[a, b] which is not Frobenius

[cobalt sonnet]

but i can't do it explicitly

like if i take a conjugacy class

C= sum_{g \in C} g

Delta( C) = sum_ {g \in C} \sum_{h \in G} gh^(-1) tensor h

But how do I express this as an element of Center tensor Center?

[cobalt sonnet]

This is my attempt so far

[lone jacinth]

So I'm not so used to the comultiplication definition of Frobenius algebra.

But for the non-degenerate bilinear form where you multiply and then take the coefficient at the identity restricts to a non-degenerate bilinear form on the center.

This should translate to the comultiplication by the bilinear form induces an isomorphism A -> A^*, then use this isomorphism on the dual of multiplication. Maybe this doesn't translate directly to restriction of comultiplication...

[cobalt sonnet]

What would the copairing be in this case?

Like the copairing for k[G]
is just

1 to sum_{g in G} g^-1 tensor g

[cobalt sonnet]

what would it be for the center?

[cobalt sonnet]

because the pairing is non degenerate there should be a copairing and if i can find that i can work out the comultiplication too

[lone jacinth]

So I guess we would need to compute it.

So for a conjugacy classes C, let C^-1 be the conjugacy classes of an inverse of one of the elements in C.

Then the pairing takes (C, D) to |C| when D = C^-1 and 0 for other conjugacy classes.

So the isomorphism A^* -> A takes C^* to C^-1/|C|.

And then the dual of the multiplication would be some horrible formula relating how many times a conjugacy class appears in the decomposition of a product of conjugacy classes

[cobalt sonnet]

yeah it should be 1 to C_{I} tensor C_{i}^{-1} where each C_{i} is weighed by the square root of the number of its elements

[cobalt sonnet]

for the last part you are saying

take A to A star then A star tensor A star and finally A tensor A, right?

[lone jacinth]

Yeah

[cobalt sonnet]

is there some standard name for how many times a conjugacy class appears in the decomposition of a product of conjugacy classe?

[cobalt sonnet]

So if I let

C_i x C_j = sigma_{h} N^{h}{i,j} C{h}

the dual map should be

C_{k}^{} to sigma_{i,j} {N_{i,j}^{k} C_{i}^{star} tensor C_{j}^{}

right?

[cobalt sonnet]

I checked, this is correct

I am exploring the properties of these constants, $N_{i, j}^{k}$ and they are interesting

$N_{i, j^{-1}}^{0}=C_{i} \kronecker_{i,j}$

Commutativity implies symmetry in the last two

and the rest i am trying to figure out

[acoustic shoal]

I think you should format your messages in latex! It will be easier to read

Enclose messages in $.................$

[broken turtle]

MINER MAN

[cobalt sonnet]

Why do we care about frobenius algebras in representation theory?

[sharp socket]

We are working with a lattice L(B), where B is a basis matrix in the lattice L(B). The Smith Normal Form of B=SAT, where S and T are unimodular matrices. Is the shortest vector in the lattice L(B) the same Euclidean norm as the shortest vector in L(A)?

I think L(A) and L(B) are isomorphic, which would hint at the answer being yes, but I don’t know if the answer is actually no.

Two matrices with the same SNF (ex: A and B) are similar, I think. Similarity is an equivalence relation on the space of square matrices.

The nonzero $d_{i}$ elements, together with the number of $d_{i}$ which are zero, form a complete set of invariants for the module. Explicitly, this means that any two modules sharing the same set of invariants are necessarily isomorphic. I think A and B are very likely isomorphic, since A and B have the same invariants in their SNF.

[broken turtle]

🌕

[sharp socket]

I think L(A) = L(B). AI disagrees with me but maybe AI is wrong

[bitter kayak]

Is the kernel of a chain homotopy equivalence an acyclic complex?

I think the answer is no but what is a counter example?

[lone jacinth]

0 -> Z -> Z
v v v
Z -> Z -> 0

maps being 0 or identity.

[bitter kayak]

Oh that’s very simple, thanks!

[fierce steeple]

The failure of this is an example of why one should consider mapping fibres / cocones instead if you care up to quasi-iso

[bitter kayak]

Right, but both of those are just the mapping cone right?

[lone jacinth]

Yeah a map is a homotopy equivalence iff the cone is contractable.

[fierce steeple]

Up to shift

The mapping cocone /fibre is more like a kernel and cone/cofibre is the kernel, but they agree up to shift

[bitter kayak]

Ah right

[limpid horizon]

Not totally sure where to post this (this is from an algebra book), but im just trying to interpret whats going on here

This is just, simplicial homology as you'd learn in alg top is it? Like, is the Ci there the free Z-module with basis of faces [vi1, vi2, ... vi(i+1)] (dimension i)

The fact that they are tensoring with a group G and calling that reduced simplicial homology i have no idea where thats coming from

[spice idol]

Tensoring with an abelian group G is simply going from a direct sum from copies of Z to a direct sum of copies of G

[lone jacinth]

You'll notice that in addition to the normal way to define simplicial homology, they have this extra term C_{-1}. This is what makes it reduced homology as opposed to just homology

The tensoring with G gives you homology with coefficients in G

[spice idol]

It's an algebraic shorthand for just the construction of homology but \bigoplus GF rather than \bigoplus ZF

You've got stuff like the universal coefficient theorem from homology that relates the homology with coefficients to the regular homology iirc

[lone jacinth]

So in the phrase
"reduced simplicial homology of Delta with values in G"

"reduced" is the C-1 thingy, "simplicial homology of Delta" you know, and "with values in G" is the tensor with G thing

[spice idol]

In the particular case that G is the underlying group of some ring R, you get a natural chain complex of R-modules, and therefore also that the homology with coefficients in R is an R-module

This is an example of extension of scalars

[limpid horizon]

when I see someone say extension of scalars the way I'm thinking of it is that you have R subring of S, M an R-module and then S(X)R M is an S-module which is like extension of scalars of M from R to S

But I guess generally u dont need R subring S

just R->S ring map?

and any ring R has Z->R map

[spice idol]

Yes exactly

[spice idol]

Lol it's in the banner

[limpid horizon]

whats the banner?

[lone jacinth]

It's the one for cohomology though

[spice idol]

Of this server

[spice idol]

Close enough, lol

[fierce steeple]

Regular homology being relative to the sphere spectrum

[spice idol]

This scares me

Omfg finally I hate the shifting lemma and modularity why can't it just be nice obvious like permutability auegaheh

[urban granite]

my profile banner has that cohomology one, special case tho

[spice idol]

I saw!!

[fierce steeple]

What are these

[spice idol]

In universal algebra, congruences play the role of normal subgroups, but are equivalence relations. As such, they can be composed. If, for all congruences R and S on some algebra A, we have R • S = R • S, then A is said to be congruence-permutable (this is analogous to the fact that NM = MN for groups, and they are indeed congruence permutable). They are incredibly well behaved and you get many nice theorems, and everyone is happy.

