advanced-algebra

[spice idol]

(usually)

For sets there really isn't that much information to be obtained lol

[forest turtle]

not familiar with universal algebras. idk how true this is in general, though. im sure you could argue that things like field extensions give you a lot of info.
i feel like in topology it can go both ways; quotient spaces can tell you quite a bit, but similarly if you know about what spaces the one you are interested embeds into

[spice idol]

Field extensions yes

But I have the intuition that quotients of an algebra in general tell more about it than the subalgebras

[forest turtle]

i can see that

[plucky arch]

Well it’s like

Given you have a first iso theorem

Quotients of an algebra tell you about mapping out of it

Subalgebras tell you about mapping into it

[spice idol]

Yes

[plucky arch]

Since every homomorphism can be decomposed as a quotient, followed by an isomorphism, followed by a subalgebra inclusion

[spice idol]

Quotients can be studied using congruence relations and that admits a much richer theory than subalgebras so I'm probably biased

[plucky arch]

Sometimes I like to visualise this as a spider diagram

[spice idol]

I know model theory has a lot to say about embeddings, so maybe it's just smt algebra

[plucky arch]

Where your algebra has a bunch of outgoing “quotient” arrows

And incoming “subalgebra” arrows

Like a spider

[forest turtle]

🕷️

[plucky arch]

And then to map one algebra to another, you just have to find a quotient of one that matches a subalgebra of the other

[spice idol]

Then it still depends on the isomorphism chosen

[plucky arch]

Mhm

[spice idol]

The quotient and subalgebra are canonical, but what actually determine the mapping is the precise isomorphism chisen

[plucky arch]

Mhm mhm

[digital parcel]

Obligatory @woven loom ping

[spice idol]

Lmao

[forest turtle]

thanks guys!

[forest turtle]

is the last condition saying that there is just any arbitrary split extension? it means the obvious SES is split, right?

i agree that if there is some split exact sequence 1 -> N -> G -> H -> 1, then G is the inner semidirect product of the image of N and the image of H in G

but is the existence of a split exact sequence 1 -> N -> G -> H -> 1 equivalent to saying that there is a split exact sequence of the same form where N -> G is the inclusion and G -> H is a left inverse of the inclusion H -> G is exact?

[golden osprey]

unironically this is what I use Obsidian for (although I'm really bad about updating my Obsidian with things I find so I still miss alot)

[limpid horizon]

I'm trying to understand how we can apply LES in homology to derived functors. after taking projective resolutions of M' M and M'' in 0->M'->M->M''->0, don't we need the Pi's to be connected in SES's too, so that we have a SES of chain complexes and then after applying the functor we can use the LES in homology?

[fierce steeple]

You use what Weibel calls the horseshoe lemma

I.e. that you can get a SES of projective esolutions from the original SES

@limpid horizon

[limpid horizon]

Do we need to repeatedly use lifting property of projectives sort of thing?

[fierce steeple]

Yeah to prove that lemma

[woven loom]

Woe

[forest turtle]

i guess what i am asking is, is there an example of a group G with two subgroups N and H such that G is isomorphic to the an outer semidirect product of N and H, but is not the internal semidirect product of N and H?

[spice idol]

I suppose you mean "an" outer semidirect product of N and H

[forest turtle]

yes, sorry

i guess like, you make the action trivial? and just take a normal N such that NH isn't G or N and H have non-trivial intersection?

[spice idol]

Z2xZ2 is isomorphic to some outer semidirect product of Z2x1 and Z2x1 but is not the inner semidirect product of em

:P

[forest turtle]

lol

[forest turtle]

does that answer this question?

[spice idol]

It means that precisely the canonical SES splits

Hm nevermind I see what you're asking

[forest turtle]

yea, like, if i have any such splitting, does the cannonical one split?

[spice idol]

No nvm that because there is no canonical such sequence

Lol

[forest turtle]

well, i mean the map N -> G is the inclusion

i don't care what the map G -> H does outside of N

[spice idol]

Yes, it answers the question, you need specifically for the projection map in the split SES to be the identity on H

Only then is G the inner semidirect product of N and H

[forest turtle]

okay, thank you

i feel like the wording on the wiki page is a bit confusing

[spice idol]

It is lmao

[forest turtle]

but the map N -> G doesn't need to be the cannonical inclusion?

or does p also need to kill the right copy of N in G?

yea, it has to be latter. otherwise, we can use the same example with Z2 x Z2 and just swap the factors that Z2 x 0 goes to

[near lantern]

The map from G to Q could be thought of (at least in some of the contexts where this arises) as a map that "forgets" part of the data stored in an element of G. So elements of G "extend" elements of Q with more data. (An OK-ish example: let G = Gal(L/K) and Q = Gal(E/K) for L/E/K and L/K finite Galois extensions. Then G → Q literally forgets what an element of G maps all the elements of L\E to.)

But this is a bit vague and at the end of the day it's just language: you can think of it as an arbitrary choice of words they made for saying "let G be a group with a normal subgroup isomorphic to N such that the quotient is isomorphic to Q".

IIRC this is actually a special case of that (but with N not normal, so Q = G/N is just a set), where G is the group of unit quaternions and N is the subgroup of unit complex numbers.

[near lantern]

The last condition should mean that there exists pi: G → H st (i) the thing they wrote (where N → G is the inclusion and G → H is pi) is an SES (ii) the SES is split by the inclusion map H → G specifically.

[signal tide]

Let ( P ) be an irreducible polynomial over ( \mathbb{F}p ) of degree ( d \geq 2 ) that, over ( \mathbb{F}{p^n} ),
for some fixed ( n > 1 ), has a root ( \alpha ). Is it true that the degree of the minimal
polynomial of ( \alpha ) over ( \mathbb{F}_{p^n} ) is ( \frac{d}{\gcd(d, n)} )?

[broken turtle]

Slomenist

[unborn rampart]

Yes, I think this follows from F_p^n(\alpha) having degree lcm(d, n) over F_p

[atomic quest]

Why Q can't be a finitely-generated Z algebra?
Meaning there is no surjective Z-algebra morphism between Z[x1,...xn] and Q, n is any natural number.

I can prove it for n=1. But the proof for n>1 isn't too neat. Is there an easier way to see this? Some elegant argument?

[fierce steeple]

Pretty sure you can do like: given a map, you can find a finite set of primes S such that x_i is sent to a_i/b_i and b_i is only divisible by stuff in S. Then the image is contained in Z[1/S]

I remember a funny proof which i think works: if Z[x1,...,xn] -> Q is a surjection then the kernel m is maximal, and by the theory of Jacobson rings m cap Z is maximal. But then m cap Z contains a nonzero prime p of Z and so p= 0 in Q

[atomic quest]

That was illuminating. Thank you.
How does one develop the mathematical maturity to come up with such proofs? How are you able to?

[limpid horizon]

Lol im thinking the same thing

I dunno if im good enough to continue it anymore tbh

[ornate atlas]

Going to go out on a limb and say : practice and experience

I’m sure when you started your UG you’d say the same about the proofs you write now, learning is a lifelong thing

It also kinda goes both ways, I regularly look back at problems I did 6+ months ago and wonder how tf did I come up with that?

You just kinda need to be thinking about these things enough

[fierce steeple]

Nah I'm just built different

[digital parcel]

holy fawk...

[fierce steeple]

But ye idk it's just some experience of playing around w these objects and things

[rare walrus]

And luck

Even if you're very knowledgeable and experienced there are some problems you just don't get, just because of chance

[ornate atlas]

Fair, I failed to consider this

[potent plaza]

i think just keep building your bag of tricks

problems that aren't just straight from the definition usually rely on a certain construction,observation or more generally a "trick". the more u see those the more u get better and u keep building this bag of tricks.. eventually you may even create your own tricks 😄

[placid flare]

Is the following result true?

Let $G_1, G_2$ be topological groups and $U \subseteq G_1 \times G_2$ an open subgroup. Then $U = U_1 \times U_2$ as sets, where $U_1$ and $U_2$ are open. Is it true that $U_1, U_2$ are (sub)groups and $U = U_1 \times U_2$ as groups? If not, does it change anything if we additionally assume that $U$ is normal and of finite index?

[broken turtle]

jp

[digital parcel]

Why is U = U_1 x U_2?

[placid flare]

Because $G_1 \times G_2$ carries the product topology.

[broken turtle]

jp

[digital parcel]

Right but not every open subset of the product topology is of the form U_1 x U_2

Those just form a basis

[placid flare]

Oh shoot, I see. I need to refresh from my topology lecture. Thanks.

[digital parcel]

Np

This is also true for groups: not every subgroup of a product of groups is a product of subgroups

Though it's certainly true that the product of subgroups is a subgroup

[fierce steeple]

Every ideal of a product of (commutative?) rings is a product of ideals though

[fossil thicket]

it also works for subrings?

me only knew about ideals, and never thought about subrings

[digital parcel]

i'm aware of a result like this for (prime) ideals

[fierce steeple]

Or ideal maybe sorry

[digital parcel]

idk if prime is necessary

yeah prime isn't necessary

[fossil thicket]

product of primes is not a prime

[digital parcel]

yeah

[fierce steeple]

It's just cause u can multiply by (0,1)

Etc

[fossil thicket]

because it's an ideal sheaf on Spec

[digital parcel]

do people study subrings?

i only hear about ideals

[fossil thicket]

alg nt is a lot about studying subrings of the ring of algebraic integers

[digital parcel]

ooh cool

[fossil thicket]

but you can equivalently say it's studying nice ring extensions of ℤ

but yee rings good

[digital parcel]

amen!

that's cool tho didn't know subrings were actually used

[fossil thicket]

"used" might not be the best word because it's the type of object we're looking at. if you study rings, then naturally you would want to study maps of rings, and thus in particular maps which are injective.

but whenever one studies a nice algebraic object, it's also nice to see if it acts on something. in this case, studying modules over a ring are also very nice, and a special class of modules are the submodules of the ring itself which gives you the ideals.

they have more structure so we can play with them a lil more.

and this is a good moment to put the shameless plug about prime factorization of ideals in a dedekind domain

:p

[digital parcel]

ah gotcha

yea my bad i guess by "used" i just meant like the word itself

i should read some alg nt sometime

[fossil thicket]

yea i just don't wanna give false hopes >.<

[digital parcel]

there's some cool geometry in it too

[sly rune]

what do you mean by this

[ornate atlas]

Sometimes you just simply don’t see the path. There could be a clever trick you just never spot, something like that

[plucky arch]

I’ve heard the name quite a lot, but what is “higher algebra” about, big-picture-wise?

