advanced-algebra

[soft parcel]

yeah fair

But yeah, I feel like the approach sort of falls short at the point where M and N might both not be flat

[fierce steeple]

But yes ok, I mean tbh I think flatness is kind of irrelevant. My hint would be to consider it all modulo maximal ideal as you do in the flat case

Like in the case that A is a field, this is standard, so try to reduce it to that

[soft parcel]

Oh really? I haven't seen that before

Thanks

[fierce steeple]

Dimension of tensor product is product

[soft parcel]

Oh yeah ok

[fierce steeple]

Or like every module over a field is faithfully flat

Cause they're all just sums of the field

[thorny fable]

can we formalize the condition of subjectivity for a "localization map" R \to S^-R

also what are the motivations to invert everything but a prime ideal (R_𝔭)

why would we want to turn a prime ideal into a maximal ideal

[past cove]

Well local rings are very rich
And the condition of being a prime ideal naturally gives you a multiplicative subset

But a lot of motivation comes from algebraic geometry as usual

Where local rings express the data about a certain subset of your variety/scheme and you want a localization operation to be able to extract that kind of information

[thorny fable]

I shouldn't be asking about non commutative cases, but is localization also a "technique" there?

[thorny fable]

might have to hold my horses and of now

[past cove]

Er good question and I think the answer is it's complicated lol
Non commutative algebra scares me ngl

[thorny fable]

have to open up Lam at some point, yeah it's scary as hell

[thorny fable]

also yeah, is there any significance for this case, as I'm completely unaware of the applications

[lofty linden]

do you study this algebra and cryptography in a undergraduate course

[past cove]

Uh idk about localizations lol

Tho ig if it's an advanced course then yeah

Cuz localization is ig more in the realm of commutative algebra

[lofty linden]

wait i never been to university before what do you typically study in an undergraduate course

[thorny fable]

whatever you can digest

idk

[ornate atlas]

Technically but it’s not super useful and it’s a fucking nightmare, assuming you can even do it

Ok I shouldn’t say it’s not super useful because it does have uses but it’s no where near as powerful as in the combative case

You can get a notion of localisation in non commutative rings under certain assumptions, but it’s no where near as simple as the commutative case of just slapping on some fractions

We only dealt with a special case in my noncom class and even that was pretty terrible lol

[spice idol]

if you take an algebraic variety V of some kind we are interested in maps that look locally like a quotient of two polynomials (these are as "nice" as possible, think of it like the algebraic variant of differentiable / analytic functions), which are called regular functions. This means that every open subset has a set of functions with some local property attached to it, and thus it forms something called a sheaf.

With sheaves we like to know the "local behavior" of the functions at each some point; that is how functions defined on some neighborhood of a point x behave as we zoom in on x. It so happens that, when V is a variety, then you can attach a ring k[V] to it and points of V correspond to maximal ideals of V, and this local behavior at some point is captured by the localisation at its maximal ideal.

So localisation by a prime is an extension of the fairly concrete idea of "looking at a point" in classical algebraic geometry

[thorny fable]

hopefully it will make sense, oneday

[spice idol]

you've got a whole concept called calculus of fractions apparently, which basically copes with the fact that things aren't commutative lmao

[spice idol]

algebraic geometry is a beast of a subject

[thorny fable]

will it teach me cool algebra?

[spice idol]

it will motivate a lot of it

[thorny fable]

would!

[spice idol]

like localisation, I understand it pretty well now purely because I went out of my way and did classical algebraic geometry

because your base field is algebraically closed magic happens

[digital parcel]

You can read up on the basics of the Zariski topology (which is the topology in algebraic geometry). Dont need to do any exercises or whatnot to just get the idea. If you want to do some exercises, see Atiyah-Macdonald ch1, ch3

[thorny fable]

nice, currently trying to prove some proposition (exercise) from the book im reading, nothing mentioned from geometry as of now

[digital parcel]

yeah

i mean even just purely algebraically, localizing at a prime gives you a local ring which are super super nice to study

[limpid horizon]

I'm not too sure how I^t+Ann(M) annihilates Ext^i(R/I^t, M)

one sec

[thorny fable]

this one is nice

[limpid horizon]

Is it because representatives of Ext^i(R/I^t, M) is given by kernel of some map between Hom(R/I^t, -), which I^t annihilates, and it is also given by kernel of some map between Hom(-, M), which Ann(M) annihilates?

[mental escarp]

Is there some theorem like, for k algebras R and S whenever the categories mod R and mod S are equivalent then the commutative algebras are isomorphic?

[spice idol]

R and S are commutative right?

if so this holds for general commutative rings; see the center of a category

the center of the category R-Mod is isomorphic to Z(R), and equivalent categories have isomorphic centers

[vague pawn]

If I have polynomials $f_1, f_2, f_3,..., f_m \in \bQ[x_1,...,x_n]$ such that there exist polynomials $g_1, g_2, g_3,..., g_m \in \bC[x_1,...,x_n]$ such that $$f_1 g_1 + f_2 g_2 + ... + f_m g_m = 1$$ How can I show that $g_i \in \bQ[x_1,...,x_n]$ or at least there are polyonomials like $g_i$ that are with rational coefficients.

[broken turtle]

ExpertEsquieESQUIE

[weak lodge]

Adding onto what enpeace music said, you might be interested in something called Morita equivalence. Two commutative rings that are Morita equivalent are isomorphic, for the same reason enpeace said

[spice idol]

goes to show how nice commutative rings are

[fierce steeple]

Morita theory is cool.

[spice idol]

does that study everything that follows from morita equivalence n such?

[fierce steeple]

You can classify when RMod is equivalent to SMod as well as the types of equivalences

[spice idol]

oh that is awesome

does it pop up "in the wild" i.e. have there been cases where it was easier using morita equivalence than showing the rings had certain properties in common?

poorly phrased question but i hope you get the idea

[vague pawn]

🙏

[mental escarp]

Wow thanks!

[fierce steeple]

more generally you can think of RMod as an invariant of R and lots of the properties are reflected by this yes

[weak lodge]

Off-topic but @spice idol as a universal algebraist you might be interested in Morita theory for algebraic theories, where instead of RMod you compare categories of models of algebraic theories

[spice idol]

does that exist?

thats pretty cool

im not very active in the category theory side of UA, tbh

and esp what im reading up on right now - tame congruence theory - is as far away from category theory as you can get (but i am hoping to change that in some way using polynomial representations)

do you have any resources for that?

[weak lodge]

borceux handbook of categorical algebra 2 discusses this i think

[spice idol]

cool, thanks!

damn they even have descent theory

[worldly zealot]

morita equivalence is most interesting in the noncommutative case imo (morita equivalent rings have isomorphic centers). theres a big field of representation theory that is only interested in rings up to morita equivalence - that is, studying module categories.

an important theorem is that every finite-dimensional associative k-algebra is morita equivalent to a path algebra modulo an admissible ideal. so this is why people look at quivers

[spice idol]

i see, because most of the properties of a ring we care about are encoded in its representation structure?

[worldly zealot]

dont know that i could say most, but many, yes

like homological invariants

[spice idol]

hmhm okay

[worldly zealot]

also schemes can be recovered from the quasi-coherent sheaves on them

this is sort of the motivation for noncommutative algebraic geometry

you can define Proj of a noncommutative ring via its module category

[ornate atlas]

Isn’t it only interesting in the noncommutative case lol

[worldly zealot]

xd

im scared of absolutes!

[ornate atlas]

It’s interesting in the commutative case in so far as knowing if rings are isomorphic or not is interesting lol

[spice idol]

an isomorphism between rings seems easier to construct than an equivalence of categories

ino

im(naive)o

[ornate atlas]

One would think

I want to learn more about morita stuff, I’d have probably done my masters thesis on it if I had stayed at edi

[near lantern]

Let V and W be vector spaces and L a 1-dimensional vector space over a given field. Suppose for every l ∈ L \ {0}, I have an isomorphism pi_l: V → W and this has the property that pi_{tl}(tv) = pi_l(v) for non-zero scalars t. Is there a neat way to explain why this induces a canonical isomorphism from Hom(L, V) (or L* (⨯) V if you prefer) to W?

This can be done very explicitly: for f: L → V, let pi(f) = pi_v(fv) for any v. But I feel it should be possible to do it more abstractly.

Never mind: obviously we have pi^{-1}_{tl} = t pi^{-1}_l, so l ↦ pi^{-1}_l is linear, giving pi: L → Hom(W, V) or pi: L (⨯) W → V which can be shown to be an isomorphism.

[willow seal]

I'm not sure why Wikipedia's definition of the tensor product of two modules is an abelian group. Other references I've read define it to be another module, which makes more sense to me. Is there an advantage to considering tensor products as abelian groups?

[near lantern]

It depends on whether your ring is commutative or noncommutative.

If R is a commutative ring, the tensor product of R-modules can be given a natural R-module structure.

If R is a general non-commutative ring, then you have to distinguish left and right R-modules, and the tensor product of a right R-module and left R-module is an abelian group. To get natural module structures on the tensor product your original modules have to be bimodules.

[willow seal]

Makes sense thanks. Is the abelian group operation addition of tensors?

[near lantern]

For example, if M is an (S, R)-bimodule and N a left R-module then the S-module structure on M "survives" onto the tensor product and M (⨯)_R N is an S-module.

[unborn rampart]

The intuition I've heard (from jagr actually) is that the tensor product M (x) N glues the right module structure of M to the left module structure of N and destroys them both in the process. If they are both bimodules then you're still left with a module structure on both sides

[near lantern]

Yes, exactly!

[willow seal]

I think I see that from the quotienting of the free abelian group

[ornate atlas]

I always remember it like matrix multiplication, for getting the dimensions right

[near lantern]

(Hom can also be thought of like this, except that the module structures on the domain are "reflected" in side.)

[lone jacinth]

Not sure this is what youre asking, but you have things such that over an algebraically closed field every finite dimensional algebra is morita equivalent to a path algebra with admissible relations. So if you want to prove something about the module category of all findim algebras its enought o think about path algebras.

Similarly End(M) is morita equivalent to End(N) whenever addM = addN, so in situations where End(M) appears naturally you can simplify them quite a bit by reduicng to End(N) for some smaller module.

[spice idol]

thats what i was looking for yes, thank you mr jagr man

ive never heard of add(M) before, huh

[zenith pagoda]

TIL that the projective dimension of C(x,y,z) as a module over C[x,y,z] is undecidable in ZFC

https://mathoverflow.net/a/2041

i can't delete it now, because you replied to it

there will just be two links

[void plank]

Ooh, very interesting!

[limpid horizon]

anyone know how to prove this one

or just even like ... where one would start with seeing why this is true

[limpid horizon]

Also, they must be isomorphic as R-modules?

[fierce steeple]

Follows from 3)

[limpid horizon]

Oh yowza

ok thanks ill take a look

[spice idol]

do we have an isomorphism H_^i_m(M) (x) R_m \cong H^i_m(M) known then?

[limpid horizon]

I havent thought about what 3) means yet

also H^i_mRm(Mm) = H^i_m(Mm) right

bro how do i write latex on here man

i been here for ages and still dont write with latex here

[fierce steeple]

Oh I misunderstood

I guess basically means it is enough to show that R \ m acts invertibly on H^i_m(M)

[limpid horizon]

whats goin on lowkey

[spice idol]

commutative algebra

probalby

[limpid horizon]

[urban granite]

communist algebra

[granite yew]

Is there a metric vectorspace (V,b) over a field F such that for some field extension K/F, Cl(V,b) is a K-algebra, but is a clifford algebra over F and not K?

There are many easy examples of the (almost) converse, e.g. Complex Clifford Algebras which aren't Real Clifford Algebras

[limpid horizon]

H_m^i(M) (x)R Rm = H_m^i(M) ?

by 3), H_m^i(M) (x)R Rm = H^i_mRm(Mm) so i see that part

[void plank]

I've seen the claim that if complexes of R-modules C and D are quasi-iso to C' and D' respectively, then C (x) D is quasi iso to C' (x) D'

Is there anywhere I can find a proof of this claim?