There is a nice characterisation of these, that being a variety of algebras (class of algebras closed under embeddings, quotients and arbitrary products) is congruence permutable (every algebra in it is congruence permutable) if and only iff there is a term p(x, y, z) such that every algebra in V satisfies the equations p(x, x, y) = p(y, x, x) = y. The variety of groups is such an example, with term xy^-1z.

Congruence modularity is whenever the lattice of congruences satisfies the so-called modular law. This seems to be the least requirement for the lattice of congruences to behave nicely in some way. There exists a similar characterisation in terms of the existence of some terms, but it is much more complicated and more pulled out of someone's ass and I hate it. Then you also have something called the shifting lemma, which is incredibly useful and equivalent to modularity but the proof of all those equivalences spans like 2 full pages of out-of-ass-pulling of terms. And specifically constructed congruences.

Safe to say I'm not the biggest fan

[fierce steeple]

Oh sure cool heh

[spice idol]

"I ain't reading allat"

[fierce steeple]

Mo i mean i did read it

Ig like i just dont know much about UA

[spice idol]

What I'm doing rn, commutator theory, was originally developed for congruence permutable algebras and everything works nicely there, but the moment you go into modular varieties everything devolves into messes of terms and I AM STRUGGLING but in the end it is quite nice

The results I mean

The theory itself is doable the proofs are just convoluted as hell because every UA person is a madman

[spice idol]

No one does here

[tidal zealot]

I am working with these formalism of peano axioms.

And I am trying to prove this theorem

But everywhere I have seen the proof and construction of addition it uses this property
$$m + \nu(n) = \nu(m+n)$$

[broken turtle]

Prïyanshü

[tidal zealot]

Like here

But I can't prove this condition from the other conditions in theorem

Heuristically it should be doabe , to show this from using (iv) and (i) of aforementioned theorem.

Can anyone show this?

[kindred maple]

Cuz the way I have seen it, its defined with $0 = {\varnothing}$

[broken turtle]

Khush

[kindred maple]

$1 = {\varnothing, 0}$

[broken turtle]

Khush

[kindred maple]

and so on

[kindred maple]

then we can show that the \mathbb{N} defined here is bijective to the \mathbb{N} I defined here with the sucessor function given by $\nu(n) = n \cup {n}$

[broken turtle]

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

[woven loom]

Horrid

[limpid horizon]

Fr

Ok like its sort of interesting but

I would much rather spend my time thinking about other things

[spice idol]

Only if you're not working in Malcev varieties!

[woven loom]

I’d like to prove I am working in one

[spice idol]

Well, just show you can construct a term p(x, y, z) such that p(x, x, y) = p(y, x, x) = y

Hi jagr!

[lone jacinth]

You would do this using induction (N1):

Notice
n + v(0) = n+1 = v(n)
assume
n + v(m) = v(n+m)
then
n + v(v(m)) = n + v(m+1) = n + ((m+1)+1)
using associative this is equal to
( n + (m+1)) + 1 = v(n + v(m)).

This means that the subset of N such that n+v(m) = v(n+m) satisfies (N1), so is all of N.

[spice idol]

And if you weren't just jesting, what variety are you talking about anyways?

[woven loom]

We don’t know it’s a variety

[spice idol]

That's okay it also works in quasivarieties and prevarieties

Just show it contains the free algebra in 3 generators

[woven loom]

Good thing this one definitely doesn’t

[spice idol]

Lmaooo

What class of algebras is this about

[supple grail]

Isn't that provided to you just in the next page (line 9 from below in pg 33)? Also this is a question more appropriate for #proofs-and-logic

[slate matrix]

I hope this is the right channel to post in. Does anyone here know how to prove that if U, V, and W are R-modules over a commutative ring R with unity, then U ⊗ V ⊗ W is isomorphic to U ⊗ (V ⊗ W) (here, U ⊗ V ⊗ W represents a trilinear map from U x V x W)?

In particular, I want to first find a bilinear map from U x (V ⊗ W) to U ⊗ V ⊗ W and first showing it is well-defined, but I have a bit of trouble doing so. Can anyone help? Thanks.

[limpid horizon]

I think the way you asked it is a bit of the wrong way round

[digital parcel]

fixing an element u gives you a bilinear map V ⊗ W to U ⊗ (V ⊗ W)

[limpid horizon]

Like im pre sure u prove tensor product is associative to justify the notation U (x) V (x) W no?

[digital parcel]

so this gives you a map U ⊗ (V ⊗ W) to U ⊗ V ⊗ W

then you can construct a map U ⊗ V ⊗ W to (U ⊗ V) ⊗ W

[rose mirage]

judging by their definition using trilinear maps, they could be defining U (x) V (x) W as the quotient of U x V x W by the trilinear relations

I do agree it's a tad odd

[limpid horizon]

Oh ok yeah

[digital parcel]

then you show that these two maps are both inverse

[limpid horizon]

Can someone help me parse whats happening there

[slate matrix]

I'm looking for a specific bilinear function to use, so given that (x, y) is an element of U x (V ⊗ W), what would it be mapped to precisely?

[limpid horizon]

Btw thm 3.5.6 says Hi_m(M) = Hi(M (X)R C) where C is modified cech complex of R

Anamono, back to cech complex localization bs again lol

I stg i need to review that (ST)^-1A = U^-1(S^-1A) exercise from a/m again

I dont know why i dont feel comfortable with it

[digital parcel]

probably something like
[ H^i_{\mathfrak m}(R) \stackrel{3.5.6}{\cong} H^i(R_{\mathfrak m} \otimes_R C^\bullet) \cong H^i(C^\bullet_{\mathfrak m}) ]

[broken turtle]

anamono

[digital parcel]

localization commutes with quotient so $H^i(C^\bullet_{\mathfrak m}) \cong H^i(C^\bullet)_{\mathfrak m}$

[broken turtle]

anamono

[limpid horizon]

commutes with quotient?

[digital parcel]

yeah

your cohomology modules are quotients

[limpid horizon]

oh ok yea I think i interpreted that wrong then. When I saw that I was like "cohomology commutes with localization?"

is that not what it is

[digital parcel]

this result tells you that local cohomology commutes with localization

actually cohomology and localization should always commute

[limpid horizon]

ya the fact this is local cohomology isnt making a difference here no

[digital parcel]

because your cohomology module is a quotient of image and kernel, both of which localization commutes with

yeah it makes no difference

hmmm actually

actually yeah i think it should be fine but idk maybe wait for someone else to pitch in

[limpid horizon]

When we are thinking about what is commuting here, is it 1) localize the chain complex then take homology vs 2) take homology then localize?