I assume it’s somehow related to higher category theory

[fierce steeple]

The term usually means doing algebra with oo-categories like derived categories (which you can think of as enhancements of abelian categories) and the oo-category of spectra (which is a homotopy theory thing)

Like there are many precise analogues between this oo-categories of spectra and the category of abelian groups, so you can consider rings, modules etc

More generally you can study oo-operads, monads and things

[plucky arch]

What is a derived category

[fierce steeple]

how much homological algebra do you know? like chain complexes or stuff

[plucky arch]

0 unfortunately 😭

[fierce steeple]

Hm i do not really know how to describe this well then sorry

but the upshot is that given an abelian category A you can associate a category D(A) and often it is more convenient to work in the latter - derived categories are super important in lots of algebra and related subjects (and have been for over half a century)

[plucky arch]

Oh ok I looked up the def

Seems reasonable enough…?

Functors out of the derived category naturally correspond to functors out of the category of chain complexes which send quasi-isomorphisms to isomorphisms

I assume

So I suppose you’re trying to capture the idea of “functors that only depend on the homology groups” without destroying the structure of the chain complexes…?

[fierce steeple]

yeah like you often want to remember more about the chain complex than just the groups so you have to use this notion of quasiiso

[plucky arch]

But still the idea that, if two chain complexes have the same homology, and moreover this is witnessed by a chain map, then your functor should give isomorphic outputs?

[fierce steeple]

Ye

[plucky arch]

That actually seems pretty well-motivated

Hm what are derived functors then

[ornate atlas]

I see on Wikipedia that derived categories are useful in the study of D modules, do you know anything about that? (I only know the basics of homalg and of D-modules so even if you do I may not follow lol)

[fierce steeple]

This sort of situation is arguably at the heart of homotopy theory

[plucky arch]

What situation?

[ornate atlas]

Basically they take functors which aren’t exact and make them so, more specifically it tells you how far your original thing was from being exact

[plucky arch]

(Idk what being exact is)

[ornate atlas]

Ah

[fierce steeple]

Like having a bunch of equivalences you want to invert

[plucky arch]

Ah sure but that’s just, like, localisation right

[fierce steeple]

Indeed but doing it in a controlled way and stuff

[fierce steeple]

Preserving finite limits and colimits

[plucky arch]

Finite limits and finite colimits, or finite lim is and arbitrary colimits?

[fierce steeple]

The former

[plucky arch]

I see

This sounds elementary topos like

[wise sedge]

Is there some paper or website with a table of second cohomology groups of irreducible modules of finite simple groups?

[tight marsh]

Anyone have any good resources on an infinite dimensional geometric algebra? I can't seem to find much thats very useful

[muted sierra]

1. You're a category theory fan, but haven't heard of derived functors before?
2. In this context, being exact means “sends exact sequences to exact sequences”. Derived functors measure how far a left (resp. right) exact functtor fails to be right (resp. left) exact.

[plucky arch]

1. yes, what’s the issue?

Also I know what exact means now dw

[muted sierra]

I said this because @fierce steeple said something about limits and colimits, but if I recall correctly, that's an unrelated notion of exactness.

[plucky arch]

Yes it’s equivalent in additive categories

You can define “exact” more generally as preserving finite limits and finite colimits

[plucky arch]

How does one compute all the “invariant tensors” of a group?

I remember one of my physics courses using the result that all the invariant tensors of SO(3) can be built up from the kronecker delta and the Levi-civita epsilon

I’m not quite sure how one would prove this

[hybrid swan]

How would one go about proving the diagram chase lemmas in homological algebra (Snake, 5-lemma, zigzag etc) for an arbitrary abelian category

The proofs I've seen in Mod_A always pick elements but thats not possible in a general category

Like for 5-lemma you pick an element of C' and chase it around the diagram until you reach a preimage in C

[lone jacinth]

There are a few different ways you can do it.

You can use something like Mitchell-embedding or the Yonedas embedding to reduce it to abelian groups.

Or you can use pullbacks and pushouts and universal properties where you would normally use preimages and images

These notes contains such a proof of the snake lemma for example
https://opperman.folk.ntnu.no/HomAlg.pdf

(page 45)

[hybrid swan]

I think I'm willing to just accept Mitchell-embeddings...

But thank you!

[lone jacinth]

Just to draw the parallel here, even if you surrender,

The crucial idea here is that, since a2 is the cokernel of a1, then a2-hat will be the cokernel of a1-hat.

Once you have this, everything is just a normal diagram chase. The map i3-hat just describes all those elements that can be chased from Kerf3 to A3 back to A2.

Then the dual situation happens at the bottom. A map to this pushout that vanishes when composed with b2-hat is really a map to Cokf1, since that's the kernel of b2-hat.

[fierce steeple]

Yeah as Pseudo says, it's the same thing in additive categories. I said that definition cause it works more generally and I thought Pseudo would like it more lol

[muted sierra]

Ah...

[plucky arch]

you were correct :>

[trail hinge]

Hi, I know this isn't exactly the proper channel for it however I'm doing a course on groups and rings and I really need help with some set of problems

if anyone got time to help me i'd be very grateful, it should be pretty elementary

[limpid horizon]

For intro algebra the channel is the groups-rings-fields one

And you can also just post whatever question u have there

[muted sierra]

If you have a Lie or algebraic group $G$, you can find the left $G$-invariant vector fields on $G$ by differentiating the action of $G$ on itself by left translation, denoted $\lambda_g : G \to G$ for each $g \in G$. Then every left $G$-invariant vector field $X$ can be recovered from its value at the identity $X_e \in T_eG$, and conversely, every vector $v \in T_eG$ can be uniquely extended to a left $G$-invariant vector field $X$ such that $X_e = v$, simply by setting $X_g = d\lambda_g(v)$, where $d\lambda_g : T_eG \to T_gG$ is the differential of the aforementioned $\lambda_g$.

Similarly, you can find all the left $G$-invariant differential $1$-forms on $G$ by taking elements $w \in T_e^\star G$ and setting $\alpha_g = (d\lambda_{g^{-1}})^\star(w)$, where $(d\lambda_{g^{-1}})^\star : T_e^\star G \to T_g^\star G$ is the dual of $d\lambda_{g^{-1}} : T_gG \to T_eG$. And, again similarly, you can find the left $G$-invariant tensor fields on $G$ by taking elements of $T_eG \otimes \dots \otimes T_eG \otimes T_e^\star G \otimes \dots \otimes T_e^\star G$ and applying the appropriately tensored $d\lambda_g \otimes \dots \otimes d\lambda_g \otimes (d\lambda_{g^{-1}})^\star \otimes \dots \otimes (d\lambda_{g^{-1}})^\star$.

If you want a concrete coordinate representation, then I guess you need to fiddle with the structure constants of the Lie algebra structure on $T_eG$, but this is more differential geometry than abstract algebra anyway.

[broken turtle]

Eduardo León

[limpid horizon]

How does this show that Hi(_ (x) C) are the right derived functors of that

and why do we need to show that its 0 for any injective R mod I

Tbh, im just trying to figure out what i even need to figure out atm

[limpid horizon]

I guess, what does it take to show a functor is the same as the right derived functor of blank

[limpid horizon]

Ok smth to do with delta functors

[rare walrus]

@rose mirage et al: is there a ring-theoretic interpretation of Clifford theory? I imagine there may be some way to extend Clifford's theorem (and other results in the theory) to twisted group algebras, or perhaps semisimple algebras with a group action. Curious if you know.

I ask you specifically, Wew, because I imagine this is how it works for blocks

[rose mirage]

well for blocks it's kind of annoying - there are nice results about restriction of Brauer characters to normal subgroups when those characters belong to a block B of G that "covers" a block B' of a normal subgroup (specifically, B covering B' is equaivalent to all irreducible constituents of chi in B restricted to N lying in a G-conjugate of B') but there's no guarantee this happens in general (it's only guarenteed to happen afaik when G/N is a p-group, then for any block of N there is a unique block of G that covers it)

I'll think more about the semisimple case

[rose mirage]

quickly scanning https://www.researchgate.net/publication/226508357_Twisted_group_algebras_and_Clifford_extensions it seems like it could be useful for you

[limpid horizon]

Why is that line true

The “and so” makes me think it has something to do with m adic topology. I havent learned that yet so idk

[past cove]

It follows from the paragraph above

like the idea is that if a family F generates the m-adic topology, you can just use the "opens" to detect when things are annihilated

[limpid horizon]

Ok, so I do need to read up on m adic topology for this?

[past cove]

I mean

do you understand the paragraph above?

[digital parcel]

Atiyah-Macdonald chapter 10 has some stuff on I-adic topology

[past cove]

ig if you don't know what I-adic rings are you should probably look into that

it's in any commutative algebra text

I recommend matsumura

[limpid horizon]

Thx guys

[limpid horizon]

Is (x^k) the ideal generated by x1^k, x2^k … etc

[digital parcel]

Yeah

[fierce steeple]

ye it seems they are using notation (s) for sequences s and then this fits into that

[limpid horizon]

and then Ry to Rxy "adds" x to the localization, and Rz to Rxy is 0. so in C2, the differential in total will result in (r1/x^n1)/y^m1 - (r2/y^n2)/x^m2 in that Rxy component ?

so Rx to Rxy is just r/x^m to ... what element of Rxy?

