advanced-algebra

[solar turret]

jagr, can you verify this one?

[limpid horizon]

what about p2 in m

[solar turret]

Oh I missed that

Thank you for pointing out

[limpid horizon]

also p1 i guess

[solar turret]

So p1 and p2 are isolated

Yes

[hallow bone]

what is r(a)?

[solar turret]

Radical of a

[hallow bone]

cant you then immediately start with the associated primes?

it feels unnecessary to do this whole dance

[solar turret]

What is associated primes?

[hallow bone]

the set of radicals of the primary ideals in the primary decomposition

this should be terminology you know, no?

or does atiyah macdonald use different terminology for that

[solar turret]

Yeah, Atiyah used associated terminology

[solar turret]

I don't get it, how?

[hallow bone]

a = √a

so applying √ to a primary decomposition of a will be another primary decomposition of a

in particular, a will be the intersection of its associated primes

[solar turret]

Yes but how does it help me?

[hallow bone]

well then you can see directly that the existence of an embedded prime is contradictory

because if there were to exist an embedded prime, then that can be removed from the decomposition, which is a contradiction

[solar turret]

I see, thank you .enpeace

I think no, 0 has no primary decomposition.

In this case, X is homeomorphic to Max(C(X))

Since X is infinite Hausdorff space, there are infinitely many irreducible components of Max(C(X))

Does it imply Spec ( C(X) ) has infinitely many irreducible component?

In simple language, if X is a topological space such that Y is subspace of X and Y has infinitely many irreducible components, does X have too?

[astral ginkgo]

Consider

X=Spec(k[x1,x2,...]), then X is a UFD thus is irreducible, but Y=Spec(k[x1,..]/I) where I is generated by all x_ix_j for i\neq j. This ring has infinitely many minimal primes, so Y has infinitely many irred components, but X of course does not.

See

https://math.stackexchange.com/questions/1542104/non-noetherian-scheme-with-infinitely-many-irreducible-components-passing-throug/1542119#1542119

For details on the quotient ring

(Also UFD is a stronger condition, any integral domain is irred cuz 0 is the only minimal prime)

[solar turret]

so in this case, is it hold?

[astral ginkgo]

You would have to check

The counterexample I gave shows there's no general result (at least without additional assumptions on the ring)

[astral ginkgo]

Oh also, idt Max(C(X)) is even closed in Spec(C(X)) (otherwise this means there's an ideal that is contained in every maximal ideal and the only primes containing it are maximal, but any element contained in every maximal ideal of C(X) is 0, and there are non-maximal primes since X is infinite)

So trying to conclude anything about irred components from the irred components of a non-closed subspace won't do you any good, since they might not even be closed in the larger space

[waxen fractal]

@hallow bone sanity check: say that an algebra is locally finite if every f.g. subalgebra is finite. is it true that V(A) is a locally finite variety iff A^k is locally finite for all cardinals k

my reasoning is that i think S and H preserve local finiteness so it just comes down to P

[near lantern]

Assuming A^k = free algebra on k generators, isn't this obvious? In fact, can't you restrict it to "A^k is finite for finite k"? After all, any subalgebra generated by k elements is a quotient of A^k.

[waxen fractal]

ok this made me realize that i didn't phrase this in a way to make it clear im talking about universal algebra varieties and not AG

the only indication being that i pinged enpeace

to be clear, by A^k i mean the direct product of k copies of A. and by algebra i mean an arbitrary algebraic structure. and by variety i mean an equational class of algebraic structures

[solar turret]

I don't know how to use exercise 7 in 8

[astral ginkgo]

Induction. For example, the ideal (x_1) is prime in k[x_1], and you can use induction to show that (x_1)[x_2]= (x_1)k[x_1,x_2], then induct

Also in general (x1,...,xi) is prime in k[x1,...,xi] as it's maximal

[solar turret]

no i mean how do i show their powers are primary?

[astral ginkgo]

Power of a max ideal is always primary

Then use induction again

[near lantern]

I mean, I got that.

[near lantern]

I see.

[waxen fractal]

ah ok so it was just confusion over what ^k meant

[near lantern]

This looks correct.

[solar turret]

If x in p for some p in D(A), how do I show x is a zero divisor?

Say (0:a) \subset p

[past cove]

show that if x isn't a zero divisor then p can't be minimal over (0:a)

[solar turret]

Yes I am thinking about how to construct a new prime ideal which contains (0:a) and strictly contained in p

[past cove]

what about (0:a) + (x)

[solar turret]

But I have to argue why this is strictly contained in p

What I am thinking take the set of ideal which is strictly contained in p and containing (0:a), and prove that minimal element in that set is prime ideal

The set is non-empty because (0:a) in that set

[past cove]

how do you know a minimal such element exists?

[solar turret]

I am trying to use zorn's lemma

[astral ginkgo]

The minimal element will be (0:a), by this construction

[solar turret]

Ah

Sorry

[past cove]

zorn's lemma doesn't tell you about minimal elements

[solar turret]

What if we take maximal one

[astral ginkgo]

Well you'll need to prove that it's not (0:a)

I guess if you prove it's prime then it will follow

[solar turret]

I don't think the maximal element has to be prime

[shy horizon]

The problem here is that the Lie algebras with Dynkin diagrams are the classical ones, and they've already had their F_q forms classified by Seligman. The remaining ones (filtered Cartan, Melikyan) don't have a root space decomposition (as far as I'm aware) and hence no Dynkin diagram.

Since these are over finite fields and their closures everything is kind of fucked up when you go out of the classical case

[summer quest]

are you able to say anything about what the automorphism groups of these remaining cases look like when working over \bar{F_q}? Probably in these cases you lose the obvious relation between outer forms and Galois actions on Dynkin diagrams but the classification you want in terms of Galois cohomology only requires understanding this automorphism group

if not the entire automorphism group then at least you can try to describe inner forms/outer forms separately

[shy horizon]

Unfortunately I don't think the automorphism groups have been completely studied or classified in any case (they're relatively huge) but I'll take a further look in the literature

[summer quest]

if you can describe this automorphism group reasonably explicitly and assuming it's not a completely horrifying mess then you are just computing H^1(Gal(\bar{F_q}/F_q),Aut) with coefficients in this automorphism group and this step should be fairly manageable

you maybe don't need complete control over these automorphism groups to perform this sort of Galois cohomology computation, since you are only dealing with a procyclic group acting on everything that gives you quite a bit of constraint on things

[solar turret]

How to prove the last statement?

If p_i is in D(A) then (0:a) \subset p_i then r(0:a) \subset p_i, i want to show p_i = r(0:x) for some x in A.

[hallow bone]

yes, because the free algebras in V(A) are subalgebras of A^k for large enough k
this is (essentially) the HSP theorem :3

[waxen fractal]

ah yay

i used this to prove something interesting

one sec tho

[solar turret]

Can I show r(0:a) is prime ideal?

I got it

[solar turret]

But I am stuck at it

[waxen fractal]

yea so i essentially wanted this cuz, i wanted to argue that, if V(A) and V(B) are locally finite, then so is V(A × B)

and this is because local finiteness is also preserved by finite products

so we use A^k × B^k = (A × B)^k

and then i wanted to use that to prove that there's no "maximal locally finite" variety (except in the case of multipointed sets) because you can always expand a locally finite variety using this method and keep it locally finite

[hallow bone]

oo i see

🔥

[waxen fractal]

[solar turret]

ii) I showed forward direction using iii) how to show converse ?

[astral ginkgo]

S_p(0) is contained in p, so we just have to prove its radical is prime if p is minimal.

The following facts might be useful to recall:

||A_p is local||

||Nilradical is the intersection of all primes.||

||Inverse image preserves radicals.||

@solar turret

[solar turret]

I see, thank you

[astral ginkgo]

Did you get it?

[solar turret]

Yes

Nilradical is prime when p is minimal prime

Because there is only one prime ideal in A_p when p is minimal prime

Last statement of 11, converse,

So it is given that every prime ideal containing 0 is a minimal prime ideal.

a = \cap S(p_i), where p_i is a minimal prime ideal, and since each p_i contains 0 so a = \cap S(q_i), where q_i is the set of all prime ideals containing 0

Now from 9 and 10, 0 = \cap S(p_i), p_i in D(A), and D(A) contains all associated prime ideals of O, since O is decomposable therefore all minimal prime ideals of 0 are associated, and all associated prime ideals of 0 are minimal.

Hence, 0 = \cap S(q_i) = a, is it correct?

[astral ginkgo]

I'm not sure how you're concluding a is the intersection over all primes, since if p_i <= q_i then S(q_i)<= S(p_i), so intersecting over all primes might result in something smaller

[solar turret]

so a = \cap S_p(0) over all minimal prime ideals of A

But if all prime ideals of 0 are minimal that means all prime ideals of A are minimal

[astral ginkgo]

Why would every prime ideal be minimal?

I might be misunderstanding since I don't know what isolated means in this context

Does it just mean the minimal primes over a specific ideal?

[solar turret]

I think here they meant every associated prime ideal of 0 is isolated

[astral ginkgo]

Yea ok

[astral ginkgo]

So the condition means that all associated primes to 0 are minimal, this doesn't mean all primes are minimal

[solar turret]

Yes

[astral ginkgo]

Yea then your proof works. 9 and 10 together give that the intersection of S_p(0) over all associated primes (which are all minimal) is 0

[lone jacinth]

Is there a nice coordinate free description of the transfer map.

I.e. for H < G a finite index subgroup if you pick representatives xi for G/H then you have g xi = xj hi and you can define G -> H/[H, H] by taking g to the product of the hi. The resulting map is independent of the choice of representatives, but is there a nice way to construct this without mention of picking representatives?

[cerulean cove]

Godtoldme

l

[near lantern]

Decompose G/H into orbits under left multiplication by G. If g maps the point xH to the point yH, it also maps the set xH to the set yH. For each orbit O = {x_i1 H, ..., x_in H}, g^n maps each x_ik H to itself in the orbit space and therefore acts on it as a set. In fact it acts on it as a right H-set, so defines some element of Aut_H(x_ik H). The latter is a group that is isomorphic to H (after picking any origin in the free transitive H-set x_ik H) but not canonically - the isomorphism is only well-defined up to conjugation. So we get an element of Conj(H). Let us compute this explicitly. If g (x_ik H) = x_i_{k+1} H and we make specific choices of representatives x_ik, then g x_{i_k} = x_{i_{k+1}} h(i_k, g) for some h(i_k, g)'s in H, and g^n (x_i1 n) = x_i1 h(i_n, g) ... h(i_1, g) h. This means that using the basepoint x_i1 (i.e. the isomorphism h ⇔ x_i1 h to identify the coset with H), left mul by g = left mul by h(i_n, g) ... h(i_1, g). So g is mapped to tr(g, O) := [h(i_n, g) ... h(i_1, g)] which as a conjugacy class doesn't depend on the basepoint. If we choose to take x_ik = g^{k-1} x_i1 then we get the same tr(g, O) for any particular orbit in O. So tr(g, O) ∈ conj(H) really does only depend on g and O.