[fierce steeple]

This is false, or otherwise higher Tor groups of modules would always vanish

(Cause taking projective resolutions would not affect M (x)_R N up to quasi-iso)

Conversely if the modules involved are all projective then you are good (by uniqueness of projective resolutions)

[limpid horizon]

I havent seen Tor in action anywhere yet. what is a good first application of it?

I have seen Ext now for defining depth and stuff

[void plank]

E.g. universal coefficient theorem in alg top as well as kunneth formula

[void plank]

This is from the book "hochschild cohomology for algebras". The claim is marked in red

Maybe the statement assumes some of the properties in the preceding theorems hold

[fierce steeple]

Yes the C_n and C_n' being flat would make it ok

[void plank]

Ah ok, this wasn't clear at all since it was stated randomly in the book

It did seem too good to be true, which is why I asked you all

[fierce steeple]

Basically the thing that has the nice property you want is the derived tensor product, whose homology groups are the Tor groups. Then when the C_n are flat this can be computed by the "usual" tensor product

(Like more generally this is cause you can use acyclic resolutions for derived functors)

[void plank]

By derived tensor product, do you mean taking a tensor product of a projective resolution of the corresponding chain complexes? Projective modules are flat, so this seems ok

[fierce steeple]

Yes

Like using projective resolutions is really the most like fundamental thing and then it's a general theorem that you could also use acyclic resolutions up to quasi-iso

Is this B. Richter's book?

[void plank]

Fortunately in my subfield, we mainly work with complexes of F_2 vector spaces

We endow them with group actions often though, in which case doing things like applying coinvariants after taking a tensor product becomes tricky (unless the group is odd order in this context)

[fierce steeple]

Curious

I now wonder what field this is aha

[void plank]

Sarah Witherspoon

[spice idol]

smt cs related?

[fierce steeple]

Ah sorry yes

[void plank]

(quantum) error correction

[spice idol]

hahaa I was right!!

..partially

[fierce steeple]

Nice

I assumed smth like enpeace said aha

[spice idol]

damn that is a surprising amount of abstract math there then, sounds really cool :0

[fierce steeple]

Well like ig finite fields are used for Hamming codes n stuff right idk

[fierce steeple]

Nice

Witherspoon is cool

[void plank]

All my stupid questions the last few weeks is because I'm doing a project in quantum error correctikn that's making me question all the hom alg I learned a while back

But it's been deepening my understanding of material I already learned, so that's always good

[spice idol]

yeah but I wouldn't've imagined chain complexes and stuff would come up, but ig i dont really know what they do there

[fierce steeple]

I haven't checked out the book tho

[fierce steeple]

No they haven't been stupid lol it's ok

But yes over a field all is good

[spice idol]

if I may ask, how does hom alg pop up there?

[void plank]

Tensor product of chain complexes were used for one of the most major breakthroughs in classical and quantum error correction a few years ago

[spice idol]

hom alg over a field.. one can dream..

[fierce steeple]

Oh very nice

[fierce steeple]

Lol yes

The study of graded vector spaces

[void plank]

It's a bit silly at first but gets a lot more interesting

A classical error correcting code is defined by a "parity check matrix" H with entries in F2 whose kernel defines your encoded information. This is just a chain complex with one arrow representing H

A "quantum (CSS) code" is an error correcting code specified by two such parity check matrices H_X and H_Z representing your Pauli X and Z operators acting on your quantum system. They must satisfy H_X H_Z^T = 0, hence it's a chain complex with two arrows

The homology in the middle of the complex (and it's dual) is your space of logical operators, which are the errors that are problematic for your encoded information.

Then you can play games like taking tensor products of chain complexes to get other codes. These constructions were essential in showing the existence of "asymptotically good quantum LDPC codes". Classical LDPC codes, for context, are used in Wifi and 5G for example

There's also lots of other tools from algebra and algebraic geometry that can appear in (quantum) error correction. For instance, QR codes use Reed-Solomon error correcting codes, which are based on evaluations of polynomials over finite fields. Algebraic geometry codes have also been used in QEC recently for certain important protocols for doing universal quantum computation

[spice idol]

hmm, that is awesome

[void plank]

I've been meaning to go through some of this book for example:
https://link.springer.com/book/10.1007/978-3-540-76878-4

[worldly zealot]

one of the letters in CSS teaches intro abstract algebra almost every year at my school and he throws in quantum error correction stuff. i took it the one rare year he didnt teach it

my professor was more commalg heavy though. big win for me personally

[limpid horizon]

That sounds cool

[void plank]

Ah that's unfortunate. It's one of those fields that's practically very useful and is mathematically quite deep

[void plank]

Fair enough!

[worldly zealot]

yeah would have been nice to brush paths with the field in any case

[void plank]

It's fairly easy to pick up if you have a good math background. The harder stuff for me is ironically the more practical and application-heavy side of things, but maybe that's just because I'm more theoretically minded

[spice idol]

me when I'm trying to prove something and there's a counterexample of an algebra on a set with three elements and two unary operations 😭

[frosty ember]

reading etingof's book on lie groups and lie algebras right now

I've been stuck on this kinda weird exercise

etingof does not require this to be part of the definition of a (g,K)-module, which is a little bizarre but ok

instead, he requires

• K is connected
• there is a map Lie(K) -> g and Ad: K -> GL(g) such that the differential of the adjoint action is the adjoint action of Lie(K) on g
• as usual, the two actions of Lie(K) on M (one by differentiating K and the other by g) coincide

[frosty ember]

I can show that the derivateives of both sides with respect to g are the same, but I can't seem to conclude from there because we are not really looking at a stabilizer ...

[near lantern]

Life lessons

[near lantern]

What book is this?

[near lantern]

Is V finite-dimensional?

[near lantern]

I would suggest the following: g and V are K-reps; the condition says that the action map g ⨯ V → V is K-linear, or that K acts trivially on it as an element of the K-rep Hom(g (⨯) V, V). But I'm unsure if in your setting you can take tensors and homs of K-reps as K-reps.

[versed swan]

I’m 80% sure this question goes here, so here goes. I’ll move it if not.

For n is a positive integer, I define a number x to be n-algebraic if it is the root of a polynomial of degree n with rational coefficients (this also works with integer coefficients, but I’m defining this with rational coefficients for reasons that will be clear in a minute). From here, I can make several assertions based on prior results:

• x is 1-algebraic iff x is rational
• All non real x are at least 2-algebraic
• x is computable by radicals if x is 4-algebraic or lower
  Etc.

Now suppose I want to try to use this to extend to transcendental numbers. I define a number to be ∞-algebraic if it is the root of a convergent infinite series with rational coefficients. For example, π is ∞-algebraic because it is a root of the sine Maclaurin series. The number e is also ∞-algebraic with the series for LambertW(exp(1+x))-x https://math.stackexchange.com/questions/4948131/power-series-where-the-number-e-is-a-root?noredirect=1&lq=1

Now, I have two questions:

• What is the accepted nomenclature for what I’m saying here, if any? I know for a fact that I’m not the first person to consider this kind of question. As a super basic example, this https://math.stackexchange.com/questions/4643919/power-series-for-functions-that-have-e-as-a-root problem calls what I’m calling ∞-algebraic an “algebraic sympathizer”
• That stack exchange question has an answer which claims that all real numbers are ∞-algebraic. Is there a way to restrict the definition of “∞-algebraic” to something like “an infinite series with rational coefficients that can be explicitly computed at each step” which would then yield a result of something like “An number is ∞-algebraic iff it is computable”?

[pearl stream]

what does it mean for a vector space to be endowed with a quadratic form?

[spice idol]

it means that you have a pair <V, f> where V is a vector space and f : V -> F is a quadratic form

[versed swan]

Which, to explain that being “a solution exists”, means there’s a “quadratic form” which essentially corresponds to some kind of inner product

(That’s not ALL a quadratic form is)

(But it’s kind of the thing to think about when relating it to a vector space)

[rare walrus]

This n-algebraic idea is just asking for the degree of the extension [Q(x):Q].
I don't know enough about the latter question to comment

[woeful lichen]

I think https://en.wikipedia.org/wiki/Field_extension might be useful here.
Given a field extension K ⊆ L (in your case ℚ ⊆ ℂ), and an element s ∈ L, write K(s) for the smallest subfield of L that contains both K and s. K(s) is a K-vector space (since in K(s) we can do addition, and multiplication by elements of K). The dimension of K(s) as a K-vector is called the degree of K(s) over K, and written [K(s):K].
If s is the root of a non-zero polynomial with coefficients in K, then the minimal degree of such a polynomial is [K(s):K].
So x ∈ ℂ being n-algebraic is the same as having [ℚ(x):ℚ] ≤ n, as @rare walrus mentions.

[woeful lichen]

The hard part of considering ∞-algebraic, I think, is that it's no longer pure algebra (concerning sets with a bunch of n-ary operations, in this case fields, i.e. sets with 0, 1, +, -, and ⋅ satisfying a bunch of laws and in which all elements have a multiplicative inverse). Instead a norm |-| gets involved too, allowing convergence of infinite sequences and series to be considered.

[spice idol]

when operations are no longer n-ary everything falls apart

[woven loom]

All computable numbers would fall under this by using your approximation to make the coefficients of the power series

So as to make evaluating it at 1 spit out 0

So, appropriately rescale and approximate, and I think you should get it?

I dunno, maybe I’m making some obvious mistake ofc, but rational approximations with some careful tweaking should probably save you?

[woven loom]

I’d think some approximating polynomials thing shouldn’t be too far from it, but trying to actually write it down idk if that one works

[full dragon]

Does someone mind checking over this to see if this is true

Let $R$ be a commutative ring, $K$ a unital $R$-algebra with unit $e$. Let $f \in \mathrm{Hom}\mathrm{UnitRAlg}(K,R)$, basically a functional that distributes over multiplication. Consider the module $M = \ker(f) / \ker(f)^2$. \
For some $R$-module $H$, define the space $D_f(H) = { p \in \mathrm{Hom}\mathrm{RMod}(K,H) : \forall x,y \in K [p(xy) = f(x)p(y) + f(y)p(x)]}$. \\

Assume $g \in \mathrm{Hom}\mathrm{RMod}(M,H)$, then by passage through the quotient we can imagine $g \in \mathrm{Hom}\mathrm{RMod}(\ker(f),H)$. \\ Now define the projected map $g'(x) = g(x - f(x)e)$ on $K$, then $g' \in \mathrm{Hom}\mathrm{RMod}(K,H)$. Now for $x, y \in K$, we have $xy - f(xy)e = xy - f(x)f(y)e = xy - f(x)y + f(x)y - f(x)f(y)e = (x - f(x)e)y + f(x)(y - f(y)e) = (x - f(x)e)(y - f(y)e) + (x - f(x)e)f(y) + f(x)(y - f(y)e)$ thus $xy - f(xy)e \equiv (x - f(x)e)f(y) + f(x)(y - f(y)e) , \mathrm{mod } \ker(f)^2$ \\
$g'(xy) = g(xy - f(xy)e) = g((x - f(x)e)f(y) + f(x)(y - f(y)e)) = f(y)g'(x) + f(x)g'(y)$, so $g' \in D_f(H)$. \
Therefore, by linearity, $\phi : g(x) \mapsto g(x - f(x)e)$ is a module map $\mathrm{Hom}\mathrm{RMod}(M,H) \rightarrow D_f(H)$
\\
Assume $\phi(g) = 0$, so $g(x - f(x)e) = 0$ for all $x \in K$, then for $x \in \ker(f)$, we have $g(x - f(x)e) = g(x) = 0$, so $g = 0$ identically on $M$, thus $\phi$ is monomorphic. \\
Now assume $d \in D_f(H)$, then we have that $d(e) = d(e^2) = f(e)d(e) + f(e)d(e) = d(e) + d(e) \Rightarrow d(e) = 0$, so $e \in \ker(d)$. Now, for $x \in K$, $d(x - f(x)e) = d(x) - f(x)d(e) = f(x)$, and that for $x,y \in \ker(f)$ we have $d(xy) = f(x)d(y) + f(y)d(x) = 0 \Rightarrow \ker(f)^2 \subseteq \ker(d)$, we have that $d = \phi(g)$ where $g(x) = d \vert_{\ker(f)} (x + \ker(f)^2)$ so $\phi$ is epimorphic. \\
Ultimately $\phi$ is an isomorphism

This is an abstract alg formulation of the fact that the dual of the zariski cotangent space is the space of derivations

[broken turtle]

miz

[void plank]

Actually, I had a small follow-up question for you about this result.