Im sorry if im all over the place. For some reason this has been very difficult for me to parse

[digital parcel]

Just as r goes to r/1 in a localizationc r/x^m goes to (r/x^m)/1

Or in this case just r/x^m

[limpid horizon]

Which can be identified with an element of Rxy? Is the multiplicativity closed subset generated by x and y? I thought it was generated by xy so denominators are like (xy)^m

[digital parcel]

Yeah but you can just clear the y^m in the top

So yeah if you want you can throw in some power of y in the numerator and denominator

Actually not even needed, since in R -> R_x you can identify r with r/x^0 = r/1

Similarly for Rx -> Rxy, use the map in up of localization

[limpid horizon]

Ok so the notation Rxy means S = {x^ay^b , a b in Z}

Not S = {(xy)^a | a in Z}

[digital parcel]

It means this but both x and y are invertible in R_xy

The inverse of x/1 is y/xy

(R_x)_y is isomorphic to R_xy

If this isnt clear, do this exercise from Atiyah-Macdonald chapter 3

[limpid horizon]

ill do it. thanks

[stray grail]

Regarding ideals and modules, I have heard that modules are a sort of generalization of ideals. So why do we still dive into the theory of ideals, and not just talk about modules? Is there a clear example where it is helpful to think about these sets as ideals rather than modules?

[spice idol]

Because ideals can be multiplied and modules cannot

[ornate atlas]

Also ideals are a special case of modules in the way that fields are a special case of rings, they’re still interesting on their own and have specific interesting properties.

It’s not a generalisation in the sense that it makes studying ideals obsolete or anything, it’s not like you’re studying a specific object as opposed to a general one if that makes sense (like studying Z/5Z as opposed to any ring, or Z/nZ)

[spice idol]

It's a generalisation in the sense that the submodules of R are ideals, and can fruitfully be considered as such

Much like how universal algebra is a generalisation of specific algebraic structures, it doesn't make sense to treat, e.g. groups, purely from the lens of universal algebra, and disregard the properties that make groups "groups"

[cunning saddle]

I’m trying to read Humphreys’ book “Introduction to Lie Algebras and Representstion theory but I’m having trouble understanding the family C_l which begins on the bottom of page 2.

to start, how is f(v, w) defined?

[rose mirage]

Standard construction sending a matrix M to a bilinear form given by (v,w) -> v^TMw

[cunning saddle]

Thanks!

What exactly is the bracket operation here? Is it [f(x(v),w)]= f(x(v),w)-f(w,x(v))

[tepid flower]

Take J a subset of S (simple reflections). I want to prove that the parabolic subgroup (S_n)J = S_n x S(n-k)

The textbook I'm reading even says it's Stab([k]) which I don't get either

[near lantern]

So consider the root system A_{n-1} with Weyl group S_n, generated by S = { (1 2), ..., (n-1 n) }. Then for any subset J ⊆ S, if S \ J = { (i_1 i_1+1), ..., (i_{k-1} i_{k-1}+1) } (1 ≤ k ≤ n, 1 ≤ i_1 < ... < i_{k-1} < n), then the subgroup of S_n generated by J is the subgroup of permutations which preserve each of the set {1, ..., i_1}, ..., {i_{k-1}+1, ..., n}, which is isomorphic to S_{i_1} ⨯ ... ⨯ S_{n - i_{k-1}}. To see this, in one direction note that each element of J preserves the mentioned subsets of {1, ..., n}; and in the other, note that (for j = 1, ..., k, where i_0 := 0 and i_k := n) J contains (i_{j-1}+1 i_{j-1}+2), ..., (i_j-1 i_j) which are a set of transpositions generating the group of all permutations of the set {i_{j-1}+1, ..., i_j}.

In terms of Dynkin diagrams, in J some of the vertices have been deleted, turning the Dynkin diagram of A_{n-1} which is a path graph on n-1 vertices into a disjoint union of path graphs, which corresponds to a direct sum of smaller A_* root systems, or in terms of Weyl groups, a product of smaller S_*'s.

[tepid flower]

Where can I read about this?

I feel like over text makes it really hard to understand stuff

[tepid flower]

Especially this

This gives a better understanding for some reason

[foggy galleon]

Suppose G is a commutative algebraic group defined over Z. Suppose that for all number fields K, G(O_K) is a subgroup of G(K) and that H^1(K, G(K))=1. Can you say G is a product of GL_1's?

[near lantern]

IDK this is something you just work out by hand AFAIK

Or rather I don't know any resource that specifically covers this properly

Perhaps one could look up: lecture notes "parabolic subgroup" A/_n

or something like that

[fierce steeple]

Let A be a DVR with residue field k. If B is a local artinian A-algebra with residue field k, is B automatically a fg A-module?

Note that this is false without the residue field hypothesis, because you could take B to be a big extension of k.

(Also, I'd be interested just in the case where A = Z_p !)

I think I have an argument if A = Z_p. If B is as hypothesised, then we have a sequence F_p -> B/p -> B/m_B = F_p. So B/p = F_p x C for some ring C, but since B/p is local (as B is), we must have C = 0 i.e. B/p = F_p. But then note that each of the F_p-modules p^nB/p^{n+1}B is finite, so B is finite and in particular finitely generated over Z_p

i.e. putting it all together, B = Z/p^n for some n >= 1

I would suspect the result still holds if the residue field of B is a finite extension of the residue field of A, though I think this follows because B is an algebra over the witt vectors of its residue field and then you run a similar argument

[lone jacinth]

If B is artinian it will have finite length as a B-module.

The length is equal to
sum_i dim_k m^i/m^i+1
where m is the maximal ideal. So it has the same (finite) length as an A module.

The argument works the same replacing the residue field with a finite extension.

You also don't need local, as long as B modulo the radical is finite dimensional

[fierce steeple]

Very nice

Does this work for noncommutative B?

[lone jacinth]

I assumed B was non-commutative

[fierce steeple]

I guess there you need to worry about what artinian means lol - let's say left and right

Very cool thanks

[lone jacinth]

You just need one side (whichever you're asking about it being fg for)

[fierce steeple]

Thankss

[fierce steeple]

Also this is of course false lol.

[soft parcel]

I don't get this exercise. Doesn't r vanish at a prime ideal p precisely if r is in p? So we simply have that r vanishes identically iff it is in every prime ideal. That doesn't seem to me to indicate that it is nilpotent (well, besides it being in (0) maybe, but then r is just 0...)

Could someone clarify?

[subtle plaza]

The nilradical is the ideal of nilpotent elements. One can show that the nilradical is the intersection of prime ideals in a commutative ring

[soft parcel]

Yes, I know that

[subtle plaza]

So r vanishes identically iff r is in every prime ideal iff r is in the nilradical

[soft parcel]

Oh

Welp

Even when you wrote it I was like what does this have to do with anything

lol

Yeah, that makes sense. Thanks

[limpid horizon]

But R_xy doesnt match the formats here does it? Because S = {1, xy, (xy)^2, …}

Its not of form ST where S = {1, x, x^2…} and T = {1, y, y^2…}

So how are we applying that exercise here

Again im assuming the notation R_xy means you localize at the subset generated by the single element xy

[digital parcel]

what's the image of T in S^{-1}R?

[limpid horizon]

Elements of the form t/1 with t in T

[digital parcel]

right

let's denote this new set U

what do elements of U^{-1}(S^{-1}R) look like?

[limpid horizon]

elements of the form (a/s)/(t/1)

[digital parcel]

yeah

can you rewrite this as an element of (ST)^{-1}R?

[limpid horizon]

Im
Guessing its a/(st)

[digital parcel]

yeah

so now this is an element of (ST)^{-1}R

so now we have a map $\varphi : U^{-1}(S^{-1}A) \to (ST)^{-1}R$ which maps some $(a/s)/(t/1) \mapsto a/st$

[broken turtle]

anamono

[digital parcel]

verify this is an isomorphism

[limpid horizon]

What im confused about tho is applying this in our situation because R_xy is not like (ST)^-1R

[limpid horizon]

What i said here

[digital parcel]

the exercise isn't saying anything about the sets themselves but the localization of the ring at the set

maybe i'm misreading your question let me see

[limpid horizon]

Sorry, yea i tend to get mixed up with the little details 😭

To apply exercise 3 we need S and T and we’re saying something about the localization at the set ST. But in the example of cech complex you said (Rx)y is isomorphic to Rxy by exercise 3

But i am confused on how we are applying exercise 3 in this case

[fierce steeple]

It's sort of clear it doesn't matter cause inverting (xy)^n for all n is the same as inverting x^m y^k for all m,k

[limpid horizon]

Yeah i can get behind that, but as it was before we couldn't directly apply exercise 3 right?

without first noting that i guess

[fierce steeple]

Idk what exercise 3 is lol

I'll look at atiyah macdonald ig

[digital parcel]

okay i see what you're saying

[limpid horizon]

it was this

[digital parcel]

@fierce steeple this is exercise 3

[fierce steeple]

What is the "it" about here?

