advanced-algebra

[hushed bone]

I’m not entirely sure what M is tbh, but I imagine that M = M_1 or something and so the filtration M_1 > M_2 > … > M_i and the regularity on M_i/M_i+1 shows that its regular on M/M_i

[ionic sequoia]

Any 8th grade American based math tutor in the house ?

[lament scarab]

I'm wondering if anyone can provide a reference for the proof that there can be two polarizations subordinate to the same dual Lie algebra element which give unitarily inequivalent induced representations (a standard construction in the orbit method). I looked up the reference shown in the screenshot and it does discuss inequivalent reps being attached to the same coadjoint orbit, but the method by which reps are being attached to orbits doesn't seem to be the orbit method.

If there is a better channel to post this question please let me know! I'm surprised there isn't a representation theory channel

[near lantern]

This is the representation theory channel (among other things). It's worked out so far 🤷.

[near lantern]

Regarding the question, can you recall the definitions (of say, polarisation subordinate to an element of g^* and the unitary representation associated with one, at least for sp_4 if that's easier to define)?

[lament scarab]

sure, this is the definition from a paper I'm writing, \mathfrak{g}_f is the Lie algebra of the isotropy subgroup of the dual Lie algebra element f under the coadjoint action.

you can also look at definition I.4.1 of https://link.springer.com/article/10.1007/BF01389744

ah I forgot to include the condition that Ad(G_f) \mathfrak{h} = \mathfrak{h} in my definition lol

the unitary rep obtained from the polarization is then the subrep of Ind^G_{G_f} \chi_f (\chi_f is a character which differentiates to if) obtained by imposing the following condition (V^{(L)}X is the left invariant vector field associated to X, \sigma is an element of Ind^G{G_f} \chi_f i.e, a section of the pre-quantum line bundle)

[wary elbow]

As a side note... I would be 100% in favor of a representation theory channel

Subordinating rep theory to algebra is just wrong lol

[woeful crane]

Did you ever find some resources on this?

[analog abyss]

Alas not. But my advisor tells me that he meant to say “presentation” instead of “representation” so it’s also no longer as pressing for me lol

(And told me of the resources he had in mind)

[limpid horizon]

Eisenbud's book is huge

only started reading it recently

W book tbh

[broken wren]

A representation is the opposite of a presentation. Maybe we should call it a copresentation

[spice idol]

or maybe we should call stuff like a coproduct a reproduct

reequaliser

relimit

resimplicial reresolution

[near lantern]

Next up: find out what a quiem is by attending two funerals.

[silver goblet]

is there a way to motivate the definition for homotopies between morphisms of chain complexes if you don't a priori know about singular homology?

[urban granite]

well idk if it would be a good idea but i'd talk about quasi-isomorphisms
it'd be reasonable to introduce homotopies between chain maps if we weaken equivalences between chain complexes to quasi-isomorphisms

like they need to have same map on homologies for quasi-isomorphisms to make sense

[lone jacinth]

Projective resolutions can be motivated as expressing a module through generators and relations (and relations between relations etc).

But taking projective resolutions is not functorial, since they're not unique. The difference between two projective resolutions is precisely a homotopy

[silver goblet]

oh wow, i didnt know about that, that's pretty remarkable

is this easy to see? if so i might request to leave it to me as an exercise haha

[lone jacinth]

It is easy to see yes

[silver goblet]

lovely

[lone jacinth]

Standard exercise

[silver goblet]

anyways thanks! that's very helpful

since it motivates derived stuff as being homotopy stuff

i guess this story is parallel to "resolving" a cgwh space by spheres?

(are projective resolutions cofibrant replacement in the usual model category?)

(nvm yes nlab has this as a remark)

[scarlet ermine]

Another motivation is homology of internal hom is chain maps up to homotopy.

If you’re not familiar with this, here’s how it works—you can consider a chain complex as a Z-graded abelian group A (or a graded object in some abelian category), together with a map
d: A->A[-1]
which satisfies d^2=0.

Now, given two chain complexes (A,d), (A’,d’), you can ask, which graded abelian group homomorphisms f:A->A’ are actually chain maps (ie respect the differentials). This is exactly the condition
d’f = fd,
which can write as
d’f-fd=0,
and then think about the assignment

f ~> d’f-fd

This is actually a homomorphism from graded abelian group homs

Hom(A,A’) -> Hom(A,A’[-1])

and the kernel of this morphism gives us only maps of chain complexes (rather than any graded abelian group hom).

So, if the kernel is interesting, how about the image? To still land in degree zero things, consider

Hom(A,A’[1]) -> Hom(A,A’)
h ~> d’h - hd

(so doing this for lower degree graded abelian group hom maps). The image of this map is exactly the nullhomotopic chain maps A->A’, and any h:A->A’[1] is a nullhomotopy

[silver goblet]

hmmm okay i think i get this

i'll need to think about the last few sentences when im more awake lol

also im a bit confused about what you mean here: "homology of internal hom is chain maps up to homotopy"

by internal hom i assume you mean the total derived hom functor right?

what is the graded structure of a homotopy class of chain maps?

[lone jacinth]

Internal hom is the Hom-complex.

I.e. the chain complex Hom(A, B) where the degree n part is homogeneous graded maps of degree n (between the underlying graded abelian groups)

[silver goblet]

ah i see

[lone jacinth]

This is adjoint to the tensor product of complexes

[scarlet ermine]

I'm trying to organize a reading group at my university on DG/infinity categories, mostly to an audience of commutative algebraists (and some algebraic geometers). Most everyone is interested in derived categories. Does anyone have any nice "goal" theorems for the seminar that show off the utility of infinity categories?

[wanton spoke]

descent and gluing for derived categories

[worldly zealot]

how much algebraic topology should i know to read about equivariant cohomology

[blissful field]

This character table does not seem to be orthonormal wrt the inner product. Neither columns nor rows. More info here:

#help-39 message

(Was told reposting here may be better)

[blissful field]

more full working - I'll do this in abit

[worldly zealot]

you need to take comlpex conjugate

[blissful field]

Thank you so much!

[worldly zealot]

inner product (a,b) is sum of the product of a(g) with complxe conjugate of b(g)

np

[blissful field]

I shouldn't attempt to do work while tired

[wicked nova]

Let A be a ring, p a polynomial over A and a \in A. Is it necessarely true that if a is a root of p then x - a | p?

I know that this is obviously true in well behaved rings A, but I don't remember if this is true in general

[muted sierra]

I'll assume that by ring'' you mean commutative ring''.
Consider the homomorphism of $A$-algebras $\phi : A[x] \to A$ that sends a polynomial $q(x) \mapsto q(a)$. Clearly, $\langle x-a \rangle \subset \ker(\phi)$, so your question boils down to whether this inclusion is actually an equality. I suggest you to use the first isomorphism theorem, which asserts that $\phi$ factors as the quotient map
$$A[x] \to B = \dfrac {A[x]} {\langle x-a \rangle},$$
followed by an isomorphism $B \cong \mathrm{im}(\phi)$, followed by the inclusion $\mathrm{im}(\phi) \hookrightarrow A$.

[broken turtle]

Eduardo León

[lone jacinth]

In general for any polynomial p you can always find a polynomial q such that
p(x) = q(x)(x - a) + p(a)

Even more generally you can always do euclidean division by monic polynomials

[wicked nova]

do you know where I can find a proof to that?

[shrewd wing]

https://math.stackexchange.com/questions/1811541/understanding-divison-by-monic-polynomial-in-rx-where-r-is-an-arbitrary-ri

[lone jacinth]

I mean, the proof is essentially: look at the algorithm for long division. It doesn't involve any division (except possibly the leading coefficient of whatever you're dividing by). Okay then it works for all commutative rings

[fierce steeple]

if $f = \sum_j b_j x^j$ then $f(x) - f(a) = \sum_{j \ge 1} b_j (x^j - a^j) = \sum_{j \ge 1} b_j (x - a) \cdot \frac{x^j -a^j}{x-a}$

[broken turtle]

Prismatic Potato

[fierce steeple]

in the special case

[kindred sphinx]

Guys please give me a small hint on this one. Let G be a group and H a subgroup such that the index 2. Then show for every g, gH=Hg

In other words show H is a normal subgroup

[silver goblet]

how many cosets are there

[neat nexus]

If g doesn’t belong to H. As a set, G on one hand is disjoint union of H and gH. On the other hand it’s disjoint union of H and Hg, so…

[kindred sphinx]

Yeah I solved it but still thanks!

[plucky arch]

What intuition do people have for the abelian category definition?

[hushed bone]

Algebra

[woven loom]

It’s modules

[hushed bone]

Derived functor :)

[plucky arch]

Could you elaborate

[hushed bone]

It’s modules

U can do algebra

[plucky arch]

Are you referring to the embedding theorem or something

[woven loom]

The intuition is modules

[plucky arch]

An overly terse answer is not very helpful to me

[hushed bone]

It works like modules

Quotients

Isomorphism theorems

Just do a proof with modules without writing “x in M” and jt works

[plucky arch]

Hm ok

[fierce steeple]

Tbh I don't feel the embedding theorem is that important

I have heard it said that abelian gives you the extra definitions you need to be able to prove the basic stuff up to the snake lemma

[plucky arch]

Me neither, it was just my attempt to interpret the terse answers given

[fierce steeple]

But it is a definition where various variants are useful

[plucky arch]

Oh which variants

[fierce steeple]

Well like often all that matters is being additive, but also often you work with abelian categories satisfying extra axioms e.g. look up Grothendieck's axioms for abelian categories, especially like the notion of a Grothendieck abelian category

[woven loom]

Important, no, but intuition yes

[fierce steeple]

Maybe to like know you can make precise the fact you aren't far off modules ye

I always have viewed mentioning the theorem as a way to be lazy with the basic proofs lol

[woven loom]

It is yes

“Why do you want all these properties? It’s what modules do”

[plucky arch]

Interesting, lemme look this up

[fierce steeple]

But not rly too important initially

[plucky arch]

Oh cool, it’s abelian category + Topos

[fierce steeple]

Idk what you mean tbh

[plucky arch]

There’s a section on the nlab saying they’re precisely Ab-enriched grothendieck topoi?