Suppose C, C', D, D' are complexes of F_2 vector spaces such that C ~ C' and D ~ D' (where ~ means quasi-iso) and let G be a group of odd order. Suppose we've also endowed C and C' with right actions of G and D and D' with left actions of G.

Now consider C (x)_F2 D and C' (x)_F2 D'. These are quasi-iso by the result we've discussed (as C and C' are trivially complexes of flat modules).

Exact functors preserve quasi-isos and taking coinvariants of an F_2 vector space endowed with a G-action with |G| odd is an exact functor.

So assuming my reasoning is correct till now, this implies that Coin(C (x)_F2 D) ~ Coin(C' (x)_F2 D'), right? And of course, Coin(C (x)_F2 D) = C (x)_F2[G] D and similarly for the other term

Maybe I'm wrong though, so I'm curious to know where my reasoning breaks down if so

[full dragon]

@fierce steeple while you exist do you mind checking over this for me :3

[rose mirage]

my eyes fell out of my head like an AWOOGA moment looking at paragraph 3

[forest turtle]

use an align environment lol

[fierce steeple]

I didnt jnow what this was for

[fierce steeple]

This sounds good to me

[full dragon]

Autism :33

[near lantern]

For the second part, I suspect - but have not tried to prove - that an ∞-algebraic number is computable iff the power series is computable.

[near lantern]

On the fourth last line, it's d(x-f(x)e) = ... = d (x). Otherwise correct.

[versed swan]

I mean, it’s easy to show in one direction: if a humber has a computable convergent power series representation, then you can compute it as precisely as you like by taking as many terms as you like and then using Newton’s method or some such

[near lantern]

I figured something like that.

[woeful lichen]

For computable numbers s that are real, the other direction isn't much harder: Pick a sequence of epsilons e_n to bound how far the partial sums p_n = sum(a_i s^i for i = 0, ..., n) are allowed to be away from zero. Pick each coefficient a_n in QQ such that you have |p_n| = |p_(n-1) + a_n s^n| < e_n. This is always possible because QQ is dense in RR. This is computable because the partial sum so far and s^n are computable.

When I tried, it looked like for non-real complex s, it's a little more tricky, but still possible. That involves using that s^n and s^(n+1) are linearly independent in CC over RR, yielding that s^n QQ + s^(n+1) QQ is dense in CC. This means that for every second partial sum, it's easy to be as close to 0 as needed. Using some budgetting and estimations, we can also get the other partial sums as close to 0 as needed.

[woeful lichen]

For the direction from computable power series to computable roots, I'd be worried about not being able to compute how many terms you need for a given precision, leading to non-termination of the computation.

[mint pine]

there's a field that, given a real number, researches the possible converging rational sequences to it, but I cannot remember the name of it. Might be of interest to you.

[woeful lichen]

Hmm, I don't actually know when a powerseries is considered computable...

[ornate kindle]

So I have a Noetherian integrally closed domain A, and I'm seeing a reference to I_A as its "ideal group"

I'm not sure how such a group would look though? It also mentions that the group is free abelian and generated by the non-zero prime ideals of A

[past cove]

Are you familiar with Dedekind domains?

[ornate kindle]

It kinda seems like the group action is products of the elements of the ideal (and hence the principal element when that's relevant), but I don't see how there's inverses with that action

Yeah that's what this is on

Well, part of what this is on

[past cove]

You can define an inverse ideal

Tho it's subtle

[ornate kindle]

Hmm alright, I remember seeing that before but I though it was a set that wasn't actually an ideal, just a subset of the fractional field

I'll reread that section, thanks 👍

Oh dammit it literally defines the ideal group in that section. Thanks lol

[limpid horizon]

for a k-module M where k is a field, the length of M = dimension of M is not trivial right

[rare walrus]

I mean it's not super hard

it's pretty easy to see that a maximal flag produces a composition series

[limpid horizon]

What is a flag?

[rare walrus]

tbh if you haven't seen a flag, it's best to just say it's a composition series lol

but yeah if you've ever tried to produce composition series for vector spaces it's straightforward

[spice idol]

the only simple k-modules are the 1-dimensional vector spaces, so a composition series must be the same length as the dimension

[limpid horizon]

I guess i havent thought about how vector space dimension and dimension of module in terms of maximum length of chain of submodules are related

[rare walrus]

if you have a basis e_1, ..., e_n then you just take 0 <= <e_1> <= <e_1, e_2> <= ... <= <e_1, ..., e_n> = V

You can quickly see that the quotients are just 1-dim (as enpeace is saying)

and so you have a composition series

done, length = dim

[spice idol]

formally you'd have to prove that a composition series implies finite dimension

but that's not too hard

[rare walrus]

Same argument in reverse tbh

[spice idol]

ts

[rare walrus]

bases extend all that

Does this make sense kiand?

[spice idol]

fuckass property I'm so jealous

[rare walrus]

Heh... 😎 guess you could say... that reps are just... too sigma 😎

[spice idol]

chagrin

[limpid horizon]

gimme a min i havent thought about this stuff in a while

[rare walrus]

I should ban you for this

[limpid horizon]

does this not show that length of V is at least the dimension?

[rose mirage]

refer to my previous message

[rose mirage]

make a longer chain I dare you. I double dare you

you can't do it! the quotients are simple!

[rare walrus]

It's a composition series so by standard results on these things, it is of the maximal length. See, uh... what's the theorem called... Jordan-Hoelder

[rose mirage]

HOElder
catching strays

[limpid horizon]

yea ive never studied composition series before somehow lol

[rare walrus]

But like if you're really wanting to prove it like that, you can do the converse argument and prove length <= dim

so you're done really

Maybe you can try forming the converse argument

Like take a chain and produce an l.i. set

[limpid horizon]

On wiki it says length of a module is just longest chain of submodules. No quotients being simple is needed there?

[spice idol]

basically a series 0 = V0 < V1 < ... < Vn = V can be seen as an inclusion { } = B0 < B1 < ... < Bn of linearly independent subsets, where Bn is a basis of V. A composition series, then, is a an inclusion of linearly independent subsets such that every step one element is added, as this is what it means for every quotient to have dimension 1. But the length of such a chain of inclusions must exactly be the size of your basis, i.e. the dimension of V.

This may help or may make it worse

[rose mirage]

if the quotients are simple you know you have the longest chain by obvious

and by obvious I suppose I mean the correspondence theorem

[spice idol]

actual sleeper agent moment

[limpid horizon]

longest chain that starts with M0 and ends with Mn right

@rose mirage

[rare walrus]

Well you have a maximal chain but not necessarily the longest. The J-H theorem proves it's the longest

[limpid horizon]

i should learn jorden holder theorem

[rare walrus]

M0 = 0, Mn = M.

So idk why you said that, that confuses me as a question

[limpid horizon]

I meant to say what ur saying here

about wew's comment

[rose mirage]

they're vector spaces the skeleton is a chain

wtf did I just say like actually read that back as if you're a non-maths person

[rare walrus]

yea it's just subtle is my point

[limpid horizon]

Lol

i like doing that sometimes

[spice idol]

what does skeleton mean here?

[rose mirage]

ok enough escaping the simulation I gotta log back in

[rose mirage]

poset of subspaces taken up to isomorphism

[spice idol]

oh hm I see

[rose mirage]

i.e. the subcategory of Skel(FinVec)

NOT true. No quotient maps

[spice idol]

I see

is that defined for any module?

[rose mirage]

it's defined for any object in any category

[spice idol]

nvm just thought about it that should be pretty well-defined for arbitrary algebras

or categories yeah sure

boo modernity

[rose mirage]

sYbaU

[spice idol]

you can't make mee

[spice idol]

you should!

honestly it's phrased pretty annoyingly usually because it does not make use of lattice theory

[rose mirage]

that's a good point I don't actually know what you need for jordan holder to hold

meets and joins?

[spice idol]

yeah and modularity

you need a modular lattice

[rose mirage]

you might need one but it remains to be seen if I do

what fails if you don't have it? I'm not immediately seeing it

oh right no I see it now. It's when you do the intersection step right?

[spice idol]

and instead of having factor objects, you've got a correspondence between so-called projective intervals, which is quite technically defined but in group terms it translates to an isomorphism that is a composition of isomorphisms of the form MN/N = M/N\cap M

[spice idol]

I do believe so

haven't gone through the proof in a detailed way myself, gotta admit

[rose mirage]

just act like you have

[spice idol]

mathematicians hate this one trick

[rose mirage]

actually we LOVE it

[spice idol]

understand ANYTHING by simple pretending to!!

[digital parcel]

Probably need her. Bc ure using jordan holder so u need her to hold. Bc holder

[limpid horizon]

ya

[limpid horizon]

Does anybody know anything about Hilbert series / functions?

My questions are just basic like definition stuff

Hilbert series of a graded R-module M has as coefficients the length of the graded pieces Ma right?

[rose mirage]

I ardly knew er!

[limpid horizon]

I want to test latex
$M = \bigoplus_{a \in G} M_a$
Does the text show up here too?
$1+1=2$

Ok cool

Lol i never asked questions using latex before

L

L imma take this L

[broken turtle]

kiand123

[golden osprey]

#latex-testing exists

[limpid horizon]

Thanks Broski

Hello everyone of MathCord. I cordially invite you to #latex-testing to cheer me on with my latex-testing journey.

[unborn rampart]

Good luck broski!

[limpid horizon]

Thank you man 🥰

[rough bridge]

@civic aurora I'm currently formalizing a bit so my question will be understandable, hold on a minute. I may not be able to finish now but withing the day.

[robust rain]

Can anyone explain why m(M/N) = M/N?

Our version of Nakayama's Lemma is as follows: Let (A,m) be a local ring, M a finite A-module. Then M=mM implies M=0.

I also don't really see how we get to the second part of the corollary - the sentence starting with 'In particular,...'

[last talon]

M/N = (N + mM)/N = m(M/N)

[robust rain]

god im silly

[last talon]

Consider the submodule generated by the s_i, and take that for N

[robust rain]

ah I see

thank you : )

[limpid horizon]

M/N is finite over A in that case?

[last talon]

M is finite over A by hypothesis, so yes

[limpid horizon]

Oh, yes, thanks

I wish i knew the proof of nak lemma. I saw the determinants and i just peaced out ✌️

Any alternative proof?

[robust rain]

It's really not that bad haha, give it a go. You can also prove it by induction on the number of generators of M

I'm sure there are plenty of other proofs

[digital parcel]

The determinant proof is actually really slick imo lol

It’s a short and sweet proof

It’s also commonly used in some proofs about integral elements

[limpid horizon]

Yeah i just dont know what a determinant is to be completely honest

Alternating bilinear map type shi? Idek

[robust rain]

adj A * A = detA * I is all u need. it's laplace expansion

[limpid horizon]

Yeah i guess what i was concerned about with it was sure i could go through the computation and see its true but that doesnt really help with intuition

Ive always seen determinant as some computation thing i didnt learn it properly theory wise

[digital parcel]

all that matters here is that, if you regard each entry of a matrix as a variable (that is, say x_{ij} gives you the (i, j)-th entry of the matrix), then the determinant is a polynomial in these x_{ij}'s

[digital parcel]

^ and this is all that you need

[limpid horizon]

Do you think about that formula intuitively or just as a computational trick

[digital parcel]

both i guess, but i think of it as the former

it's how you give things like GL and SL and other subgroups of GL a variety structure, for example

actually in the proof of nakayama which uses determinants you do use the fact that the determinant is a polynomial in the coordinates

that's how you reach your conclusion

[rough bridge]

@civic aurora I guess after thinking for a bit I run into problems because of the way I asked my question here a couple times before. In general I tend to reach a bit with math.
I really want a reference to study this material better. I tend to have endless chains of problems that develop from previous problems until I run out of theory to accurately assess what I'm looking at. Is this like, the "hidden subgroup problem", or what exactly? I can explain my direct confusion in more detail if you prefer.
Here's an image of what I'm referring to:

It's a structure with commutative diagrams that relates to integer factorization / multiplication

[limpid horizon]

I met a new post-doc and he’s like why am i using bruns and herzog

That book is universally hated man 😂

[spice idol]

skill issue

[spice idol]

what is "this material"

without context your image looks like nonsense (I'm sure its not- but youre giving no context as to what anything means here)

[rough bridge]

I have given context, I am looking for a reference request to discuss e.g. the pictured math. Could you read the text in the image?