This is maybe not thaaaat representative-independent, but it shows how you could encapsulate the parts that need a choice to define: namely, prove that there is a canonical isomorphism between Conj( Aut_H(X) ) and Conj(H) for any group H and free transitive H-set X.

Anyway, once you have this, {tr(g, O) : O an orbit of g on G/H} is a well-defined multiset in Conj(H). One could keep this datum, which is maybe more refined than the transfer product. Or one can argue that the product of a multiset in H is well-defined in H/[H,H], and in fact we can then do it for elements of Conj(H) (or even of H/[H, H]).

[lone jacinth]

Hmm, I think I actually have an idea at least for H abelian.

ZG is a free (right) ZH module of rank [G:H]. Then left multiplication by g is a H-linear map, so you have a homomorphism
G -> (ZH)^x
given by taking determinant (which can be defined coordinate free)

[fierce steeple]

I remember seeing this funny overflow post a few years back

[lone jacinth]

It's a bit annoying that the codomain is (ZH)^x and not H, but close enough I guess

[fierce steeple]

I guess yeah is what jagr describes

[near lantern]

A more "topological" flavour of what I said is that for a diagram of free transitive H-sets of shape "n-cycle", the "monodromy" of the diagram (what is the action of the loop) is independent of basepoint up to conjugacy, and the automorphism group of the fibre is H (by an isomorphism well-defined up to conjugacy).

[fierce steeple]

Reminds me of how like most definitions of the sign of a permutation are somewhat unsatisfying to me (though it is possible to produce many definitions that are clearly devoid of choices)

[lone jacinth]

Determinant of permutation matrix is not too bad

[near lantern]

I think if you do this with H-sets instead of H-modules, you get Aut_H(H) = H instead of Aut_H(ZH) = ZH^⨯.

The only thing to define is "determinant".

[fierce steeple]

Sure ye

[lone jacinth]

Yeah, not that easy I would guess

[near lantern]

The H/[H, H] thing also seems similar to how trace over non-commutative rings is usually taken to be valued in R/[R, R].

[fierce steeple]

The fun one I saw was that you look at orientations of the complete graph on n vertices

It is just rephrasing but is cute lol

[near lantern]

Just take determinant over F1

Does that actually work?

[lone jacinth]

Determinant over F1 is the sign of a permutation usually

GL(n, F1) = Sn, SL(n, F1) = An

[near lantern]

I mean determinant over the group ring F1[H], rather.

As the determinant was over Z[H] before.

[lone jacinth]

Well, I don't know that F1H is

[near lantern]

Maybe the regular representation of H in F1-Vec

[waxen fractal]

i blinked and then ppl started talking about

[near lantern]

So if F1-Vec = pointed sets, then IG it's H?

H is a pointed monoid, so it can be an F1-algebra maybe

(pointed monoid = monoid)

[lone jacinth]

That's reasonable enough. But it doesn't seem to help with defining determinant

[fierce steeple]

Sheaves

[near lantern]

Is ∧^k subsets of size k?

with an orientation?

[lone jacinth]

I guess it should be

[near lantern]

Well ∧^k is not the way to define det over a non-commutative ring IG

IDK how to actually do that though. I only vaguely remember trace.

[near lantern]

Maybe this multiset is the set of elementary divisors of g on the F1[H]-module G.

[buoyant fox]

"PIDs have dimension 1"

what does this mean? what is the dimension of a ring

[digital parcel]

supremum of lengths of chains of prime ideals

[buoyant fox]

thanks

[digital parcel]

when you define chains you start with like p_0 \subsetneq p_1 \subsetneq ... \subsetneq p_n, and you say n is the length

[astral ginkgo]

The reason this is called dimension is because this is equivalent to a topological notion of dimension, which is the supremum of lengths of decreasing chains of irreducible closed subsets. The dimension of R is the same as the dimension of Spec(R) since prime ideals correspond to irreducible closed subsets

[buoyant fox]

is that something that comes up in like manifolds

i havent really studied manifolds yet

[astral ginkgo]

And this corespondence is inclusion-reversinh

[astral ginkgo]

This is algebraic geometry

[fierce steeple]

Amusingly, with this definition ("Krull dimension") all nonempty manifolds have dimension 0 so it is very much adapted to situations more like algebraic geometry than manifolds

(I guess empty manifold has dimension -oo in some conventions lol)

[lone jacinth]

The notions of dimension agree whenever you have a variety that is also a manifold though

[hallow bone]

when you look at the ring of global sections?

I'd assume so lol because hausdorff spaces aren't known to have many irreducible closed subsets

[lone jacinth]

That's one way of phrasing it, but what I'm saying is take some polynomial equations in R^n (or C^n). If the variety carved out is a manifold then the dimension (with Zariski topology) equals the dimension as a manifold

But there are two different topologies at play

[last talon]

The notions of dimension (of the manifold) and krull dimension (of the variety) agree
Just being a little more clear (because krull dimension of the manifold doesn’t agree)

[hallow bone]

ah, yeah, that is very nice

[hallow bone]

krull dimension of the manifold is like the 10th dentist

[last talon]

it doesn’t exist?

[hallow bone]

how does one begin to prove this even lol

[lone jacinth]

I guess something about measuring the dimension of the tangent plane at a point using derivative properties of the defining polynomials

[hallow bone]

right, I guess the dimension of a manifold is the dimension of the tangent space at some point, because you can just look at a neighborhood isomorphic to R^n

[digital parcel]

do you think the 10th dentist likes being a contrarian? or do you think they hate it but feel like it's a necessary evil

[hallow bone]

I think its the latter

[digital parcel]

i pity their poor soul then

[hallow bone]

and then they all draw straws to see which one of them will be the 10th

[digital parcel]

☹️

[astral ginkgo]

That's what they said about reviewer 2

[lone jacinth]

I think the first 9 dentists are paid shills, and there's only one trustworthy dentist

[hallow bone]

this is 10th dentist propaganda

[digital parcel]

Ohhhh

[astral ginkgo]

Oh oops I read your questions as rhetorical, I'm sure that was not your intention

[digital parcel]

Nah it wasn’t rhetorical

[narrow kraken]

Take R=C[x,y]/(y^2-x^3-x), i.e, the coordinate ring of an elliptic curve; then, K(R-Mod_f.g) is Z \oplus C/Z^2. my professor gave this example of the Grothendieck group in class but didn't explain it, so I was wondering if there is some geometric explanation for this?

[distant harness]

Can you perhaps start by stating the "this" you desire an explanation for, separate from your guesses?

[eager hound]

Yes the geometric explanation is that if C is the compactified curve (an elliptic curve) then by general theory of elliptic curves Pic^0(C) \cong C, and Pic is this \oplus Z. Then Pic(R) fits into an exact sequence Z \to Pic(C) \to Pic(R) where the incursion of Z sends 1 to the divisor of the thing we are removing, which is the degree one divisor corresponding to the identity. Finally because R is dedekind any vector bundle is a sum of line bundles by the fundamental theorem, thus K_0(R) is just the group generated by Z linear combinations of elements in Pic(R)

So I think actually it should be something like Z[C(k)] or something unless I am making a mistake

where C is the projective closure of that cubic

[hexed tangle]

I am trying to figure out why (\lim \textrm{Ext}^1_{\mathbb{Z}}( \mathbb{Z}/p \overset{p}{\to} \mathbb{Z}/p^2 \overset{p}{\to} \cdots, \mathbb{Z}) = \mathbb{Z}_p)

[broken turtle]

mh_le

[limber wharf]

I was wondering if anyone has read eisenbuds commutative algebra with a view towards AG, I was told that it covers almost all the theorems that hartshrone would use.

[hexed tangle]

I've managed to confuse myself, the Ext^1's are the cokernel of...?

[wary elbow]

This is false. Eisenbud is a good book but it doesn't have any sheaf cohomology, which is a big part of Hartshorne.

[lone jacinth]

If F is free and
0 -> K -> F -> A -> 0
is exact then Ext^1(A, B) is the cokernel of
Hom(F, B) -> Hom(K, B)

I'm a little unsure what limit you mean, but you can show
Ext^1(Z/n, Z) = Z/n

[hexed tangle]

right

[limber wharf]

Yes, I am not really seeing much geometry motivations in that text

[wary elbow]

There's actually a decent amount of geometry in Eisenbud -- he gives geometric explanations for many of the algebraic constructions -- but it is first and foremost a commutative algebra book, so it is not a replacement for an algebraic geometry text

[hexed tangle]

I am trying to solve this problem and have the following (see photo) since there can't be a 0-homotopy s that is an inverse to *p in Z, id_Z is not zero in D_+, but how does that make it a morphism from Z/p in D_+?

ah because it's a morphism from its prjective resolution

[eager hound]

I think the point was “statements hartshorne will use that are not explained in Hartshorne” e.g. that eisenvud supplies the commutative alg input you need to start Hartshorne

[wary elbow]

Ah yeah, in that case I agree that it has the required commutative algebra for Hartshorne. It also has some derived functor stuff that Hartshorne assumes.

I don't know if I would suggest it as a first course in commutative algebra, though, bc it's very long. You don't have to know everything in Eisenbud to get started in Hartshorne.

[lone jacinth]

So you've found a morphism from [Z -p-> Z] to Z[1], so if you can also find a morphism Z/p to [Z -p-> Z] you would be done.

Now remember the definition of morphism in the derived category.

[hexed tangle]

If P is the projective resolution, then there is a homotopy equivalence P -> Z/p

[lone jacinth]

It's a quasi isomorphism, not a hmtpy equivalence, but yes.

And then what are morphisms in the derived category?

[hexed tangle]

they are morphisms between chain complexes of projective objects modulo homotopy

[lone jacinth]

Well, if that's the definition you're using then Z/p isn't even an object in the derived category.