[digital parcel]

yeah

okay i guess i shouldnt say localization commutes with local cohomology

i mean it's true but it doesn't follow from this

you have to do a little more work

"localization commutes with local cohomology" means if i have some multiplicative set $S$, then
[ H^i_{S^{-1}I}(S^{-1}R) \cong S^{-1}H^i_I(R)]

or something like that

[broken turtle]

anamono

[limpid horizon]

hm ok

[digital parcel]

the point of this is to show that computing local cohomology at a maximal ideal is the same as computing the cohomology of the modified cech complex

[limpid horizon]

hm

at a maximal ideal referring to what

Rm?

[digital parcel]

at this maximal ideal

okay actually i guess bruns-herzog only does local cohomology with support in the maximal ideal m

you can do local cohomology with support in any ideal

so when i say "local cohomology at m" i just mean "local cohomology with support in m"

[vapid axle]

Could we also see this via $(x \otimes m) \otimes (y \otimes n) \mapsto xy \otimes (m \otimes n)$ and $x \otimes (m \otimes n) \mapsto (x \otimes m) \otimes (1 \otimes n)$? Granted I haven't checked if they're well-defined yet (or even linear)
but I know that they're inverses

[broken turtle]

okeyokay

[limpid horizon]

tbh im not too sure what that terminology means. I know that for an R-mod M, Supp(M) is the set of prime ideals of R so that Mp is not 0

[digital parcel]

when you defined local cohomology as you know it now, you took the limit $H^i_{\mathfrak m}(M) = \varinjlim_n \text{Ext}^i(R/\mathfrak m^n, M)$ right

[broken turtle]

anamono

[digital parcel]

you can do this for an arbitrary ideal I, taking the limit over powers of I instead of powers of m

this limit, denoted H^i_I(M), is called the "local cohomology with support in I"

[limpid horizon]

Yeah ive seen that defn but the way i was looking at it atm was as the right derived functors of I_m(M) = {x in M st m^kx=0 for some k}

[digital parcel]

yeah that's fine replace m with I

all these definitions are equivalent

[limpid horizon]

yea

[digital parcel]

which is the nice thing about local cohomology

[fierce steeple]

P cool

[limpid horizon]

so the right derived functors of that guy is "local cohomology with support in m"

[digital parcel]

yeah

$\Gamma_I(M) \cong \varinjlim_n \text{Hom}_R(R/I^n, M)$

[broken turtle]

anamono

[digital parcel]

so then you take the right derived functor and you get $H^i_I(M) \cong \varinjlim_n \text{Ext}^i_R(R/I^n ,M)$

[limpid horizon]

how does this relate to support of module in the sense of Mp is not 0 then p is in support of module

[broken turtle]

anamono

[digital parcel]

this im not too sure about actually, i asked this before to some prof but forgot the answer

https://homepages.math.uic.edu/~bshipley/huneke.pdf

from here

[limpid horizon]

Ok now im confused because: we want to compute H^i_mRm(Rm). So for that, we are considering Rm as an Rm-module? But then 3.5.6 using the tensor product thing is tensoring over R not Rm

The statement in 3.5.6 is:

M an R-module. Then

Hi_m(M) = Hi(M (x)R C)

This was one of those annoying ass days where i sat here for like 2 hours and understood barely anything

[digital parcel]

R_m as an R-module

[limpid horizon]

But mRm is not an ideal of R

[digital parcel]

pull it back to R

[limpid horizon]

Ok so we want to compute Hi_m(Rm)

It was using some theorem that says if M is an R-module and (R,m) is local then M is CM iff Mm is CM

But it didnt say if its considering Mm as an R-module or an Rm-module

[digital parcel]

likely M_m as an R_m module

this is a pretty common form of a statement

see eg this

[urban granite]

atiyah

[digital parcel]

i mean actually either way

you have that $H^i_{\mathfrak m}(M) \cong H^i_{\mathfrak m R_{\mathfrak m}} M_{\mathfrak m}$

[broken turtle]

anamono

[digital parcel]

for any R-module M

[limpid horizon]

Ok so we tryna compute Hi_mRm(Rm). How does 3.5.6 work here now?

Ill show u 3.5.6

[spice idol]

"were tryna compute himmrmrmrmmrm"

What is math even about

[digital parcel]

i mean anyway you can consider R_m as an R-module

[limpid horizon]

Idek bruh

[digital parcel]

yeah

so just consider R_m as an R-module

you have a map R -> R_m which is the localization map

this is a ring hom

now m is an ideal of R

okay let me state this more generally for just modules

there's probably some noetherian assumption needed here

if i have a ring hom $A \to B$, $I$ an ideal of $A$, and $M$ an $S$-module. Then I have an isomorphism
[ H^i_I(M) \cong H^i_{IB}(M) ]

[broken turtle]

anamono

[digital parcel]

in this case replace A with R, B with S, and I with m

to be honest i think his notation should've been H^i_m(R_m) instead but whatever

see here, in the proof of theorem 3.5.7

[limpid horizon]

Thx a lot i will look into it

[digital parcel]

Np

https://homepages.math.uic.edu/~bshipley/huneke.pdf

See here as well

[vapid axle]

Let $A$ be a local ring and $\mathfrak{m}$ its maximal ideal. Let $k = A/\mathfrak{m}$ be its residue field, and let $M$ and $N$ be $A$-modules. I am trying to verify that $k \otimes_A (M \otimes_A N) \cong (k \otimes_A M) \otimes_k (k \otimes_A N)$. First, I am trying to show that the map $f: k \otimes_A (M \otimes_A N) \to (k \otimes_A M) \otimes_k (k \otimes_A N)$ given by $x \otimes (m \otimes n) \mapsto (x \otimes m) \otimes (1 \otimes n)$ is well-defined (I have found an inverse for $f$, but I will postpone verifying its well-definedness until I receive confirmation that $f$ is well-defined) as follows: fix $x \in k$. I want to show that the map $\mu: M \times N \to (k \otimes_A M) \otimes_k (k \otimes_A N)$ given by $(m, n) \mapsto (x \otimes m) \otimes (1 \otimes n)$ is bilinear, in order to get a well-defined map out of $M \otimes_A N$ which I can use to construct the desired map. It should be bilinear; for instance,
\begin{align*}
\mu(m_1 + m_2, n) &= (x \otimes m_1 + m_2) \otimes (1 \otimes n) \
&= (x \otimes m_1 + x \otimes m_2) \otimes (1 \otimes n) \
&= (x \otimes m_1) \otimes (1 \otimes n) + (x \otimes m_2) \otimes (1 \otimes n) \
&= \mu(m_1) + \mu(m_2)
\end{align*}
however I am not sure how to verify, say $\mu(am, n) = a \mu(m, n)$ for $a \in A$. The issue is that $(k \otimes_A M) \otimes_k (k \otimes_A N)$ is a $k$-module, whereas $a \in A$. If $s \in k \otimes_A M$ and $t \in k \otimes_A N$, am I to interpret $a \cdot s \otimes t \in (k \otimes_A M) \otimes_k (k \otimes_A N)$ as $\pi(a) \cdot (s \otimes t)$, where $\pi: A \to A/\mathfrak{m} = k$ is the canonical projection?