My handwriting is bad

[spice idol]

not really sure what I'm looking at

like I can somewhat get it but "finitely supported group"? "lift"?

"F"? "Lift = \varnothing"?

[rose mirage]

just because you can describe something with a commutative square doesn't mean that you should

I also don't really know what you want. You can express properties of algebraic operations diagrammatically through a standard process if that's what you're after

[spice idol]

also, if I'm correct in what I'm thinking, the top diagram doesn't even commute, as group multiplication may not be commutative

and the second diagram is just a mystery

to me

[rough bridge]

The diagram was made with the intention of saying that the group should commute. I agree that it doesn't need to. Here G would need to contain a copy of the integers, and be homomorphic with multiplication

[spice idol]

I agree with wew's sentiment, don't express stuff with commutative diagrams if you don't need to

unnecessary convolution is how you make mistakes lol

[rough bridge]

Sure. This doesn't answer my question. Do you have a reference for studying this kind of math (commutative diagrams, the Ring of integers, etc?) I'm beyond basic field theory, but before Galois Theory, although I do have a reference for Galois Theory I trust and could bridge that gap

[rose mirage]

category theory?

[limpid horizon]

no

[spice idol]

commutative diagrams and rings of integers are wholly different things

one is algebraic number theory which you could look into

and the other is just category theory

[limpid horizon]

ur handwriting bro 😢

[foggy galleon]

What's known about the finite dimensional division algebras (not necessarily associative) over real closed fields? In particular, is it true that any such algebra must have dimension 1,2,4 or 8?

There's this article In which dimensions does a division algebra over a given ground field exist? by Darpö and others, but I couldn't find it

[limpid horizon]

well i mean what do you want to learn? you have choices here

[rough bridge]

I want to start with a reference discussing Group Factorization methods (factorizations of groups into subgroups), in a structured way. Do you know of a textbook?

[foggy galleon]

oh okay it is indeed true. This is spelled out in Dietrich "A general approach to finite dimensional division algebras". Given the theorem over R and the first order theory of those fields the rest is elementary

[rough bridge]

wait just so I'm following is this a parallel convo or why did you answer your own question in 5 minutes

[foggy galleon]

I had a question and found the answer, so I reported it

I wonder, is it possible to prove by purely algebraic means that the dimension must be a power of 2 (say, just for the case of R)?

[rough bridge]

This is a theorem, you want an algebra proof?

It does exist and there's a proof involving matrices I think

[civic aurora]

are these supposed to be commutative diagrams in groups?

I dont think there are any group homomorphisms Z -> Z×Z that look like integer factorisation

[rough bridge]

Hureitz' Theorem you mean, it's provable with matrices I think (had to google the name)

[foggy galleon]

mmh probably not, otherwise we would be proving that there are no nontrivial odd dimensional field extensions of the reals

wait idk what you mean by Hurwitz

he was the first to prove what I asked for, but he also has another theorem were he introduces extra normed structure

[rough bridge]

I don't profess to understand "Hurwitz'" theorem

https://en.m.wikipedia.org/wiki/Cayley–Dickson_construction

but it's there

[civic aurora]

there are only two group homs that look like they factorise 1, one sends 1 to (1,1) and the other sends 1 to (-1,-1). and for both of these, the only number they factorise is 1, and every other n gets sent either to (n,n) or (-n,-n) neither of which factorise n

[rough bridge]

So you don't think it could form a homomorphism, and hence the diagram is flawed?

For example, maps that factor integers exist, 15 -> (3, 5), but it may not be consistent in a meaningful way

and primes / units would need to be like, (p, 1), (u, 1)

[foggy galleon]

although I'm guessing it should follow from C being algebraically closed

[ornate atlas]

I was going to say that’s rude before I saw the actual writing, fuck me

[foggy galleon]

or R admiting only power of 2 field extensions. So if k is a field that only admits field extensions of degree p^k for some prime p does this imply that every finite dimensional division algebra over k must have degree some power of p?

[molten vale]

F_2, F_3, F_3\times \prod_{i\in I} F_2 are examples of commutative ring where everything except 1,-1 are zero divisors. What other examples are there and is there a classification?

[spice idol]

"not necessarily associative"

matrices are, famously, associative in their multiplication

also "can be solved by matrices" is like saying "we can solve this equation by doing algebra" lol

[spice idol]

really, model theory comes into play?

that's awesome

[foggy galleon]

I think the statement of "there exists a ... algebra of degree n over k" can be turned into a system of polynomial equations. If it has a solution, then you are good. This is first order

Tarski says that the first order theory of real closed fields is equivalent to the first order theory of R. Idk how the proof works tho

[spice idol]

ah right

you've got a wacky ass theorem that says that says a field is elementary equivalent to R iff some conditions hold

[foggy galleon]

isn't it what we are saying already (the Tarski thing)

[spice idol]

yes yes

the theorem is wacky

[foggy galleon]

why

it seems reasonable. But I should look closer into it

[spice idol]

yeah I guess they are somewhat reasonable but it's still cool that the algebraic closure being a finite extension is such a strong property

[foggy galleon]

oh yeah

but is that first order?

[spice idol]

yes

hmm,

it is equivalent to the extension being of order 2

[foggy galleon]

but this isn't obvious

[spice idol]

no, that's why the theorem is weird

[foggy galleon]

I mean that the statement "[ bar K : K] is finite" is not first order

[spice idol]

sure

but that statement is equivalent to K and R being elementary equivalent

[foggy galleon]

First you use Artin-Schreier

then Tarski

Tarski is more recent I think

Artin-Schreier is more surprising to me than Tarski

the Artin-Schreier paper is from 1927, Tarski's paper is from 1931

[rough bridge]

https://kconrad.math.uconn.edu/blurbs/linmultialg/hurwitzrepnthy.pdf

Again, I never said I understood Hurwitz' Theorem. I said I knew there was a proof somewhere involving matrices.

[molten vale]

wait actually are those the only ones?

[limpid horizon]

For a graded R-module M over a graded ring R, we have
$M = \bigoplus_{a \in G} M_a$ decomposition as Z-modules.

The Hilbert function is defined as
$H(M,a) = l(M_a)$

[broken turtle]

kiand123

[limpid horizon]

M_a is not necessarily an R-module, right? So is this the length of M_a as a Z-module (or abelian group?)

ok yall get the point

[limpid horizon]

My question is here

[spice idol]

no, else R_i M_j wouldnt be included in M_ij

(because of the direct sum)

for example R[X] is a graded module over itself

but the components are mere R-modules

from the definition, every component is an R_1 module though - R_1 M_i \subset M_1i = M_i

[olive schooner]

it's the length of M_a as an R_0-module where R_0 is the zeroth component in the grading of R

[limpid horizon]

0 in group G yea

[limpid horizon]

What is this?

Not sure what u wrote

[spice idol]

1 is the identity of G here

so by the definition of a graded module we have R_g M_h \subset M_gh, which means that R_1 M_g \subset M_g, i.e. each M_g is an R_1-module

[limpid horizon]

Ok cool

So we are indeed looking at the length of these guys as R1-modules

I dont know where in the definition ive seen for this it specifies that tho. Ill look again ig

[digital parcel]

Usually for hilbert functions/series/polynomials one assumes R is a standard graded ring and that M is a standard graded R-module

If you're willing to use a standard grading then Atiyah-Macdonald has some stuff on this

(They don't call their grading the "standard grading" but that's what it is)

[limpid horizon]

What is a “standard” grading

[digital parcel]

G = non-negative integers

[limpid horizon]

My purposes uses Z^n-grading on ring k[x1, … xn]

[limpid horizon]

I didnt know a/m had stuff tho thx

[digital parcel]

Roberts' "Multiplicities and Chern Classes in Local Algebra" deals with this stuff in Z^n

If you want to look there as welll

Everything for the Z^n case ends up being a generalization of the Z case

Well I should be precise and say Z_{\ge 0}^n and Z_{\ge 0} since that's what Roberts and A/M and p much everyone else does

[limpid horizon]

\ge 0?

[digital parcel]

non-negative grading

so like M_i = 0 for i < 0

this is from Roberts

[limpid horizon]

Ok thanks

Why does that read nicely

[digital parcel]

Wdym? Like why do we want the Z_{\ge 0} grading?

[limpid horizon]

Nah im just saying the exposition is nice

[digital parcel]

Oh yea

[limpid horizon]

Why cant all advanced math texts be so nice like that

Some authors just love being like yo this is serious business im so serious with this shit

Ig personal taste tho

[digital parcel]

I wouldn't be so quick to gas up Roberts lol it's pretty encyclopedic

It has some nice exposition/historical remarks but they are very short from what i've read

[limpid horizon]

Hahahaha

[digital parcel]

You could also look at Serre's "Local Algebra"

Though there's not much meat to it in the relevant section B.3

[void plank]

Actually, I think there's a gap in my reasoning here that I wanted to double check with yall.

As a reminder, let C, C', D, D' be complexes of F_2 vector spaces. Endow C, C with a left G action and D and D' with a right G action, where |G| is odd.

My overall proof strategy is to show that two quasi-iso tensor product complexes remain quasi-iso after taking coinvariants.

I started off assuming C ~ C' and D ~ D' as F_2 vector spaces and then concluding C (x)_F2 D ~ C' (x)_F2 D'

But to then apply coin, I need to interpret them as complexes of F2[G] mods since coin: F2[G]-Mod --> F2-Mod. Am I doing this by interpreting e.g. C (x)_F2 D as a complex of left or right F2[G] modules where

g (v (x) w) = v (x) gw

or

(v (x) w)g = vg (x) w?

If so, what's the guarantee that C (x)_F2 D ~ C' (x)_F2 D' as complexes of F2[G] modules?

[lone jacinth]

The usual way to give a group action on the tensor product would be
g(v (x) w) = vg^-1 (x) gw

Then the tensorproduct preserving the quasi-isomrphism should follow from it being exact. But thats assuming C and C' (and D and D') are quasi-isomorphic as F2G-modules

Otherwise if you just have C=C' but with a different action, of course there will be no relationship

[void plank]

Ok, so the right proof would be to assume C ~ C', D ~ D' are quasi-iso as F2[G] modules (and my condition that |G| odd implies by Mashcke's theorem that C and C' are complexes of projective F2[G] modules, hence flat ones). This then implies they're quasi-iso as F2-modules since restriction of scalars is an exact functor

Then C (x)_F2 D ~ C' (x)_F2 D' as F2[G]-modules with the group action you described

Then I can take coinvariants of both and conclude they're still quasi-iso.

Is all this correct?

[lone jacinth]

I guess its not really about them being projective/flat, since the tensor product over F2 is always exact, dont need G odd for thar (you need it for taking coinvariants being exact)

But you could just tensor over F2G to begin with I guess

since tensoring over F2 then taking coinvariants is the same as tensoring over F2G

[void plank]

I guess the only thing I need to convince myself of is that

C ~ C' and D ~ D' as F2[G]-modules

implies

C (x)_F2 D ~ C' (x)_F2 D' as F2[G]-modules (with the module structure you described in your previous message). Is this obvious?