The category you're describing is equivalent, but the usual definition is that you take the category of chain complexes and invert quasi-isomorphisms.

[hexed tangle]

invert, how?

take cohomoly instead of homology?

[lone jacinth]

Introduce a formal inverse

I.e. localize the category

[hexed tangle]

Ok

so it induces an quasi iso, P -> Z/p

and since we have P -> Z[1] is non-zero, then there is a morphism Z/p ~ P -> Z[1]

is this correct?

[hexed tangle]

@lone jacinth thank you 😊 🙏

[astral ginkgo]

To be fair you usually do this in practice by passing through the homotopy category

[fierce steeple]

It is a shame that "homotopy category" is ambiguous lol

[plucky arch]

yeahhh

[lone jacinth]

What is "this" in your sentence here?

[astral ginkgo]

I just meant you define the derived category in 2 steps, but I now reread and saw they were talking about the category of projective chain complexes mod homotopy

[fierce steeple]

Insert comment about (truly) unbounded complexes

[lone jacinth]

Like acyclic complexes the are not contractable, or what kind of comment we inserting here?

[fierce steeple]

I mean like using projective complexes is only really enough to get bounded above/below (depending on convention) derived cat

[lone jacinth]

I think you in general have an essentially surjective adjunction
K(Proj A) -> D(Mod A)
But yeah, not faithful

[fierce steeple]

Hm I do not know why it should be essentially surjective but I mean in any case LHS is wrong thing for unbounded case innit

[lone jacinth]

Let
i: K(Proj A) -> D(Mod A) :p
be the adjoint functors.

Then for X in D(Mod A)
ipX -> X
we want to show the couinit is a quasi-iso.

You have
Hom(iA[-n], ipX) = H^n(ipX) = H^n(pX) = Hom(A[-n], pX) = Hom(iA[-n], X) = H^n(X)

[fierce steeple]

:p

(Tbh idk what this adjunction is)

I must admit I never work with K(Proj) and things lol

[lone jacinth]

It's a magical thing that exists because of Brown representability

[fierce steeple]

I forget lol I am sure K(Proj A) has another nice description hm

[worldly zealot]

K(Proj A) comes up when the scheme is not smooth at A because it isn't D(Mod A)

[fierce steeple]

Lol wdym

[worldly zealot]

i.e. the categorical quotient D^b(mod O_X,x)/K^b(proj O_X,x) is nonzero iff x is singular

[fierce steeple]

Sure but I mean that doesn't sound like a reason it comes up

Ok I guess you mean it detects it

Presumably this is related to IndCoh?

[lone jacinth]

I'm looking at some old notes now and it says if you take the localizing subcategory of K(Proj A) generated by A, then the resulting subcategory is equivalent to D(A) (through the usual localization map)

[lone jacinth]

I guess homotopy theorist usually use D^perf in place of K^b(proj) here

[worldly zealot]

tru

[fierce steeple]

Ah lol that makes sense

Well D^perf is more intrinsic I would say

But also with Db is this DbCoh

[worldly zealot]

yez

[fierce steeple]

And then it is this lol

As in Perf vs Coh (or QCoh vs IndCoh)

[fierce steeple]

Ok that was bad of me, add D if u wish

[fierce steeple]

I would just say Perf tbh lol

[lone jacinth]

Well whatever the notation, you would think of it as compact objects in the derived category. Instead of complexes of projectives

[fierce steeple]

Yes

Or equivalently for rings like it is generated by A under retracts and fiber sequences / triangles and so on

[lone jacinth]

thick(A)

[fierce steeple]

Yes

[tacit mountain]

Hey is anyone here familiar with with this result by Chevalley? I'm trying to find a precise reference to inspect (ultimately, I'm trying to find a graded analogue to his result).
Let R be a 2-dimensional Noetherian domain and let P be a height 2 prime of R. Then by a result of Chevalley, there exists a discrete (rank 1) valuation ring V containing R with maximal ideal M, such that M \cap R = P.

[lone jacinth]

It's theorem 3.11 here
https://homepages.math.uic.edu/~marker/math512-f18/valued_fields_3.pdf

Oops this only gives you that it's a valuation ring

So I guess you need to use the hypothesis's to show Northerianness

[steady lance]

I say Noetherianness and feel bad about it every time

secret 3rd option

[hallow bone]

Noetherianity?????

[hidden shore]

is it not possible to just use "Noetherian" without any suffixes

[astral ginkgo]

This is about when you want to say "by noetherianity/noetherianness". Sure you can avoid this by saying "By the ring being noetherian" or w/e but that means you are not the poll's demographic

[exotic lark]

its grammatically unpleasant

[astral ginkgo]

I've only heard noetherianity until very recently

[ornate atlas]

Notherianity is what I would use, but I would also try to avoid it because it feels clunky lol

[hidden shore]

noetherianity sounds cooler

[past cove]

I always just say "being noetherian"

[undone idol]

yesetherian

[fierce steeple]

"Smol"

[hallow bone]

"nice"

[distant harness]

"Which adjective do you use?" (Asks to select between two nouns).

[fierce steeple]

Lol

And I don't know why you would say "Both" to refer to being Noetherian

[ornate atlas]

https://tenor.com/view/iq-to-high-iq-to-high-dumbass-gif-8703840328342123411

[fierce steeple]

Thanks

I'm used to being asked this

[distant harness]

@Moderators automod circumvention.

[digital parcel]

I love using “By virtue of A being Noetherian”

[last talon]

@ornate atlas there’s a moderation problem, see Troposphere’s comment

[fierce steeple]

Yes

[ornate atlas]

https://youtu.be/n9gRbhMugD8?si=a6qSmA-UGMhfKer6

[astral ginkgo]

Oops

[silver goblet]

"Noether would tell us that ________"

[astral ginkgo]

Can't believe I confused adjectives and nouns. That's almost as embarrassing as the time I misunderstood the temporal scope of the paronomastic infinitive

[tacit mountain]

thanks for the help 🙂

[light badger]

Is there a nice way to compute H^i(G,F_p(j)) for i=2?

Where G = Z/pZ \rtimes (Z/pZ)^{\times} and F_p(j) is the 1-dim F_p space on which G acts by jth power of the primitive character?

Using Hochschild-Serre, it boils down to H^2(Z/pZ, F_p(j))^{(Z/pZ)^{\times}} and I can do that explicitly but its very ad hoc and annoying

[lyric rapids]

Quick question, is the following fact is true? I'm not sure if it is a typo or just something that is completely left out and not proven yet. R is any arbitrary ring btw. (It only talk about the case for it is the tensor functor of R-module with R being naturally isomorphic to the identity functor, not when we treat it as tensor product over Z-module)

[hushed bone]

What does - take as input?

Wait sorry

[lone jacinth]

It's supposed to be tensoring over R yes.

Same goes for the top sentence.

Hom_R( -, Hom_Z(R, D)) is isomorphic to Hom_Z( - (x)_R R, D)

[hushed bone]

Oh thats composition

Sorry if it is composition isn’t that equality true?

[lone jacinth]

The equation is true, but not relevant to anything written above or below

[hushed bone]

But it’s not true that tensoring with R over Z is identity on R-modules

Concretely like, if R is Z[x] then tensoring with R is the countable infinite direct sum functor

For example

[lyric rapids]

got it, thought i was going crazy.

[hexed tangle]

what is an example of two complexes that have the same homology groups but are not quasi-isomorphic?

[distant harness]

Does "tensoring over Z" even produce a well-defined R-module? There seems to be two different ways to define a scalar multiplication for the result.

[hushed bone]

Yes

[distant harness]

Yes to which of my sentences?

[hushed bone]

Well defined

It is given the structure coming from base extension

[lone jacinth]

Consider the ring R = k[x]/(x^2) and the two complexes
R -x-> R
and
R/x -0-> R/x

[distant harness]

I don't understand that description, sorry.

[hushed bone]

You consider R as a Z-module in the natural way

For any algebra B over a ring A, the functor - (x)_A B gives a natural structure of B module, this is base extension

This is the process applied here

[hexed tangle]

what does (x^2) mean here?

[lone jacinth]

I'm missing a /

[hexed tangle]

ah

[distant harness]

What I mean is, if we have $R \otimes_{\bZ} R$, then how do we define $r\cdot(s\otimes t)$? Unless I'm missing something, $rs\otimes t$ and $s\otimes rt$ are different elements of $R\otimes_{\bZ} R$ in general, so they can't both be the value of $r\cdot(s\otimes t)$.

[broken turtle]

Troposphere

[hexed tangle]

do you mean 0 -> R -*x-> R -> 0 ?

[hexed tangle]

thank you!

[distant harness]

Perhaps what I should be asking is, what is the source category of that functor?

[hushed bone]

A-modules

[lone jacinth]

But fun fact: over Z (or any hereditary ring) then two things are quasi-isomorphic iff they have the same homology.

(At least if quasi isomorphic means isomorphic in the derived category and not the existence of a quasi-iso between them)

[distant harness]

Okay, that makes sense then.

[distant harness]

But then "tensoring with R over Z is identity on R-modules" is not merely false but a type error, right? It can't even aspire to be the identity on R-Mod when it is actually a functor Z-mod -> R-mod.

[hushed bone]

Ugh

You are first applying restriction of scalars

There’s a fixed map from Z to R that lets you first consider R-modules as Z-modules

[lone jacinth]

I think it would be reasonable to say a functor Z-mod -> R-mod is the identity on R-mod if it did restrict to the identity there.

But either way it's not what the author meant

[distant harness]

Since that restriction corresponds to precomposing with a forgetful functor which is not faithful, the result still has no hope of being (isomorphic to) the identity.

[lone jacinth]

It is faithful though

[distant harness]

Yeah, the argument I had in mind doesn't quite seem to hold together when I think it through.

[lone jacinth]

It's not (typically) essentially injective. So that would cause a problem

[steady lance]

Say R is a Noetherian ring and I is an ideal. How “far” can I be from being cohen macaulay. AKA is there a standard example of an ideal with the “maximal” difference between its height and its grade?

This might have a very simple answer but I just realized I don’t know it

I have a feeling the answer is “arbitrarily far”, but I have trouble constructing ideals of certain heights without just using regular sequences

[lone jacinth]

I guess take
R = k[[x, y1, ..., yn]]/(x^2, xyi)
and I the maximal ideal.

Then (x) = R/I, so I has grade 0.

[steady lance]

Perfect thanks!

This is exactly what I was looking for

[signal moon]

Does U_lambda have a name?