[fierce steeple]

Ah sure like Shv(X, Ab)

[plucky arch]

Mhm

[fierce steeple]

Ye many of the nice "big" abelian categories one meets are Gothendieck abelian

[plucky arch]

Makes sense

[latent crypt]

You might also like Godement's definition of an abelian category. He kinda gives it as a single axiom.

Well not really but it kinda highlights all the things we want. There's a zero object, every map admits a kernel and cokernel, and the first isomorphism theorem holds (the natural map coker ker f → ker coker f is an isomorphism for all f).

[spice idol]

I see them as objects which are very convenient for computation and measuring "stuff", and the axioms kinda give all the properties needed for that

this is very vague I'm sorry lol

[near lantern]

Hello

[latent crypt]

🌊

[fierce steeple]

An abelian category is a heart of a t-structure

[rocky hatch]

Interesting actually. I can't appreciate it as much cuz I haven't done much hom alg unfortunately

[rocky hatch]

Hullo

[limpid horizon]

Hi.

[north stream]

Hola/ola

[broken wren]

I thought that was the standard definition. Indeed it is the definition in nlab. But it is not the definition in wikipedia

[ornate atlas]

It’s the definition in Weibel I think?

But he does waffle about preadditive categories and stuff so maybe I’m misremembering

[broken wren]

How about this: a map that has trivial kernel and cokernel is an isomorphism. Is this equivalent to abelian? (in the presence of additive and finite limits and colimits)

[lone jacinth]

Yes

I've also always seen the preabelian + first iso definition.

Not that wikipedias definition is terribly different

I guess wikipedias definition lends itself to the question, are there preabelian categories where every epimorphism is normal, but not every monomorphism?

I don't think that's possible.

The map from the coimage to the image should always be a bimorphism, so if it is a normal epimorphism it would just be the cokernel of 0 which is an isomorphism.

[hushed bone]

Tf is a bimorphism

[lone jacinth]

morphism both epi and mono

[spice idol]

morphism that swings both ways

[lone jacinth]

So bijection - morphisms basically

[hushed bone]

Hello isomorphism

This meme is brought to you by Abelian Category

[lone jacinth]

Yeah, so preabelian + bimorphisms are isomorphisms should be another definition of abelian

[fierce steeple]

Oh lol why isn't it just stated like this usually

Tbh I have to google what preabelian means every time lol

[hushed bone]

Additive and all finite limits and colimits exist?

I never have to think about preabelian lol

[fierce steeple]

Same lol

[hushed bone]

Idk who is actually working with these

[fierce steeple]

If your category doesn't have finite limits and colimits then cry

Jk manifolds exist

I have seen some interest in quasiabelian categories for like analytic things

But that is again different lol

[broken wren]

The axiom I gave is very appealing, but that doesn’t make it a good definition, particularly for first exposure. The definition in Wikipedia is about kernels and cokernels because that’s what you use abelian categories to work with

[fierce steeple]

Yee sure

[broken wren]

The purpose of abelian categories is to do homological algebra. Homology is the kernel of the cokernel. You need to the first isomorphism theorem to make that definition symmetric

[lone jacinth]

People do work with topological groups at least, so someone works with at least one of these

[near lantern]

IDTS

[near lantern]

That's exactly how far preabelian supposes.

[lone jacinth]

IDontThinkSo?

[scarlet prairie]

Let $\mathfrak{a}$ be a commutative unital ring which has additive group that is free abelian on ${x_i}_{i \in J}$ where $J$ is an infinite indexing set. Let $\mathfrak{X} \subseteq J$ and define $\langle \mathfrak{X} \rangle := \langle x_i \mid i \in \mathfrak{X} \rangle$. Suppose that $\langle \mathfrak{X} \rangle$ is an ideal. My claim is that $\mathfrak{a}/\langle \mathfrak{X} \rangle$ is free abelian with basis ${\overline{x}i}{i \in J \setminus \mathfrak{X}}$.

[broken turtle]

Former Rank 7 LLORT AJNIN

[scarlet prairie]

Might be better phrasing but would my claim above be right? I am just taking a some portion of the basis, assumign their linear span forms an ideal, then claiming that the quotient is free abelian

[lone jacinth]

Yeah, it's not really much to do with rings or ideals.

If you take a free abelian group and mod out a subgroup generated by part of the basis you get a free abelian group.

[scarlet prairie]

hmm yes, i thought about it a little more. actually the ring structure doesn't even matter here right? If I just take any part of the basis and take the subring generated, this might not be an ideal so i can't take quotient but since I am assuming I do get an ideal, then I just have to check what you have wrote

[lone jacinth]

I mean you can take the quotient either way, it's just the quotient might not be a ring

[scarlet prairie]

ah okay, thanks for the help, I think i was overthinking the multiplication part

[golden osprey]

I keep seeing the words "Grobner degeneration / Grobner deformation" thrown around in papers

I have genuinely zero idea what is going on here, and these sources neither give background nor elaborate on anything nor give references

Googling has just pulled up more papers with the same issues

Does anyone have any good sources for this stuff about what a Grobner degeneration is?

I guess its supposed to be a geometric interpretation of a Grobner basis? Not really clear what's going on here

[lone jacinth]

Seems to be on page 571 here
https://link.springer.com/article/10.1007/s00233-023-10351-4

[golden osprey]

oh cool

oh and it gives a pointer to a thing in Eisenbud incredible

thanks

[spice idol]

are there any ways of showing that a grothendieck category has enough projectives? Besides just showing it directly, of course

[fierce steeple]

Hmm well they don't even have enough projectives in general so i am unsure what the question is

[spice idol]

like are there sufficient conditions for a grothendieck category to have enough projectives?

[fierce steeple]

Or do you mean possible methods to try to prove it in some cases

Hm

[spice idol]

yeah

[buoyant fox]

my proof here is bunk right

because k isnt necessarily just a constant?

[fierce steeple]

f and m are irreducible so it's fine

[buoyant fox]

but I think I assumed k is some constant

but it might not be right

[fierce steeple]

I mean that being irreducible forces them to be constants

[buoyant fox]

oh im tripping yeah you are right

[fierce steeple]

Indeed an irreducible monic poly is the min poly of its roots

[foggy galleon]

is there some kind of classification of (non-associative) finite dimensional real division algebras over R?

[worldly zealot]

i mean you could just try to find a projective generator? idk to what extent this is easier

for presheaves of abelian groups on a small category C let P_c be the presheaf Z[Hom(-, c)], can show this is projective, the coproduct of all the P_c is a generator

[broken wren]

Grothendieck gave methods for showing enough injectives. Apply these to the opposite category

[wise sedge]

Yes

There are the octonions

And that’s it

[ruby otter]

Can't you just directly say that f' divides f but then f=kf' etc. Like f IS the minimum polynomial of alpha since f is irreducible

[lone jacinth]

The opposite of a Grothendieck category isn't necessarily grothendieck though. E.g. ModR

[foggy galleon]

This is wrong

Oops sry I meant to acually respond to the message below lol

[lone jacinth]

That's the only normed one

Or normed unital I guess

[lone jacinth]

This MO gives an argument for there being uncountably many at least

https://mathoverflow.net/a/475954/157483

So I would guess there isn't really a classification

[foggy galleon]

Ohh nice

Maybe it's possible to say something more profound about the topology/geometry of D/P?

[foggy galleon]

this is from a 1999 paper

[buoyant fox]

F may not be not be monic

But yeah I think that detail is minor

[buoyant fox]

Is this statement correct as stated? Or do I also need the condition that beta_j is a root of the minimal polynomial of beta over K?

[broken turtle]

Sunny — ping when reply plz

[charred kestrel]

I've been told that this is true, but I want to know if there is a relatively easy one to prove this. One direction is relatively easy, the other less so. I'm basically a complete outsider to this field, but can follow anything if spoon-fed enough.

[charred kestrel]

i need help 😭

[cursive crag]

Any tip for starting representation theory, I mean i need direction, I can't think of any way to get in, where to begin so my books and resources, where to begin carving out the first piece

[coral bolt]

finite groups

[lone jacinth]

Why do you want to get into it / what kind of representation theory do you want to do?

The most befinner friendly is probably rep theory of finite groups and character theory. Which then can branch into things like compact lie groups, more combinatorial things or modular / integral rep theory.

If you want something more homological algebra flavored there's commutative noetherian rings or quivers / finite dimensional algebras.

In any case there's probably lots of good books out there. Something I like to do is find a the University website for some course that follows a book / has good online resources and just follow that.

[cursive crag]

After finite groups , for compact lie group

And also I want to know about books or other sources? Can you recommend any

[lone jacinth]

Open coursewear has a course at least
https://ocw.mit.edu/courses/18-757-representations-of-lie-groups-fall-2023/

[cursive crag]

Is hall enough as a prerequisite for representation theory

[lone jacinth]

Well idk what hall is

[cursive crag]

Brian Hall: lie algebra and lie group book

[lone jacinth]

The one with representation theory in the title? Then I would hope so

[cursive crag]

Yeah

[lone jacinth]

I don't think this is true unless I'm misunderstanding the statement.

Like sqrt(3) is totally real, but 3 is not a sum of two squares in Q

[distant harness]

You can apply the functions multiple times. Set h=sqrt(1²+1²); then you can make sqrt(3) as sqrt(h²+1²).

[charred kestrel]

^

[lone jacinth]

I thought we were in some fixed extension? Are we in the algebraic closure of Q?

Okay, I see what was meant

[distant harness]

S is the closure of {1} under the five given operations.

[lone jacinth]

Right, but not the closure in something, but an arbitrarily constructed closure

Got it

[distant harness]

(On the other hand, I'm not sure what a "totally real number" is -- google insists on showing me stuff about totally real fields).