[limpid horizon]

(Rx)y is isomorphic to Rxy

[fierce steeple]

I mean like why can't you use exercise 3

[limpid horizon]

Because for Rxy the subset is {1, xy, (xy)^2, ..} which wasnt something like ST with S = {1, x, x^2..} and T ={1, y, y^2..}

[fierce steeple]

Sure but then this

Anyway I should say like this is all much clearer if you think of localisation in terms of what it does i.e. universal property

[digital parcel]

yeah i mentioned that here but i wasn't sure if that was getting through

[fierce steeple]

Like for any ring $B$, restriction along $A \to S^{-1} A$ induces a bijection ${$ maps $S^{-1} A \to B} \to {$ maps $A \to B$ sending $S$ to units$}$

[broken turtle]

Prismatic Potato

[fierce steeple]

And then if you unwind what maps out of (R_{x})_{y} or (UT)^[-1} R look like etc it's fine

[digital parcel]

yeah that's my bad i shoudlve phrased this in terms of the UP

[limpid horizon]

Oh thats what u meant with up lol

[digital parcel]

yeah

[limpid horizon]

I aint even gonna lie i need to now parse how youre using the UP

I havent thought in terms of up much yet

New for me

[warped crypt]

Let f : A^n -> A be a function where A is the universe of discourse of a p-structure G (where p is a signature).
I'll call a function f p-representable iff there exists a p-term q such that G(q) = f (where G is also the interpretation function of the p-structure G)
Notice that there exist functions that are not p-representable in general (there may exist countably many p-representable functions, but |P(N)| functions)
Are there any resources on anything like this? Has anyone seen anything like this?

[digital parcel]

Ignore the blue line

[ornate atlas]

https://tikzcd.yichuanshen.de…

[limpid horizon]

Should be Rx to Rxy right

U can use up here to show (Rx)y iso Rxy?

[digital parcel]

Yeah sorry

[digital parcel]

lol I was in a car and didn’t feel like doing that but ty

[limpid horizon]

for cech complex, the map from like Rx -> Rzy is just 0?

is it even possible to have a map Rx -> Rzy since x is not invertible in Rzy

oh these are R-mod homs not ring homs

[digital parcel]

the map is 0 by definition, see the last line

[limpid horizon]

ok thx

[vapid axle]

I'm trying to prove iii. Is it easiest to show injectivity and surjectivity instead of constructing an inverse? The inverse is not immediate to me

[fierce steeple]

Well it is easiest in all of them to use the universal property of the tensor product

Then you will see more intuitively why they are true

[vapid axle]

I used the universal property to get c)

[fierce steeple]

But in iii) the inverse is actually not bad. To give a map out of a direct sum, you just need to show where each summand goes, and then it comes from te,soring the maps M,N -> M (+) N by P

[vapid axle]

Well I guess I can construct some bilinear map from (M \otimes P) \oplus (N \otimes P) to (M \oplus N) x P

[fierce steeple]

What I mean is you want to also show the maps are isomorphisms using universal properties

[vapid axle]

Oh I think I see

[fierce steeple]

Like for example in i) the point is that a bilinear map M x N -> P is the same as a bilinear map N x M -> P

So M (x) N and N (x) M have the same universal property

[vapid axle]

So something like consider the injections $\iota_M, \iota_N: M, N \to M \oplus N$, then there are maps $(\iota_M \otimes 1), (\iota_N \otimes 1): M \otimes P, N \otimes P \to (M \oplus N) \otimes P$. So we can consider the coordinate wise map $(\iota_M \otimes 1) \times (\iota_N \otimes 1): (M \otimes P) \times (N \otimes P) \to (M \oplus N) \otimes P$ which is (hopefully) bilinear and induces a map out of the tensor product (which is hopefully an inverse to the given map)

[broken turtle]

okeyokay

[vapid axle]

Never mind

We don't even need the universal property here I think, we can just map it coordinate wise out of the direct sum

Nvm

Oh we add them?

[fierce steeple]

Ye

[vapid axle]

Fascinating

Thank you

Am I doing something wrong? $(m \otimes p_1, n \otimes p_2) \mapsto (m, n) \oplus p_1 + p_2 \mapsto (m \otimes p_1, n \otimes p_1) + (m \otimes p_2, n \otimes p_2) = (m \otimes p_1, n \otimes p_2) + (m \otimes p_2, n \otimes p_1) = (m \otimes p_1, n \otimes p_2)$ if and only if $m \otimes p_2 = 0$ and $n \otimes p_1 = 0$

[broken turtle]

okeyokay

[fierce steeple]

Idk what ur first step is there

[vapid axle]

Well we have $(m \otimes p_1, n \otimes p_2) \mapsto (\iota_M \otimes 1)(m \otimes p_1) + (\iota_N \otimes 1)(n \otimes p_2) = (m, 0) \otimes p_1 + (0, n) \otimes p_2 = (m, 0) + (0, n) \otimes p_1 + p_2 = (m, n) \otimes p_1 + p_2$ right?

[broken turtle]

okeyokay

[vapid axle]

I think this is the right map since the other order of composition works

I must have done something wrong in my calculations

Do you see where I went wrong in my first step Potato?

[fierce steeple]

Ig what I mean is like it is hard to pass what is going where to me cause the domains aren't written

[vapid axle]

Sorry

[fierce steeple]

Dw

I'll read it more carefully

[vapid axle]

The first map is from $(M \otimes P) \oplus (N \otimes P) \to (M \oplus N) \otimes P$

[broken turtle]

okeyokay

[fierce steeple]

That second equality doesn't hold

Also like some bracketing is missing

[vapid axle]

Okay wait I'm very bad at tensor products

Let me try to understand why

[vapid axle]

Are you saying the second equality doesn't hold here?

[fierce steeple]

Ye

[vapid axle]

Oh is it because of the definition of a free module

They're just formal linear combinations or something

So we don't add them component wise

I have thoroughly confused myself

[fierce steeple]

Yeah (x) is bilinear

[vapid axle]

Never mind I'm dumb

We can just apply it to (m, 0) (x) p_1 + (0, n) (x) p_2 straight from there

Bruh 😂

Alright thanks you saved my life

[fierce steeple]

Np

But yeah the argument is had is basically like

Consider what a map out of each side means in terms of bilinear maps

[vapid axle]

Ye, I feel like it's good practice with working with the universal property

[sand pelican]

what book is this from?

[vapid axle]

Atiyah-Macdonald

Commutative Algebra

[sand pelican]

any recommendations for prerequs before talking this? I’ve had Undergrad Group, Ring, Field, and Galois Theory

[vapid axle]

You could probably jump right in if you've taken a course in abstract algebra

I'm sure knowledge of some point set would be good for intuition in the exercises which is where you'll do most of your learning

Since a lot of exercises involve Spec

But I'd say you're good to go

[sand pelican]

Ive had some pointset from analysis and reading Munkres first two chapters but admittedly would like to go further in munkres

[sand pelican]

oh boy

[vapid axle]

Honestly I'd just say you could start yeah

[sand pelican]

Where do you see commutative algebra the most?

[vapid axle]

Algebraic geometry and algebraic number theory

I'm mainly reading through Atiyah as a prereq for Hartshorne

[sand pelican]

I getcha

Any application to Alg Top?

[vapid axle]

I have no clue tbh lol

I consider myself very new to algebra

My guess is probably?

[fierce steeple]

I would say should know the stuff in at least the first few chapters of atiyah macdonald for alg top

[limpid horizon]

I havent learned alg geo yet either i wonder if ill end up liking it

i still really like commutative algebra so im hoping so

[limpid horizon]

man i did point set topology like last sept-december but its been so long since i thought about topology

im taking diff geo this upcoming semester. should be nice cuz i know nothing about that

[sand pelican]

I feel like I dont ever remember the top stuff but when I look at a textbook Im like “I dont need to look at this” so Im conflicted if I should move on or review

[limpid horizon]

haha i feel u

for me there is like an endless list of things i want to review

in algebra for me rn cuz thats what i study the most

i want to review like all those ring theory proofs that would be found in like the first chapters of dummit and foote and be able to just know them off hand

that would be nice

i feel like for most of them i probably remember the gist of it though

[spice idol]

Pro tip: think of integers, but noncommutative!

[limpid horizon]

noncommutative?

i aint even gonna lie one thing that i still like cant see proprely is why if order of a in G is n then order of a^m is n/(n,m)

there are a lot of little things like that that got lost on me a bit along the way

[fierce steeple]

idk if you actually want a proof but the funny way is WLOG assume G = Z/nZ and then it becomes the statement that mn = gcd(m,n).lcm(m,n)

[limpid horizon]

This was nice thank you

[hard kite]

Dope that is a funny way lol

[limpid horizon]

ya that did the trick for me

i think usually they apply division algorithm to prove it right?

[spice idol]

I mean that's just basically the proof right

[fierce steeple]

Yes lol

[fierce steeple]

Yeah my joke is more just like

Psychologically easier to think about Z lol

[spice idol]

Ah right fair

Me when the forgetful functor is representable

[tepid flower]

Let J = S \ {s_k}. I'm trying to understand the minimal coset representatives of (S_n)^J. Why do you get the set {x in S_n | x_1 <...<x_k and x_(k+1) <... <x_n}

Even then what happens if J was something else other than S \ {s_k}? Say we take S_6 and J = {2,3,5} then how does (S_6)^J look like?

[fierce steeple]

I am confused by what your S_6^J means

[tepid flower]

Symmetric group

But the minimal coset representative

[fierce steeple]

And what is S \ {s_k}

Idk just there seems to be some notation I am missing or something.

[tepid flower]

The set S without the s_k

A set of simple reflections

[golden osprey]

The fundamental theorem of symmetric functions says that for every $f_{\text{sym}} \in \mathbb{Z}[x_1, \ldots, x_n]^{\mathfrak{S}_n}$, there is a unique $f \in \mathbb{Z}[x_1, \ldots, x_n]$ such that $f(e_1, \ldots e_n) = f_{\text{sym}}(x_1, \ldots, x_n)$.
My understanding is that Waring had an incomplete proof in 1770.
Is the first complete proof due to Gauss in 1876 or is there an earlier one known?