[broken turtle]

okeyokay

[hushed bone]

Too complicated

(k (x)_A M) (x)_k (k (x)_A M) = k (x)_A (M (x)_k k) (x)_A N = k (x)_A M (x)_A N

[vapid axle]

thanks bud

I'll try to work it out

[hushed bone]

All it requires is proving tensor product associativity over different rings

And that A (x)_A M ≈ M

[limpid horizon]

Now im confused on why all this was relevant (i.e my question now is: why did we first localize R at m?). Goal: Determine when R = k[x1, ... xn]/I is cohen-macaulay. R is a local noetherian ring with maximal ideal (x1, x2, ... xn). We can detect if its CM by considering local cohomology H^i_m(R).

By thm 3.5.6, if we have M an R-module, and C is the modified cech complex on x = x1,x2,..xn where x is a system of parameters of R (i.e, the x1, ... xn generate an m-primary ideal? I have not looked into systems of parameters much yet, not sure how they are relevant) then H^i_m(M) = H^i(M (x) R C)

so in our case R = k[x1, ... xn]/I, we cannot simply do H^i_m(R) = H^i(R (x)R C) = H^i(C)?

[digital parcel]

this is my gripe with his notation though i may also be missing something

he uses $H^i_{\mathfrak m}(R)$ to denote $H^i_{\mathfrak mR_{\mathfrak m}} (R_{\mathfrak m})$

[broken turtle]

anamono

[limpid horizon]

yea only once he started talking about stanley reisner ring example

in thm 3.5.6 he doesnt do that

[digital parcel]

replace M here with R_m

okay i should say though. that theorem assumes you're taking a map between local rings

though the stanley-reisner ring isnt necessarily local

[digital parcel]

hence where this comes to play

[limpid horizon]

doesnt it say it is a local ring in the blurb

[limpid horizon]

*local

[digital parcel]

it's *local but idk what that means

[limpid horizon]

i looked it up i think its just a graded version

[digital parcel]

oh okay

then yeah you can just use that theorem then

[digital parcel]

lol okay

never seen that used before but sure

[limpid horizon]

like this?

[fierce steeple]

Worst possible naming scheme

[limpid horizon]

why did he localize and do all that mumbo jumbo

[digital parcel]

here you need to replace R with R_m

which is again why i think his notation sucks

you have an isomorphism $H^i_{\mathfrak mR_{\mathfrak m}}(R_{\mathfrak m}) \cong H^i_{\mathfrak m}(R_{\mathfrak m})$

[broken turtle]

anamono

[limpid horizon]

ok forgetting his notation just the usual notation, why cant given R = k[x1, .. xn] / I stanley reisner ring, H^i_m(R) = H^i(R (x) R C) = Hi(C)

why do i need to consider Rm at all

[digital parcel]

it's often nicer to reduce to showing the localization of a ring has some property

i dont know anything about stanley reisner rings so i cant comment on that in this specific case

but in general it's easier to check things locally (i.e., localized at prime or maximal ideals)

[limpid horizon]

whats the difference here, is the cech complex difference?

[limpid horizon]

does that formula i wrote hold?

[digital parcel]

the difference is that $C^\bullet$ is a complex of $R$-modules. but you want to look at it as a complex of $R_{\mathfrak m}$-modules

[broken turtle]

anamono

[digital parcel]

hence why you take $R_{\mathfrak m} \otimes_{R} C^\bullet \cong C^\bullet_{\mathfrak m}$

[broken turtle]

anamono

[hushed bone]

No these are the same thing

Well

Okay no they aren’t but like

Nvm

[digital parcel]

yeah i mean if you're just working over R and not R_m then sure this formula holds

but it's not very useful

because that's basically the definition of local cohomology of R

[willow nebula]

If the decomposition of a k[G] module into indecomposables yields some non -projective modules in that decomposition is there something one can say about the k[G] module itself

Basically I want to know if there are consequences to having non-projective summands in the decomposition

And references to those kinds of results

[lone jacinth]

Then that kG module is not projective either

[willow nebula]

Is it because of some tor ext computations?

[lone jacinth]

Not really

[willow nebula]

Can you gimme an outline of why that's the case

[lone jacinth]

A direct summand of a projective is always projective

Easy exercise just using the definition of projective

[willow nebula]

Right

Projectives are direct summands of free

So it follows

I think

[lone jacinth]

Yes, you can prove it like that as well

[willow nebula]

Ohkay so is there something that happens when k[G] modules are not projective?

Like from a representation theory standpoint

[limpid horizon]

Example; Rm (x)R Rx1 = (Rx1)m, as an Rm-module?

[digital parcel]

Yes; for any $R$-module $M$, there’s an isomorphism $S^{-1}R \otimes_R M \cong S^{-1}M$ as $S^{-1}R$-modules

[broken turtle]

anamono

[digital parcel]

This is, eg, Atiyah-Macdonald chapter 3

[lone jacinth]

So it's in some sense "rare" for modules not to be projective (and in some sense not).

When G is a finite group and the characteristic of k doesn't divide |G|, then every module is projective.

In the other cases, most modules are not projective. So I would say the special thing is what happens when the module is projective

[limpid horizon]

Thats cool

[willow nebula]

Yeah what you're saying is the semisimple case of representations right?

In the case that k[G] becomes semisimple when char k dosent divide |G|

[solemn thunder]

If you guys are so smart with your algebra, then what is x if 2x=4?

[spice idol]

Yes, everything is nice and everyone is happy

Even happier when k is alg closed

[willow nebula]

I'm assuming alg closed in the background

So I'm interested in when the characteristic is bad

Do the non projective summands tell us something in that case

[spice idol]

Something something Brauer stuff (I'm not at that part of Serre's book yet)

[willow nebula]

Brauer characters?

[spice idol]

That's how you can study it I think, but I'm not sure what happens with the representation theory of the group algebra

[willow nebula]

Ohkay thanks
Btw is it possible to tell anything about a ring about whose modules we know the following statement to be true:
A module is projective iff it's injective

Apparently it holds when the ring is a group algebra (and I think always even in bad characterestic)

[fierce steeple]

Depends on the ring

[lone jacinth]

Yes

[spice idol]

Or group, I ain't seeing no addition here

Also depends on what meanings you give the symbols '2' and '4'

[fierce steeple]

So true

https://tenor.com/view/jordan-peterson-gif-9239594

[spice idol]

Now wheres my nobel prize

[limpid horizon]

im reading about some simplicial homology stuff, and for simplicial cohomology (with values in a group G) they define it as $H^i(Hom_Z(C, G))$ (C is the chain complex)

[broken turtle]

kiand123

[limpid horizon]

why do we apply Hom(-,G) for? My only intuition rn is that its contravariant so the arrows reverse to get "co"homology ..