[distant harness]

Noetheritude.

[undone idol]

this result is ridiculous someone must have botted the vote

[astral ginkgo]

Common sense wins again ❤️

[hallow bone]

no.

[indigo wing]

Why is algebra so hard

😭

Maybe it is just skill issue

[ornate atlas]

Because it’s maths and maths is hard, tis the struggle

[indigo wing]

but no joke

algebra is the hardest among analysis, algebra, geometry

smh

[hallow bone]

the mathematician's favorite state of mind is confusion

[hallow bone]

depends who you ask

algebra is far easier for me than analysis

and geometry, depends on the geometry

[indigo wing]

oh really

wow

[hallow bone]

but you probably will get a biased answer on mathcord, its very algebra pilled

analysis people are the ones with jobs

[indigo wing]

I guess in algebra I am introduced more to foreign ideas

[last talon]

Geometry is algebra 🙃

[hallow bone]

geometry = arbegla

[indigo wing]

and analysis usually builds up and ties well with geometric picture and previous concepts

that is my impression at least

and when I am handed with foreign stuff I get lost easily at first

[last talon]

No, algebra = geometry
We’re actual geometers here, not algebraic geometers

[hallow bone]

geometry is when locally representable sheaves on sites

[indigo wing]

Anyways

I have no idea how to solve this

💀

[last talon]

Geometry is when isometry or homeomorphism groups

[lone jacinth]

Do you know about seperable degree and the relation with homomorphisms?

[indigo wing]

In terms of

the number ways you can embed E into algebraic closure of k is separable degree of E over k

is \emph{the} definition we have

[lone jacinth]

And then an extension is seperable iff the degree equals the seperable degree

[indigo wing]

For finite ones?

[lone jacinth]

Yes

[indigo wing]

Yea we had that result

I guess then for finite case we can apply that?

[lone jacinth]

Yup

[indigo wing]

$[E(\alpha):k]_s = [E(\alpha):E]_s[E:k]_s = [E(\alpha):E][E:k] = [E(\alpha):k]$

[broken turtle]

Kintiru

[indigo wing]

I see

[lone jacinth]

And then in general something is seperable if all the finite sub extension are

[indigo wing]

Ah unfortunately I never saw that result

[lone jacinth]

I mean, that's just kinda the definition

Or how are you defining seperable?

[indigo wing]

element in a field is separable if its minimal polynomial is separable (factors into distinct linear polynomials in its splitting field)

a field extension is separable if every element in the field is separable

[lone jacinth]

Right so then E/k is seperable if k(alpha)/k is seperable for all alpha in E

That's what that's saying

[indigo wing]

Ah

I was somewhat trying to relate minimal polynomial of \alpha over k and E, but it didnt work well

[indigo wing]

sadge

[hallow bone]

ill ask it here too, may be more fit for the channel

[hushed bone]

I find this separable degree stuff so fascinating

Like I’ve worked through the proofs, but I think it’s magical these things are true

[grave pumice]

i haven't really heard of this

but i've seen "number of embeddings into C" described for some number fields as being identical to degree

i guess that's by char 0 fields being perfect

[hushed bone]

Yes

[ornate atlas]

This is a very vauge question, but is there some standard notation of what is meant by (kG-mod)_\sigma where \sigma is a family of subgroups of G closed under subgroups and conjugation? I think k could be any field and G any (finite) group, but lets say im interested in the modular case.

This is based off of a conversation I had, so I cant really give much more in the way of helpful context, but its in the real of like modular rep theory and model categories, but im a bit confused about what this category should actually be. It could just be adhoc notation that ive forgotten the meaning of, so this could be somewhat meaningless, but yeah

[ornate atlas]

Like obviously this is just some subcategory of kG-mod, but I dont know if theres some standard way that its restrcited. Possibly just to modules which are Ind^G_H(M) for some H, but that feels like itd be very small, I get the feeling its something a bit larger than that (but this is probably impossible to answer without knowing more if this isnt a standard construction/notation lol)

[lone jacinth]

Things that are direct summands of Ind^G_H stuff would be reasonable.

That's not really that small either necessarily

It would only be interesting in the modular case though

[ornate atlas]

Hmm yeah it could be that, ill play around with it and see if it seems to match up with what im thinking about. Only interesting in the modular case is fine though, its what im interested in (and I guess expected)

[ornate atlas]

This is cocomplete right (if so I think its what I want)? This might be nonsense, but as a quick sanity check
Suppose that all the $M_i$ are direct summands of some $\mathrm{Ind^G_H}$, then $\bigoplus_{i\in I} M_i = \bigoplus_{i\in I} \mathrm{Ind}^G_H(M_{i_j})\bigoplus_{i\in I} N_{i_j} = \mathrm{Ind}^G_H(\oplus_{i\in I}M_{i_j})\bigoplus_{i\in I} N_{i_j} $ since $\mathrm{Ind}^G_H$ is left adjoint to restriction. Then since kG-mod is cocomplete, so is $(kG\mathrm{-mod})_\Sigma)$

[broken turtle]

Nope

[fierce steeple]

Notation smh

But also i mean I think like this is not cocomplete unless you already close under direct sums cause can't H vary

[ornate atlas]

Hmmm yes

[fierce steeple]

like considering things like $\bigoplus_{i} \mathrm{Ind}^G_{H_i}(V_i)$

[broken turtle]

Prismatic Potato

[ornate atlas]

No you make a good point

[fierce steeple]

omitting some direct sums makes me sad

reminds me of hatcher lol

but it's okay if you choose that lifestyle

[ornate atlas]

wdym lol

[fierce steeple]

Well like

[ornate atlas]

But also this makes me sad because im now at a loss again, rather unsure what this category should be

[fierce steeple]

looks like you're writing $\bigoplus_i A_i \bigoplus_j B_j$ instead of some variant of $\bigoplus_i A_i \oplus \bigoplus_j B_j$

but it's ok

[broken turtle]

Prismatic Potato

[fierce steeple]

i am not sure but this looks familiar lol

[ornate atlas]

Ah, yeah lol

[fierce steeple]

hmm

[ornate atlas]

This is somewhat hopeful, I had hoped one of you wew or jagr wouldve come across this before. Its something my advisor was talking about, and I need to put a model structure on it, but ive either entirely forgotten what the actual notation meant (or never really knew)

[fierce steeple]

One guess would be like the stabiliser has to be in the set of things or smth idk

cause i have seen smth similar for variants of BG and things

[ornate atlas]

I do remeber we spoke about fixed points a bit, so something like that could be involved. I may honestly just need to rock up on friday like "ive not done anything because I forgot what we spoke about and couldnt work it out again (sorry please dont kill me)"

But yeah Ive tried to find a refernce for this for the last few hours with no luck, and I cant seem to come up with a reasonable seeming definition myself

I wonder maybe if I just actually want to work with kG modules and deal with the family of subsets condition elsewhere, like weak equivalences are given by H_* isos but like WRT fixed points or something

Back to the thinking mines I guess (but if anyone has seen this before please do let me know!)

[fierce steeple]

i'm just confused so like were you just shown this notation by someone in person?

hm

it seems like what Jagr said is reasonable, just maybe you wanna close under some operations

[ornate atlas]

Yeah, but im a dumb dumb and have either entirely forgotten what it meant or just never understood in the first place (this was a couple weeks ago, im just only getting around to looking at it now). I took a note that ive to work out what the model structure is on this and that its probably cofibrantly generated and very similar to the usual one on ChR, but nothing else, we then just waffled about different but related stuff for the rest of the time

[fierce steeple]

sure hm

[hallow bone]

whats a good place to read on bi/hopf algebras and their representation theory?

[near lantern]

Just email now and ask what it meant

[balmy pollen]

https://math.aalto.fi/~kkytola/files_KK/lectures_files_KK/Hopf-lecture_notes.pdf

i like montgomery for hopf algebras in general

[hallow bone]

thank you so much!

[void hare]

idk if this is the best place to ask this, but i want to find a structured way of decomposing RU(C_p)into indecomposable components (after p-completion and maybe inverting p) so that adam’s operations behave nicely (i want them to be stable under most of the summands), i.e. RU(C_p) = Q_p[C_p] = Q_p[g]/(g^p-1), but i will just write Z for now and “neglect” the problems with p-completion and inverting. i have a few examples for p = 2 and 3, namely the way it works there is finding a set of idempotents and try to work with that, there’s a general formula that gives you this for e.g. C, but i don’t necessarily have sufficiently many roots of unity, for p = 2, we do have nice idempotents 1+g/2 and 1-g/2, so (up to working 2-locally) have Z[C_2] = Z[1+g] and Z[1-g], same thing can be done for p = 3, but this is a bit more complicated, the only non-trivial i can think of is the one corresponding to the regular representation e=1+g+g^2 and thus 1-e must also be one and playing around a little and doing some check, we get a decomposition Z[C_3] = Z[1] + Z[1+g+g^2] + Z[g-g^2]. is there a way to generalize this somehow or has this been done in some paper? (note here: that the adam’s operations on RU(C_2) are trivial and on RU(C_3), they are stable under the first two summands, but not the third)

[narrow kraken]

is it true that hdim(R) = sup{pdim(M) | cyclic R-modules M} for any ring R? wikipedia says its true for any ring R, but i thought it's only true for noetherian rings. here hdim is global dimension of the ring

[lone jacinth]

It's true yes

[lone jacinth]

To see why:
First if the supremum is infinite you're done. So say pdim(R/I) <= n for all. Then we want to prove that Ext^n+1 vanishes.

Ext^n+1(-, M) = Ext^1(-, Sigma^n M) where Sigma is cosyzygy. So if we can show that Q := Sigma^n M is injective we're done.

As Ext(R/I, Q) = 0 we have that Hom(-, Q) is exact on the sequence
0 -> I -> R -> R/I -> 0.

This is known to characterize injective modules. Exercise if you haven't seen it before

[narrow kraken]

ic

is there any relation between global dimension of a ring R and a subring S of R

[drowsy niche]

definitely nothing that holds in general

consider e.g. k and k[x] and Z and Q

[drowsy niche]

if R is finite projective as an S-module we should have gldim(S) \leq gldim(R)

further I'm fairly certain that this works for left global dimension if R is flat as a right S-module and projective as a left S-module

ok yes (edit: see below)

proof left as an exercise uwu

[lone jacinth]

Restricting to commutative Noetherian rings you have that the finitistic dimension equals the krull dimension.