[lone jacinth]

Number x such that any homomorphisms Q(x) -> C maps x to a real number

I.e. Q(x) is totally real

[distant harness]

I see.

[distant harness]

Hmm, "totally real" aside, the claim seems to imply that, for example $$\sqrt{(1+\sqrt2)^2+1^2} = \sqrt{4+2\sqrt{2}}$$ can be written without nesting square-root signs inside each other, and it's not obvious how that can be done.

[broken turtle]

Troposphere

[spice idol]

ended up needing something similar; I remembered my category was the module category of some ringoid (small Ab enriched category), so i can just use the Yoneda lemma to get a free-ish object, and that will be projective

[charred kestrel]

yah, numbers such that all galois conjugates are real

[charred kestrel]

which claim?

[distant harness]

The one you're trying to prove.

[charred kestrel]

you mean that rhs can always be transformed into something that looks like lhs?

-# i apologize, im going to sleep very soon; do ping me and i will respond

[distant harness]

The LHS is just there to show that number on the RHS is in S.

[charred kestrel]

yah, thats kinda the idea i guess

[distant harness]

But I don't see how to write sqrt(4+2sqrt2) without nesting of the roots.

[lone jacinth]

How is it claiming that?

Like
Q < Q(sqrt(2)) < Q(sqrt(4+2sqrt(2))
is given by adjoining square roots

[charred kestrel]

oh, no, if you mean the field extension, you can adjoin √2 and then adjoin √(4+2√2)

[distant harness]

Hmm, I read the specification as saying every number in in S is is some extension where we have adjoined only square roots of rationals ("existing elements of Q").

[charred kestrel]

sorry. im not very literate in the field*, so i dont know how to phrase is clearly. radical extensions, maybe? what's the right way to phrase this
-# *pun intended

[lone jacinth]

I don't think that's what it's saying

[charred kestrel]

yah i meant this
is there a better way to phrase this?

[distant harness]

If we can adjoin square roots of elements that were not in Q to begin with but were added later, then my objection falls away.

[charred kestrel]

again, is there a better way i could have phrased this?

[north stream]

I feel like you can proof that for every rational polynomial whose degree is 2^n has its roots in this field iff the roots are totally real.

[charred kestrel]

counterexample... $x^4-x=0$?

[broken turtle]

Sunny — ping when reply plz

[north stream]

This is basically just saying that every 2-group is solvable and you reduce to showing that you can always adjoin a square root of a positive element

[broken wren]

That’s not a verbatim quote. It doesn’t say “of Q,” but “to Q.” So I read the existing as meaning at the intermediate stage

[north stream]

x^3-1 roots are not totally real

[charred kestrel]

fine, but surely this counterargument still holds... . . . apparently not?
just ignore me

[north stream]

For what?

There is like some weird asociation with like proving the cube with volume 2 can't be drawn with a rule or something like that.

So I don't think you can prove it without like using some Galois theory machinery or at least some field theory

This is what I'm talking about if you want to know more about
https://en.wikipedia.org/wiki/Doubling_the_cube

So basically your problem is saying is describing all constructible numbers

[vague pawn]

I have a rank 2 free $\bZ$-module $M = \langle e_1, e_2\rangle_\bZ$, and integer matrices $A,B$ with nonzero determinant. Is $AM \cap BM = ABM$?

[broken turtle]

ExpertEsquieESQUIE

[north stream]

Take A the matrix which switches coordinates and multiplies one coordinate by 3, (e_1,e_2)->(3e_2,e_1), and take A=B

[fierce steeple]

Also no reason for the RHS to be symmetric in A and B i guess

[vague pawn]

If B = adjA?

they commute in that case

[north stream]

The switching is not neccesary lol but yeah

[fierce steeple]

If they commute then you have one inclusion since ABx = A(Bx) = B(Ax)

[vague pawn]

this gives \supseteq

[fierce steeple]

uhh

I think I thought this was the harder direction lemme see

Oh lol

[north stream]

Even if they conmute, if you send 3e_1 to e_1 then the product send it to 9 e_1 which means like 3e_1 is not in the image of the product

[fierce steeple]

Ye only one inclusion

Not sure what happens with adjugate though

[north stream]

This feels like it is only true for scalar multiplication or sl(Z)

[fierce steeple]

i suspect it is false in general

[vague pawn]

sad

I need to show an isomorphism here

[north stream]

not even for scalar now thinking about it

[vague pawn]

I thought that the homomorphism given by inclusion works

[north stream]

What is A'?

[vague pawn]

adjA

[fierce steeple]

I guess you just consider M -A-> M/(det(A)M) which has image Ax/det(A)M and you need to compute the kernel. If Ax = det(A)y then det(A)y = A adj(A) y = Ax so adj(A)y = x -- here i used the fact that A is injective

and conversely if x = adj(A)y then Ax = A adj(A)y = det(A)y

So the kernel is precisely adj(A)M

[vague pawn]

the map is just multiplication by A modulo det(A)M?

ok, thanks potato and kev

[fierce steeple]

yeah like multiply by A on left and then project

[charred kestrel]

yah i am aware — its specifically ||all constructible numbers in google slides||

im from the ||powerpoint construction|| community lol

[limpid lagoon]

is what they described not the Pythagorean closure of Q? https://math.stackexchange.com/questions/116716/is-the-pythagorean-closure-of-mathbb-q-equal-to-the-field-of-constructible-nu

@charred kestrel

[charred kestrel]

im not very familiar with the terms — what is this?

[limpid lagoon]

its essentially just what you defined in your question https://en.wikipedia.org/wiki/Pythagorean_field

[charred kestrel]

oh ok

its probably the same, yah

[limpid lagoon]

if i may ask, why are you trying to prove that in your question?

the specific formulation of your question seems weird

[north stream]

Yes, I think constructible numbers also includes the square root of -1. So this is the name for what I have in mind

But now that I think about, the question is a little bit more nuances than I think about

[charred kestrel]

the part i need help is that, in an informal way, all totally real numbers with just square roots can be made with the five operations

[north stream]

Well, not really but the questions is really confusing. I mean who writes a÷b??

[limpid lagoon]

could you phrase this more clearly?

[north stream]

No really, there are polynomials with totally real roots which aren't of even degree.

[charred kestrel]

sorry i guess, but its exactly the same as a/b

[north stream]

So can't be generated by square roots

[low orbit]

I think what you’re really claiming is: every totally real number that lies in a field obtained from \mathbb Q by successive adjunctions of square roots can be built using the five operations

[limpid lagoon]

from what I understand, this is wrong

[charred kestrel]

yah, this — i cant tell whats the difference

-# im sorry im really stupid you'll have to be patient with me 😭

[spice idol]

nvm spitting bullshit

[spice idol]

what are the five operations here

[limpid lagoon]

see above

[limpid lagoon]

.

[spice idol]

oh i see

yeah seems unlikely

just at a glance

[limpid lagoon]

@charred kestrel is this what you mean?

$R=\bigcup_{n\ge 0}​E_n​,\hspace{1em}E_0​=\mathbb{Q},\hspace{0.5em}E_{n+1}​=E_n​(\sqrt{a}:a\in E_n​, a>0)$

[broken turtle]

c2b7
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[charred kestrel]

yah, and we would want to prove $S_{\min}$ to be all the totally positive elements in $R$

[broken turtle]

Sunny — ping when reply plz

[limpid lagoon]

yes, im pretty sure this is false

this would be strictly larger than the pythagorean closure of Q

[charred kestrel]

why

explain more

[charred kestrel]

bump

[limpid lagoon]

if you denote P as the pythagorean closure of Q, and R as above, then to say P = R would be to say that every positive a \in P can be represented as b^2 + c^2 for b, c \in P

[north stream]

Just fourth squares are not totally real!

Like the solutions of x^4-2

[charred kestrel]

every totally positive a \in P can be represented as b^2 + c^2 for b, c \in P

[north stream]

I think you are confusing the fact that the pythagorean field F of Q contains the square root of every interger with the fact that it contains every single one of its square roots, that last one isn't true

I think troposphere put it already, try writting $a^2+b^2=\sqrt{2}+1$ or even $a^2+b^2=\sqrt{2}$

[broken turtle]

KevLee

[limpid lagoon]

https://en.wikipedia.org/wiki/Pythagoras_number @charred kestrel

p(Q) = 4, i think this is all you need

[charred kestrel]

how do i come to prove this?

[limpid lagoon]

https://en.wikipedia.org/wiki/Lagrange's_four-square_theorem#The_classical_proof

[charred kestrel]

no, im trying to state that it contains every single one of its square roots, provided the thing inside is totally positive

[charred kestrel]

oh god 😭

[north stream]

What is totally positive?

[charred kestrel]

all galois conjugates are positive

[north stream]

that would work

[limpid lagoon]

yeah, here is an explicit proof from Martin's Geometric Constructions

[north stream]

Nice

[limpid lagoon]

so @charred kestrel , i believe you are wrong

this book might be interesting to you in general since you are doing a very similar thing investigating powerpoint constructions lol

you should probably check it out

[charred kestrel]

i dont understand why this matters — do you mean that √(1+√5) is constructible qith √(a^2+b^2)?

[limpid lagoon]

that number is not in the pythagorean closure of Q

[charred kestrel]

yah, but the prediction is that its not constructible, since √(1+√5) is not totally real

for instance, its galois conjugate √(1-√5) is complex

maybe im expressing bad :-(

[limpid lagoon]

i understand what you're saying, yeah that example doesn't apply for your question. however, i think this paper (see the example after 1.2) shows that your more specific question is also not true.

[charred kestrel]

√(2+√3) can be constructed

[limpid lagoon]

oh i see that nvm

ignore that

[charred kestrel]

in fact, $\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}$ so its trivial lol

[broken turtle]

Sunny — ping when reply plz

[charred kestrel]

ok lemme see your example

are the examples even numbered

after proposition 1.2?

do you mean the $1+\left(\sqrt[3]2\right)^2$ example?