[digital parcel]

Read Hatcher chapter 3

He gives good motivation for cohomology

In short you get to study the topology of a space, not by looking at the space itself, but via the maps from the space, ie Hom(-, G)

There’s some theorem which says that H^i and H_{n-i} are isomorphic for some nice conditions

I can’t remember the name but it’s on the tip of my tongue

Poincare

Anyway there’s many such duality theorems which relate homology and cohomology under various nice conditions

Hatcher has some metaphor about homology and cohomology and paths on a mountain or something like that

It’s nice motivation

[willow seal]

yoneda's lemma right

[hasty flume]

so the slogan here is that whereas maps f: X -> Y induce maps H_i(X) -> H_i(Y) on homology they induce maps H^i(Y) -> H^i(X) on cohomology. In particular the diagonal map D: X -> X x X induces a map H^*(X x X) -> H^*(X). theres a natural map H^*(X) tensor H^*(X) -> H^*(X x X) so composing this together gives you a notion of multiplication in cohomology which doesnt exist in homology

[plucky arch]

oh I didn’t realise that’s how it worked

[hasty flume]

Of course there are natural maps X x X -> X (the projection onto each factor) but the induced map on homology there is not very interesting. intuitively those maps are "destroying too much information"

[plucky arch]

that’s pretty neat

presumably homology has some notion of comultiplication?

[spice idol]

https://mathoverflow.net/questions/415/does-homology-have-a-coproduct

Short answer: homology is not as nice as cohomology

[limpid horizon]

Tbh im really not sure on what most of what youre saying is

Chain maps f: X ->Y induce maps on H_i(X)->H_i(Y), i know that part

Im also not sure of what youre saying about this notion of multiplication on cohomology is relating to what my question was

[spice idol]

We apply the Hom(-, G) because that turns the coefficient groups Z^(X) into the abelian groups of functions from X to G

And it's the easiest way to get a cochain complex, i.e. get the maps the other way

[limpid horizon]

what is Z^(X)?

[spice idol]

Direct sum of Z over the set X

[limpid horizon]

size of X should be like # of simplices of dimension i or smth right

[spice idol]

Yeah

Unless you're doing like singular homology lol

[limpid horizon]

why do we care about turning Z^(X) into functions from X to G

[spice idol]

Then it's some set of maps

[limpid horizon]

so its "valued" in G?

or smth?

[spice idol]

I phrased it badly; the easiest way to turn the maps around is using the Hom(-, G) functor, and that gives you the set of maps to G

[limpid horizon]

yea thats what i originally thought i guess

Hom(-,G) turns maps around alright sure

[spice idol]

But other than that I don't know enough Abt it

Google is being unsurprisingly useless

[fierce steeple]

Yes good post

[fierce steeple]

(Also worth noting (for others) that the same argument works on the level of cochains)

[digital parcel]

Sort of, I suppose

I cant give a good answer for this one

Unless what youre asking is if what I said is the content of yonedas lemma

Then not really, I meant the duality theorems

[digital parcel]

This is said in more detail in hatcher if ure curious @limpid horizon

[limpid horizon]

Thanks, ill prob look at it some time

[spice idol]

Hatcher is cool

[limpid horizon]

Lots of random bits and pieces i need to study properly

[spice idol]

Pretty pictures

[digital parcel]

It’s very nice

Yeah you can just read the intro section

[limpid horizon]

Most of what i hear is bad reviews

[digital parcel]

P much all the intuition for cohomology is laid out there

[limpid horizon]

Ya actually ill do that bc last time i looked at it i didnt know anything at all abt homology stuf

[digital parcel]

Yeah some ppl don’t like it but for the expository stuff (which I am suggesting you read, not necessarily the technical stuff), I think Hatcher is nice

[spice idol]

Cuz it assumes post-rigor

People don't like that

[limpid horizon]

Lol

Ya i mean prob at level of alg top ppl still want full rigor

Intro alg top

[spice idol]

That's fair, but at a 3rd year undergrad / grad level hatcher is completely fine

[limpid horizon]

Ya i have a physical copy of it

Its a real big textbook

[digital parcel]

Yeah

[digital parcel]

Wdym post rigor

[fierce steeple]

I feel like post rigour is a bit generous

[rare walrus]

I think it just assumes you kinda already know what it's talking about

[digital parcel]

Oh

[rare walrus]

I had a terrible time with hatcher some years ago

[digital parcel]

I didn’t get that impression

But maybe it’s bc I did Hatcher w a prof

[rare walrus]

I think it is a Bad™ book, but I will say that the section on the Galois connection for covering spaces is Good™

[spice idol]

Galois connections 🤤

[digital parcel]

Though I only did ch 1-3 so idk if it got worse after that

[rare walrus]

The homology chapter was really bad

and I read further than that but it's blurred now

Hatcher slander over, bedtime

[fierce steeple]

This is a thing i think due to (and at least popularised by) Tao where he talks about prerigorous, rigorous and post-rigorous stages of learning math. I think the point is roughly that at a post-rigorous stage you are able to use analogies and be slightly less pedantic with everything because you are able to fill in the details and make things rigorous

But this is a book for people learning, so i dont think it is great to skim over important details

[digital parcel]

Ahh I see

[limpid horizon]

Bruh

[digital parcel]

Okay yeah I agree

[spice idol]

They're cool

[limpid horizon]

You are salivating

[fierce steeple]

Chapter 4 is the homotopy theory one right? It was fairly standard, though i would have liked more details here

For the homotopy theory bit i would probably suggest reading elsewhere though

[spice idol]

It's exaggerating, for the funis

[digital parcel]

My school doesnt have a regularly offered AT course so I had to do a one semester reading course so idk

[fierce steeple]

Chapter 5 did not exist in my days

[spice idol]

I would do that if it were stuff like UA but oh well

[digital parcel]

This guy is obsessed with UA

[fierce steeple]

I will write a book on algebraic topology and sort everything out

[limpid horizon]

How much did u get through in the reading course

[digital parcel]

Ch 1-3

[digital parcel]

I should prob read homotopy theory anyway

Suggestions?

[spice idol]

Relevant xkcd about new standards

Too lazy too look it up

Just imagine it

Yk what I'm talking about

[fierce steeple]

Lol idk it but i will find maybe

[spice idol]

https://xkcd.com/927

xkcd: Standards

Fortunately, the charging one has been solved now that we've all standardized on mini-USB. Or is it micro-USB? Shit.