[broken turtle]

Spamakin🎷

[hushed bone]

Yeah I proved it in 1875

[plucky arch]

when can we get a fundamental theorem of fundamental theorems

[sacred sentinel]

If y is invertible, then xy = 0 implies x = 0. It follows that the resulting ring is just the localization of k[y] at y

Ah i misread your question, my bad, in that case x is invertible which implies y is 0

Yes i believe so

[past cove]

Localization commutes with quotients so this is the same as quotienting k[x,y,y^-1] at (xy) but y is a unit it in the ring so you're just quotienting by (x) which gives you k[y,y^-1]

[sacred sentinel]

Yeah just another way to express localizing at a single element

[past cove]

Oh wait

You're not localizing at y you're localizing away from y lol

[sacred sentinel]

Also this should hold for any ring

[past cove]

Wait are you sure you're localizing at the prime (y)/(xy) or at the set {1,y,y^2,...}?

[sacred sentinel]

Wait no I think im talking nonsense

You actually get K(x) as in the fraction field I think

[past cove]

Oh well it doesn't matter I think

[sacred sentinel]

I think it does, because you're also inverting x-a for every a

So you get like k adjoint x

I think this might not be true, because the ring localized away from (y) contains 1/(x-1) which is not in k[x] localized at x

[past cove]

Then your claim isn't true I think

[sacred sentinel]

Okay, that makes more sense

[limpid horizon]

The context of this is showing how a ses of modules 0->C->C'->C''->0 gives rise to a ses 0->Hom(B,C)->Hom(B,C')->Hom(B,C'')->Ext1(B,C)->Ext1(B,C')->Ext1(B,C'')->...

it does this by taking projective resolution of B and then forming that diagram there, then using the LES in homology theorem

my question is how do we know that diagram commutes. I understand the rows are exact because you can get them from applying Hom(Pi,-) where Pi is projective

[sly rune]

going up is precomposing and going right is postcomposing

in more categorical language Hom is well-defined as a bifunctor

[pastel shoal]

What are some examples of the restricted product of topological groups/rings other than adeles and ideles?

[digital parcel]

left R-module

[hushed bone]

You just see that each square commutes by definition of the maps

[lone jacinth]

Seems to me it should be isomorphic to k(x)

[pliant sonnet]

Could anyone please help me with this problem:
Let D_n= { a,b| a^n, b^2, abab} be the nth Dihedral group. If n is odd, prove that D_2n isomorphic to <a^n> × <a^2,b>.

[lone jacinth]

I don't know exactly what you know about groups, but a sufficient condition to showing that the product of two subgroups is a direct product is that both are normal and they don't intersect.

So if you can show these two things you just need to show that a^n, a^2 and b together generate D_2n

[pliant sonnet]

Thanks! I ended up using a similar method to solve it just now

[rare walrus]

This question I asked in #algebraic-geometry also could work here.

N.b. I am actually meaning to ask about finite groups of Lie type here

[sterile leaf]

Would probably be groups that satisfy conditions from lie groups but are not literally a lie group

Finite groups*

[rare walrus]

OK, maybe I can be clearer, I mean finite reductive groups, i.e. fixed points of a Steinberg endomorphism of a reductive algebraic group over F_p

[sterile leaf]

In the case of this I don't know exactly

But I might point out that lie type probably means something like a finite reductive group that satisfies the lie (reductive algebraic) group conditions

Again I don't know this branch of rep theory haha

I just placed my thoughts here

[rare walrus]

My issue isn't about the definition of a finite reductive group... I do know this 😭

[sterile leaf]

Lie type you mean?

[rare walrus]

"groups of Lie type" is a term of art which arguably does not have a precise definition, see e.g. https://mathoverflow.net/questions/136880/definition-of-finite-group-of-lie-type

I avoided it deliberately by using the term "finite reductive group" instead

[broken turtle]

mh_le

[tepid flower]

.

Bump

[cloud karma]

For that J I believe you get x_2 < x_3 < x_4, x_5 < x_6

[tepid flower]

hmm

how about the map x -> x^J? What if x = 7125346 and J = {1,2,4,6}

@cloud karma

[cloud karma]

Well you arrange x_1 x_2 ... x_k in an increasing order and similarly for x_{k+1} ... x_n

so x^J = 1273546

[tepid flower]

uhh how

[cloud karma]

noggin

[tepid flower]

[cloud karma]

In J, 2 and 3 correspond to the first 3 numbers {7,1,2}, 4 is (4,5) so you have {5,3} and finally {4,6} for 6

now you just rewrite the sets in increasing order

[tepid flower]

oh that's it?

[cloud karma]

yeah lol

[tepid flower]

this so easy idk why it wasn't obvious

[cloud karma]

noggin failed

[limpid horizon]

So this is just exploiting a nice property of Hom functors basically?

[fierce steeple]

Yes

[hushed bone]

Yes it’s that it’s a bifunctor which someone said

[limpid horizon]

Ok yea thanks

[limpid horizon]

How exactly do left exact functors preserve kernels and right exact preserves cokernels?

Still am just not exactly sure what is being preserved when say, a ses 0->A->B->C->0 gives 0->F(A)->F(B)->F(C) upon applying F

And same for right exact

[limpid horizon]

(Also, exactness at C here isnt necessary right)

[past cove]

you have a left exact sequence like 0 -> ker(f) -> A -(f)-> B, so when you apply a left exact functor you get 0 -> F(ker(f)) -> F(A) -(F(f))-> F(B) which is also left exact
then you can show by the axioms of an abelian category that F(ker(f)) = ker(F(f))

for right exact you have a right exact sequence A -(f)-> B -> coker(f) -> 0

and you do the same thing but with a right exact functor

[limpid horizon]

A-(f)?

[digital parcel]

$A \xrightarrow{f} B$

[broken turtle]

anamono

[limpid horizon]

Ty

[digital parcel]

that's what they mean by A -(f)-> B

[limpid horizon]

Im not really sure what this showed

[digital parcel]

okay so you have an exact sequence

$0 \to \ker(f) \to A \xrightarrow{f} B$

[broken turtle]

anamono

[limpid horizon]

Yea

[digital parcel]

when you apply your left exact functor F you get an exact sequence
[ 0 \to F(\ker(f)) \to F(A) \xrightarrow{F(f)} \to F(B) ]

[broken turtle]

anamono

[limpid horizon]

yea

[digital parcel]

when we say the "functor preserves kernels" it means $F(\ker(f)) \cong \ker(F(f))$

[broken turtle]

anamono

[digital parcel]

so that's what milky milk milk said

in an abelian category left exact functors preserve kernels

same thing holds for a sequence $A \xrightarrow{g} B \to \operatorname{coker} g \to 0$

[broken turtle]

anamono

[limpid horizon]

ok yea so it just comes from the definition of this being exact

thx

[past cove]

who tf is hchan

[digital parcel]

oh shit sorry knee jerk reaction

[limpid horizon]

thats ok hchan helped me earlier in groups rings field

[digital parcel]

he's active here too so i get the blues mixed up

[digital parcel]

anyway this just comes from the fact it's left exact

[digital parcel]

this is what the real statement is

this is what it means to "preserve kernels"

[limpid horizon]

and this left (right) exactness property of functors is what makes 0th homology equal to the functor evaluated at the object right

in the derived functors setup

[past cove]

yeah I should've been more straightforward with it but the left exact sequence is exactly the definition of the kernel of a morphism so you get your result

[past cove]

do you mean the 0-th derived functor?

[limpid horizon]

and thats a very nice thing because it what makes us continue exact sequences

[digital parcel]

not quite, but also not sure what youre asking here

[limpid horizon]

yea

[digital parcel]

oh i misread

yes left exact functors admit right derived functors

[limpid horizon]

yeah i said 0th homology which i guess is being unspecific but these derived functors are always still some homology right

[digital parcel]

right derived functors basically tell you how far away a left-exact functor is from being right-exact

derived functors should be cohomology tho

#prealg-and-algebra

[last stratus]

kk

[past cove]

this depends on if it's left or right

[digital parcel]

oh yeah true

sorry i was thinking about left-exact functors lol

[limpid horizon]

i havent bothered thinking about if something is homology or (co)homology yet lol

probabl worth doing

[past cove]

what you do is (for right derived) you take an injective resolution of your object, and you take the cohomology of the complex attained by applying the functor to this resolution

[digital parcel]

(assuming your category has enough injectives)

[past cove]

I always like to imagine we're in R-Mod 🤩

[limpid horizon]

this is also assuming the functor is covariant tho right. If its left exact contravariant you would need a projective resolution?

this is something i was thinking about. Left exact functors would have right derived functors, but constructing them via projective or injective resolution just depends on variance of functor?

[digital parcel]

yeah

[limpid horizon]

Ok, i feel like finally a lot of this is clearing up for me. Last thing I have no idea about is why computing Ext^i(A,B) gives the same result whether you do so by taking projective resolution of A and computing right derived functors of Hom(-,B), or taking injective resolution of B and computing right derived functors of Hom(A,-).

[digital parcel]

i dont really remember the proof but this can be found in any standard textbook

[hushed bone]

This is called balancing Ext

I believe with Ext specifically (maybe all bifunctors?) you can do some ad hoc argument

But in general I believe you use a spectral sequence and the two different horizontal / vertical filtrations to compare

[limpid horizon]

thanks

seems kinda dense

[lone jacinth]

Typically everything is framed in terms of functors being covariant. And then contravariant functors C -> D are just covariant functors C^op -> D

[fierce steeple]

Or just ur global convention

[fierce steeple]

I think the "balancing" argument is just the spectral sequence proof in a special case lol

[hushed bone]

Yeah but you can do it without ever saying the word

[fierce steeple]

Ye

I feel like hm

There are a few things in hom alg where ad hoc proofs are given but a spectral sequence thing would be quicker rlol

But idk the best way to deal w this

I mean also like UCT for example and stuff

[dark widget]

for any natural number n, an n-category is also a 1-category right?

[plucky arch]

hm if you take equivalence classes of 1-cells then probably...?