[broken turtle]

Sunny — ping when reply plz

[charred kestrel]

if so — my field extensions are all like this

[limpid lagoon]

@charred kestrel i had to work on some other stuff, but i finally figured this out. you are correct, look at theorem 2.2 of https://arxiv.org/pdf/math/0408159, and https://arxiv.org/pdf/math/0407174 for the general statement and proof

in particular, this is the general proof from the latter paper

[charred kestrel]

is https://en.wikipedia.org/wiki/Lagrange's_four-square_theorem#The_classical_proof still useful for me? i assume not?

[limpid lagoon]

all of the previous stuff i was talking about does not apply when you have the totally real restriction

[charred kestrel]

ok

[limpid lagoon]

why did you want to prove this again in the first place?

just knowing that a proof exists is enough right

for what you're doing

[charred kestrel]

i want to know how it works for fun, i guess

well, what we're currently currently trying to do is to get 11th roots or something in powerpoint, but i kinda want to understand more about this

i want to understand more

i don't understand why the step can follow like this:
Hence, each of the conjugates of (β+c/2)^2 are positive and (β+c/2)^2 is a sum of squares of elements in K_(i-1).

[low orbit]

I believe we use the fact that in a totally real field $K$, any element that is positive in every ordering (equivalently, “totally positive,” i.e., all real embeddings give $ > 0$) lies in the cone generated by sums of squares.

[broken turtle]

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

[low orbit]

The Hilbert-Landau-Siegel theorem.

[low orbit]

How funny. I literally skimmed through the first paper after I saw your question and somehow missed the key theorem.

[lime python]

If I have an algebra $(\mathbf{1}, \mu_{\mathbf{1}}, \eta_{\mathbf{1}})$ is it true that \mu_{\mathbf{1}} \lambda_{\mathbf{1}}^{-1} = id$, where $\lambda$ is the left unitor

[broken turtle]

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

[charred kestrel]

is there some simple way to prove this theorem 😭

(that i can understand)

[low orbit]

I believe the standard proof uses the local-global principles for quadratic forms over number fields.

Hasse–Minkowski

For quadratic forms over number fields, a value is represented over K iff it’s represented over every completion K_v

[charred kestrel]

what the fuck even is that

[low orbit]

(meaning all real/complex places and all p-adic places)

[charred kestrel]

what the fuck even are forms 😭

3 days ago, i spent 3 hours going from knowing what a field is to knowing what a galois conjugate is

and now im here 😭

that's what google slide does to a person

😭

stupid capitalism

[limpid lagoon]

homogenous polynomial

[charred kestrel]

whats homogenous

[limpid lagoon]

all of the terms have the same degree

so like x^2 - xy - y^2 is a quadratic form

[charred kestrel]

oh ok

what are local-global principles? and whats Hasse–Minkowski? 😭

im going to diiiiiiie
-# not serious, like at all.

[limpid lagoon]

https://en.wikipedia.org/wiki/Hasse_principle, https://en.wikipedia.org/wiki/Hasse–Minkowski_theorem

[charred kestrel]

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

how the fuck do i even begin to understand this.

[limpid lagoon]

the good thing is you don't need too lol

[charred kestrel]

wdym

[limpid lagoon]

what do you need to understand this proof for?

it's kind of beyond what your main motivation is, no?

[charred kestrel]

for fun

well

my main motivation is to learn more mathsTM

plus, that's not how autism works 😭

[buoyant fox]

shouldn't this be m and not n?

in the box

[low orbit]

What box?

[limpid lagoon]

they posted a question in an image, then deleted it

[vapid axle]

Can somebody help me see why 5.21 implies this highlighted part?

Ah this is just another application of Zorn's lemma huh

[vapid axle]

Also does anybody know why x^{-1} maps to 0? I am assuming that they are referencing the extended map from B -> \Omega. If so, why is x^{-1} in B?

[zenith pagoda]

holy shit what is your preamble

[spice idol]

g a y

[astral ginkgo]

Because it's in the maximal ideal m' that you're quotienting by

So after mapping to k' it's already 0

[astral ginkgo]

What is Sigma here

It seems they're doing a zorn's lemma construction starting from A', then just highlighting the fact that B contains A

But it contains A' by design thus x^-1

[low orbit]

,tc

[broken turtle]

Personal LaTeX configuration.

​ ​ ​ ​ ​ ​ autotex: Disabled (may be overriden by guild settings)
keepsourcefor: Don't delete source (may be overriden by the guild)
​ ​ ​ ​ ​ ​ ​ colour: Using the trans_black colourscheme
​ ​ ​ alwaysmath: Disabled
​ ​ ​ alwayswide: Disabled
​ ​ ​ ​ namestyle: DISPLAYNAME
autotex_level: WEAK

To see more detailed information use ,texconfig <option>.
To set an option use ,texconfig <option> <value>.

Preamble

Using a custom personal preamble with 4554 lines!
Use ,preamble to view or modify your preamble!

Dec 13, 2025, 5:01 AM

[wise sedge]

Can you always quotient a loop by its nucleus like how you can always quotient a group by its center?

[buoyant fox]

Can someone help me with galois theory problem?

[north stream]

drop it

[buoyant fox]

So x^n - 1 is solvable by radicals, it has galois group equal to the group of units of the integers mod n

The galois group has order phi(n) , totient function

let alpha be a root of x^n - 1. If kth roots are required in a minimal expression of alpha, show that k divides n.

for instance, galois group of x^10-1 = C_4 = phi(10)

primitive 10th root of unity can be expressed like so (it only involves square roots, and square roots divide 4)

My approach is to try and use tower law

The problem I am running into is the following: If I have a tower of extensions up to L, where L is the final extension, and I could show that this is a tower of simple radical extensions, I would be done. But I don't know how to prove this

L is splitting field of x^n - 1

in fact, I could just take the tower L over Q

[summer quest]

(Z/nZ)* is a finitely generated Abelian group and is in particular solvable, what more is there to say

[buoyant fox]

" If kth roots are required in a minimal expression of alpha, show that k divides n." why does this follow

The nuance is, it may be possible that the splitting field does not sit atop a tower of radical extensions

For instance, see the following case

[summer quest]

write x^n-1 as a product of cyclotomic polynomials

[summer quest]

n-th roots of unity are primitive d-th roots of unity for a unique d|n and x^n-1=\prod_d|n \Phi_d(x)

[buoyant fox]

but how does this show splitting field of x^n-1 is at the top of a tower of radical extension

I understand all about the galois group

but this isnt necesasrily true. See above example, K splitting field of x^3 - 3x -1 is a solvable galois extension, but its not a radical extension

and I understand that L/Q is actualy radical, since Q(z_n) = L over Q is a radical extension

but just viewing it like this doesn allow us to invoke tower law to conclude naything

we would need to construct a radical extension tower with more intermediate fields if we want to concldue anyhting about divisibility

[summer quest]

right yeah sorry I am sleepy and not thinking straight

[buoyant fox]

and when I try to construct such radical extension towers, I dont know how to guarantee that they are still ending in splitting field L

[north stream]

In this case it is a radical field extension since just droping the 10th primitive root of unity will work.

[buoyant fox]

you mean the tower Q(z_10) > Q right

[north stream]

It is not only square roots allowed, you just need something in the form of x^n-a and in this case, this just gives a Galois extension when you add the first root.

[summer quest]

maybe you want to use the fact that the Galois closure of a simple radical extension of degree n is the extension by a further primitive nth root of unity to deal with this possible issue around splitting fields/Galois closures

[buoyant fox]

hmm

I havent learned about galois closures

so Im assuming there is some kind of other approach

[summer quest]

the Galois closure is a very basic thing though

[buoyant fox]

yeah, its just its for a practice for an exam I am taking soon, and I feel like professor is intending for us to use ideas we have learned in the class to solve these kind of problems

If I am needing to learn extra theory, I feel like I am missing some key idea

[summer quest]

if you adjoin a new element to form a field extension, this may not be a Galois extension and instead you might want to force this to be separable and pass to the minimal possible extension which includes this new element and which is also Galois

intuitively the minimal way to do this is to adjoin this new element and also its Galois conjugates (that is what the splitting field is for after all)

[buoyant fox]

yeah we actually used such an approach, the thing is that when doing this approach, it no longer guarantees that L is the top of hte tower (or even in the tower at all) right?

[summer quest]

hmmm

[buoyant fox]

you are talking about such a procdure right

So there is something about L for x^n-1 that appears when we do such a procedure, we actually get the splititng field somewhere in it

and like repeating this procedure explicityl, we actually get it immediately as K_0,

but thats not what we want, cause that doesnt give us anything insightful via tower law about divisiblity

[buoyant fox]

Maybe if the following is true, ti could be used: "Let K be a field. Let L be a radical field extension. If we have a tower from L < M < K, and L to M is radical, then M to K is radical

i dont know if this is a true statement however

[north stream]

You mean if L=K(alpha) with x^n-a being the minimal polynomial of alpha?

[buoyant fox]

Well my strategy is

[buoyant fox]

yeah

so thats obviously radical right

its Q(z_n)/Q, which is radical

[north stream]

By definition, yeah

[buoyant fox]

lets say we need to adjoin kth roots, then there is some radical extension in between that and the splitting field, or the splitting field right

so "If M/K
and L/K
are both radical extensions then L/M
is a radical extension"

if this statement is true

it gives what we want right?

cause then we can just get the tower law

[north stream]

I mean, can you prove what is the Galois field of x^n-a?

[buoyant fox]

and show that k divides degree L over K

[buoyant fox]

in our case x^n-1 always has galois field group of units of Z/nZ

so its phi(n)

order

[north stream]

I mean, can you explain what you are trying to understand?

[buoyant fox]

second sentence here

i udnerstand setnecne 1, why its solvable by radicals, also what the galois group is (group of units Z/nZ is abelian so solvable)

I will write someting out

[north stream]

Well, this is just like, some Galois correspondance

[buoyant fox]

If L/K has cyclic galois group of order p, does L(z_p)/K(z_p) have cyclic galois group of order p?