I stopped being lazy

[spice idol]

I'm UA #1 fAn

https://tenor.com/view/electric-fan-wind-turn-hot-fan-gif-4860088

[digital parcel]

Just the other day werent u talking abt how u hated some theorem in UA

The real #1 fan would love that theorem

[fierce steeple]

I think it is good to use Hatcher as a scaffold as it has useful topics but for the proofs/details I remember Spanidr and Bredon's texts being good

[digital parcel]

Oh I have a copy of Bredon

We were gonna do bredon but then the prof decided to switch to hatcher lol

[spice idol]

Not really a theorem but just how utterly unmotivated some things are

[digital parcel]

Ah okay

[fierce steeple]

Ulimately there are some results which are kinda tricky here and I remember Spanier and maybe Bredon being particularly precise

[spice idol]

I've got a copy of Bredon too!! I like it although trying to learn point set using the first chapter wasnt fun

Especially because it was self study

I may be stupid

[digital parcel]

Some metaphor about how it’s good to learn to swim by diving in the deep end or something

Probably some xkcd about that too

[spice idol]

On god they tryna drown me in this h03, I cannot see

Dunno how obscure that reference is

[digital parcel]

Not what I had in mind but enjoy

Yeah I dont get the reference

[spice idol]

Omg I remember this from the geography exam

[digital parcel]

I am utterly sorry

[spice idol]

So glad that's over

Well the physics part was fun

[spice idol]

It is alright I am somewhat chronically invested in meme culture

[spice idol]

But we're cool now I'm slowly learning the motivations behind everything

🔥 🔥

Turns out things feel unmotivated when you take the shortcut from decades of the field existing

[digital parcel]

math as it should be

Oh wrong one

💯

Lol

[spice idol]

LMAO

I love the 50 emoji

[digital parcel]

This is why I peddle ppl to do some AT while doing homological alg

Or at least to look at an AT text and see where it all comes from

[spice idol]

Doing the tiniest bit of singular homology helped me understand quandle homology

[digital parcel]

TIL quandle homology is a thing

[spice idol]

Awful to calculate though would not recommend

[digital parcel]

I thought this was some quandale dingle joke

[digital parcel]

At a point isnt every (co)homology theory awful to calculate

[spice idol]

Quandle homology is uniquely awful

[digital parcel]

[spice idol]

It exists for any left self-distributive magma

[digital parcel]

I see

[spice idol]

Cuz that means that left-multiplication acts as some kind of translation, in the expected way

And from that you can construct hypercubes in the quandle, and construct the homology from that geometric intuition

[digital parcel]

“In the quandle”

Where did “quandle” even come from

[spice idol]

No clue

It literally doesn't mean anything

[digital parcel]

[spice idol]

But treat a quandle as some kind of topological space, because it really is just an abstraction of a knot

[digital parcel]

Oh that’s interesting

[spice idol]

It's a faithful knot invariant

But fucking hard to work with, of course

We can't have nice things

[limpid horizon]

Lol

Memories of 2024

[spice idol]

Hello guys, quandale dingle here

I should go to bed soon, I feel my grip on sanity slipping

That's why I'm so rambly

Apologies

[digital parcel]

Yeah mine would too if I had to write “rack” and “quandale” in my papers

[spice idol]

I should be nonchalant

[digital parcel]

I appreciate the rambling anyway, it’s nice to hear abt other fields of math

[spice idol]

Did you know that quandle homology is a special case of the quandle homology for quandle representations in some ring R

[limpid horizon]

Why do u always learn the most weird stuff @spice idol

[spice idol]

Fu nni

[limpid horizon]

Lol

[digital parcel]

Ohhhh it all makes now sense now!

[spice idol]

I like weird algebraic structures, gives me hope that UA isnt totally useless

[spice idol]

Basically every group has a canonical quandle structure:
x • y -> xyx^-1
So a quandle representation of Q in an R-module M is a quandle homomorphism f : Q -> Aut_R(M)

You've also got quandle-sets analogous to G-seta but their homology is isomorphic to the homology of the linearised version

(i.e. take the induced permutation representation on the free ab group generated by your set)

I wonder if you can define cohomology of a G-set

Similar to that of a group

Okay so the simplicial resolution stuff can def be generalised

[fierce steeple]

Quandale dingle

[spice idol]

Spindles are also cool

[digital parcel]

Hmm i see

Is there any geometry in these aside from knots

[cobalt sonnet]

I am trying to compute the invariant associated with the TQFT coming from the center of a group in terms of the structure constants (describing product of class sums) does anyone know if this simplifies?

Here alpha(k) denotes the inverse conjugacy class

For genus 1 i get number of conjugacy classes

for genus 2 it's a mess (unless G is abelian)

[cobalt sonnet]

can you please help with this?

[lyric berry]

How is this defined?

[digital parcel]

idk lol

see the conversation following that message

[weak lodge]

See Section 3 of this paper (which afaik is the paper that introduced quandle homology) for a construction, and cf. Section 2 of this paper

[hot relic]

Can one view quandle homology as a derived functor? If so, how?

[spice idol]

Very good question! No idea, I'm not that good at hom alg yet

I found a paper comparing quandle homology to quillen homology but ?? It's only talking about cohomology??

This is the second quandle homology paper that feels a bit sketch

[hushed bone]

That shit sounds so fake

[spice idol]

It could be true for all I know

I'm honestly too lazy to go and fact check it all though

[hushed bone]

I mean the word quandle homology vro

[spice idol]

😔

[fierce steeple]

Heap homology

[spice idol]

That's just group homology

Heaps are just depointed groups

[hushed bone]

Torsor type shit

[spare solstice]

Am I stupid

What's the Jacobson ideal of F_p[C_(p^k)]

[rose mirage]

If you’re associating a simplicial complex to them at any point (quick scan of the first paper indicates that you can do this) then you’ll be able to do it via a standard construction for simplicial nonsense

[spice idol]

What paper are you reading?

There are a couple different definitions for quandle homology

[rose mirage]

The first one that came up when I googled “quandle homology”. They don’t actually give a simplicial complex but state two homology theories, one of which very clearly arises from simplicial homology of some complex

https://arxiv.org/abs/1012.2923

[vapid axle]

Can $\implies$ be proved as follows? Let $M_i'$ be the subspace of $M$ where $x_j = 0$ if $j \neq i$. Then $M_i' \cong M_i$. Suppose that $f: N' \to N$ is injective. Since $M$ is flat, $f \otimes 1: N' \otimes M \to N \otimes M$ is injective, and the restriction to $N \otimes M_i'$ must be injective. It follows that $f \otimes 1: N' \otimes M_i \to N \otimes M_i$ is injective.