[lone jacinth]

Obvious is in the eye of the beholder I guess, but you might notice that if D is concentrated in degree 0, then C(x)D ~C'(x)D because -(x)D is exact. Then by 5-lemma, the category of D for which C(x)D ~ C'(x)D is a thick subcategory, so every bounded complex.

then same argument gives C'(x)D ~ C'(x)D'

Im assuming this is the same argument you used to see it is true as F2-modules....

I guess if you want an actually obvious solution. Since its semisimple, quasi-isomorphic is just the same as homotopy equivalent.

and tensor product is additive

[void plank]

You mean C ~ C' and D ~ D' as F2[G] modules implies C (x)_F2 D ~ C' (x)_F2 D' as complexes of F2-modules? I was just applying restriction of scalars to C ~ C' and D ~ D' before taking tensor products over F_2 for that argument

But it wasn't a priori clear to me that after giving the terms in C (x)_F2 D and C' (x)_F2 D' the structure of F2[G] modules, I'd get C (x)_F2 D ~ C (x)_F2 D as complexes of F2[G] modules

[lone jacinth]

I mean the argument that C(x)D ~ C'(x)D' as F2-complexes yeah. Like restriction of scalars turns them into F2-complexes, but thats not the argument for why theyre quasi-isomorphic

unless you mean that the main part of the argument is "its obvious" I guess

In which case I guess "its obvious" is an approriate argument for F2G-complexes as well

[void plank]

Ah sorry, I'm assuming/applying the result that if

C ~ C' and D ~ D' are quasi-iso and C and C' are complexes of flat modules, then C (x) D ~ C' (x) D'

[lone jacinth]

Isnt that like the defintion of flat?

or whats the definition of flat?

Either way I guess it would just come down to proing C and C' are flat then

anyway, a homomorphism of F2G-modules that is an isomorphism of F2-modules is also an isomorphism of F2G-modules.

So if you have a morphism of F2G-complexes that is a quasiisomorphism of F2-complexes it is also a quasiisomorphism of F2G-complexes

[void plank]

Yeah, that's why I assumed C and C' were complexes of F_2[G] modules where |G| is odd, so that C and C' would consist of flat F_2[G] modules by Mashcke's theorem

[void plank]

Ahh ok, this is the missing thing I was looking for

[lone jacinth]

yeah so if you want to skip the whole coinvariants step and just tensor over F2G, this would work.

But by proving it this other way I guess you get something a little more general since you dont need G being odd untill the very end

[void plank]

Yeah, this was mainly just an exercise for fun I was thinking about to show how the two coincide

I guess now the last thing I'm missing is why we're guaranteed that the quasi-iso C (x)_F2 D ~ C' (x)_F2 D' (as F2-modules) is a morphism of F2[G] modules

Say f: C --> C' is quasi-iso (for complexes of F2[G] modules) and similarly for g: D --> D'.

Then the map h: C (x)_F2 D ---> C' (x)_F2 D' is certainly an F2 quasi-iso. But I suppose the fact that h is built out of f and g (something like h = f (x) g or direct sums of them?) means it is also a morphism of complexes of F2[G]-modules with the action g(v (x) w) = (v g^-1 (x) gw)? Should be easy to check assuming h = f (x) g is correct

[lone jacinth]

Yes
a(x)b [g (c (x) d) ] = a(x)b [ cg^-1 (x) gd ] =
a(cg^-1) (x) b(gd) = a(c) g^-1 (x) gb(d) = g [ a(c) (x) b(d) ]

[lone jacinth]

Z/4 is another

[void plank]

Thanks once again!

[molten vale]

oh yeah misse that, but is there anything else?

[lone jacinth]

Well you can do Z/4 times F2 stuff aswell.

Feels like there might be more to me, but that would need to be somewhat complicated rings if they exist

[molten vale]

yeah make sense

[near lantern]

Is Grothendieck's simultaneous resolution a pullback square?

I am referring to the following: let G be a semisimple algebraic group over ℂ with Lie algebra g, h the (universal/abstract) Cartan algebra, and W the Weyl group. Let F the flag variety of Borel subalgebras of g and g' the "incidence" variety g' = {(x, b) ∈ g ⨯ F : x ∈ b}. Then the resolution is the commutative square g' -> h -> h/W, g' -> g -> h/W (the last map is given by Chevalley's Theorem ℂ[g]^G = ℂ[h]^W, and g' -> h is analogous to "taking eigenvalues").

[low obsidian]

so I have to show "If V is an n dimensional irreducible representation of algebra A, then End(V) is isomorphic to the direct sum of n copies of V"

So I thought about it in terms of A modules, and took the following map:

but I never needed irreducibility of V to show this was an isomorphism

so where is irreducibility of V needed? Or is my proof wrong

[near lantern]

Isomorphic as an A-module? What's the A-module structure on End(V)?

[low obsidian]

V is an A module, so define the obvious structure on End(V):

(f+g)(v)=f(v)+g(v)

(af)(v)=af(v)

[near lantern]

Right. Then indeed you don't use that V is irreducible. In fact, for any A-module V and vector space W, Hom(W, V) is an A-module isomorphic to the direct product of dim W copies of V, by exactly your argument. It's just a coincidence here that W = V as vector spaces.

[low obsidian]

I see

Thanks

[near lantern]

Actually, this is obviously not a pullback square: g, h, h/W are affine but g' is projective over g. Rather the question should be whether this holds restricted to the regular semisimple locus, i.e., whether g'^sr -> g^sr is the pullback of h^reg -> h^reg/W as principal W-bundles.

[low obsidian]

https://math.stackexchange.com/questions/2317020/classification-of-modules-over-complex-matrices

In this SE post, he establishes M, as an M_n(R) module, is isomorphic to e11M+...+e_nnM. I get that, but then he says eiiM is isomorphic to an R module for all i, in fact they are all isomorphic to the same R module. Ok fine, but then later he uses this to claim that any M_n(C) module is isomorphuc to (C^m)^n, but I dont understand this. He knows that each eiiM is isomorphic to C^m as a C module, so how did he deduce that it's isomorphic to C^m as an M_n(C) module?

[lone jacinth]

eii M is not an Mn(R)-module

So it can't be isomorphic to anything "as an Mn(R)-module"

The point is you think of an Mn(R)-module M as an R-module (you can think of R acting as multiples of the identity). Then M splits as a direct sum of eiiM, which are all isomorphic R-modules. And multiplication by eij gives isomorphisms between them.

Hence M can be described as (e11M)^n acted on by matrix multiplication.

[kindred mauve]

anyone has some resources that cover solely tensor products? Possibly that make use of categorical language.

[pastel shoal]

Have you looked at Keith Conrads notes?

[kindred mauve]

Thanks (for other courses but not for tensors)

[pastel shoal]

I don't think it uses a lot of categorical language, but obviously covers the universal property

[kindred mauve]

Ok that is probably enough

[low obsidian]

hmm ok, so if he says as R modules each Mn(R) module is isomorphic to L^n, where L is some R module, how does he use this to conlcude that the only simple Mn(C) module is C^n. I get we can replace L by C^m in the case of R a field, but don;t see how his argument then says anything about simplicity of Mn(R) modules

[lone jacinth]

So the Mn(C) modules are exactly V^n where V is a C-module.

The only simple C-module is V=C

So then C^n is the simple Mn(C) module

[low obsidian]

Wait so the only simple C module is C. Each Mn(C) module as as a C module is isomorphic to (C^m)^n for some m. In order for the C module structure of the Mn(C) module structure to be simple, we need m=1?

Oh, or is it: as an R module, Any Mn(R) can be written as the direct sum of n isomorphic R modules, call this sum R^n. But each R^n can be made into an Mn(R) in the obvious way, and doing this will give us an isomorphic Mn(R) module to what we started with?

[lone jacinth]

So if I understand your question the answer is yes.

The Mn(R)-modules are exactly given by M^n for an R-module M where the action is matrix multiplication.

So for example for n=2 the matrix
[r1, r2; r3, r4]
acts on the element
[m1; m2] as
[r1m1 + r2m2; r3m1 + r4m2]

This are exactly what the Mn(R)-modules are and for R modules M and N, then N^n is a submodule of M^n as Mn(R) modules exactly if N is a submodule of M.

So similarly M^n is simple iff M is simple

Using scary words the two categories of modules are equivalent, making R and Mn(R) Morita equivalent.

[low obsidian]

interesting

[fierce steeple]

Say M is a free module over a commutative ring R. Is there a name for elements which can be written as linear combinations of basis elements where all non-zero coefficients are in invertible?

Maybe this is too niche lol

[distant harness]

Are you sure that doesn't depend on which basis you choose?

[fierce steeple]

It does depend on the choice of basis, e.g. (2,1) and (3,2) form a basis of Z^2. This is a good point, but in my case I'm fine fixing a basis and indeed working with the standard basis of R^n

I should probably just say what I want though aha rather than expect there is a common notion

[distant harness]

It sounds like "oh, those are surely well-behaved elements" for the first ten seconds.

[lone jacinth]

If it depends on the choice of basis I would assume there is not a name for it

[fierce steeple]

Yeah this is a good point lol

[spice idol]

what could possibly lead to this question 😭

[fierce steeple]

Basically I have a chain complex of free modules and I wanted conditions under which the homology was free. I convinced myself that certain conditions on the matrices could imply this, but then it ended up being too complicated and not working

Fortunately the case at hand is simple enough that i think i can get around it lol

[spice idol]

oh that's good

good luck o7

[fierce steeple]

thank

[fierce steeple]

Lol turns out this stuff being shot down helped me find a better path ( I think)

[spice idol]

many such cases

if it doesnt work nicely often its not the correct path

[fierce steeple]

Maybe ye only thing is hm

Well it's like in problems made for exams or whatever this is usually the case but I feel in research stuff can often just be ugly

Or at least is until someone finds the theory later on or something

[spice idol]

yes yes

but sometimes it just works out and when it does its beautiful

[lone jacinth]

The MO trauma washes over me

[fierce steeple]

Oh lol why

This does make me think of Qiaochu saying like everything should be clear from the right theory or something / there should be no tricks

[lone jacinth]

I post to MO
People say meh
I figure out this approach is useless for my problem

[fierce steeple]

Sorry oop

I love spectral sequences.

I ran one today and the universe conspired to make it manageable which is a miracle

[lone jacinth]

I'm joking, but also... What I'm saying is true

[fierce steeple]

Basically like ended up being concentrated in two columns

[mental escarp]

Does it make sense to define the rank of any multilinear form?

For bilinear forms there is an associated matrix I think so you could take that rank

[honest maple]

I'm not sure if that make sense (defining rank for multilinear form), but maybe what you need is called tensor

[soft parcel]

Could someone give a hint as to how to show that any p-primary factor must contain the kth symbolic power of p?

I believe this is equivalent to: if q is a p-primary factor in a minimal primary decomposition and if x is in q, then there exists some t in R - p such that xt is in p^k

[round seal]

why is there a notion of an acyclic complex. is it not the same as an exact sequence. i feel like i have it be missing something

[limpid horizon]

Just terminology, no?

[fierce steeple]

They are equivalent but I would say they are different points of view. I think of a complex being acyclic like a topological space being acyclic (and indeed they are very related notions(

[round seal]

thanks. seems like this is another instance that would be solved if i knew some algebraic topology

[limpid horizon]

What are you trying to learn rn?

[round seal]

(i still have no idea why they are called boundaries and cycles)

[limpid horizon]

Yea look up simplicial homology

[round seal]

i’m learning homological algebra

[limpid horizon]

Its not too hard to understand

[limpid horizon]

For what?

[round seal]

but i havnt see algebraic topology so i have no intuition behind some of the stuff we’re doing

[limpid horizon]

Oh ur in a class?

[round seal]

yes

[limpid horizon]

What have u been doing

I learned a little bit over the summer

I dont know much alg top either

[round seal]

we just constructed tensor products

[limpid horizon]

Oh ok yea that part is more just everyday algebra

[round seal]

i had already seen that though from my masters research. we went through ext the week before last and it was cool

[limpid horizon]

Yeah i like ext i learned about that stuff over summer

The process for constructing the derived functors was nice

[round seal]

it’s really cool

[limpid horizon]

I agree

[round seal]

we’re gonna do tor this week i’m sure since we have tensor

[limpid horizon]

Have u not seen tensor products before? If i was introduced to all of that all at once id die

[round seal]

i wish i wasn’t taking other classes right now so i could dedicate all my time to learning this stuff very well bc it’s so cool

i have bc i’ve seen commutative algebra and they also cane up in my masters thesis

[limpid horizon]

What was your masters thesis about?