[last talon]

Yes
L(z_p)/K has degree at most p(p-1), by L(z_p)/L/K
It has degree divisible by the coprime numbers p, p-1 as it contains L, K(z_p), so the degree is exactly p(p-1)
Therefore by tower law again, L(z_p)/K(z_p) has degree p, and as we know the minimal polynomial of z_p in L is the same as in K (by the degree computation we did), it follows that every automorphism of L/K extends to one of L(z_p)/K(z_p)

(assuming p is a prime)

[lone jacinth]

K(zp)/K doesn't necessarily have degree p-1, but it does have degree relatively prime to p which is what matters I guess

[last talon]

Ok yeah
Same argument goes through with [K(z_p) : K], which divides p-1

[buoyant fox]

So would this be a valid argument

[last talon]

What’s the question that’s solving?

[buoyant fox]

really just the second sentence, the first sentence is straightforward

[lone jacinth]

Do you have a question here?

[vapid axle]

Right, I read some notes and they mistakenly did Zorn's Lemma involving rings containing A, but with A' makes more sense

[astral ginkgo]

Yea otherwise you really can't guarantee that x is inside

[limpid horizon]

I’m trying to show:

For M a module over noetherian R and x in R a nonzerodivisor on M:

If p in Ass(M), and q is a minimal prime over p+(x), q is in Ass(M/xM)

If someone has a hint would be nice.

I have a map R/(p+(x)) -> M/xM, and q is an associated prime of the domain. So was wanting to use facts about how Ass and Supp transfer through SES’s

But If kernel is all of R/p+(x) then i dont have a nonzero quotient of the domain as a submodule of M/xM

If kernel is not everything then it works out

I just did a silly example of:

M = Z[x]/(x), R = Z[x], 4 is a nonzero divisor, and (x) is an associated prime of M

(x)+(2) is a minimal prime over (x)+(4), and yeah (x,2) is an associated prime of M/(4)M = Z[x]/(4,x) ( (x,2) is prime and the annihilator of 2…)

idk if that example helped me with anything. Also hoping i didnt even mess up that example

[buoyant fox]

So L over K has degree p. But how do we know anything about degree L(z_p) over L? What if L contains z_p or some other expression involving pth roots of unity?

[last talon]

It’s a standard fact of cyclotomic extensions that for any field M [M(z_p): M] divides p-1 for any prime p (as every pth root of unit has the same degree minimal polynomial)

(Cf Jagr’s comment)

[lone jacinth]

To see this fact: not that
M(zp) = M(zp^k) for any 0 < k < p. This they all have the same degree.

So when factoring (x^p - 1)/(x - 1) into irreducibles, they all have the same degree. Thus the degree divides p-1

[buoyant fox]

I see

Thank you so much yall

[wise sedge]

Is every irrep of H over C 2-dimensional?

[summer quest]

What do you mean by representation here

I assume you mean as modules in which case yeah Artin Wedderburn explains this

[broken wren]

An R-linear representation of H over C extends to H tensor C, which is isomorphic to 2x2 matrices over C

[buoyant fox]

Can someone help me identify an error in a proof?

[ruby otter]

yes

[limpid horizon]

I love the just ask!

[buoyant fox]

I can’t see what the mistake is, except for perhaps the statement boxed in stars

But my professor says the starred statement is probably true, whereas the overall statement is false, so I am confused

[lone jacinth]

I'm confused what it is you actually want to show.

Obviously nth roots of unity are expressable in radicals as you can express them as the nth root of 1.

And what exactly do you mean by a radical of index k being required?

[buoyant fox]

For example, to express the 3rd root of unity as a tower of simple radical extensions, you need radicals of index 2, (1 +sqrt(3))/2)

You could also write that as 4th root 9, but it’s not needed

Only square roots are NEEDED

[lone jacinth]

So what is the overall statement that is false?

[buoyant fox]

My professor claims that “If kth index radicals are needed, then k divides phi(n)” is a false statement. But I’m confused as it appears I have a proof that shows its true

The idea of my proof is contra positive, I Show that there is a tower of simple radical extensions containing the splitting field of x^n-1 where the radical index of each extension is something which divides phi(n).

This shows that obviously things which don’t divide phi(n) are “not needed”

Do you have any idea what might be wrong with the proof, if at all? Is it the statement boxed in stars? My professor says she thinks the star statement is true, which leads me to believe there might be some technical error on the last page I am not seeing

[lone jacinth]

I mean you're also adjoining phi(n)th roots of unity, so I guess phi(phi(n)) roots might possibly be necessary.

[wanton spoke]

a cyclic extension of prime degree q is obtained by adjoining a q-th root that’s false in general it’s only true if the base field already contains the q-th roots of unity

Kummer condition

[lone jacinth]

I think they are adjoining qth roots of unity, but maybe I'm not understanding the notation.

But yeah if you're adjoining roots of unity, then the point is kind of moot anyway

[buoyant fox]

But the tower does contain the qth roots of unity

Look how I constructed it

[buoyant fox]

So where is the step of the proof which is in error

[lone jacinth]

Like at the end you say you have a tower where every step is degree qi, but it's just not true

Because you're also adjoining roots of unity

[buoyant fox]

Not degree q_o

I say “radical index”

[lone jacinth]

Them I guess I have no idea what you're trying to say

[buoyant fox]

As in it’s not clear what radical index is?

Radical index of an extension K(alpha) over K is the smallest integer k such that alpha lies in K

[wanton spoke]

even if the final field contains the q-th roots of unity that’s not enough Kummer needs the q-th roots of unity to be in the base field at the moment you take the q-th root step

[lone jacinth]

But then no radical index is ever needed.

You could just say
Q < Q(z) has radical index n, so no other radical index is needed

So that can't possibly be what that statement is supposed to mean

[buoyant fox]

But n does not divide phi(n)

[buoyant fox]

Yes they are

[lone jacinth]

So then the statement is obviously false then

[buoyant fox]

No because just by showing there is a tower with radical index n, it doesn’t necessarily mean that nth roots are needed

[lone jacinth]

Or I guess vacuously true

[buoyant fox]

For instance cube roots are not needed to express 3th root of unity

[lone jacinth]

Well what is your definition of needed? Because square roots can't be needed if it can be done with cube roots

[buoyant fox]

Interesting point

Let me think about that

[lone jacinth]

Like at one moment you're saying
Q < Q(z_q) has radical index q, so we're using qth-roots. Next you're saying we're not using cube roots for Q(z_3)

It can't be both

[buoyant fox]

Okay, how about the following reformulation of the question: it’s possible to form a tower of simple radical extension with radical indices only dividing phi(n)

[lone jacinth]

Then that's what you've proven I guess.

Though it seems radical index only makes sense for an element, not an extension (Q(z_3) = Q(sqrt(-3))

[buoyant fox]

Well perhaps the radical extension of a degree is not unique

So you don’t see any error with the proof I’ve attached?

Even with the star box statement?

[lone jacinth]

Looks fine yeah.

Just not really proving what was claimed I guess

[buoyant fox]

Yeah my professor removed the problem from the problem set, and said it was incorrect/vaguely stated

But I am still curious about it

[glacial tangle]

is it true that mR_m neq R_m if m is maximal and we see the inequality in R-Mod,
proof is ig trivial using a charterisation of elements in the jacobson radical, just wanna ensure

[broken wren]

First of all, for any ideal I, RI=I
Indeed, for any module M, RM=M. That is kind of the definition of a module. A (left) ideal is the same as a (left) submodule of R

So RI=I so if you ask RI=R, you might as well ask I=R
If I=R, then 1 is in I. And conversely.
By definition, maximal means excluding the possibility of the whole ring

[lone jacinth]

What do you mean by seeing the inequality in R-mod?

[glacial tangle]

just making presice that this isnt in R_m Mod (which wouldnt make much sense)

[lone jacinth]

Why wouldn't that make sense?

Besides if you're asking for equality why would it matter which category you're in?

They're not equal whether you think of them as R modules or ideals on R_m or just sets

Are you working with localizations of noncommutative rings?

[glacial tangle]

i just wanted to proof that in the category of R_moduls the equation (m maximal ideal )mR_m = R_m doesnt hold using jacobson radical and wanted to know if this is a standart procedure to proof this._.

[lone jacinth]

I guess I'm still a little confused about what you want. Like a proof that only uses that R_m is an R-module and no properties of localization?

[glacial tangle]

ye i have to just use elementary calc

[lone jacinth]

Like it's very elementary to show, but if you don't have a definition for R_m I don't see how you could

I mean you don't need to use anything more than the definition.

What is "the approach"?

[glacial tangle]

use

[lone jacinth]

This is a characterization of the Jacobson radical

Like you can just say
x/y = 1/1 means there exists s such that
sx = sy

But that can't be because m is prime

[glacial tangle]

ah i just looked at x from ideal, y from the ring and z from R setminus m x y/1 = 1/1 and this gives z(1-xy) = 0 which also gives a contradiction

[jade valley]

I am doing algebra 1 and 2

[digital parcel]

#prealg-and-algebra

[jade valley]

Done with algebra 1 going to algebra 2

[digital parcel]

i have a sneaking suspicion that channel might be more relevant for you

[spice idol]

beyond where the light touches...

[last talon]

(The light touches #advanced-algebra and #alg-top-geo-top only
Maybe #groups-rings-fields and #category-theory on good days)

[spice idol]

I bring the light with me whenever I enter #foundations

[jade valley]

Is there graphing there?

Graph comparisons?

Looking to find this

[spice idol]

#prealg-and-algebra

this is a channel for algebra in the sense of algebraic structures and representations

[jade valley]

1/50 x 50/1=1

[spice idol]

so true, man

[jade valley]

100%

[buoyant fox]

Would need to consult an expert about this one

[jade valley]

Hi

Nice to meet you

I am getting a scale today

going to learn how to use it

I am good but not 0

I am going to do that!