[past cove]

what definition of n-category are you using

[dark widget]

idek, this is a dumb question sorry idk highter category theory. ig there's no natural way then

like i assumed for 2-categories the objects are functors and functors are natural transormations so it's a 1-category

[fierce steeple]

n-categories can be viewed as more general as there is more structure. You can neglect some of it to get a 1-cat, but can also view as a 1-cat as an n-cat

[fierce steeple]

Not sure what you mean

"The objects" of what

[dark widget]

the 2-category

[fierce steeple]

Idk you seem a bit confused as to what these are, like you seem to be wording this as if there is only one 2 category or smth

You are probably thinking of Cat as an example of a 2-categpry, where for any two objects (i.e. categories) you have an associated functor category, which is a 1-category

[dark widget]

ah yes

[limpid horizon]

Yeah, i continue to mull over contra vs covariance and every time people here tell me “its just always assumed covariant from Cop”

So then, “canonically”, right derived functors are always constructed using injective resolutions ?

[scarlet ermine]

maybe canonically isn’t the right word (ie there are several choices of injective resolutions hehe) but yes.

for example with taking the right derived functor of the contravariant functor Hom(-,A): Ab^op -> Ab, to compute
Ext^i(B, A),
you take an injective resolution of B in Ab^op, which is a projective resolution of B in Ab

[digital parcel]

you can take your favorite injective resolution since the derived functors you get are all isomorphic

same for projective resn

[scarlet ermine]

yup no objections there, but you’re making some initial (very large) choice and you only got your hands on your derived functor up to isomorphism. Are there occasions where you can choose things canonically? I guess for sheaves you got something like the godemont resolution (not injective but is acyclic)

[digital parcel]

oh i see what you're saying, idrk then

i guess there's natural choices of resolutions in some settings but idk if there's a universal "canonical" choice

[fierce steeple]

Well I should say like there is a lot of canonicity here: given any two injective resolutions M -> I*, J** there is an induced homotopy equivalence I* -> J* unique up to chain homotopy, and so in particular a canonical isomorphism H^*(F(I*)) -> H*(F(J*)

And yeah then you need choices to construct them as a functor but there are canonical isomorphisms between any two choices

This is very common in category theory though - for example taking limits of a certain shape (provided they all exist) is a functor defined up to canonical isomorphism, in the same way

[fierce steeple]

Generalising the Godement resolutions, there are very often (co)monadic resolutions, the basic example being how you can resolve a module in a stupid way by using R^M -> M and continuing (and the Godement resolution is another one)

Generally these will just be acyclic resolutions

[scarlet ermine]

oh nice!

[limpid horizon]

Isnt Hom(-,A) contravariant along Ab->Ab?

wouldnt Hom(-,A) on Ab^op->Ab be covariant

[scarlet ermine]

oh yes that’s right

I usually will say a contravariant functor and write A^op -> B the associated covariant functor, though maybe the language is slightly bad

I guess in my head there are only covariant functors and I call a functor contravariant if it’s from a familiar category with an op on it lol

[limpid horizon]

yea it is very clear to me how no one thinks about this later on

[scarlet ermine]

but then I’ll run into trouble if I’m defining functors out of CRing/AffScheme hehe

[limpid horizon]

im still kind of new to using category theory in algebra

[scarlet ermine]

yup for sure

[fierce steeple]

This is a funny thing. In smth I'm writing I need to write Spec as a functor {commutative rings} -> {schemes}^op and it feels wrong

[scarlet ermine]

haha

[near lantern]

The point is just that you can use the equivalence to get the others from one. There are right-derived functors for covariant lim to lim preserving and contravariant colim to lim preserving functors, and left-derived functors for covariant colim to colim and contravariant lim to colim functors. Hardly anyone wants to deal with all four of them when most arguments for one apply to all of the others by adding op's to the domain and/or codomain category.

[limpid horizon]

If F is not left exact, its possible that even for A-g->B, F(kerg) is not a subobject(?) of F(A)?

I guess so, right? Because it doesnt preserve kernels

[limpid horizon]

I set up my question weirdly, i should just be asking that if F is not left exact, subobjects may not be preserved under F?

[fierce steeple]

Ye and then that is sort of immediate by this

[limpid horizon]

Yea it may not preserve subobjects immediately by the fact that it may not preserve kernel

[fierce steeple]

your original question is a bit stronger but also correct: the functor (-) (x)_Z F_p is not left exact, and sends Z to Fp and Q to 0

[limpid horizon]

I feel like any intro to left/right exact functor should mention this stuff. It makes it way more intuitive to understand the point of it

[limpid horizon]

Isnt the stronger question if F preserves subobjects in general?

[limpid horizon]

Yeah this is nice example of functor not preserving subobjects

[hushed bone]

Actually I guess, can you do this purely in ZFC? Like to pick a specific injective resolution for each module can’t be done with just normal choice right?

[sly rune]

don't you have the same problem with defining the limit functor that Potato mentioned

for abstract complete categories

[hushed bone]

I guess so

[fierce steeple]

Hm

[woven loom]

Well, consider how that needs you to pick a class worth of em

Namely, resolutions for things of arbitrarily large sizes

(Which you’re screwed by regardless of being a skeleton or not)

But you’re fine for some like, skeleton of bounded dimensions involved or wtv

Now, if there were some highly canonical choice, you might be saved, but I’m not aware of one

[near lantern]

You can pick a projective resolution for all modules M simultaneously by P_n = Free(...(Free(M))...) (n+1 Free's). Essentially this is because "every module has an epi from a projective" can be done canonically. If there's a similar "canonical" injection of any module into an injective module, it can be used to canonically construct an injective resolution (albeit often not a convenient one).

And I believe if we write M* := Hom_R(M, ℚ/ℤ) for any ring R and left/right R-module M, then (Free(M*) → M*)* ∘ (M → M**): M → M** → Free(M*)* is such a canonical injection.

[sly rune]

that doesn't sound exact

[lone jacinth]

It's a composition of two injective maps, so is injective (and the maps are injective because Q/Z is cogenerating).

There aren't really any other exactness conditions to consider

[sly rune]

i meant the P_n thing

[lone jacinth]

Ah, there's just missing a ker in there, i.e.

Free(ker(Free(..)))

[sly rune]

hm, i guess so

[dusty plaza]

Hey guys. I tried asking a question about linear algebra on the corresponding channel, but nobody answered. Do you mind that I ask it here?

[spice idol]

Depends on the linear algebra

[dusty plaza]

Can you check out the question in the linear algebra forum and tell me if I should post it here or leave it there?

[unborn rampart]

I think it belongs in #linear-algebra , but you should write it in latex, because it's kinda hard to parse what you've written

[dusty plaza]

will do

[hasty flume]

Suppose i have a 1 dimensional Noetherian local ring A (in fact it is a localization of a k-algebra of finite type with k alg closed but im unsure if this will be relevant). in this case there is a bijection betwee the minimal primes of A and the minimal primes of its completion. can we upgrade this to a full correspondence of the primary decompositions of 0 in each ring, respecting multiplicity?

[hushed bone]

Huh, and I guess this canonically gets you chain maps on these resolutions right? So you don’t have to pick specific ones.

[near lantern]

I feel like it ought to be functorial, so yes, but don't feel like verifying that.

[hushed bone]

I think it’s clear, send generator on m to generator on f(m) iteratively

[hushed bone]

I don’t think so. If A is reduced but not analytically reduced, then the primary decomposition of 0 should just be the intersection of all the minimal primes. This should (if I understand correctly your setup) correspond to the nilradical of A^ which isn’t 0, so you didn’t get the primary decomposition of 0 in A^

And I think you can have this? Idk, I guess I don’t know for certain if there are examples of non-analytically reduced rings of dimension 1

[hasty flume]

I think something like this must be true because analytically isomorphic singular points on curves are equivalent via embedded resolution

[hushed bone]

what is “something like this”

[hasty flume]

So their irreducible components are in bijection with one another and that preserves multiplicity

[hushed bone]

The statement about primary decomposition?

[hasty flume]

Ya

Maybe with more assumptions

[hushed bone]

So what is multiplicity here?

[hasty flume]

Perhaps the finite type ness over k is important

[hushed bone]

That is confusing me a little

Is the primary decomposition of I always like

Via symbolic powers of the minimal primes?

Cuz otherwise I can’t really think of what you mean by multiplicity

[hasty flume]

Uhh so geometrically what's going on is this is corresponding to a curve embedded in a surface and I'm really studying the ring O_X,P/(f) where f is a local equation for C

So I have two such curves C, C' in two surfaces X, X'

Now the completions of the local rings of each curve are isomorphic

So in this case there is a bijection of the minimal primes = irreducible components

[hushed bone]

Yup

Although I’m actually not even familiar with this result lol

It super isn’t true in general I think, but I guess the dimension 1 does a lot of work for you?

[hasty flume]

Yeah

[hushed bone]

But then what’s going on after?

[hasty flume]

Ummm maybe this notion of multiplicity is only relevant after the blow ups

[hushed bone]

Like what’s the multiplicity here

[hasty flume]

But anyway what has to be true is that successively blowing up these singular points on each curve, the irreducible components of the total transforms are in bijection and that bijection preserves their multiplicity as divisors

[hushed bone]

Okay so this is what I don’t understand

[hasty flume]

So perhaps multiplicity is not meaningful for the base

[hushed bone]

If your thing is sufficiently fucked up, the whole divisor thing falls apart

[hasty flume]

I was thinking of this in terms of like factorizing f

[hushed bone]

Okay okay I see

But… I need to think more but like

Something feels funny to me

Like say A is your ring

And say it’s nonsingular in codim 1, that’s just saying it’s a DVR almost

Then I think the ring is super nice and probably things push through

[hasty flume]

Isn't A already a local ring of dim 1

[hushed bone]

Yeah but if it isn’t reduced then it will have some shitty xodim 1 singularity

And then the whole divisor thing kinda falls apart

[hasty flume]

Ic

[hushed bone]

Or wait codim 1

So uh

No that’s the closed point lol

So okay yes

[hasty flume]

Maybe we can assume the curve is reduced

[hushed bone]

Uhhh

Okay no I confused myself

But okay, let’s say X is nice and whatever

Then f definitely can be written as like, a bunch of stuff with multiplicities

But…

🤔

Like okay here’s where I’m confused lmao

In a general dim 1 local ring, I don’t see how the notion of multiplicities of the irreducible components cutting out 0 makes sense like…

The way you do it before wuotientinf is like

Push f through the valuations of O_X,P where P is the generic points of the components of V(f)

But then once you quotient by f like, how do you even see that anymore?