[round seal]

but they were confusing at first

[limpid horizon]

I was uncomfortable with tensor products for ages

[round seal]

it was in the area of noncommutative invariant theory

[limpid horizon]

Nice idek what that is

[spice idol]

the idea is that you take some space (think any subset of the n-dimensional euclidian space) consider C_n the free abelian group generated by all simplices in your space. That is, all ways you can embed an n-simplex into your space (an n-simplex being the n-dimensional version of point, line, triangle, tetrahedron, etc). Then you have a boundary mapping d : C_n -> C_n-1, that takes a simplex to its "boundary". Here is how you'd do that for the triangle spanned by the vectors (a, b, c):

d([a, b, c]) = [a, b] + [b, c] + [a, c]

Then, as one would intuit, the boundary of a boundary is 0 (like, a sphere is a boundary, so it does not have a boundary itself). This means you get a chain complex. The cycles of this complex can be seen as actual "cycles" of n-dimensional simplices that in a sense form a "closed loop". It is justified in the n=2 case, where 2-cycles are exactly those (up to boundaries) given by a linear combination of actual loops in the space. Then the boundaries are, as the name would suggest, the linear combinations of boundaries of simplices.

whoops 😭

[round seal]

looking for as regular algebras that a certain family of hopf algebras act on

[spice idol]

ignore this for now

[round seal]

why

[limpid horizon]

Youve lost me now idk this stuff

[round seal]

i didn’t either i had to learn a lot

but it made me decide i will like commutative algebra better lol

[limpid horizon]

Yea im doing combinatorial algebra

I knew i didnt like it from the start

lol

But im pushing througg

[round seal]

i wish i studied as much as you

i feel like i’m not doing enough bc i can’t find time to study outside of completing my weekly psets

[limpid horizon]

Im just taking one class and trying to finish my masters thesis

And i feel like im barely working enough too

Im just tired and its all really hard tbh

[round seal]

it will feel worth it once you defend though

[limpid horizon]

Yeah probably, once i get there

[soft parcel]

Could someone help me understand what is being said here? Are they claiming that the map R -> R[e^-1] x R[(1-e)^-1] by r -> (r,r) is an isomorphism? If so, I can see that it is surjective, but I don't see why it necessarily is injective. If r maps to (0,0), then e^k r = (1-e)^l r = 0, so r is in I_1 and I_2, hence in the nilradical. Can one conclude that r=0 from there?

Or does anyone know a reference for this lemma?

[lone jacinth]

Say k>=l

Consider
r = 1^(2k) * r = (e + (1-e))^2k * r

Expand with binomial theorem and you get r=0

[soft parcel]

So clean

I like it

Thank you

[lament scarab]

Is there a version of Mackey’s irreducibility criterion that applies to locally compact groups? The versions I seem to find online all pertain to finite groups https://dec41.user.srcf.net/h/II_L/representation_theory/12

For the particular case I’m interested in G isn’t compact, the induced representation is infinite dimensional, and I’m inducing from a character

[last talon]

Does this do a good job of what you want? https://www.jstor.org/stable/1969423

In particular theorems 12.1, 12.2, 13.1

I don't know enough about the theory of LC groups to be able to tell if this converts into what you want in the same way Mackey for finite groups does, but it seems in the right direction

[last talon]

(I'll also leave part II here https://www.jstor.org/stable/1969786)

[lament scarab]

awesome thanks! I'll check those papers out

[low obsidian]

sorry to bother u again, Im trying to prove that again and am totally stuck

this is what I have so far:

and then I get stuck at the very last line

if we can eliminate the e_{1i} we are done

but I dont see a way to do it

[lone jacinth]

So have are you defining phi_i exactly?

Because now it seems like you're saying they're just arbitrary / unrelated...

[low obsidian]

so we know the e_{ii}V are all isomorphic

so pick one of them as W, say e_{11}V, and let phi_i be any isomorphism from e_{ii}V into e_{11}V

or is that too general to proceed

[lone jacinth]

Well if it's just any random isomorphism it probably won't match with the Mn(F)-module structure

There is however a very natural choice of isomorphism

(how do you know they're isomorphic in the first place)

[low obsidian]

Is e_{ii}V to e_{jj}V: e_{ii}v mapsto e_{jj}v an isomorphism?

probably not actually

[lone jacinth]

Well, what happenes if v=eii?

[low obsidian]

ok so I think

e_{ij} and e_{ji} will invert each other

multiplying by them

e_{ii}V to e_{jj}V: e_{ii}v mapsto e_{ji}e_{ii}v seems to work

no

lemme think about the indices

yea ok

it works, e_{jj}e_{ji}=e_{ji}

ok, youre saying replace the $\varphi$ with the map into $e_{11}V$?

[broken turtle]

kerchooboi

[lone jacinth]

Something that simplifies verification here also is that all the maps are additive, so you really only need to check for A = a eij

Just a little easier not messing up all the sums and indicies

[low obsidian]

hmm, but at the last line here

for example the first coordinate

if I replace phi 1 with multiplication by e11

I still have the same issue of there being e1i

[lone jacinth]

So what is the issue?

e1j ejj eji = e1i, so exactly what you want yeah?

[low obsidian]

Yea I might be tweking, let me try to formally write this up

thanks

[low obsidian]

ok that was insane for indexing, shouldve done this

but I got it

thanks a lot

So defines a functor from Mn(R) modules into R modules right

[elfin ice]

I am learning about Lie groups and Lie algebras using Representation theory. A first course of Fulton & Harris.
I have already proven that the Lie algebra of Aut(g) is Der(g) in its previous problem 8.27;
However I have difficulty in proving Aut(G)->Aut(g), φ=>dφ is one-to-one and onto...
For (b) I know that using the hint in the end of this book, which uses a universal cover.

[weak lodge]

use the theorem that if G is a simply connected Lie group, then every Lie algebra endomorphism f of g lifts to a unique Lie group endomorphism F of G, namely f = dF
(side note: this is actually a special case of the fact that if F: C -> D is an equivalence of categories, then for every object X in C, F induces a group isomorphism Aut(X) -> Aut(F(X)) )

[elfin ice]

The problem is, this is like the "ultimate goal" of this chapter which isn't proved before, hence one technically shouldn't use this result

[soft parcel]

Does anyone know of a way to finish this off?

If the last isomorphism were k-linear, then I think one gets a contradiction as the sides have different dimensions a k-vector spaces, but I'm not sure about the details here... the last isomorphism is also only assumed to be a ring isomorphism
(p.s. k is the number of primes assumed to be lying above p)

[lone jacinth]

I mean you can argue by dimension or just, since A is artinian the sum is finite.

[soft parcel]

But I need to argue that the dimensions don't match up, right?

To get a contradiction

[lone jacinth]

Why do you want a contradiction?

[soft parcel]

I want to show that there cannot be k > n primes lying above p. So in the above, I want to get the contradiction k<=d < k

[lone jacinth]

I mean

A has length <= d so there are at most d factors.

If we assume there are more than d factors that contradicts that there are at most d

[soft parcel]

What is length?

[lone jacinth]

The number of simple modules in its composition series

It's essentially the same as dimension, so you can just use that if you're not familiar with it

An n-dimensional vector space cannot be the sum of more than n non-zero vector spaces

[soft parcel]

But we do not know that the last isomorphism is an isomorphism as vector spaces, a priori

I am looking at the proof of 13.25 and I believe it actually shows that it should be kappa-linear, but I'm curious whether your argument works assuming only ring isomorphism

[lone jacinth]

I mean, the k-vector space structure is just induced by the ring homomorphism

[soft parcel]

Yeah

[lone jacinth]

So then ring homomorphism implies linear, since that's just how the vector space structure is defined

[soft parcel]

Yeah, I guess to me it wasn't so clear that the k-vector space structure was actually induced by the ring homomorphism

But in the proof of that lemma, the map is given by R -> R_1 x R_2 by r -> (r,r), so I think you are right

Anyway, thanks for helping clear that up

[near lantern]

I'm not sure but I think you can go in the other direction using the exponential map?

[elfin ice]

I used the exponential map to prove that φ=>dφ is one-to-one, but I can't prove that it is onto

[void plank]

Let A be a k-algebra. Does the functor sending A to its underlying k-vector space have any adjoints?

[lone jacinth]

Yes, the left adjoint is the tensor algebra

[spice idol]

yeah, any functor sending an algebraic structure to some reduct has an adjoint

(reduct meaning you remove some operations)

[lone jacinth]

It does not preserve coproducts so does not have a right adjoint

[void plank]

Thanks jagr!

[golden osprey]

What are some sufficient conditions for a polynomial to have n distinct roots?

I know GCD(f, f') = 1 but this is annoying when you don't have a specific polynomial (rather a family of polynomials with some parameter)

Univariate polynomials ofc, can even say we're over a field if you have some field specific conditions

[distant harness]

Hopefully the field is algebraically closed. Otherwise even the gcd(f,f')=1 condition is not sufficient.

[lone jacinth]

Irreducible over a seperable field would work

[golden osprey]

Oh sure, good point lol

[lone jacinth]

Having nonzero discriminant

That might be computational tractable with some parameters even

Or I guess it's about the same as computing gcd(f, f')...

But that seems tractable to me anyhow

[near lantern]

For g a finite-dimensional (semisimple) Lie algebra over an algebraically closed field of characteristic 0, the regular (semisimple regular) locus is a Zariski open and dense subset. For the regular locus this can be made very explicit: let char_{ad(x)}(t) = t^n + a_{n-1}(x) t^{n-1} + ... + a_r(x) t^r for x in g, where n = dim(g) and a_{n-1}, ..., a_r are polynomial functions on g; then x is regular iff a_r(x) ≠ 0, so the regular locus is the complement of the zero set of a_r. Is there a similar explicit description of the semisimple regular locus?

[distant harness]

Well, the discriminant being a (degree-dependent) polynomial in the coefficients makes it feel more attractive as a symbolic condition than calculating a gcd.

[lone jacinth]

What is the semisimple regular locus?

[near lantern]

{x in g : x is regular and semisimple}. I defined regular above; semisimple means that ad(x) is diagonalisable.

[lone jacinth]

I see.

Well, the polynomial being seperable is certainly sufficient. Maybe if it's rank 1 that would give a characterization in some cases.

Idk

[near lantern]

Unfortunately I think that is not necessary (I assume you are talking about char_{ad(x)}(t)/t^r to remove the t^r that is always there). For sl_n, one can show that semisimple regular <=> distinct eigenvalues, i.e., that char_x(t) is separable (so in this case there is an explicit description: the complement of the zero sets of the discriminant of char_x(t) p_x(t) := char_x(t)/t^r and its constant coefficient). But this does not imply that ad(x) has distinct (non-zero) eigenvalues: one can consider x = diag(0, 1, -1, 0) ... and this does not have distinct eigenvalues so it's not semisimple regular 💀. I need to think more then.

[lone jacinth]

I mean you would just tack the condition of being regular onto it, no?

[near lantern]

I'm not sure what you mean: being semisimple is not a (Zariski) open or closed condition.

[lone jacinth]

I guess I'm not following what you're saying in the "but this part".

Unless you meant to say that x in that example is semisimple regular. I.e. that there are examples of semisimple regular elements with repeated eigenvalues.

[near lantern]

I started saying something then realised it was wrong, but thought the rest of the message was still useful information, so decided to send it. 😅

I'm not sure now whether "char_{ad(x)}(t)/t^r is separable" is necessary or not (turns out it is sufficient).

[lone jacinth]

Yeah, I wouldn't really expect it to be necessary in general, but maybe in some cases.

And are you assuming the algebra is semisimple to get the sufficient part?