So if for example R/S is an integral extension and S has finite global dimension, then the global dimension of R is either equal to that of S or infinite.

[lone jacinth]

Hmmm,
So it's not too hard to see
||pd_S R(x)M <= pd_R R(x)M <= pd_S M||
So one would think the clue is in showing
||pd_S R(x)M = pd_S M||

[drowsy niche]

I was wrong
consider R = M_2(k) and S the ring of upper triangular 2x2 matrices over k

[lone jacinth]

I see.

Yeah here R is fg projective on both sides

[drowsy niche]

if we add in something like S is a direct summand of R as a left S-module it should work since then it's not hard to see that pd_S(M) <= pd_S R(x)M

[lone jacinth]

Yeah, but that's kindof a big ask

[narrow kraken]

is there a reference for different methods (or ways to simplify) of computing global dimension (or an overview of the common methods}? are there any sort of bounds? for ex, all I know right now are essentially, based on various properties of R, reduce to computing sup{pdim(M)} for some class of modules M (i.e, cyclic or if R is artinian, irreducible}. also, hdim(R)=0 iff R is semisimple, and hdim(R)=1 iff R is hereditary but not semisimple. also, the other definition of hdim(R) = {max d for which there exists M,N s.t Ext^d (M,N) is nonzero} seems kind of unusable, since ultimately you still need to enumerate projective resolutions right. in particular, without any assumption of R being artinian/noetherian

[lone jacinth]

I used the Ext characterization in the proof that it's enough to consider cyclic modules, so I wouldn't call that unusable.

Ext has many nice properties that often makes it easier to compute then directly dealing with projective resolutions

[narrow kraken]

yea i've seen that M is injective iff Ext^1 (I, M) for every ideal I of R (baer's criterion i think?), but i haven't learned about (co)syzgies yet so i will have to look into that

[hallow bone]

The module category of a cocommutative bialgebra is symmetric monoidal right? because you can give it the identity as universal R-matrix

[lone jacinth]

It's Ext(R/I, M), which corresponds to the lifting property with respect to the inclusion I -> R.

Anyway, a cosyzygy Sigma M is just the cokernel of
M -> E
an embedding into an injective module.

You can see immediately from the long exact sequence that
Ext^n+1(X, M) = Ext^n(X, Sigma M)

[lone jacinth]

Should be yeah.

You just make M (x) N into a module through comultiplication. Then cocomutativity ensures it's symmetric

[hallow bone]

alright yeah,
so cocommutative bialgebras are boring

imagine not having a nontrivial braiding

[lone jacinth]

All in the eye of the beholder

[hallow bone]

well I can imagine they're less useful for knot invariants lol

[lone jacinth]

There's a nice little table here
https://ncatlab.org/nlab/show/triangular+Hopf+algebra

[hallow bone]

ah yeah tannaka duality, which definitely isn't just enriched yoneda

[lone jacinth]

If you want braided monoidal you should reach for the quasitriangulated bi algebras

[hallow bone]

yisss

[fierce steeple]

Lol

I remember like

I think some original things in this theory are just Barr–Beck and Yoneda

[hallow bone]

representations of these get you braid invariants, though knot invariants need a little extra conditions

because the closure of two different braids may yield the same knot

[hallow bone]

what's Barr-Beck? I think I've heard of it

[lone jacinth]

I would have done my master on this Young-Baxter hopf algebra stuff if it wasn't for covid

[hallow bone]

well I've definitely seen the two names

[lone jacinth]

I think maybe it's a good thing I didn't 🤷‍♀️

[hallow bone]

oh, lol

[fierce steeple]

There are many variants but provides conditions for a functor to be monadic

Very useful

[hallow bone]

I'm thinking of doing my bachelor's on this, or something related

maybe how it relates to like racks and quandles (I've been told there is some very deep stuff there)

[fierce steeple]

One of my favourite theorems ig lol

[hallow bone]

ohh that one

[fierce steeple]

Often just called Beck's theorem too lol

[hallow bone]

poor Barr

[fierce steeple]

Though.there is also Barr–Beck–Lurie

[hallow bone]

how so?

[fierce steeple]

I am sure someone calls it Beck–Lurie

[lone jacinth]

It just feels a little meh

And then my life would be different, and I kinda like my life

[hallow bone]

to you maybe 😔
I for one am having a blast

[hallow bone]

Lurie-Beck?

[hallow bone]

that is valid

[fierce steeple]

Idk ig alphabetically should be Beck–Lurie

[lone jacinth]

Well, I don't know if I would be doing what you're doing exactly

[fierce steeple]

And chronologically lol

[hallow bone]

I guess I'm going very much into the knot invariant side of things

so I'll be reading up on vassiliev stuff too

the book I'm reading has a chapter about monodromy too which is cool

Another thing is that the alexander polynomial was initially defined super topologically and I'd be interested in seeing how that specifically connects to the super algebraic/combinatorial ideas of quantum and vassiliev invariants

[hallow bone]

right right

why do we insist on calling theorems after the person who "found" them

lol

[fierce steeple]

i thought we insisted on naming them after the 2nd European who found them

[hallow bone]

hence "found" in quotations

[fierce steeple]

yeah lol

[hallow bone]

so if I do this strategically I should just wait for someone to discover something and then immediately write a paper clarifying some minor things

[distant harness]

I don't think there's all that much expectation in practice that the person whose name is attached to a result was the first one to prove it. That's somewhat often the case, but there are enough prominent and well-known exceptions that nobody ought to take such naming in itself to be a claim of priority.

[narrow kraken]

let's take the triangular matrix ring R = (Z Q | 0 Q) and a left ideal I = (0 B | 0 0) of R, where B is any additive subgroup of Q. i want to show that pdim I <= 1. any ideas? i was trying to construct a Z-resolution for B and lifting it to a R-resolution but idk if that works

also does anyone have references on this "lifting" of resolutions from one ring to another?

[near lantern]

Wait WTH

I thought you were older

[narrow kraken]

im stuck on proving [A:K]_r = [A:K]_l here

[lone jacinth]

So I is isomorphic to just [B, 0]^T. So for a Z-resolution
F1 -> F0
you can just do
[F1, 0]^T -> [F0, 0]^T

I guess if you need some kind of pattern here I is just B tensor the projective module [Z, 0]^T

[lone jacinth]

How old do I appear?

[lone jacinth]

Here's a hint: start with a basis for Z(D) over Z(D)nK and work from there

[narrow kraken]

i'm having trouble showing that for F: = Z(D)nK, the F-basis of Z(D) is K-linearly independent (I showed it's K-spanning)

oh does this work || suppose {z_i} is an F-basis for Z(D), and suppose sum k_i z_i = 0 for k_i in K. then multiply on the right by arbitrary z in Z(D): this gives sum k_i (z z_i) = 0. but since {z_i} in an F-basis for Z(D), the map Z(D) -> Z(D) that sends x -> x z_i is like an invertible F-linear map, and then you can isolate each of the k_i to get they all have to be 0||

[lone jacinth]

You don't need to show it's K-linearly independent

[narrow kraken]

hm i was trying to show that the F-basis of Z(D) gives a K-basis of A

[lone jacinth]

Any spanning set contains a basis

[narrow kraken]

oops so true lol

[lone jacinth]

Also you probably shouldn't expect it to be K-linearly independent in general.

Like if you had something like
Z(D) = Q(a real cuberoot 2) and K = Q(a complex cuberoots of 2), then Z(D) is 3d over the intersection, but Z(D)K is 2d over K

[narrow kraken]

wait sorry im confused acc, why does this show the left and right K-basis of A have to coincide

[lone jacinth]

Well you found something that is both a left and right basis

[narrow kraken]

oops i forgot the {z_i} are central 🐭

thank you!

[marble cove]

Enpeace Cohomology

[hallow bone]

yes

[scarlet prairie]

Let k be a commutative ring and R be a k-algebra. As an exercise, I wanted to show that the bar resolution of $R$ as a left $R^e$-module is free. $R^e=R \otimes_k R^{op}$ denotes the enveloping algebra

[broken turtle]

Former Rank 7 LLORT AJNIN

[scarlet prairie]

but im fairly confident now that the bar resolution isn't free

im not even sure if the resolution is even projective, does anyone know?

[lone jacinth]

It is important that R be free as a k-module.

For example if k=Z/4 and R = k[x]/(x^2, 2x), then
Just as abelian groups
R^e is Z/4 (+) (Z/2)^3, but R(x)R(x)R is Z/4 (+) (Z/2)^7, so cannot be a free R^e module.

And it's not projective either

If R is free you get a free resolution, and if R is projective you get a projective resolution

[rose mirage]

I think your R there takes the crown for my least favourite ring I've ever seen

[lone jacinth]

Smallest ring that isn't principal ideal ring

Anyway, the bar resolution is most useful for k a field, because then every R module is free over k and you can use it to construct free resolutions.

[narrow kraken]

why do we need k to be algebraically closed here? like don't we just have [Pj] = sum over i of C_{i,j} [Li], and then dim_k Hom(Pi, Pj) = <[Pi], [Pj]> = sum over k of C{k,j} <[P_i], [Lk]> = C{i,j}

this thing below is also stated for any R that's artinian and of finite homological dimension, and i don't see the problem with it. the only thing that the proof uses for this is that every term in the projective resolution of each irreducible L_i is finite length and hence can be written as a direct sum of the indecomposable projectives P_j. but then that would suggest the cartan matrix C in GL_n (Z) is always invertible, i.e det(C) = \pminus 1 always, but that's not always true

[astral ginkgo]

Oops I assumed k was a field

Even though it says commutative ring

[narrow kraken]

i realized here that we should instead have dim_k Hom(P_i, P_j) = C{i,j} dim_k (End(L_i)) oops. when k is algebraically closed, End(L_i) = k so thats when we get the equality

[narrow kraken]

ok this is also fine lol, the cartan should always be invertible for hdim finite and artinian R, i was getting confused

[near lantern]

Is there a simple combinatorial formula for the order of a fundamental weight of a crystallographic root system modulo the root lattice?

IIRC if you take the fundamental chamber of the affine root system, its vertices are 0, li/ni, where the ni's are the coefficients of the simple roots in the highest root. Moreover, the vertices in the weight lattice, i.e., 0 and the fundamental weights corresponding to the simple roots appearing with coefficient 1 in the highest root, form a complete system of representatives for the weight lattice modulo the root lattice.

[lone jacinth]

Consider for example k = the real numbers and R = the complex numbers.