Ya feel me?

[hushed bone]

But also like doing this gives you the decomposition of f? But I guess it might not be the primary decomposition because you’re taking powers and not symbolic powers, but I think things are nice enough they agree anyway

Anyway I am just kinda puzzled atm

But it might be Skissue

[hasty flume]

well i thikn it should be detectable by the primary decomposition where i want to hope that everything is the same as a prime power decomposition

and that it corresponds upstairs to a factorization of f

into irreducible components

[hushed bone]

Yeah so that’s the thing, idk abstractly how you’d recognize that the primary decomposition is into powers of primes

[hasty flume]

[hushed bone]

Like admittedly you do know A is the quotient of a like

Smooth 2-dim k-algebra

(Localized)

And that does something at least

But like, f can be all sorts of fucked up

[hasty flume]

tbc

[hushed bone]

So I think you should be able to make A as totally fucked as you want

[hasty flume]

this is the notion of equivalence that im trying to show holds if analytic isomorphism between the two points holds

so maybe the preserving multiplicities bit theyre referring to only holds for the reduced total transform and we can just assume C is reduced

[hushed bone]

Yeah I think here everything is always reduced

[hasty flume]

maybe it just works out idk

[hushed bone]

But that’s also weird cuz

Then the multiplicities should all be 1?

But umm

[hasty flume]

Yeah im realizing i dont understand what they mean by preserving multiplicites

Like

[hushed bone]

So in my brain

[hasty flume]

do they mean that the non reduced total transforms have corresponding components with the same multiplicity

or are they literally saying to throw that information away

[hushed bone]

Okay no I’m now convinced

Reduced means all multiplicities are 1

[hasty flume]

😵‍💫 😵‍💫

[hushed bone]

If not you absolutely are not reduced

You have introduced non reduced structure

[hasty flume]

I just odnt understand what "preserving multiplicities and incidence" means then

[hushed bone]

I feel like that’s about multiplicities of like

Self intersections

Of the singularities

[hasty flume]

Oh my god they mean multiplicity of the singular point

[hushed bone]

https://en.m.wikipedia.org/wiki/Multiplicity_theory

[hasty flume]

im stupid

[hushed bone]

Like this

And this can be read off the Hilbert polynomial

Which won’t care about completion

[hasty flume]

aaaaaaaaaaaaaaaa

okay i get it now

[hushed bone]

I’m the best

[hasty flume]

Chreww

thanks chmonkey

[hushed bone]

Chmonkey W again

[limpid horizon]

That sounds interesting but i cant understand it yet lol

[digital parcel]

i did my bach thesis on mixed multiplicities

waow

teissier had some minkowski-type inequality and he proved it and used it in ag

smth to do w milnor numbers but unfortunately i can't read french

[limpid horizon]

Tbh i think homological algebra is really interesting

I read somewhere that someone said its dry on its own

[proper frost]

I just joined. Haiii. I understand nothing here and I love it 🥹. I've found my place

[limpid horizon]

Welcome dopamine giver

[limpid horizon]

GIVE ME DOPAMINE NOW GIVE IT NOW

The fact that sending that message is what gave me dopamine 🤯

Mind = Blown

(Ok ill stop)

[last talon]

I don’t see how a direct product of sofic groups is sofic?
Like it doesn’t seem immediate
I’d want to do something like F -> S(n) x S(k) -> S(n + k) where the first map is induced by soficity of the two product groups, but that doesn’t satisfy (iii)

[last talon]

S(n) x S(k) -> S(nk) works

[lone jacinth]

What is S(n) here?

[last talon]

Symmetric group on n elements

[lone jacinth]

Then what does d(x, y) refer to in point (ii)

[last talon]

d(x, y) = |{x(i) =/= y(i)}|/n

[lone jacinth]

Shouldn't that make d(phi(g), e) = 0 then if there are no fixed points

[last talon]

Fixed

[lone jacinth]

I see

[lone jacinth]

You have
(a+b) / (m+n) <= a/m + b/n
so for a product of groups GxH and a finite subset F you can pick phi for epsilon/2 and the projection of F onto G and H.

Then combine them into a map -> S(m)xS(n) -> S(m+n)

[last talon]

That’s what I thought, but it doesn’t work because if you have some (g, e) then that’ll have fixed points
If you do -> S(mn) instead then it works

[lone jacinth]

How are you doing S(mn)?

Oh, I see

[solar linden]

how is t transcendental?

[foggy galleon]

by definition?

[solar linden]

wdym

[foggy galleon]

it simply is

it's a variable, no?

[solar linden]

yes

[foggy galleon]

it doesn't satisfy a polynomial equation over C. Note also that C is algebraically closed

[foggy galleon]

they are simply stipulating that t is transcendental

[solar linden]

doesn't it satisfy t-t = 0? isn't t like, already the variable of the polynomial?

I think I'm confused with the definition of transcendence

[foggy galleon]

right. Transcendence is relative to a base field. They are talking about transcendence over C

[solar linden]

oh ok I was being stupid

wait do the meaning of a "transcendental function" is different?

like, just the term "transcendental function"

[foggy galleon]

idk, I think people might mean different things by that phrase, I don't think it's very standard terminology

[solar linden]

oh ok I was mixing up transcendence wrt a field and transcendental function

[foggy galleon]

yes so in that case they are talking about transcendence over the field C(x)

[solar linden]

OHHH

damn my brain fried so much over this I wasn't able to eat

I'm 1 hour late

thx crocodile

[prime vigil]

Does Ado's theorem (on finite-dimensional Lie algebras admitting a faithful representation in some gl(V)) generalise to infinite-dimensional Lie algebras?

[lofty marlin]

Are Milne's notes a good intro for comm alg?

[scarlet ermine]

https://math.stackexchange.com/questions/3031/cayleys-theorem-for-lie-algebras
this might answer your question

[prime vigil]

Thanks, this is what I was looking for

[plucky quest]

Can i ask about sets here

Empty ahh channel💀

[ruby otter]

i think you better ask your questions about sets in discrete math/proofs and logic

[fathom fox]

#proofs-and-logic

[vapid axle]

Can I have a hint for this problem please? We can tensor with the right half of the sequence to get $A \otimes M / \ker (\pi \otimes 1) = A \otimes M / \text{im } (\iota \otimes 1) \cong A/\mathfrak{a} \otimes M$, but I'm a little stuck trying to show that $A \otimes M / \text{im } (\iota \otimes 1) \cong M/\mathfrak{a}M$

[broken turtle]

okeyokay

[swift cove]

Or you can just category theory sledgehammer, namely left adjoint preserves colimits

but well exactness is a weakened continuity condition so this is prolly what ultimately happens anyways

[hushed bone]

There’s an isomorphism A (x) M -> M given by multiplication. Under this isomorphism the image of that map is exactly aM

[hushed bone]

No, you need to recognize aM as the image of a map. Can’t just completely category theory that one away unless you define aM in some other way

[north remnant]

isnt aM supposed to be the kernel of M -> M (x) A/a

and not the image

[hushed bone]

Okay but that’s the same thing cuz of exactness?

[north remnant]

oh

[hushed bone]

I see what you mean actually

Hmmm

I would need to think more if it’s obvious that this kernel is aM.

You can say it contains it easily, but then…

Probably clear, but saying a specific tensor is not zero is always kinda MonkaS

[sacred sentinel]

I mean a (x) M is just aM and then the map aM -> M is just inclusion so the sequence is exact after tensoring ( exactness on the right/middle comes from right exactness of tensor product). Maybe a bit handwavy

[sacred sentinel]

Maybe this is not quite right and you have to instead identify the image of the composition a (x) M -> A (x) M -> M with aM

[sacred sentinel]

Yes this is definitely wrong

[hushed bone]

No a (x) M is not aM in general

Okay sorry you realized that

It requires something like M being flat

[digital parcel]

can't you just use this diagram?

[hushed bone]

that’s what I was saying lel

[digital parcel]

and then there's some hom alg lemma which says there's a map from A/a (x) M to M/aM which is an isomorphism since the left two are

oh

[hushed bone]

.

[fierce steeple]

Ye lol

[digital parcel]

oops

[fierce steeple]

Chmonkey is right and only thing needed lol

[hushed bone]

Of course I am right

I’m chmonkey

[fierce steeple]

A more interesting question is when this is an iso

[hushed bone]

When what is an iso

[fierce steeple]

I (x) M -> IM

But it is easy

[hushed bone]

I mean it’s exactly a Tor group being zero

[digital parcel]

isn't this actually equivalent to M being flat

[hushed bone]

No

It’s true when you enumerate over all I

[undone idol]

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

[fierce steeple]

If it holds for all ideals

[digital parcel]

oh yeah i was thinking holding for all ideals

[hushed bone]

You don’t need that, it just follows because cokernels of isomorphic maps are isomorphic

Which… actually I guess follows from 5 lemma

[digital parcel]

yeah

[hushed bone]

But I don’t like to think of it that way

[digital parcel]

i mean 5 lemma doesn't tell you about the existence of a map A/a (x) M to M/aM, but that's not hard to see

[hushed bone]

Yeah

[digital parcel]

anyway in this case i think you need four lemma and not five lemma

but i guess it doesn't matter too much

[hushed bone]

Five lemma works

You extend the diagram with two 0s

Lol

[digital parcel]

oh yeah on the right

true

[hushed bone]

It’s super hacky

But it’s useful

[fierce steeple]

Just put the chips in the bag bro

[hushed bone]

Just put the kernel in the bag bro

[digital parcel]

you're literally a potato i'd have to kill you to make chips

sorry potato that was very toxic of me. lapse of judgement i just got super defensive and threw my phone and broke it. wont happen again

[hushed bone]

Bro took 2 mins to follow up cuz he had to grab his laptop to type that

[digital parcel]

it's hard to find my laptop when all i see is red

heh..