[near lantern]

Yes. I don't know if it's sufficient in general. (Anyway semisimple element is not so useful if the algebra is not semisimple because different definitions become inequivalent etc. And I already know that {regular} is open for any fd Lie algebra.)

[near lantern]

OK so for sl_n the eigenvalues of ad(x) (at least for x diagonalisable) are all pairwise differences of the eigenvalues of x. In particular x = diag(1, 0, -1, -2, ...) has distinct eigenvalues hence is semisimple regular, but ad(x) has 1 as an eigenvalue multiple times (for 1, 0; 0, -1; etc.).

So it's not necessary for sl_n for n ≥ 3. It is necessary for sl_2 (rank 1, as you guessed).

[near lantern]

It seems I got confused by two different definitions of regular.

Let r = min { dim ker(ad x) : x in g} and say that x is regular if it's an argmin, i.e., dim ker(ad x) = r.
Similarly define R as the minimum generalised (ad x) 0-eigenspace dimension (i.e. minimum dim ker(ad x)^∞ for x in g) and x to be strongly regular if it achieves it.

Now, for any Lie algebra, for any 0 ≤ k ≤ dim g, the sets
A_k = {x : dim ker(ad x) ≥ k},
B_k = {x : dim ker(ad x)^∞ ≥ k}
are Zariski closed subsets of g.

To see this for the latter, use the characteristic polynomial: if char_{ad x}(t) = t^n + a_1(x) t^{n-1} + ... + a_n(x), then a_i is a (degree-i homogeneous) polynomial of x, and B_k = V(a_n, ..., a_{n-k+1}).

To see it for the former, use that the rank is Zariski lower semicontinuous on the vector space gl(g), i.e., {T in gl(g) : rank(T) ≤ k} is Zariski closed as the zero locus of all (k+1)⨯(k+1) minors. Thus A_k = V(all (n-k+1)⨯(n-k+1) minors of ad x, as polynomials of x).

(Incidentally, since a_i(x) = tr(∧^i (ad x)) = ∑ {all i⨯i principal minors of ad x}, A_k ⊆ B_k, as it should be.)

Thus we have g = A_0 ⊇ A_1 ⊇ ... and g = B_0 ⊇ B_1 ⊇ ... The numbers r resp. R are the last indices i where the set A_i resp. B_i is equal to g, and the regular resp. strongly regular locus is A_r \ A_{r+1} = g \ A_{r+1} resp. B_R \ B_{R+1} = g \ B_{R+1} and hence Zariski open (and non-empty hence dense); in fact
{regular} = ∪_{each (n-r+1)⨯(n-r+1) minor} g \ V(that minor)
and
{strongly regular} = g \ V(a_R).

So far, this just amounts to the classical facts that geometric and algebraic multiplicity are Zariski upper semicontinuous (which just means that {mult ≤ k} is an open subset of the space of matrices for any k).

However, if ad x is semisimple then obviously geometric and algebraic multiplicity coincide. So for all k, A_k^ss = B_k^ss, where ^ss means to intersect with the set of ad-semisimple elements (which is not really a variety - it, is in the usual cases, neither open nor closed (nor locally closed) in the Zariski topology, although it contains an open dense subset - I won't assume these properties though).

In particular, if the ad-semisimple elements are Zariski dense, then they meet every non-empty open set, so {regular}^ss = A_r^ss \ A_{r+1}^ss is non-empty. But if A_r^ss \ A_{r+1}^ss = B_r^ss \ B_{r+1}^ss is non-empty, then B_{r+1} is a proper subset of g so r = R.

So (if ad-semisimple elements are Zariski dense in g) we can define a single number r as the minimum value of either multiplicity. However, note that this does not imply that the regular and strongly regular elements are the same - every strongly regular element is regular but the converse is only guaranteed for ad-semisimple elements.

Now, for g a semisimple Lie algebra, any strongly regular element is semisimple (over alg closed char 0). This is true because it can be shown that the automorphism group of g acts transitively, not on the strongly regular elements themselves, but on the set of generalised 0-eigenspaces of strongly regular elements (Cartan subalgebras), and it can be shown that one such subspace consists entirely of semisimple elements - so they all do.

So semisimple regular = strongly regular, which is Zariski open. If regular is used to mean "strongly regular", then actually the adjective semisimple is redundant.

[worldly zealot]

@lone jacinth what sorts of things are people interested in wrt the complement of the g-vector fan (of a tame algebra)? vague question so i can give more context if needed, but i really mean it in as much generality as possible

[lone jacinth]

I'm not sure. Whether it is finite / discrete would be a natural question I guess.

Wether it contains only rational vectors / whether they correspond to 1-parameter families.

I guess these are more things I'm curious about off the top of my head as opposed to "things people are interested in"

Why do you ask?

Mousavand and Paquette have a recent paper going over a lot of natural questions.

https://arxiv.org/abs/2508.11789

Framed more in terms of bricks than g-vectors though maybe

[worldly zealot]

was just reading a paper which describes the complement in terms of flows through the fringed quiver and certain nonkissing strings, so was interested what implications that may have but that ig depends on the way and extent to which people want to understand the complement

[lone jacinth]

hm, whats the paper?

[worldly zealot]

https://arxiv.org/abs/2507.12688

[low obsidian]

so Im stuck on pt 2 rn

I have this rn, if I could find the inner product of the characters in that sum I would be done by S churs orthogonality relations

but I dont see how to simplify to that point

[spice idol]

prove that each ψ_i is an endomorphism of the regular representation (by left-multiplication) and look at what the image of it may be

[bronze surge]

can anyone give a crash course explanation in L_\infty algebras and possibly how they are relevant to Lie theory?

[plucky arch]

I’ve been watching Borcherd’s lectures on Galois theory and he mentions the notion of an “absolute Galois groupoid”, which is analogous to the fundamental groupoid in terms of getting around choosing a particular algebraic closure. Is this… actually useful in any way?

[fierce steeple]

Pls don't crosspost

[bronze surge]

sorry, didn't know that was a rule. Won't happen again :)

[zenith pagoda]

ngroupoid was just talking about their use in p-adic geometry in neam’s thread

[ornate kindle]

I'm seeing mention to one module being a "direct factor" of another, but can't seem to locate a definition for that

Would it be like:

If $V = U\oplus W$ then $U$ and $W$ are direct factors of $V$

[broken turtle]

NotABot

[spice idol]

direct factors are usually in reference to factors or a direct product

[spice idol]

this would be a direct summand

though in this case they are the same ofc, biproducts and allat

[ornate kindle]

So would that be an appropriate definition do you think?

For context, I'm trying to understand what it means to be "relatively projective" here

[spice idol]

ah yeah they mean direct summands here

im pretty sure

[ornate kindle]

Awesome, ty

[summer quest]

yes this is often useful to consider

[fierce steeple]

Explicitly or in the back of your mind lol

[summer quest]

sometimes very explicitly otherwise you can make some mistakes

[fierce steeple]

Ye sure

[summer quest]

I am reminded of any arguments in l-adic cohomology where you define weights using a choice of embedding \iota:\bar{Q}_l->C

[fierce steeple]

What I mean is like ok you should not usually fix an embedding

Or if you do then fix forever and be careful or smth

But I haven't seen like idk the groupoid used as a groupoid lol

[summer quest]

it's more like you sometimes need to keep track of this choice and what happens if you change it

[fierce steeple]

Ye

Schemes.

[summer quest]

of course this picture is much more obviously useful when you're looking at \pi^et_1 for things that aren't just Spec of a field

in that case yeah the groupoid situation is incredibly useful

I guess another reason why you might want to consider this perspective is that you can often model these kinds of absolute Galois groups in terms of objects which are not 0-dimensional things anymore, and then you do need to keep track of basepoints

[hybrid swan]

Just want to sanity check if my proof works

By applying $S^{-1}$ and $\hom(-,N)$ in either order we get that the two rows are exact. The rightmost square commutes since the vertical isomorphisms are natural. The red arrow composed with $f^\ast$ is zero so it factors through the dotted arrow and by the five lemma thats an isomorphism

[broken turtle]

Jussari

[hybrid swan]

now that I've written it down I'm not even sure why I felt unsure

[limpid horizon]

what is the Koszul complex?

[novel nimbus]

@limpid horizon is there a certain context you need to understand in, because the definiton can be found anywhere

[vapid glen]

The first line in the last paragraph seems to suggest that all left exact presheaves on Ring^fin are representable. I can see how to extract a ring from such a functor (recover the underlying set by looking at F(Z[x]), recover the additive/multiplicative identity by looking at F(Z[x] --> Z) given by x goes to 0/1, recover addition/multiplication by looking at F(Z[x] --> Z[y,z]) given by x maps to y+z/yz). Let R be this ring. How does F being left exact help in showing that F is represented by R?

Any finitely generated ring A can be written as a quotient of $Z[x_1] \otimes ... \otimes Z_[x_n]$. Since the tensor product is the product in $Ring^{op}$ and quotienting is philosphically a limit in $Ring^{op}$, F should commute with everything (of course, the problem here is that we are quotienting by an ideal here, not a ring so it doesn't make sense to evaluate F at the denominator)

[broken turtle]

Finitely Many Bananas

[lone jacinth]

The fact that you're quotienting by an ideal is an easy fix. Just take Z[x1, ..., xm] where the xi maps to the generators of your ideal.

[vapid glen]

I see. Will my ring then be the coequalizer of this map and the map that sends all the x_i to 0?

[pastel agate]

Very basic question, but it doesn't necessarily go in the groups-rings-fields channel I guess. I'm trying to wrap my head around free Algebras. In particular I'm having trouble connecting the "universal property" definition to the "explicit construction" (polynomials in words on an alphabet) definition. I've gotten some hand-wavy explanations of how a free algebra is supposed to contain every possible polynomial, and how evaluation corresponds to the universal property, but in all honesty it really just flies completely over my head and the lecture notes I've been given haven't really been helping me. This is the first time I've really had to deal with "free" structures (there was something short about free groups in my group theory class but we moved on very quickly from it and never really revisited it, so I came away with basically nothing).

Any tips on how to approach this? Analogies/explanations that made it click for you? Anything is appreciated

[spice idol]

noncommutative algebra right?

[pastel agate]

Correct

Well the course I'm taking is actually on modules, but this is included and it's the first time I've worked with algebras as well

[plucky arch]

My suggestion would be to carefully examine what it means to define a homomorphism out of a free algebra

[lone jacinth]

So take for example the free algebra
k<x, y>
a monomial is some multiplication of xs and ys for example
x^2 y^3 x y^5 x^7
the elements are linear combinations of monomials, and a homomorphism can map x and y to whatever you want where you extend the homomorphism by for example
f(x^2 y x) = f(x)^2 f(y) f(x)

Which part is it that gives you trouble. The fact that f is determined by the value of x and y? The fact that you can map x and y wherever? Exactly what the elements of k<x, y> is?

It might be beneficial to think about the case of commutative algebras first if that's more familiar.

There the free algebras are just the polynomial rings (for example k[x, y]) and maps are still given by evaluating the polynomials

[spice idol]

you can think of a free structure as being as "generic" as possible. This means that your generators (which you can see as variables of some expression) cannot have any relations. Just like how you wouldn't expect the equation xy + 3z = 0 to hold for all x, y, z.

So, what might a "generic" algebra look like? Well, it must be generated by its generating set, so every element must be written as some combination of the generators, i.e. some linear combination of the monomials jagr mentioned, which are all the ways you can add multiply and scale some set of variables noncommutatively. But we also want no nontrivial relations between these variables, as our free object must be "generic", so two of these expressions must be the same iff all the coefficients before the monomials are the same (i.e. if one can show they are the same using only basic algebra).

To connect this to the universal property, convince yourself that any function f : X → A where A is some algebra, can be uniquely extended to an algebra homomorphism from your free algebra "assigning" to each variable x in X the value f(x), and then evaluating the expression. For this reason these maps are commonly called evaluation maps.

[pastel agate]

I think I have a good idea of what the actual free algebra itself looks like (since I've been given the explicit construction), what I'm still not really getting is how this is equivalent to the existence of a unique morphism from K<X> to some K-algebra A given just a regular function f:X-->A (this is how it's been defined to me)

[plucky arch]

So, given the explicit construction, can you see how such a unique morphism is made?