Then for R = P1 = L1 you have C11 = 1, but dim End(P1) = 2.

[lone jacinth]

If you don't have finite global dimension the sum wouldn't be finite.

[lyric rapids]

Can someone explain how I have the matrices formula for the differential (pic on right side) from the pic on the left side?

[past cove]

write $\partial^Q_n = \begin{pmatrix} a & b \ c & d \end{pmatrix}$

[broken turtle]

Irony

[past cove]

then by the commutativity we have that $\begin{pmatrix} a & b \ c & d \end{pmatrix} \begin{pmatrix} p \ 0 \end{pmatrix} = \begin{pmatrix} p' \ 0 \end{pmatrix}$

[broken turtle]

Irony

[past cove]

this immediately implies that c = 0

and that $a = \partial_n^P$

[broken turtle]

Irony

[past cove]

restricting to the other summand gives you $d = \partial_n^R$

[broken turtle]

Irony

[past cove]

and $\mu_n$ is some homomorphism

[broken turtle]

Irony

[lyric rapids]

I get that for (p, 0), what about (0, r)? Does that suppose to give (0, r')?

[lyric rapids]

Mu is just some random homomorphism? Do we need to know anything about mu, or we dont need to at all?

[past cove]

now you go in the other direction

you project to the second coordinate instead of embedding it

i.e. you have a commutative diagram like this

[past cove]

I mean it's the homomorphism such that $\partial^Q_n (0,r) = (\mu_n(r), \partial^R_n(r))$

[broken turtle]

Irony

[past cove]

but it's not super important what this is concretenly

[lyric rapids]

Ahh made sense

Okay thank you a lot.

[balmy pollen]

If u have a valuation on a division ring and define a new valuation by shifting by some constant, the valuation rings are isomorphic right

[lone jacinth]

What do you mean by "shifting" here?

[balmy pollen]

v'(x) = v(x) + c for some fixed c in Z (disc val in my case)

im 99% sure it's true

[lone jacinth]

But that won't be a valuation anymore

[balmy pollen]

oh ur right

whoops

[balmy pollen]

I have a (noncomm) algebra $A$ which is a right Ore domain of GKdim 2, and a disc val on the right ring of quotients $Q(A)$. The context for the above is that I want A to sit inside $\mathcal{O}_\nu$, I think I can pass to $S^{-1}A$ where $S={a\in A:\nu(a)\leq 0}$ to do this -- when does this preserve GKdim

[broken turtle]

Sara

[balmy pollen]

A is fg also

[blissful field]

Any good resources on Finite Gorenstein Algebras?

[fierce steeple]

Wassup

[steep pivot]

someone sent some spam about crypto or something should I delete my message?

It was deleted

[fierce steeple]

Dw but thanks for pinging. Just usually mod pings get thumbs up by the mod who deals with stuff but this wasn't so I thought I'd follow up lol

[balmy pollen]

turns out almost never, cor. 6.5 here shows every non-PI complex domain of GK2 has localizations of infinite GK dimension

neat

[analog schooner]

are lie rings/modules a thing

they probably are but havent yet bothered with those

[analog schooner]

maybe modules over a lie algebra idk

lie algebras are skew symmetric right which mean the transpose of the lin operator A is the same as -A i think

using the bracket operation, where [X,Y] = -[Y,X] or something

[weak lodge]

Yep, nothing in the definition of a Lie algebra breaks if you replace the ground field with a commutative ring. So in particular picking Z gives you a Lie ring. These come up pretty naturally in group theory, as there's a way to associate a Lie ring to any strongly central series, particularly the lower central series. There's also the Malcev and Lazard correspondences which are pretty useful if you want to study (locally) nilpotent groups

[analog schooner]

malcev/lazard correspondences??

what are nilpotent groups

[weak lodge]

Yeah, but it's better to define the Lie bracket as being alternating (i.e., [X, X] = 0) since this implies skew-symmetry when the ground ring doesn't have characteristic 2

[weak lodge]

There are a few definitions, but one is that G has a central series of finite length

[analog schooner]

seems familiar

are the composition series and central series different

never done anything with central series yet

[weak lodge]

Yeah, but you're right that they're similar in concept (both types of subnormal series)

[analog schooner]

when the ground ring has characteristic 2

does that mean [x,x] = -[x,x]?

by anti symmetry

[weak lodge]

Yeah

[analog schooner]

i still dont follow what central series is

lower central series mean like descending central series with subalgebras instead of normal subgroups

not entirely sure

arent abelian groups an example of being nilpotent

which is determined by whether its central series length is finite or not

[weak lodge]

Yeah

[weak lodge]

Yeah, those are the groups of nilpotency class 1

[lyric rapids]

Can I ask what is the canonical map of A -> cone(id_A)? I was thinking it might be the inclusion map from An to A(n-1) (+) An or to An (+) A(n+1) but i got stuck tbh. Wondering if someone could help me out.

[lone jacinth]

That's what it is yeah

[lyric rapids]

The inclusion of An -> A(n-1) (+) An?

Ahh okay

Ill get back to you if im still stuck

[sudden mountain]

whatbook is this

[astral ginkgo]

Actually

I found the notes but this might dox @lyric rapids since these seem to be from an ongoing course

[signal moon]

Why does the definition of the Pontryagin dual of a locally compact abelian group G consider only homomorphisms from G into the circle group, and not the entire multiplicative group of C? If we allow for these extra homomorphisms we get a different notion (e.g. if my example works, in the case of G=Z) that gives a perfectly good abelian group.

Is there a name for this object?

[astral ginkgo]

Some good answers here

https://mathoverflow.net/questions/294019/why-the-circle-for-pontryagin-duality

One reason not to take C^x is that taking a non-compact group you get that the resulting dual won't send LCA groups to LC Groups

Which is already not great

[signal moon]

I think I'm making some stupid mistake here. Do I not get a homomorphism between Z and C by sending k to 2^k?

[fierce steeple]

You get a cts hom Z -> C* for each element a of C* by sending k to a^k

[signal moon]

I found some slides which say that for an abelian group G, any homomorphism from G to C maps into S1

[fierce steeple]

Are you sure that G is not assumed compact or even finite

[last talon]

For any finite abelian group that holds

[signal moon]

But a^k need not lie in S1? Have I misremembered the definition of C*? I thought that was C\{0}, the group of units of C under multiplication

Sorry to put this in this channel, I wasnt sure where else it would go...

[fierce steeple]

Yes

There is no problem with what you said

[fierce steeple]

Except this

[signal moon]

Ah okay yes I figured it was probably something I was reading wrong there

[fierce steeple]

Well okay I think your message is correct in that I do believe you found such slides

[lone jacinth]

Your suggestion also wouldn't be a duality.

You can of course define Hom(A, B) for any abelian groups A and B, but there's nothing special about C^* in that regard.

[fierce steeple]

I think initially people often define it as C* for finite abelian groups which adds to the confusion potentially lol

[signal moon]

Yes, I saw that somewhere else and my confusion was added to, lol

[fierce steeple]

Bear in mind for finite abelian groups you could even work with the set of roots of unity in C (can be written as Q/Z) so you will see that too lol

[hushed bone]

Never write the category of finite abelian groups using just letters

Acronym

[fierce steeple]

Yeah this comment came to mind lol

It's okay in the UK as long as you draw a cigarette next to it

[hushed bone]

It’s okay if ur writing it in French

[signal moon]

Well you can't have any homomorphism from a finite abelian group to S1 that doesn't lie in Q/Z can you?

[hushed bone]

Just like FGA

If u translated it and then acronymized it

[signal moon]

FinAbGrp

[hushed bone]

If only…

[fierce steeple]

That's what I mean

For finite ab groups it doesn't matter if u pick S^1 or C* or Q/Z

[hushed bone]

I pick R* just to be cool

[fierce steeple]

Ig more generally discrete torsion groups lol

[fierce steeple]

U r cool

[hushed bone]

😎

[fierce steeple]

I use Z* to be cool

[hushed bone]

Damn

Actually I use 1*

[fierce steeple]

Sheaves

[signal moon]

Well thanks all, this was very helpful

[hushed bone]

You’re welcome

My comments were very helpful here

I will take the majority of the credit and thanks

[signal moon]

I will take care when writing down the names of categories of groups of any sort

[hushed bone]

👍

It actually pisses me the F off that the obvious abbreviation for spectral sequence is SS

But like… bro

At least you can write GSS

[signal moon]

What does the G stand for?

[hushed bone]

Grothendieck

The most important spectral sequence

[hushed bone]

I haven’t read it but I imagine Tohoku isn’t that hard to make sense of

[lyric rapids]

Oh no it is just some random textbook i picked up to learn homological algebra on my own lol. The material in my own course was lowk quite lacking.

They don't even teach left, right derived functor 💔

[astral ginkgo]

Oh ok. The reason I said this is because the notes I found (which seemed to contain this exact section verbatim) were last updated 3 days ago

[astral ginkgo]

Oh wow that's awful

That's like

One of the major points of homalg

[lyric rapids]

They only taught like Tor, Ext functor. Which is like whatever i guess but no where there is more general treatment of just left, right derived functor in general.

[narrow kraken]

does the the projective cover of a simple module over an Artinian ring have to be finitely generated

[rose mirage]

tbf that does require a lot more machinery

[lyric rapids]

Which is would need for algebraic geometry.

[lyric rapids]

Well yeah, i suppose. Which is why I got the textbook to begin with. Very nice for someone like me that is tryna self study

It give proof of theorem, (unlike my uni material) and example of stuff. And a good enough amount of calculations so I could do it on my own and check thing through.

[fiery wasp]

Whats a slope

[rose mirage]

https://ncatlab.org/nlab/show/tangent

[ornate atlas]

Does it?

I guess if you dont know any cat theory and just know what like modules are

[rose mirage]

that is usually what the prereqs to hom alg at unis are

[ornate atlas]

Fair, feels odd not to just introduce abelian categories but maybe this is just me having learned it from Weibel

[last talon]

Tbh abelian categories feel like overkill for most times I care about homalg

[lone jacinth]

Yes.

The ring itself is a projective mapping onto it, so the projective cover will be a direct summand of it.