[broken turtle]

okeyokay

[vapid axle]

Suppose that $M$ is flat. Let $M_i'$ be the subspace of $M$ containing elements of the form $(0, \dots, 0, a_i, 0, \dots)$. Then $M_i' \cong M_i$. If $f: N' \to N$ is a monomorphism of $A$-modules, then $f \otimes 1: N' \otimes M \to N \otimes M$ is injective. In particular, the restriction $f \otimes 1 \mid_{N' \otimes M_i'}: N' \otimes M_i' \to N \otimes M_i'$ is injective. The following diagram commutes:
[
\begin{tikzcd}
{N' \otimes M_i'} && {N \otimes M_i'} \
\
{N' \otimes M_i} && {N \otimes M_i}
\arrow["{f \otimes 1 \mid_{N' \otimes M_i'}}", from=1-1, to=1-3]
\arrow["{\varphi^{-1}}", from=1-3, to=3-3]
\arrow["{\varphi'}", from=3-1, to=1-1]
\arrow["{f \otimes 1}"', from=3-1, to=3-3]
\end{tikzcd}
]
where the vertical arrows are isomorphisms. In particular, $f \otimes 1$ is injective, so $M_i$ is flat.

Is this valid?

[broken turtle]

okeyokay

[fierce steeple]

Yes, though i wouldnt worry so much about direct sums like thay

[vapid axle]

Wdym

[fierce steeple]

Well just distinguishing internal and external direct sums like that feels unnecessary beyond like intro lin alg

But can add if needed ig

[vapid axle]

Oh I guess so

[fierce steeple]

More important is justifying why like you can resfrict f (x) 1 like that

As in why it lands in N (x) M_i

You are implicitly using that tensor product commutes w direct sums

[vapid axle]

True

I thought that was trivial but things can be deceiving with the tensor product ig

[plucky arch]

could someone eli5 exact sequences to me?

i still don't quite appreciate their significance

[digital parcel]

maybe one example is excision in algebraic topology

this is from Hatcher

so you have a short exact sequence 0 -> A -> X -> X/A -> 0

and you apply reduced homology

and this lets you study the reduced homology of X by looking that of A and that of X/A

in the context of math phys i think an example of chain complexes/exact sequences is the de Rham complex

and then you have that notion of exact differential forms

so like if you look at
[ \Omega^{p-1}(X) \xrightarrow{\partial^{p-1}} \Omega^p(X) \xrightarrow{\partial^p} \Omega^{p+1}(X), ]
then you always have that $\text{im} \partial^{p-1} \subseteq \text{ker} \partial^p$, so this forms a chain complex. now every $\omega \in \Omega^p(X)$ is exact when you have exactness at this step, ie $\text{im} \partial^{p-1} = \text{ker}\partial^p$

[broken turtle]

anamono

[digital parcel]

a special case is short exact sequences, ie exact sequences of the form
[ 0 \to A \to B \to C \to 0 ]

[broken turtle]

anamono

[digital parcel]

now these are quite nice because sometimes you can study properties of B by looking at A and C (e.g., additivity of hilbert polynomial, I think there was some discussion about this in #algebraic-geometry)

[digital parcel]

actually maybe this is why some people call the maps in chain complexes "differentials"

TIL

Ext/Tor pop up everywhere which are defined via injective/projective (respectively) resolutions, which are exact sequences

[plucky arch]

I see…

perhaps if I knew more algebra this would be more helpful

[digital parcel]

Maybe. I think the de rham example doesnt need much algebra for motivation tho

You can look at this purely from a diff geo perspective

If your dR complex is exact, then that says “every closed form is exact”

[plucky arch]

sure but im trying to understand homology more broadly

[digital parcel]

Ah okay

Homology/cohomology give you invariants of spaces (or more generally, algebraic objects) basically

Eg in AT, homology classifies “holes” in your space; the objects in your chain complexes are Abelian groups

Cohomology usually appears when you’re asking about certain obstructions

So like the cohomology of the dR complex basically tells you what issues you face that stops a closed form from being exact

Hopefully that helps, if not someone else can give a better answer

[fierce steeple]

Wouldnt really call this a short exact sequence without caveats

[digital parcel]

how so?

[fierce steeple]

Well short exact sequence is almost exclusively used for abelian categories (or slightly more generally, e.g. groups), so this is a more general use. In particular in topological spaces there is not a zero object, so that already makes it abusive as written

[spice idol]

Idk what the first 0 is doing there anyway

Second 0 I can somewhat get

[digital parcel]

ah yeah that's true

[spice idol]

Pff imagine not being congruence representable

So cringe

*crying

[plucky arch]

I’ve heard you can define exact sequences in Set*, for example

[spice idol]

Yes, although they do not mean much

[fierce steeple]

I guess I could be a lil negative and say they are "just" special cases of pushouts/pullback squares, and they are particularly nice in pointed categories. Also in additive categories you can turn pullbacks/pushouts and equalisers/coequalisers into exact sequences, so it is a convenient way to deal with stuff

[spice idol]

If you mean thats the category of pointed sets

[plucky arch]

wait pushout pullback squares??

[fierce steeple]

Wdym

[plucky arch]

what is a pushout pullback square

[fierce steeple]

A square which is simultaneously a pushout and pullback

[plucky arch]

right that makes sense

are you saying you can phrase exact sequences like this…?

[fierce steeple]

Yes

From the definitions of kernels, cokernels

[plucky arch]

could you spell it out explicitly? I’d really appreciate it🥺

[fierce steeple]

given a map f: A-> B, the kernel is 0 x_B A

And cokernel is 0 \coprod_A B

Here i assume we are in a pointed category

[plucky arch]

mhm

[fierce steeple]

Well that is mostly it aha but yeah point is kernel of M-> M/N is N essentially

[plucky arch]

now im confused

how does this give exactness

[fierce steeple]

To say 0 -> A -> B -> C -> 0 is an exact sequence is equivalent to saying that A-> C vanishes and
A -> 0
| |
B -> C
is a pushout/pullback square

[plucky arch]

what about just saying A -> B -> C is exact at B?

[spice idol]

Any exact sequence can be split up into SES

Short exact sequence

[plucky arch]

so how does that work here...?

[spice idol]

Extend A -> B -> C to a longer exact sequence by padding it using the kernel and cokernel, and then apply the decomposition

[plucky arch]

hm...

[fierce steeple]

Explicitly 0 -> im(A -> B) -> B -> coker(A -> B) -> 0 is a SES

[plucky arch]

wait where's C

[spice idol]

In the other SES

[plucky arch]

@.@

[spice idol]

0 -> im(B -> C) -> C -> coker(B -> C) -> 0

Exactness says that these two can be conjoined together

[lone jacinth]

Might be better to write
0 -> im(A->B) -> B -> im(B->C) -> 0
then this is exact iff A->B->C is

[plucky arch]

i'm very lost...

oh

[lone jacinth]

In the case it's exact you have im(B->C) = cok(A->B)

[plucky arch]

so it's like

[spice idol]

Yeah was about to say lol

[plucky arch]

hm

https://math.stackexchange.com/questions/141215/what-are-exact-sequences-metaphysically-speaking

i'm trying to see if i can understand it in terms of this

so B is "built up" from im(A -> B) and im(B - > C)?

maybe combined with this?