[mental escarp]

If A:V->V is a linear map and the induced linear map on the k blades $\wedge_k A: \wedge_k V\to \wedge_k V$ is not full rank does that mean A is not full rank?

[broken turtle]

HausdorffT1

[limpid horizon]

blades? dafok

[mental escarp]

I don’t know the lingo. Whatever elements of the form $v1\wedge…\wedge vk $ in the exterior algebra of A are. There is a vector space worth of these elements.

[lone jacinth]

That should follow from \wedge_k being functorial

Since being full rank means the map splits

Oops sorry, that would be the converse of what you asked.
Edit: wait no it is what you asked. I read it wrong the second time

[fierce steeple]

Or just like A is an iso so it induced an iso

Which is the contrapositive of ur ting

[mental escarp]

Oh there is an exact sequence $0\to V\to_A V\to 0 $ and the functor mapping this exact sequence to another implies that the induced map of A is also an isomorphism. Nice that’s a cool application of exact sequence.

[lone jacinth]

Sure, the exactness isn't really the relevant part. But the fact that functors preserve (one sided) inverses.

In this case it's actually an isomorphism, but the argument would also work for a map V -> W

[digital parcel]

Vakil only uses this "ideal of denominators" construction in one other place: the proof of Algebraic Hartog's Lemma. Are there other cool uses of this construction?

In CA and/or AG

I found something on stacks (02P1) called "ideal sheaf of denominators" https://stacks.math.columbia.edu/tag/01X1

Basically the same construction

[lone jacinth]

It's a special case of ideal quotient
https://en.m.wikipedia.org/wiki/Ideal_quotient

Relevant to fractional ideals, which is something that comes up as an invariant

(e.g. ideal class group)

[digital parcel]

oh yeah that's true i forgot it's an ideal quotient

okay thanks

yeah ive seen that before a lot actually, eg local cohomology

[hushed bone]

You know local cohomology ??

[hushed bone]

Ideal quotient……..

Real ones call it colon ideals

[limpid horizon]

i am learning about local cohomology

[fierce steeple]

Real

[limpid horizon]

Im still not sure what the motivation for it is though

i know some properties of it now like how it depends on the ideal but only up to radical

its an algebraic geometry thing right because if a prime ideal does not contain I then the localization is 0 or something right

when you localize the submodule l-torsion(M) at P

[digital parcel]

not well by any means but my prof taught it in a topics course

[hushed bone]

Well its formation commutes with localization, and I becomes all of A. And clearly A-torsion is nothing

[limpid horizon]

Does that first part mean, localize M at S then take I- torsion elements of that is the same as taking I-torsion elements and then localizing at S?

Do you mind spelling out a bit more what youre saying?

[hushed bone]

Yes

S^-1H^i_I(M) = H^i_IS^-1A(S^-1M)

[limpid horizon]

ok cool thanks, whats the benefit of this?

[limpid horizon]

were you trying to explain why what i said here was true or smth?

[hushed bone]

I mean, to know how something behaves at each point essentially tells you how it behaves entirely. Something something A -> Prod A_p over all p is faithfully flat

[hushed bone]

Yes

[hushed bone]

So you can reduce almost everything to the local case which is highly desirable

[limpid horizon]

oh ya localizing at a prime ideal produces a local ring, i forgot about that

ik yall would go mad at tht

is that a particularly nice property for algebraic geometry?

[fierce steeple]

No i mean it is valid

[limpid horizon]

Lol

[fierce steeple]

It's a nice property in comm alg in general tbf

[limpid horizon]

Yea i know it is but i dont really remember or have seen why yet

[fierce steeple]

often common to reduce to local considerations

[limpid horizon]

I know there are local properties so that if smth is true in all localizations then its true in whole thing

is that the vibe ?

[fierce steeple]

Ye and also like lots of simplifications occur like

Ye

[limpid horizon]

so true

Ye

[vapid axle]

What lemma is this? I was aware of the five lemma

Can we implicitly use the five lemma by extending the diagram with a few more squares of 0s lol

[digital parcel]

yes that's what chonkey said

chmonkey*

sorry keyboard lagged out

[vapid axle]

Oh did he say that after your message

I didn't scroll down lol

[digital parcel]

technically five lemma doesn't tell you about the existence of the map

but that can be shown by diagram chasing using the fact there's surjectivity

five lemma tells you that it's an isomorphism

[vapid axle]

Yeah I'm just trying to find out what the middle map even is lol

[vapid axle]

I have no clue if I'm on track here but I'm thinking about constructing a bilinear map from A/a x M -> M/aM then using the universal property

Okay I think that works, I showed it's well-defined and bilinear

Now if commutativity holds then the five lemma applies...

Let $g: A \otimes M, \mathfrak{a} \otimes M \to M$ be the isomorphism $a \otimes m \mapsto am$. The following diagram is commutative with exact rows, by Proposition 2.18:
[
\begin{tikzcd}
{\mathfrak{a} \otimes M} && {A \otimes M} && {A/\mathfrak{a} \otimes M} && 0 && 0 \
\
{\mathfrak{a} \otimes M} && M && {M/\mathfrak{a}M} && 0 && 0
\arrow["{\iota \otimes 1}", from=1-1, to=1-3]
\arrow["1"{description}, from=1-1, to=3-1]
\arrow["{\pi \otimes 1}", from=1-3, to=1-5]
\arrow["g"{description}, from=1-3, to=3-3]
\arrow[from=1-5, to=1-7]
\arrow["h"{description}, from=1-5, to=3-5]
\arrow[from=1-7, to=1-9]
\arrow[from=1-7, to=3-7]
\arrow[from=1-9, to=3-9]
\arrow["g", from=3-1, to=3-3]
\arrow["\pi", from=3-3, to=3-5]
\arrow[from=3-5, to=3-7]
\arrow[from=3-7, to=3-9]
\end{tikzcd}
] The map $h$ is defined as follows. There is a bilinear, well-defined map $A/\mathfrak{a} \times M \to M/\mathfrak{a}M$ given by $(a + \mathfrak{m}, m) \mapsto am + \mathfrak{a}M$. Such a map induces an $A$-homomorphism $A/\mathfrak{a} \otimes M \to M/\mathfrak{a}M$ given by $h(a + \mathfrak{a} \otimes m) = am + \mathfrak{a}M$, which we take to be $h$. Since all vertical arrows save the middle one are isomorphisms, the five lemma shows that $h$ is an isomorphism, as desired.

How does this look?

[broken turtle]

okeyokay

[digital parcel]

a (x) M is not isomorphic to M

the quotient in M/aM comes from the fact that a (x) M has image aM in M

[vapid axle]

Oh okay

That makes more sense lol

[fierce steeple]

i do think this appproach with five lemma etc is overkill

[sly rune]

lol

actually if you look at this double complex and flip some pages...

[digital parcel]

[vapid axle]

What would be easier?

[sly rune]

universal property of quotient

[fierce steeple]

Well you have I (x) M -> A (x) M -> A/I (x) M -> 0. You can identify A (x) M with M, and also replace the left term with its image in M (as that doesnt affect exactness), which is IM

And then yeah youve exhibited A/I (x) M as the cokernel of IM -> M

[hushed bone]

Lots of things can be written down as “X is true if M is 0” where M is some module. Like you are flat if all Tors are 0 against stuff or like surjective iff cokernel is 0. If you can write it that way, you can say M is 0 if and only if M_p is 0 for all 0, so if the formation of M commutes with localization you immediately get that X is true if and only if it’s true at all P

[hushed bone]

FLIP SOME PAGES

So good

[spice idol]

Who out there flipping them pages

[limpid horizon]

Thats great, thanks

[digital parcel]

another nice reduction to local case is that an integral domain A is integrally closed <=> A_p is integrally closed for each prime ideal p <=> A_m is integrally closed for each maximal ideal m

[limpid horizon]

Tbh, I still am not sure in what sense something like Ext^i(A,-) is a functor

a functor from RMod to RMod?

Given B->C what is the map Ext^i(A,B) -> Ext^i(A,C)?

[drowsy compass]

Let's say I have some nested root formula, is there a way to compute whether or not it can't be the root of a polynonial of degree N

[golden osprey]

IDK off the top of my head what algos are out there, but I think programs such as mathematica can find minimal polynomials of algebraic numbers

So perhaps trying to find algorithms for computing minimal polynomials would be a start

https://arxiv.org/pdf/1001.0649 like some googling found this

and perhaps someone more knowledgeable than me would know of actual reference texts

[lone jacinth]

Take injective resolutions of B and C and lift the homomorphism to a map up complexes
I(B) -> I(C)
(such a lift is unique up to homotopy).

Then apply Hom(A,-) and take ith homology

[limpid horizon]

Ah ok, how does this relate to the long exact sequences for Ext?

[lone jacinth]

Or if you're working with Yoneda-Ext:

For an i-extensions
B -> E1 -> ... -> Ei -> A
take the pushout along B->C

[fathom fox]

i've always wondered if there was something like the yoneda Ext for tor

[lone jacinth]

Given a short exact sequence
0 -> B -> C -> D -> 0
the horse shoe lemma says you get a short exact sequence of complexes
0 -> I(B) -> I(C) -> I(D) -> 0
which is pointwise split because the objects are injective.

Then apply Hom(A,-) and take the long exact sequence in homology

[fierce steeple]

I think there shouldn't be just cause yoneda ext is so connected to maps basically