[narrow kraken]

oh i see

i somehow ended up with the fact that P must be cyclic, but this doesn't seem like it should be true? take our projective cover pi: P ->> L, then take x in P such that pi(x) is nonzero, hence pi(x) generates L i.e R pi(x) = L, so pi(Rx) = L, where Rx is a submodule of P. Since this is an essential surjection we need Rx = P

[lone jacinth]

So the indecomposable projectives are exactly Ae for primitive idempotents e and they are the projective covers of the simples
Ae/Je
(J Jacobson radical)

[lone jacinth]

Yes, that's correct

[narrow kraken]

oh wow interesting

[lone jacinth]

So this isn't really limited to artinian rings either, except that you don't usually have projective covers in general

[narrow kraken]

if Q is injective, why does any surjection Q ->> L have to split? i thought we only knew that every SES 0 -> Q -> A -> B -> 0 splits when Q is injective

[indigo wing]

How do you remember semidirect products

I always forget which by which

Extension of H by N or Extension of N by H

💀

[lone jacinth]

Every surjection splitting is equivalent to Q being semisimple.

Injectives are usually not semisimple, but they can be...

[narrow kraken]

ah ic

[lone jacinth]

Are you just asking whether it is called H by N or N by H?

It's not particularly important.

You usually think of it as N being a normal subgroup and H a quotient group. So it would make sense to say N is extended by H.

But the notation Ext(H, N) has lead many to say H extended by N instead. So doesn't matter to much as long as you're consistent / clear

[narrow kraken]

let's say I have an SES 0 -> K -> P -> L -> 0, where P is injective and projective. then does this have to split (?)

[lone jacinth]

No

This would again only be the case (for all such sequences) if P was semisimple

Consider for example
R = P = Z/4

[narrow kraken]

oh right. i was basically trying to prove that R being Artinian self-injective with finite hdim must be semisimple, i.e hdim(R) = 0. equivalently, pdim(L) = 0 for every simple R-module L. we have here that every f.g projective must be injective. so then I took a projective cover pi: P ->> L, i.e an SES 0 -> K -> P ->> L -> 0, and we know P must be f.g projective, hence injective. so then I was somehow trying to establish that L must be projective

[lone jacinth]

Start by assuming L has a finite projective resolution. How does that resolution end?

[narrow kraken]

with a surjection P0 ->> L

[lone jacinth]

That's how it starts

[narrow kraken]

oh

with an injection P{n} -> P{n-1}?

[lone jacinth]

And what do you know about such injections from your assumption?

I.e. what do you know about Pn?

[narrow kraken]

they have to split?

each Pn is injective

[lone jacinth]

Yes, Pn is injective so it splits

So then what must the minimal resolution of L be?

[narrow kraken]

oh ok so now we ended up with a shorter proj resolution

0 -> N -> P{n-2} ... -> P0 -> M -> 0

but that contradicts the fact that we took a minimal proj resolution so we're done

so L has to be projective

[lone jacinth]

Yeah, so the general fact being proven here is that any module is either projective or has infinite projective dimension (for self injective artinian ring)

[narrow kraken]

makes sense tyty

[indigo wing]

It is just because I have to memorize for test

My professor just uses H by N or whatever is correct (he has been doing this in homework) and I dont wanna be confused mid test

[lone jacinth]

Well, then you'll just have to stick with whatever convention your professor uses

[indigo wing]

Oh does different literatues/professors use different convention?

I thought H by N is the convention

[lone jacinth]

.

Both are in use.

N by H is the one that makes the most sense / historically the first.

But H by N is also in use, presumably because it matches the order in Ext(H, N)

So if you want a rule of thumb, I guess just think about what would make sense and then do the opposite.

[indigo wing]

I see

I guess I will just have to write it down ten thousand times and make sure I get it right or something

engrave in my brain

[lone jacinth]

I think your time is probably better spent and understanding the concept so that you wouldn't be confused if you encounter the terminology swapped. But idk what kind of problems you'll be meeting.

[indigo wing]

When we have external semidirect product

I am pretty sure you can have both N -> Aut(H) or H -> Aut(N)

or maybe I have misconception

[lone jacinth]

Yeah, so I guess if all you're given is something like
"Describe all semidirect products of H1 by H2 for two given groups"

Then there's no way to tell which is which.

So if you expect to be given exercise with absolutely no context then I guess you just have to memorize what is being asked of you.

But for example if you're asked
"Describe all semidirect products of H by N where H=.. and N =..."

Then it would be reasonable to assume N was the normal group (since normal starts with the letter N)

[worldly zealot]

a normal subgroup will be invariant to conjugation so you ought to have Aut(N)

[indigo wing]

Well yea if you are given with a larger group

but as above, something like compute all semidirect products of S_n by S_m would be troublesome, since you dont have a larger group to realize which one should be normal

It's okay I just realized the convention is just we write s.e.s. and it is the opposite of the order of its appearance in s.e.s

0 -> N -> G -> H -> 0

and reverse the order : H by N

[broken delta]

What’s the difference between “Advanced Algebra” and “Groups, Rings, and Fields”?

[hushed bone]

Advanced algebra is for stuff like homological algebra, noncomm algebra, commutative algebra, Lie algebras

Groups rings and fields is more like undergrad level or beginning graduate level algebra stuff

[broken delta]

Ah okay

Thanks

[hushed bone]

Just FYI

If you click the channel’s name you can read a description of the channels

[verbal panther]

Do algebras with uncountably many elements need infinite generators/n ary operations/operations?

[hushed bone]

What

Is this a universal algebra question

Because an algebra has a fixed number of operations if you are using it in the usual way it is in like, ring theory

And the answer is no, as long as the base ring is uncountable, and necessarily yes if it’s countable

[verbal panther]

Mb

Erm I think that channel got removed

[waxen fractal]

what channel

[waxen fractal]

yea it comes down to the countability of the language, if you are referring to algebras in general, and not R-algebras

[verbal panther]

Universal algebra

[waxen fractal]

what chair monkey said applies to the general case tho

[waxen fractal]

nooo this server could have been peak

[verbal panther]

Base ring?

[waxen fractal]

i think they might have interpreted you as talking about R-algebras

admittedly. an R-algebra is an algebra

and R is sorta the language in a sense

so it ends up being the same answer

[astral ginkgo]

There was never a universal algebra channel

[hushed bone]

https://tenor.com/view/there-is-no-war-in-ba-sing-se-avatar-the-last-airbender-atla-no-war-there-is-no-war-gif-15884087966518847999

[upper beacon]

there is no secret test factoid in ba sing se

[upper beacon]

[static plaza]

Given the rotation group $SO(3)$ and its irreducible representations (irreps) $\mathbf{V}^{(l)}$, what are the most computationally tractable methods for computing the Clebsch–Gordan decomposition of the tensor product $\mathbf{V}^{(l_1)} \otimes \mathbf{V}^{(l_2)}$? Specifically, I am looking for algorithms that avoid the $O(l^3)$ complexity of traditional summation and focus on maintaining the structural integrity of the resulting direct sum $\bigoplus_{L} V^{(L)}$

[broken turtle]

Canvas123

[hushed bone]

[near lantern]

Is a cocomplete abelian category with a finitely generated (i.e., compact) projective generator P equivalent to End(P)^op-Mod or do you have to assume something else?

[rose mirage]

those are the only conditions for the mitchell embedding theorem so you're good to go

if memory serves correctly you don't even explicitly need cocompleteness, just all small coproducts

[near lantern]

Same thing since we have cokernels.

[near lantern]

Will that give me that it's an equivalence though?

[rose mirage]

I meant the full embedding theorem not the "it's a full subcategory" theorem

[near lantern]

OK, neat.

If P is an R-module in an abelian category A (i.e., an object with a ring homomorphism R → End(P)^op) is there a functor P (⨯)_R -: R-Mod → A?

I mean you should just map presentations of R-modules to the cokernel of the same map of direct sums of P's but has someone already checked that this is well-defined?

[lone jacinth]

It is well defined.

Basically just that a morphism of projective presentations is well defined up to a diagonal lift and if you have such a lift the map on cokernels is 0.

[near lantern]

Ah, so like

For any abelia category and projective object P, let C be the category whose objects are morphisms of direct sums of P's, and morphisms are commuting squares up to a diagonal. Then the cokernel functor embeds C as the full subcategory of objects which have a presentation using P's.

This is the argument basically?

[lone jacinth]

Yes exactly

[near lantern]

And in particular if Hom_{R-Mod}(R^I, R^J) = Hom_A(P^I, P^J) for all I, J (true iff End_A(P) = R^op and P is fg), then the first category is the same for R and P.

(^I = direct sum)

[lone jacinth]

Depending on your definition of generator you might need some form of well powered-ness.

For example it should be possible that for any pair of maps M => N there exists a map P -> M that seperates them without a direct sum of Ps mapping onto M. (Essentially when M has a proper class of quotient objects).

[near lantern]

Should be taken care of if I define generator as there is a direct sum of P's mapping onto M, right?

[lone jacinth]

Hmm, actually I think you should always be able to do
P^(Hom(P, M)), so nvm

[near lantern]

Although I think if the category is locally small then TFAE (for projective P): (i) Hom_A(P, M) is non-zero for non-zero M (ii) P^{Hom_A(P, M)} surjects onto M.

(Because Hom_A(P, M/all images of P) = 0.)

[lone jacinth]

Yeah so the issue only comes if you don't have all coproducts

[waxen fractal]

currently thinking about how many binary operations there are on a 2 element set, up to isomorphism and swapping the input order, which depend on both inputs

i think there is ∧, +, ¬∧, ¬+, →

[distant harness]

if your + and ¬+ are what I think they are, they are isomorphic, by exchanging the two elements of the set.

[waxen fractal]

xor and nxor to be less terse

[distant harness]

Yeah.

They are De Morgan duals of each other.

[waxen fractal]

oh yea ur right

im a silly bean

∧, +, ¬∧, →

semilattices, F_2 vector spaces, boolean algebras, ???

i have deep seated unearthed trauma from the last one cuz in my first logic course it was the only midterm exam question i couldn't answer

i still dont know the answer

[distant harness]

1. If the two results on the diagonal are the same, then it must be either xor/nxor or implication.
2. If they are different, then either f(x,x)=x for both x or f(x,x)!=x for both x. The requirement that no input must be irrelevant then fixes it as either and/or or nand/nor.

[waxen fractal]

to be precise, what operations are generated by → ?