[fierce steeple]

"Metaphysically speaking" lol

[lone jacinth]

Yes, every element in B maps to one in im(B->C), and two things map to the same if there difference is in im(A->B).

So if you pick some noncannonical preimages of im(B->C), then every element is uniquely a sum of one of those preimages and something in im(A->B).

So it's not as nice as a direct sum, but these two contain a lot of information about B.

In particular they determine things such as B being Noetherian/artinian/finitely generated/finite length

[lone jacinth]

This visual representation as things being build by exact sequences is kinda nice

https://www.3blue1brown.com/blog/exact-sequence-picturebook

[plucky arch]

ooh ok, i'll take a look!

[digital parcel]

Had no idea 3b1b did that

Fun

[limpid horizon]

Why are graded objects hurting my head so much

Doesnt seem like it should but it is

[digital parcel]

Just use polynomial rings as your prototype

[lone jacinth]

Technically Ravi Vakil made the post.

[spice idol]

Though the polynomial ring is deceptively nice

The homogeneous components being rank 1 free modules

[lone jacinth]

Pfft, it's not even graded commutative

Not nice at all!

[spice idol]

Graded commutative?

[lone jacinth]

https://en.m.wikipedia.org/wiki/Graded-commutative_ring

[limpid horizon]

Not sure where im going wrong but i was looking at k[x,y,z] with Z^3-grading. The ideal (xy,yz) should be a graded ideal right. What is the direct sum decomposition into homogenous components?

[spice idol]

Is such a ring nicer than a commutative ring?

[limpid horizon]

Graded ideal is also the same as a graded submodule of R as an R-module right

[spice idol]

I would honestly set the world on fire if it wasnt

[limpid horizon]

Lol

[lone jacinth]

Just the homogenous elements of each degree.

Like of degree (1,1,0) it's just the span of xy, for (2,1,0) it's the span of x^2y etc

[digital parcel]

Yeah I meant more the blog

But true haha

[lone jacinth]

Oh yeah, there's a blog

[limpid horizon]

So its (1,1,0) and (0,1,1) but isnt the direct sum of those components just giving you the k-linear combinations

[lone jacinth]

I'm not sure what you mean by "it's (1,1,0) and (0,1,1)"

For every possible element in Z^3 you get some component

[limpid horizon]

In my example (xy,yz) the degree of those geneators are (1,1,0) and (0,1,1)

[lone jacinth]

Being the homogenous elements in your ideal of that degree

Sure

Those generators are in particular homogenous elements yeah

[limpid horizon]

So (xy,yz) = direct sum of (1,1,0) component plus (0,1,1) component?

Ok graded components of k[x,y,z] are the sets {qx^a1y^a2z^a3 | q in k}

[lone jacinth]

Plus all the other components

[lone jacinth]

The guy who wrote the wallis product video also has a blog I quite like
https://sridharramesh.github.io/HowSridharThinks/3b1bsineproduct/

[limpid horizon]

The other components are the ones with degree (a1,a2,a3) with either a1 and a2 at least 1 or a2 and a3 at least 1?

Like idk if im understanding this properly cuz it seems weird

[lone jacinth]

Yeah, (I mean you have components for smaller degrees, but they're just 0)

[limpid horizon]

Ok thanks this is probably gonna make more sense now

[plucky arch]

unfortunately this might be the first time in my life where a picture seems to hinder my understanding of a concept

[limpid horizon]

Oh idk he had something on exact sequences

I didnt know *

[plucky arch]

Maybe it’s just beyond me at the moment…

I’ve tried a few times but really cannot seem to grasp exact sequences

[digital parcel]

Nice

[limpid horizon]

What kind of understanding are you looking to have? Like idk im pretty beginner but to me theyre useful algebraic tools to study because lots of properties are like transferred between the objects if theyre connected in an exact sequence

I dont have some deep understanding of it tho

[plucky arch]

to me exact sequences are still a googit

[limpid horizon]

Idk how deep it gets really

[spice idol]

Chain complexes are just amazingly diverse

[digital parcel]

they get pretty deep

i should probably learn spectral sequences

i've only been putting it off because i've never had to use one

okay well vakil doesn't use it until chapter 9

[lone jacinth]

I usually try to reach for some examples.

Trying to translate some examples to this visual metaphor might be helpful even though it seems to be hindering you now. But idk ¯_(ツ)_/¯

[digital parcel]

so i'll learn it when i get to chapter 9

[plucky arch]

I guess even the examples themselves feel too hard for me

[subtle plaza]

I like spectral sequences. I read through the constructions and proofs about 10 times before it began to stick with me. At a certain point I was ok with just knowing how to use them

[digital parcel]

yeah lol the only time i've ever really seen them were in a few talks

[plucky arch]

Yeah I guess I don’t really use exact sequences in my work

[digital parcel]

and they didn't write out the spectral sequences either they just drew boxes and arrows

[spice idol]

Spectral sequences are scary but I've heard they are really powerful

[digital parcel]

oh wait i've seen it one other time

i was trying to understand the proof of some result that i had discussed my prof

so i went to the original paper and it was done with spectral sequences

so i emailed him and said "sir i dont know spectral sequences" he said "you don't need to i don't care about that proof anyway" lol

or something along those lines

[lone jacinth]

I've only used spectral sequences once. It was to make diagram lemmas with funny names
https://www.researchgate.net/publication/341029857_The_Anaconda_lemma

[spice idol]

I love the anaconda lemma

Citations 0 references 0

[hushed bone]

U wrote that?

Dox

[lone jacinth]

jagr2808 is also the name of my homepage, so I'm already doxed

[hushed bone]

Dox

[lone jacinth]

It's okay, there are two authors so no one will ever know which is me

[spice idol]

I will keep in mind both

And when you next share a paper of yourself i shall see which of the two is shared

[lone jacinth]

I will have to start only writing papers with him from now on I guess

Hi guys, new to server. Looking forward to talk about algebra and stuff

[spice idol]

Hi.. jagr

Didn't know you were new

[lone jacinth]

I'm "Not jagr"

[rare walrus]

Who?

[lone jacinth]

You must have me confused with someone else

[rare walrus]

Never heard of you before

[spice idol]

Ohhh right

[rare walrus]

Welcome to mathcord!

[spice idol]

Hai!!

[digital parcel]

#prealg-and-algebra

[spice idol]

You must be soo confused

Guess lil old me will have to do

[plucky arch]

ok i think i understand exactness now :D

[spice idol]

See here? That's your SOUL, the culmination of your being