Close! But could be better!

Drinking a milkshake

Doing math equations

[ornate atlas]

Yo, as people have pointed out, none of this is really appropriate for here, you’re looking for #prealg-and-algebra

[limpid horizon]

W

[limpid horizon]

W

[scarlet prairie]

could someone help me with 2.14?

i dont see the link at all

ah okay, is it just because an ideal does not contain invertible elements and so its closure won’t contain either(by 2.12 we can find a ball around every invertible element that has empty intersection with ideal)

[jade valley]

Yall doing anything interesting?

[limpid horizon]

I am watching youtube then gonna go sleep soon

[neat nexus]

Yeah, combine with the fact that closure of an ideal I, denoted by cl(I), is still an ideal. (x_n in I ->x then rx_n -> rx). I<=cl(I)<=the algebra and your cl(I) can’t be the algebra by your reasoning

[limpid horizon]

M fg over noetherian R. p in Ass(M) and x is a nonzerodivisor on M. If q is a minimal prime over p+(x) then q is in Ass(M/xM)

Any help would be appreciated

[neat nexus]

How is p+(x) defined though, p a prime ideal of R, (x) is a submodule of M

Or I misunderstood, x is in R?

[lone jacinth]

x is in R yes

[neat nexus]

if you can show q is in Ass(M) then done, though I don’t know whether it’s true, if true, Ass(M) is contained in Ass(xM) cup Ass(M/xM), and q is not in Ass(xM), thus q is in Ass(M/xM)

[limpid horizon]

Not sure what ur referring to when “though i dont know whether its true, if true …”

[neat nexus]

Whether q being in Ass(M) is true

[spice idol]

i am a mature adult
i am a mature adult
i am a mature adult

[limpid horizon]

p in Ass

[neat nexus]

But q is in Ass(M) or not I don’t know. If q is in Ass(M) then done.

[spice idol]

ghost ping

[limpid horizon]

Lol

[limpid horizon]

Yes q not in Ass(xM) because q contains (x) , and x cant kill some xm’?

[neat nexus]

Yeah.q=Ann(xm) then x(xm)=0->xm=0 contradiction

I don’t know whether we can show that: we take elements in Ass(M) that is minimal and contains p+(x), then this minimal among Ass(M) may can be shown to be the minimal among all primes

Anyway I provided too much unverified guess…

[limpid horizon]

Nw its appreciated

[hushed bone]

No way that q is in AssM in general

If p isn’t an embedded prime this is easy

Because M has a filtration where one of the quotients is isomorphic to A/p, and then when you mod out by x you get A/(p + x) which has q as a minimal prime so it must be associated

Well, I guess you have to argue this is minimal among all the new primes appearing in this filtration blah blah blah

It’s clear if M is A/p

[limpid horizon]

I was originally trying something like looking at the map A/(p+x) -> M/xM

[hushed bone]

And there should probably be a way to reduce to this

Oh this works

Hurb

This is an injection

And so associates primes of the former are associated to M/xM

And q is a minimal prime of the former so it’s associated

[limpid horizon]

This is injection?

[hushed bone]

I don’t know why I threw away this map when I was reasoning about it earlier

[hushed bone]

I mean it should be, I think

[limpid horizon]

Thats what i was getting hung up on but i may have missed smth

[hushed bone]

I mean the kernel is isomorphic to Tor_1(M/(A/p),A/x)

[limpid horizon]

O shit bringing out tor

[hushed bone]

I mean

lol

We’re asking if it’s injective and Tor tells you kernels

[limpid horizon]

I actually havent used tor at all really yet

I probably should look into it more

[hushed bone]

I think the kernel is isomorphic to like

[lone jacinth]

What is the map exactly?

Like sending 1 to m + xM where m has annihilator p?

[hushed bone]

(p:x)/p as an ideal of A/p

Errrr

[lone jacinth]

Because if x is not a zero divisor then surely m and xm both have annihilator p, but
xm + xM = 0

[hushed bone]

Okay yeah

[lone jacinth]

It might be that you can pick m to make this injective, but I don't think it's trivial

[hushed bone]

Yeah I think maybe not

But if you can make it so the kernel stays in q or something you win

Grrrr

I was trying to be really cute with something with local cohomology

[lone jacinth]

Is there some kind of relationship between Ass(M) and primary decomposition of Ann(M)?

[hushed bone]

I forget if it’s with Ann(M) but Ass(M) is related to primary decomposition yes

[lone jacinth]

If so I guess looking at the annihilator of Ann(M/xM) would be helpful

[hushed bone]

Oh I do have a proof

P being associated to M with fg is equivalent to having H^0_PA_P(M_P) nonzero

because for a local ring having the maximal ideal associated is the same as degree 0 local cohomology being nonzero

also note that in degree 0 local cohomology is fg because it's a submodule of M

take the SES 0 -> M -x> M -> M/xM -> 0

Note that local cohomology commutes with localization, and so H^0_P(M)_P is nonzero

Which also means that H^0_P(M)_Q is nonzero because this can localize even further to H^0_P(M)_p

Note also that H^0_P(M/xM) = H^0_(P + x)(M/xM) because M/xM is supported over A/x

Since Q is minimal over P+(x), in the ring A_Q Q is the only prime ideal minimal over P + (x) and so sqrt(P + (x)) = Q, thus H^0_(P + x)(M/xM)_Q = H^0_Q(M/xM)_Q

We thus have the SES
0 -> H^0_P(M)_Q -x> H^0_P(M)_Q -> H^0_Q(M/xM)_Q

Since H^0_P(M)_Q is nonzero and fg, and x is in the maximal ideal, the first map is not sujrective by Nakayama

so H^0_Q(M/xM)_Q is nonzero and Q is associated to M/xM

There is most definitely most certainly most absolutely a simpler proof without local cohomology

I like local cohomology :)

[hushed bone]

Says SES, there is no 0 at the end though. SMH, epic fail, what the helly

[limpid horizon]

Looks cool

I did see that section in eisenbud about local cohomology and primary decomposition but havent read it yet

[jade valley]

Yall = today?

[digital parcel]

hang up the cleats bro ure done

[hushed bone]

More like amanono

[digital parcel]

idr if i mentioned this but the proof (at least, one of them) that Schubert varieties (at least in the grassmannian) are cohen-macaulay uses local cohomology

[hushed bone]

Probably

[digital parcel]

jumpscared me

[hushed bone]

You can maybe avoid it but being CM is the statement that local cohomology is concentrated in a single degree

[digital parcel]

i didn't read the proof tbh i just noticed it used local cohomology

sorry I should've said arithmetically CM

(ie the cone is CM)

because the proof takes the cone and checks the local cohomology at (0)

[hushed bone]

Sure

[digital parcel]

yea

[limpid horizon]

yea

[limpid horizon]

Hello beautiful plant grower

[spice idol]

what is local cohomology even about

[hushed bone]

Uh

Cohomology with support

[spice idol]

⊕ is ∏ with compact support ahh

[limpid horizon]

Lol

put "ahh" at end = me chuckle and giggle

[spice idol]

"ahh" ahh

""ahh" ahh" ahh

[limpid horizon]

😂 😂 😂 😭 😜 😮 😝

[digital parcel]

closed captions at a haunted house:

[limpid horizon]

im DYING bro 😂

[spice idol]

average yt shorts comment

son im crine 💔 🤣

[spice idol]

[hushed bone]

Stop this tomfoolery. You are acting in a way that denigrates the noble spirit of algebra.

[digital parcel]

"denigrates" shakespeare not dead ahh

[spice idol]

excuse you i am very honorable
..
being i guess

[hushed bone]

https://tenor.com/view/blue-role-blue-aqua-role-aqua-aqua-roles-gif-12242037083989466285

[spice idol]

pipe down? can anyone put it back up?

[digital parcel]

i think they got medication for that

[spice idol]

oh yeah im gonna need that

esp in my old age

[digital parcel]

use my promocode

[fierce steeple]

Pipe down gold role

[worldly zealot]

read 24 hours of local cohomology

it’s really good

[spice idol]

i was kinda serious though, whats the intuition for it / what does it measure?

[digital parcel]

i had a prof recommend huneke's notes for motivation

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

[spice idol]

ah tyvmm

i will absorb this information happily

[hushed bone]

I gave you an actual answer

[spice idol]

not to me but i scrolled up and now i see

"cohomology with support" is just as helpful as saying that universal algebra is the representation theory of small categories where each object is some finite power of a fixed generic object

lmao

[hushed bone]

I mean it’s the way you tend to model cohomology with support for schemes

Affine schemes are a local model and the algebraic gadget there is local cohomology

You globalize to all schemes and end up with what is cohomology with support

[broken wren]

For a space X and a point x, you can define the local cohomology as H^*(X,X-x)
If your space is locally the cone on another space, you get the shifted cohomology of that other space. In particular, a manifold is locally the cone on a sphere, so the local cohomology is the cohomology of a sphere, ie, rank 1 and in a single degree. For a space that is singular it gives you a measure of the singularity

For schemes you can do this with etale cohomology. But usually when people say local cohomology, they mean coherent cohomology
H^*(X,X-x;O)
Which is related to topological cohomology the way coherent cohomology is related to topological cohomology. Eg, a node has deviant local cohomology

[jade valley]

I am awesome!

[toxic trout]

Very humble indeed

[near lantern]

Let G be a semisimple (or reductive) algebraic group over ℂ with Lie algebra g. We know that there are finitely many conjugacy classes of nilpotent elements of g. Fixing a maximal torus T, is every conjugacy class (as above) represented by a weight space for T?