[waxen fractal]

that makes sense

(i think i just did process of elimination)

[hallow bone]

post's lattice might be helpful

[waxen fractal]

i keep getting distracted irl by my class lecture

[waxen fractal]

good idea

although i should probably just overcome my fear and systematically investigate it myself

i should think about what the following operations are
x → (x → y)
x → (y → x)
(x → y) → x
(x → y) → y
(x → y) → (y → x)

the others are obviously trivial

oh the first two are also trivial via currying

the next one is less obviously trivial

(x → y) → y = x ∨ y ???

and the last one is trivial

wtf why is there exactly one way to escape

no wonder my soul left my body that fateful day

this is so evil

[waxen fractal]

the funny thing is that i can come up with a categorical justification for why x ⊔ y always maps into [[x, y], y] in a cartesian closed category

as a sort of way to check im not making a mistake

ok I've mentally recovered from the shock that ∨ is in the clone of →

the question is, what's next

[waxen fractal]

tbh im demoralized imma just check posts lattice

nvm this thing requires a manual to decipher wtf

ok i went on Wikipedia and i found a useful fact

apparently x → y has the property that y ≤ x → y, and this is preserved by composition

that is, the operation is bounded from below by a projection

yeah i finally know the answer now

[waxen fractal]

the proof is quite beautiful to me

[waxen fractal]

there's exactly 6 operations that have this property, x, y, x ∨ y, x → y, y → x, and 1

@hallow bone my inner turmoil is finally resolved

sorry yall for the chat spam

[hallow bone]

YIPPEE

wow

I didn't mean caps

[waxen fractal]

that is how i feel tho

very all caps yippee

[hallow bone]

that's good :3

why did it

I didnt mean to do winky fac

e

[waxen fractal]

because a long time ago i had an exam question that stumped me and now i know the answer after over a year

[hallow bone]

🔥

peak

[waxen fractal]

the exam asked to prove ∧ is not in the clone of →

actually i still have a copy of the exam hold on

time to check my memory

[waxen fractal]

yes i just bullshitted apparently and said the only things were 1, x, y, x → y, and y → x

and i got half credit cuz the instructor said there was one more

well i think half the credit was for the ∧ doesn't generate → part

the ∨ that got away...

take that martin zeman

i win

for the record i took this class in fall 2024

but it's been a thorn in my side ever since that i did not know the clone of →

[waxen fractal]

btw does it generate a weird reduct of heyting algebras

like if you downgraded distributive lattice to semilattice

[hallow bone]

I'm gonna be honest idk

[waxen fractal]

nooo

dw it doesn't matter much

[late frost]

What is the use of the lattice theory (order theory)? outside of distributive lattices, I do not see any application

[exotic lark]

the concept of an ordering is pretty general and poset theory is the language we have for that

subset inclusion, partition refinement, divisibility, etc

theres a lot of examples

discussion is probably better suited for #combinatorial-structures

[hallow bone]

the language of lattices and closure operators (in particular Galois connections) is one that pops up very frequently in algebra

think of the lattice of submodules, or subgroups, or normal subgroups, or ideals, for example

from what ive seen though its not necessarily the lattice theory itself thats applied but moreso that abstracting to arbitrary lattices makes proofs a lot more elegant

for example, the jordann-holder theorem for modules has a nice lattice-theoretic analogue in modular lattices, or the equivalence between every ideal being finitely generated and the ascending chain condition

if you can stomach it, lattice theory is absolutely instrumental in a lot of universal algebra and related fields

[waxen fractal]

@hallow bone btw i have a question about post's lattice

Wikipedia's description reveals that, apparently, you can use → to create any operation f on 2 such that there exists a projection x such that x ≤ f

im wondering if you have an idea how to prove this, i thought about it for a bit but im getting stuck

also it's kinda bizarre to me I've never heard of this construction before, it seems really interesting

[hallow bone]

seems like some induction typa thing lol

hmm

Suppose xi ≤ f(x1, ..., xn)
=> f(x1, ..., xn) = (xi -> 0) -> f(x1, ..., xi-1, 0, xi+1, ..., xn)
then apply induction

im not sure if i see a way to do it without 0

[waxen fractal]

if we had 0, then we could get any operation

[hallow bone]

lol im stupid

[waxen fractal]

i did just get an idea

(x ∨ (y → z)) → z = (x ∨ ¬y) → z = (x → z) ∧ (¬y → z)

is this true?

from there i think we can get a CNF of any operation f such that z ≤ f

[waxen fractal]

rubber duck i guess

[hallow bone]

idk youre the one doing a master in logic stuff

im too lazy to check

[waxen fractal]

im a cs bs major

😭

[hallow bone]

bullshit major

[waxen fractal]

so im thinking about the universal property of ∨

(x ∨ y) → z = (x → z) ∧ (y → z)

this is gonna sound sad

i remember logic rules by corresponding category facts

the issue was i got stuck how to put ¬y in there

until i realized i can put y → z

[hallow bone]

this doesnt even work i should go to sleep ugh

[waxen fractal]

(y → z) → z = y ∨ z

i.e. ¬y → z

yea ok i believe Wikipedia now

but anyway it's kinda wild this is a clone on. any poset

the proof is quite beautiful

given a term t(s1, s2, ..., sn), there exists i such that si ≤ t, and by induction on complexity we can assume x ≤ si for some projection x, so x ≤ t

Wikipedia didn't list a proof but the proof isn't hard thankfully

it's interesting that this forces all constants to the the top element of the poset if it exists

thus the top element is also fixed by all operations

[lusty galleon]

maybe im thinking abt this completely wrong, but i dont know how to find splitting field without finding the roots, but the roots of this thing kinda suck, and im thinking what if i have an even worse polynomial with terrible roots? is there a better way to do it?

[hushed bone]

It’s a polynomial in x^2

Write y = x^2 and it’s a quadratic

Apply quadratic formula to find solutions of y

Take square roots to get all roots

This one is actually p easy to do directly

[lusty galleon]

i found the roots of it, i just dont like them very much lol
i need to do some complex number shit to simplify
but regardless, my question is, is there a better way to do it that isnt just find the roots and then adjoin?

[hushed bone]

Just find the roots

[lusty galleon]

ok lemme rephrase

what if it was like degree 5

and the roots were not obvious

[hushed bone]

You just get fucked lol

[lusty galleon]

so there is no other way to do it?

[hushed bone]

I think there’s not really a nice way to go about it

If you know extra information you can maybe try and figure something else out, but I don’t think there’s anything very general

Maybe a NTist can say more, but I don’t know how to look at the polynomial and know if the splitting field is generated just by a single root for example

[lusty galleon]

hm ok

[shy cargo]

https://math.stackexchange.com/questions/45893/how-to-find-the-galois-group-of-a-polynomial is fairly comprehensive on a range of methods but this is not easy to do in general

[signal moon]

Let me know if this does not belong here. If a finite group G has l representations, and we know the characters for the first l-1, we can always find the character of the last. This is because there are (number of conjugacy classes) variables and as many equations as ways to get the inner product of the trace of the remaining irrep with a different irrep (namely l-1 = num conjugacy classes - 1). Is that right?

[rose mirage]

Yeah you just use row orthogonality

Or column orthogonality if it’s too hard

[signal moon]

Is this the usual statement of column orthogonality?

[rose mirage]

Yeah Thats it although it’s normally written as |C_G(g)| instead of the fraction

[signal moon]

Maybe I'm being silly but you need to use row orthogonality at least once, at which point you can use column orthogonality, right? Since if you don't know any values of the remaining character, then all column orthogonality can tell you is the absolute values of the remaining character (using part (i) in the screenshot)

Well after thinking about it it seems that this may not always work; you may get some tautological equations. But just using row orthogonality should always work (?)

Excuse me, this is nonsense. You know chi(e).

[signal moon]

I'm wondering if this is correct: to show phi is injective, suppose phi(x)=phi(y). That means that x-y is in the kernel of every irrep rho. Hence, by Mashke's theorem, x-y is in the kernel of the regular representation. Yet by uniqueness of inverses in groups, this means that x-y=e, or x=y. Next, since G is finite abelian, we can use the structure theorem for finite abelian groups to write it as a direct sum of finite cyclic groups. By modifying the argument that |V|=|V*| for vector spaces (we showed this in class. I'm thinking of irreps of G as Z-module homomorphisms G -> C), we find that |G|=|G dual|. Since we have an injective between groups of the same size, we have an isomorphism.

Well I think it would do to go into more detail where I show that G and G dual have the same size, but we have other ways of showing this from class. I'm wondering specifically about where I show that phi is injective.

Also, if the trick with the regular representation works, then I think I need to use both the left and right regular reps to guarantee x-y=e.

[undone idol]

you are overthinking the injectivity - if x is in the kernel, then rho(x) = 1 for all rho

[signal moon]

Is your rho an irrep?

[undone idol]

rho is any arbitrary element of Ghat

[signal moon]

Ok, yes

But why does rho(x)=1 for all x imply x=e?

(Just making sure, when you write 0 you mean the identity map on C?)

[undone idol]

it is enough to show that for all x != 1 there exists a rho such that rho(x) != 1

You can do an explicit construction

[signal moon]

I guess you can identify $G\cong\mathbb Z/{n_1}\mathbb Z\times\cdots\times\mathbb Z/{n_t}\mathbb Z$, write $e_i\in G$ for the tuple with a 1 only in the ith place, write $x=\sum a_ie_i$, pick $a_i\neq 0$ and let $\rho$ send $e_i$ to the principal $n_i$-th root of unity, while $\rho(e_j)=0$ for $j\neq i$? Maybe I'm completely in the wrong ballpark

[broken turtle]

person2709505

[undone idol]

Yeah that’s roughly the right idea

[signal moon]

Did I get something wrong?

[undone idol]

No it looks right but I’m a bit tired to check the details

[signal moon]

Thanks for the help

[hollow briar]

some stupid question but why both the trivial module and the zero module have their own articles on Wolfram?

It seems to me that they are essentially the same.

[urban granite]

because wolfram is stupid

[hollow briar]

lmao

[solar turret]

I am not sure about i), what does it mean by reducing this equation mod b?

[sly rune]

the ring map B → B/𝖇 sends 𝖆 to 0, so you get an induced map A → A/𝖆

[solar turret]

but we have already natural map A -> A/a, right?

[sly rune]

yeah what i mean is that this diagram commutes, where the right vertical arrow is the one you want to show is integral

essentially all the coefficients go to A/𝖆 when you reduce mod 𝖇

[median halo]

what's crazy is that they both have the same author

[solar turret]

so we are identity A/a with subring of B/b?

[lyric rapids]

Can anyone help me on this or give me a hint on how to do this btw.

This is the cone differential btw

[dark widget]

have you got any direction