[pastel agate]

Not right now, thinking through all the responses I just got lol

[plucky arch]

Mhm mhm

I think it’s like - you define $\psi(x^2 y x y^2) = f(x)^2 f(y) f(x) f(y)^2$ and stuff, given your map $f : X \to A$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[lone jacinth]

Right, so then the point is that given any function X -> A, it should extend to a unique homomorphism from K<X>.

So in the case k<x, y> -> A if you know the value of f(x) and f(y) can you determine what the homomorphism should do with other elements?

[limpid horizon]

If an object satisfies a universal property P does it usually also mean that any other object that satisfies P would be isomorphic?

[pastel agate]

I mean I can definitely see that I can construct a homomorphism in this way, but I still can't really convince myself that it's unique. Like, why does it have to be that evaluation is the only way to pass from K<X> to A? I can definitely see that evaluation of a polynomial is a way to do it, but I don't really see why it is the canonical, unique way

[limpid horizon]

i think the point is that given a function X -> A that function gives a unique way to map K<X> to A

where K<X> to A map restricted to just X is the f map

[spice idol]

yes it does, thats why universal properties are nice

[pastel agate]

K<x,y> is a ridiculously massive set right, it's all possible elements in K in front of all possible combinations of x and y

[spice idol]

unique in that it makes a diagram commute

[limpid horizon]

is that a category theory proof or something? I remember for tensor products, later on there was a proof that showed that if an abelian group L satisfies those properties, its isomorphic to a tensor product. But we needed a whole separate proof for that

[spice idol]

its unique with the property that it sends x to f(x) for x ∈ X

[spice idol]

yeah, universal properties can be defined in a couple different ways

one of them being as a universal object of some functor, which has a particularly nice relationship with adjoints

[limpid horizon]

so something would only be called a universal property if it also satisifies this nice thing

[limpid horizon]

oh ok cool

[limpid horizon]

yea

maybe just write out some elements in K<x,y> to see what they look like

[pastel agate]

It's actually even bigger than what I stated lol

It's also all possible linear combinations of those

[limpid horizon]

yeah and sums and stuff

[pastel agate]

Yeah

[spice idol]

the important part to note is that a homomorphism is totally determined by where it sends the generators

[limpid horizon]

a classic

[pastel agate]

Ok I think I'm starting to get this direction better

[limpid horizon]

it took me some time to get used to this stuff especially for tensor products

[pastel agate]

Writing it out I get that I have to eventually evaluate a word letter by letter essentially

And obviously then if I already have that then there's only one way to do it

[spice idol]

like this is what the uniqueness part of the universal property says, that the free algebra is generated by X

[pastel agate]

Still seems quite cloudy but it is clearing up a bit

[spice idol]

yes thats it

[pastel agate]

Haven't struggled this much with something that's supposed to be so basic in a long time

[limpid horizon]

idk its kind of abstract at first

[pastel agate]

Everything in this fkn field is 💀

[limpid horizon]

yeah the thing is with universal properties tho they do come up a lot so you're kind of just forced to get used it

[pastel agate]

I can see that now

[torn harbor]

I think if you're used to vector spaces you can kinda think of it that way

To specify a linear map you can send the basis anywhere

[pastel agate]

When I first encountered it in group theory I was just like "eh this was barely half a lecture I doubt it's that important"

[torn harbor]

and there's always a unique linear map

[pastel agate]

<---- clueless

[limpid horizon]

lol

[torn harbor]

but in general when your objects aren't free relations get in the way of that

[limpid horizon]

shoot yeah idk why im thinking now why doesnt just finitely generated modules not have that property

[torn harbor]

for example there's no (nontrivial) homomorphism from the cyclic group Z/3Z to the integers precisely because 3•1=0 so you would need 3•f(1)=0 for any homomorphism (which of course wouldn't work)

[pastel agate]

What I kinda hate about this is that I can already tell that once I get it fully it'll feel super obvious and if anyone asks me for help with it I'm almost certainly gonna give them answers that sound just as unhelpful as many I've already received

[torn harbor]

well finitely generated modules can have relations

[spice idol]

because they arent free and can have relations

[torn harbor]

the special thing is that any module over a field is free

[spice idol]

which corresponds to the fact that the ring has nontrivial ideals

[limpid horizon]

yeah i got kinda stuck on thinking what that ruins

[ornate atlas]

This is exactly how I feel about universal properties lol, I did not understand the tensor product for the longest time and found all of peoples comments so unhelpful but when it finally clicked I was like oh yeah it is just that

[torn harbor]

each relation on generators imposes a restriction on the possible homomorphisms

[pastel agate]

Anyways thanks for the help everyone, I still don't really feel like I fully understand what's going on (although I can't point out exactly why) but I definitely have a better idea at least

[torn harbor]

For example if x^3=1 in your group, then for any homomorphism you must have f(x)^3=f(1)=1

That is the image of f must have order 1 or 3 which is a restriction

and other relations impose more restrictions

[limpid horizon]

i see

[spice idol]

also, if we have a field, any relation between elements of a vector space can be rewritten as one element being a linear combination of the others

[torn harbor]

with no relations, you can send a generating set anywhere and get a unique homomorphism extending it

[spice idol]

so this means that any relation on a free structure again just gives a free structure

[torn harbor]

also if you respect the relations you always get a homomorphism

[limpid horizon]

oh wow so you cant send the generators of a fg module anywhere? i certainly did not have the right understanding of this before then.

i see what ur getting at though

[torn harbor]

easy example, both Z/Z2 and Z are Z modules that are f.g.

but you can't send 1 in Z/2Z to anything other than 0 in Z

since it's image must have order 1 or 2

but everything in Z has infinite order except 0

[limpid horizon]

the relation in question here is the fact that 1+1 = 0 (ik obvious just wanted to say it)

[torn harbor]

Yeah

If you've done group presentations it only really clicks once you see that

or at least to me

[limpid horizon]

do u mean that, if u send the generators in a way that respects the relations, then u get a homomorphism

[torn harbor]

yeah in the sense of group presentations basically

[spice idol]

yes

[limpid horizon]

this makes a bit mroe sense to me now because i did some field theory and thinking about homs of fields send solutions to solutions

over a polynomial yk

[spice idol]

this is what the universal property of quotients says!! lol

[torn harbor]

so say you have a group <x,y | x^2=y^3> generated by two elements x and y subject to the relation x^2=y3

if H is some other group with elements a and b satisfying a^2=b^3

then there's a unique homomorphism from the group given as a presentation to H

sending x to a and y to b

[limpid horizon]

theres a unique hom from that to H (not sure what u mean by given as a presentation to H)

[torn harbor]

I just mean let G be the group generated by x and y subject to that relation

[plucky arch]

I think the key point is that the homomorphism has to agree with f on the generators

[torn harbor]

So G is quite large

[limpid horizon]

sorry can u elaborate just a lil :p

[torn harbor]

but you can define a homomorphism on G by only knowing where the generators go

provided that the images of the generators obey the relations

[limpid horizon]

yea

[torn harbor]

It comes from the fact that a group presentation is defined as a quotient of a free group

[limpid horizon]

i thought enpeace is citing universal property of quotients in general

[torn harbor]

Universal property of the quotient says if you have f:G -> H and a normal subgroup N of G such that f(n)=0 for any n in N, then there's a unique map from the quotient G/N to H

[limpid horizon]

which, is.. R->S factors uniquely through R->R/I->S i guess (im just thinking in rings cuz i like them)

[torn harbor]

where f is given as the projection followed by that unique map

but yeah there's an analogous statement for most algebraic structures

[pastel agate]

Yeah I can see exactly that, but it's still a bit unclear to me that this would only correspond to polynomials right. Like going off of the universal property alone why should I expect the free algebra to be a set of polynomials? I can see that if we accept that it is to begin with then there's a unique homomorphism there, but I guess I'm just uncomfortable that we just sorta immediately assumed it was that way.

[torn harbor]

So if you want to define the group generated by x, y subject to the relation x^2=y^3

[pastel agate]

Like how would I eventually deduce that it did in fact look like that were I only given the universal property

[limpid horizon]

do we also need to cite the factorization of maps like in the diagram? Like do universal properties always come with some factorization of maps in that way>

[torn harbor]

the precise way to do that is to take the free group on x and y and then quotient by the normal subgroup generated by the relation x^2=y^3

[torn harbor]

yeah f factors uniquely through the quotient in this case

[spice idol]

say you have a quotient map p : A → A/R. Then a homomorphism f : A → B can be factored through p if and only if R ⊆ ker f

[plucky arch]

Ok this is a cool question

Here’s one way to think about it

Suppose you have a set X, and a function X -> A

[spice idol]

i tried to motivate this in my initial response to you

[torn harbor]

(the normal subgroup generated by x^2 = y^3 is just the normal subgroup generated by the element x^2y^{-3})

[spice idol]

.

[plucky arch]

Given that you only have access to X, this function, and the operations in A

What are all the possible elements of A you can reach?

[limpid horizon]

jagr is back too

wow active chat

lol

[lone jacinth]

The way to think about this is that a free structure should consist of all the different ways to combine your operations.

So a free group should consist of all possible products and inverses of your generators.

A free algebra should consider of all possible ways to multiply and take linear combinations of your generators, and that's exactly what a polynomial is (some linear combination of multiplications of variables)

[pastel agate]

Ok this is making more sense now

[lone jacinth]

Like k<x, y> would need to contain x and y.

Then it also needs to contain x^2 and xy and so on of course. And what makes it free is that we never impose any relations. So for example xy shouldn't equal something we had before, it should be something new

[plucky arch]

Elements of a free object always tend to be built out of syntax

[limpid horizon]

Jagr is the goat for understandable explanations

[plucky arch]

Each element corresponds to a way of taking a function f : X -> A, and producing an element of A

In a way that makes for “all” A

[pastel agate]

Ok I see

And then if we wanted commutativity for example then we start imposing relations like xy=yx

[spice idol]

yes

[pastel agate]

This is making sense

Thanks everyone 🙏🙏🙏

It's still probably gonna take some getting used to but at least I'm not completely lost like before lol

[plucky arch]

Mhm, this would correspond to only considering commutative algebras A

[plucky arch]

to be honest i don't really understand what you mean by "as generic as possible"

[limpid horizon]

generic in the sense that you're imposing the fewest algebraic conditions on the elements ig

[plucky arch]

why does that correspond to "generic"

[limpid horizon]

maybe cause its like a blank slate or something

just a generic canvas

idk

[plucky arch]

@.@

[fierce steeple]

I think generic is not really the right word lol because this is maximally simple

[spice idol]

because it's generic in the sense that doing highschool algebra is treating the variables as "generic numbers"

[ornate atlas]

Yeah it’s sort of the minimal set of assumptions so that your object actually is that object. I agree it’s not really any more generic than a given ring but I do kinda see what he’s saying

[fierce steeple]

which is often like extra structure rather than generic lol

[spice idol]

just like you're treating the variables in a free algebra as "generic elements" of an algebra in your class/variety

[plucky arch]

oh no this confuses me even more 😭

[lone jacinth]

In alggeo you have generic points that special to more specific things.

So something generic should map onto something specific

[fierce steeple]

there is a sense though too in that these are universal rings so like dealing with these is often as good as dealing with everything

[plucky arch]

"the" universal ring?

[fierce steeple]

idk but there it seems the naming is more just cause those points are dense

[lone jacinth]

For example you might make a variety of algebras
k<x, y> / (axy = byx)
where (a:b) is a point on the projective line.

Then the generic point of the projective line would correspond to k<x, y>

[lone jacinth]

Well yeah, a specialization is a point in the closure. So being dense is the same as specializing to a bunch of stuff

[fierce steeple]

ye

[plucky arch]

maybe i'll just stick with my existing categorical intuition...

[fierce steeple]

Oh yeah Jagr I have a funny question

[lone jacinth]

Of the word generic?

[fierce steeple]

do you know much about reps of (say) sl_2 mod p lol