Nvm this is trivially false for SL_n (all weight spaces have JCF of type (2, 1, ..., 1) and so represent that same conjugacy class of nilpotent matrices) I messed up the calculation

Love it when I solve a question when trying to write it up here

[broken wren]

A nilpotent conjugacy class in a semi simple Lie algebra has 0 in its closure

Consider the class functions, the functions on G invariant with respect to conjugation. For a compact real Lie group, it is easy to see that this is functions on T/W. But this is an algebraic construction that doesn’t really change when you pass to complex coefficients. So it’s still functions on T/W. in the compact case G/ad = T/W. in the complex case G/ad is not a proper quotient. But the invariant functions are the same as on T/W. so continuous class functions cannot detect unipotent elements, or equivalently the identity is in the closure of the unipotent classes

I think that’s relevant to weight spaces

[cyan osprey]

Solved it yet?

[limpid horizon]

In b) you couldve also just said for all maximal associated primes right

Which i guess is better cause theres less to check maybe

[dull cedar]

Yeah you can just check the maximal associated primes if it’s zero there it’s zero everywhere using all associated primes is just the standard way to say it

[limpid horizon]

Not actually sure how this proof goes

I figure we use correspondence theorem on submodules to get the chain

But not sure how to show say M2/M1 = R/P1

Also why do the submodules satisfy ACC

M fg over R noetherian implies M is noetherian module?

[spice idol]

yes

direct sum of Noetherian modules is Noetherian

so R^n is Noetherian

and quotients of Noetherian modules are Noetherian

so M is Noetherian

[limpid horizon]

Ah ok cool

[spice idol]

that is by definition

oh lol the book has an indexing error

it should be Mi+1/Mi = R/Pi+1

that way M1/0 = M1 = R/P1

and let P2 be an associated prime of M/M1. Then we define M2 to be the (unqiue) submodule of M such that M2/M1 \cong R/P2

repeat this process and you get an ascending chain 0 < M1 < M2 < M3 < ... where each Mi+1/Mi is isomorphic to R/Pi+1 where P is a prime ideal.

But M satisfies the ascending chain condition, so this chain must stabilize somewhere. The only way this can happen, though, is if it eventually reaches M. Else we would be able to find another prime to continue the chain

[neat nexus]

Which book is it from?

[limpid horizon]

Eisenbud’s Commutative Algebra book

Commutative Algebra (with a view towards alg geo) is the full name

[neat nexus]

Thank you

[last talon]

How does the hint for 2 help at all?
Like, the field of fractions of A_{x} is literally just the field of fractions of A, unless 1/x is in A or x = 0, in which case it’s 0

[hushed bone]

The localization isn’t finite type over K anymore

It will, I think, be finite type over K(x)

So the transcendence degree goes down by 1

I mean, I’m not entirely sure what this exercise is shooting for, I imagine it’s for the “dim A = tr deg_k Frac(A)” when A is a finite type domain over k

so this is where this could end up being important

[last talon]

It’s for dim k[t1, …, kn] = n

[hushed bone]

Okay then this will work I think

I mean you’re trying to establish a bound on the dimension so I guess the finite typeness doesn’t matter

But the point is, I think, you’re incrementing the dimension by 1 and also the transcendence degree by 1 because you’re changing the base field

What I said is for x transcendental btw

Like I think what you want to do is take a transcendental element over k, then if tr deg_k A = n you have that tr deg_k(x) A_{x} <= n-1 (I imagine it’s honestly equal but whatever)

And now you need to just establish dim A_{x} <= n-1

By 1 c)

[last talon]

For x algebraic it’s just the 1/x case that I already showed

[hushed bone]

And that’s the induction

[weak lodge]

See channel description

[limpid horizon]

Why don’t you go to #discussion ? If you just want to chat with people

But these algebra channels are about the study of “algebraic structures”

[limpid horizon]

What happened to that guy

[spice idol]

extermination

[limpid horizon]

bro 😢 🥺

rip BeautifulPlantGrower

[humble plover]

I have a question: How do I begin to solve this?
Translation of question: Let E be a finite 'body' (I think this means field but I am unsure) with 3^6 elements. (1) Let {theta}: E -> E be an isomorphism of the body (ring) E. Define E^{theta} = {a element of E | {theta}(a) = a}
Show that E^{theta} subset or equal symbol (idk how to write this) E be a subring of E and that E^theta is a field (or body, whatever it may mean)

[hushed bone]

Yes, it means field

I mean do you see how to show it’s a subring first?

After you do that you just need to show that it’s closed under inverses

I think these verifications are frankly, really straight forward. Just literally show the axioms hold

[humble plover]

Wait, I think I get it

It's nonempty because we can just say like theta(0) = 0

Closed under subtraction because theta(a-b) = a - b = theta(a) - theta(b)
and use the same reasoning for multiplication
theta(ab) = ab = theta(a)theta(b)

And since these properties hold it is a subring?

Showing that it's closed under inverses is basically the same thing, no?
theta(a^-1) = a^-1 = theta(a)^-1

I'm a bit new to this whole algebra thing, so the language still confuses me somewhat, thanks for the guidance!! :D

[hushed bone]

Yeah

[limpid horizon]

chmoney

[primal bramble]

Hey y'all - trying to work on 9.8 (1) from this exercise. I've shown the <= direction and have been told I should be able to use 9.9 (which I proved already) to show the >= direction, but not really seeing how it fits in. Any suggestions?

[hushed bone]

What I cd_R

Cohomological dimension?

[primal bramble]

Yep!

[hushed bone]

You don’t need 9.9

Let’s say B is an A-algebra and M is a B-module

Then M is naturally an A-algebra by restriction of scalars

Are you familiar with this?

[primal bramble]

Yep sounds good so far

[hushed bone]

Let I be an ideal of A

Then H^i_I(M) is equal to H^i_IB(M), where we make the latter module an A-module by restriction of scalars

For this, it suffices just to show that Gamma_I(M) = Gamma_IB(M) (natural isomorphism)

But this is obvious, like, the former is elements of M killed by a power of I, but uh, how does I act on M? As IB, cuz you need to push I through into B to even have an action on M

Is that good?

[hushed bone]

Also sorry this should say A-module

[primal bramble]

Gotcha gotcha

I think I'm good with everything you've said so far, just trying to see how to "plug in" this problem into that setup

[hushed bone]

Yeah yeah

I’m almost there

So if IM = 0, then M is naturally an A/I module. Two ways to see this or phrase it

Just by elements

The difference by something in I doesn’t change what rm is, because well, IM = 0

Or, A/I(x) M = M (this = is kind of a lie tbh)

Cuz A/I (x) M = M/IM = M

So what I really ACTUALLY want to say

There’s an adjunction from restriction and extension of scalars

So for any A-module M and A algebra B there’s a map M -> M (x) B (where you do restriction of scalars back to A on th right)

[hushed bone]

So the really key thing here is that if IM = 0, and you take B = A/I this is an isomorphism which is what this says

So the point is, every M with IM = 0 arises via restriction of scalars of “itself” from A/I

[hushed bone]

So let’s return to the problem at hand

Ann(M) is the largest ideal I where IM = 0

And so by what I said above M is gotten by doing restriction of scalars along itself

So H^i_a(M) = H^i_(aR/Ann M)(M)

So the vanishing over R and over R/Ann M are exactly the same

The explanation I gave was admittedly a bit over the top, but i wanted to illustrate some concepts and put things in a way that would help you later

[primal bramble]

Yeah for sure! I appreciate it, will definitely go over all of this in detail

[hushed bone]

Sorry the other reason I was kind of talkin about this in such generality

Im going to assume you have done some amount of AG?

[primal bramble]

Yeah some!

[hushed bone]

Okay, well if F is a sheaf with support in Z

Say i:Z -> X is the inclusion

Then F is isomorphic to i_*i^*F

this is literally saying the exact same thing as what I said before about the push-pull for rings when IM = 0

Well when you have an i_*F and i is the inclusion of a closed subset you have H^k(X,i_*F) = H^k(Z,F)

So what this says is, because of Grothendieck vanishing, which says if F is a sheaf on a Noetherian space of dim n, that H^i(X,F) = 0 when i > n

If F is supported on an even smaller space than X, say X is dim 10 but F is supported on a dim 2 thing

Then H^3(X,F) = 0, because this is H^3(X,i_*i^*F)

Which is just H^3(Z,i^*F) which is 0 cuz Z is dimension 2

This is basically exactly what was going on in the proof I presented for local cohomology, the cohomology is the same even when you look over the “right space” that the module or quasi coherent sheaf lives over

That being its support or well, R/Ann M

I hope this makes some lick of sense

[primal bramble]

I'll have to think about it a bit more but I believe so 🙂‍↕️

one quick question: it's possible I'm overlooking something really obvious, but isn't what we've argued then that cd_R(a,M) = cd_(R/ann M)(a,M) rather than cd_R(a,M)=cd_R(a,R/ annM)?

[hushed bone]

hmmm

Okay, what’s the definition of cd_R(a,M) you have?

Oh wait wtf, I totally misread the thing HUH

Okay, so what I did do was turn that problem into this

[primal bramble]

Hey no worries, definitely a worthwhile discussion regardless!

[hushed bone]

If M is such that Ann M = 0, show that cd(a,R) = cd(a,M)

Because I showed you can replace R with R/Ann M lol

[primal bramble]

Makes sense yep

[hushed bone]

Hmm

Okay I think I see

lol

[primal bramble]

Nice 😎

[hushed bone]

So now you can apply 9.9

Since the support is all of Spec R

And so I think it wants you to do something like take H^i_a(R) to be H, and relate H^i_a(M) to H^i_a(R) (x) M

I don’t think these two are equal though…

[primal bramble]

Ooooh by replacing with R/Ann M sure sure

[hushed bone]

But the other problem gets you the latter of the two is nonzero

So maybe you can show that this proves H^i_a(M) is nonzero

If you do that you get >=

[primal bramble]

Okay that does sound like a very reasonable direction to go in

I'll give it a shot. I really appreciate the suggestions!

[hushed bone]

I feel like this is harder to do when you take H^i_a to be derived functor of Gamma_a

Because you take injective resolutions and that gets destroyed when you tensor

But there’s also a definition as colim Ext^i(R/a^n,M)

And I think Ext tends to play nicer with tensor

[primal bramble]

Interesting

[hushed bone]

I still don’t see it tho

[hushed bone]

Idk if you’ve seen this, but this is just because Gamma_a(M) = colim Hom(R/a^n,M)

Any Hom(R/a^n,M) is just the a^n-torsion of M, and the colimit is the union of those

And then because direct limits are exact the derived functor of this is just the direct limit of the derived functors which are the Exts

[primal bramble]

Yeah I've just looked at it for a bit and I'm a little perplexed myself hmm

I kind of feel like there's gotta be some random lemma from earlier in the book I'm supposed to use lol

[hushed bone]

@primal bramble okay sorry so I had a slight error here. One thing you also need to do is verify that if I is an injective B-module that it is Gamma_a-acyclic. This can be done, and I’m just gonna ignore that fact

You can more directly prove the equivalence of H^i_a(M) and H^i_aB(M) if you wanted to avoid that

[hushed bone]

@primal bramble I got it

So for the <= direction which you got, you ended up showing the following, if n = cd_R(a,R), then H^i_a(M) = 0 for all i > n

This means that H^n_a is right-exact, because the cokernel would be an H^n+1_a which is gonna be 0

Now let M be a finitely generated (presented) module and take a presentation
0 -> F’ -> F -> M -> 0 where F,F’ are finite free modules

Now look at the natural transformation H^n_a(R) (x) - -> H^n_a(-), (do you see why there’s a natural transformation? This is similar to how you show Ext commutes with tensor when the first guy is finitely presented)

This natural transformation is an isomorphism when you input a finite free module, because both sides distribute the direct sums so it just reduces to checking that H^n_a(R) (x) R -> H^n_a(R) is an isomorphism… yeah. Clear

[hushed bone]

Now take this and apply this natural transformation to get

H^n_a(R) (x) F’ -> H^n_a(R) (x) F -> H^n_a(R) (x) M -> 0 -> 0
As the top row mapping down to the following:
H^n_a(F’) -> H^n_a(F) -> H^n_a(M) -> 0 -> 0

The outside 4 arrows are isomorphisms so the middle arrow involving M is an isomorphism by the 5-lemma

So this shows that H^n_a(R) (x) M = H^n_a(M) for finitely generated M, now take direct limits to see this holds for all M

So actually the two functors are naturally isomorphic at the cohomological dimension of R!

[hushed bone]

Now this proof works

[kindred mauve]

Best software to implement algorithms about elliptic curves (Shanks-Menstre, baby step giant step,...)?

[golden osprey]

Python via SageMath if you just want to practice implementing things

If you want something performant for large scale application then you shouldn't implement these things yourself most likely

[dull cedar]

<@&268886789983436800>

[fierce steeple]

I remember thinking about this a while ago, but not sure if there is an easy answer. Fix a field k (say char not 2 or 3 if you want). Is the abelian Lie algebra functor the unique (up to (unique?) iso) functor Vect_k -> Lie_k which is the identity on underlying vector spaces?

I guess an easier way to phrase it is to show that such a functor lands in the category of abelian Lie algebras lol (in which case it's obviously unique). Then we can probably do this just by checking k (which is obviously sent to an abelian Lie algebra) and then considering filtered colimits and direct sums

I guess yeah Lie_k -> Mod_k should preserve all small colimits (and is easily conservative) so your functor Vect_k -> Lie_k preserves colimits and sends k -> abelian lie algebra k, so ur done

[near lantern]

The forgetful functor from Lie algebras to vector spaces preserves coproducts?

[limpid horizon]

Can I ask what a lie algebra is or is that too basic a question to ask here lol

I just dont know any lie theory

[urban granite]

tangent space at identity element of a Lie group

or just a vector space over a field equipped with a Lie bracket

[weak lodge]

A vector space equipped with an alternating bilinear form [-, -], called the Lie bracket, that satisfies the Jacobi identity (meaning that [X, -] is a derivation for all X)
Like man1fold said, the usual examples are T_eG for G a Lie group or algebraic group

[near lantern]

Unfolding your argument I think it amounts to the claim that F(V (+) W) = F(V) (+) F(W), but the latter is the coproduct of F(V) and F(W) modulo [F(V), F(W)], so this means that F(V) and F(W) commute in F(V (+) W). And indeed given this the Lie bracket has to be 0.

[near lantern]

Equivalently you want a Lie algebra structure on every k-vector space such that every linear map V → W is a Lie algebra homomorphism. Taking W = k we see that every hyperplane contains [V, V], so it must be 0.

Surprisingly short.

[fierce steeple]

Very nice

I was actually unsure if there are any other kinda good examples of this phenomenon lol

I guess you could potentially play a similar game with G-representations, using the same argument you just used

i.e. every map arises from a G-equivariant thing, which sounds very unlikely unless the structures are all trivial

[near lantern]

Let k be mapped to the 1-dimensional character chi. Then every map k -> V is equivariant, which is the same as saying that every v in V is a eigenvector with weight chi.

[broken wren]

Right, it could be any 1d character

[near lantern]

Another idea: the functor N must be left-exact (the map from N(lim_i V_i) to lim_i N(V_i), where N: Vect → Rep(G) is a map in Rep(G) but is the identity on underlying vector spaces; so we can use the conservative property of the forgetful functor). So(?) it is of the form Hom(V0, -) for a (technically right, but it's all the same) kG-module V0. Now dimension reasons imply V0 is 1-dimensional.

We could also show right exact and do V0 (⨯) - since the forgetful functor also preserves colimits here.

Also by exact I meant including infinite limits/colimits.

[fierce steeple]

oh wait yeah just tensor by any character lol

[fierce steeple]

lol

[primal bramble]

wonderful! Thank you again 😎

[wanton heron]

Why are the two kernels equal again

[last talon]

ker(p) is a subset of L’ = p^{-1}(N’) as it’s just p^{-1}(0)
And L’ -> N’ is just the restriction of L -> N, so it’s kernel is ker(p) \cap L’ = ker(p)

[wanton heron]

I'm so stupid 😭

Thanks

[wanton heron]

I cannot figure out the diagram chasing part of the argument

Starting with an element a in the kernel of M' \otimes N to M \otimes N . I can lift a to M' \otimes L since beta is surjective

Now a' in M' \otimes L pushed by the maps M' \otimes L to M \otimes L to M \otimes N is 0 as well

Now Idk what to do

[wanton heron]

Okay i think this step is not needed
Lift a in kernel in M' \otimes N to a' in M' \otimes L
Then push a' to M \otimes L to M" \otimes L

a' maps to 0 under this push as the middle sequence is exact

so from injectivity of alpha only 0 maps to 0 in M" \otimes L. Then we can drag this element back to an element in M \otimes K and the. Again to an element in M' otimes K

Now all I need is this element in M' \otimes K maps to a' in M' \otimes L, then from the composition and exactness it would end up as a is 0 which is what we wanted

[wanton heron]

Okay this is because the middle sequence starts with an injective map

[limpid horizon]

Where are these notes from?

can u link them here

[wanton heron]

It's qing liu 's algebraic geometry book

[limpid horizon]

ok thanks

[limpid horizon]

Tbh that was kinda tricky to me too

[wanton heron]

I have wasted enormous amounts of time just to understand two pages, i thought I'll get to excercises 😭

[limpid horizon]

it be like that

[lone jacinth]

If you know snake lemma then you can just apply that.

M''(x)K -> M''(x)L is injective because M'' is flat

Otherwise I guess the steps are just the same as proving the snake lemma more or less (I didn't read everything you wrote yet)

[wanton heron]

I guess.

But it was fun to chase the elements and waste an hour 😭

[lone jacinth]

Okay, you figured it out. Good

But then yeah, if it took you an hour then probably practicing some diagram chasing is a good thing

[wanton heron]

The injectivity in the middle sequence is the key actually

[lone jacinth]

Certainly a necessary step

[wanton heron]

That's what leads to the element that was dragged back end up mapping to the element

That was the lift of an element in the kernel

Then the element in the kernel is just the image under the composition

Of an exact sequence so it's 0

[spice idol]

Liu? nice

[neat nexus]

Q.Liu? No wonder looks familiar

[polar hawk]

Hello! Maybe this is a question that beginners ask quite often, but I have just proved the Snake Lemma while I study an introduction to Homological Algebra. What I was wondering about is an interpretation that arises from the common use of this result as I was surprised by the final relation between kernels and cokernels. I would like to provide an example of what I mean by this, in order to clarify myself on such a vague question:

From the factor theorem (or first isomorphism theorem) one could conclude that all relevant information about a homomorphism is encoded in the kernel of the morphism given that we know the "ambient object" (group, ring, module, etc). As one would expect from this interpretation, it extends to the dual cokernel as containing the "residual information" of no need, as it could be identified in most cases as the object that annihilates the image of the morphisms.

Finding such relation between those concepts in the Snake Lemma surprised me and perhaps as a way to procrastinate I started to think in all of this. Has anyone thought about or read something along these lines

[worldly zealot]

maybe im not understanding the question but the cokernel should feel highly important in that you should already care about the codomain, and thus what part of it isn't in the image

[limpid horizon]

Wdym by it can be identified in most cases as the object that annihilates the image of the morphism

just that the annihilator of the cokernel is the image?

[polar hawk]

Well, i was meaning the usual quotient... M \xrightarrow{f} N, then coker f = N/imf. Here the image of f is "annihilated". The "in most cases" comes from the exception in the Grp category (where one can not take the quotient group that easely)

[polar hawk]

Yeah, I get that. Maybe calling that "residual information of no need" was not the best. Haha. But I was trying to know if something analogous can be extracted from the Snake Lemma. The clearest thing is that we can "unfold" such two exact sequences throught the kernels of the first and the "vertical" morphism's cokernels. But what eludes me is if there is a deeper implication from that unfolding results in an exact sequence as it involves such distinguished objects (relative to the morphisms involved)