groups-rings-fields

[rocky cloak]

Fp don't have many subfields

[somber bluff]

oh yeah lol

[south patrol]

So basically this factors as Fp -> K -> Fp

Where all maps are injective

[south patrol]

Or just this lol

[somber bluff]

ok ok so for example K -> F_3 exists if K has 3 elements and is iso right

all are iso i think

right?

[rocky cloak]

There's only one ring with 3 elements yeah

[tardy hedge]

What is [R : Q]?

Its prolly not a finite extension of Q eh

lol

e moment pi moment

[thorn jay]

Its transcendental

Finite extensions are always algebraic

[rocky cloak]

It's the continuum, |R|

[somber bluff]

does there only exist at most one field homomorphism to the field of rationals?

[thorn jay]

Yes, that would have to be an isomorphism, and Q has trivial automorphism group

[coral spindle]

Can you say that in more words

are you saying that If F is a field, then there is at most ring homomorphism F -> Q?

Enpeace is also correct if this is what you are saying

[thorn jay]

I dont see any other interpretation

[coral spindle]

They may be trying to ask which fields embed in Q

in more words

Kinda the same question

[thorn jay]

That's still 1 lol

(Up to isomorphism)

[coral spindle]

Yes, but we ought to approach the question differently :)

[thorn jay]

💥

[somber bluff]

how does one prove that. i kno whow ot prove q has trivial automorphism group

[crystal vale]

Third line, by the Euclidean algorithm and row operations, we may assume that f_1 has leading coefficient 1, degree d, and that deg f_i < d for j ≠ 1.

I don't get it

[thorn jay]

Q has no subfield other than itself, and any ring homomorphism to a nonzero ring must be injective, hence by the FIT any ring homomorphism from a field into Q must be an isomorphism

[crystal vale]

are they divide f_i, i =2,3..,n by f_1?

[frigid epoch]

We can also be hella pedestrian: let F be a field of char 0, x in F irrational over Q. Then x can't map to 0 (else what happens with x^-1?), so it maps to a rational a/b. So bx maps to an integer a. So bx/a maps to 1. Now x is irrational, so bx/a is still irrational, so bx/a - 1 is nonzero. But it maps to 0.

[glad osprey]

Isn't Q having no subfields equivalent to the statement that if f : F -> Q is a field hom then F = Q? Or are there easier ways to show that Q has no subfields?

[frigid epoch]

I should have phrased it positively. If x in char 0 is nonzero, it maps to a rational a/b, so bx/a - 1 maps to 0. Hence x must already be a/b.

[glad osprey]

My proof idea was this: there is a unique field hom from Q to any char 0 field F, in particular id : Q -> Q is unique. Then if f : F -> Q there is a hom g : Q -> F, and since id is the only hom Q -> Q we know that f o g = id, so f is surjective, therefore F = Q

[thorn jay]

There are easier ways, namely by the fact that any subfield of Q must contain 1, hence Z, hence Frac Z = Q

[thorn jay]

Nice

[crystal vale]

can i get a finitely stably free module which is not a free?

[thorn jay]

alternatively
any subfield of Q must contain every integer a, hence 1/a as it is a subfield, thus b/a for all integers a, b

[crystal vale]

hint?

i don't know much about modules, which is not free, one is Z/nZ over Z but not sure about stably free

[rocky cloak]

(2, 1+sqrt(-5)) in Z[sqrt(-5)] should be an example.

[crystal vale]

is it trivial example?

[rocky cloak]

Trivial in what way?

[crystal vale]

i mean i just know the definition of stably free but i don't know how to come up with example

[rocky cloak]

Well, it's not so easy to come up with examples in general.

But to check an example you can just come up with a finite free resolution

[crystal vale]

okay thank you jagr

[rocky cloak]

I think maybe, I made a mistake and this example is not stably free

[crystal vale]

i understand this

but not this, what does it mean by resolution here?

i got it

[crystal vale]

how R noetherian helps here?

[tardy hedge]

What are u learning rn? ^

[crystal vale]

Quillen-Suslin Theorem

[analog hemlock]

Does anyone know the classification of all subgroups of $\mathbb{Z}\times\mathbb{Z}_p$, ex. $\mathbb{Z}\times\mathbb{Z}_2$?

[cloud walrus]

lacchu

[candid patrol]

p prime ?

[thorn jay]

I'd assume so

[candid patrol]

It doesn't change much anyway

[crystal vale]

why?

[analog hemlock]

Meant to type Zn, but I don't think it matters

Will there always be n+1 classes?

[candid patrol]

what do you mean by n+1 classes ?

[analog hemlock]

In the case of Z times Z2 for example

We can classify the subgroups by taking a subgroup of Z, which is kZ, then {0}, or {1}, or {0,1}

[candid patrol]

Make sentences plz I d'ont get you

[analog hemlock]

I'm asked to give a classification (and proof) of the forms of subgroups of Z times Z2 after having proved that all subgroups of Z are of the form kZ for k in N.

[candid patrol]

So you think we have n+1 distinct subgroups in Z x Zp ?

[analog hemlock]

I'm not so sure about the generalization but I think there are 3 types of subgroups of Z x Z2

[candid patrol]

Fact

[analog hemlock]

There are n+1 distinct subgroups of Zn x Zn

[velvet hull]

Could you describe them for me? I’m not sure that sounds right

[acoustic igloo]

doesn't Z2 x Z2 have 4 subgroups?

[velvet hull]

Should be 5 total including the trivial ones - {e}, Z2xZ2, Z2x{e}, {e}xZ2, and the diagonal {(0,0), (1,1)}

[velvet hull]

https://math.stackexchange.com/questions/485512/subgroups-of-a-direct-product

The gist of it, is that subgroups of a direct product either look like a direct product, or are almost like a direct product in the sense that they are one after you quotient out some kind of algebraic “resonance” between the two factor groups

iirc if you tack on the extra condition that the two groups are algebraically independent (i.e. there does not exist any quotients of the two that are isomorphic), then the subgroups should all be direct products (someone verify this for me please)

Intuitively, its because algebraic independence forbids any “resonance”, so any subgroup collapses into a direct product

[grizzled spindle]

Is a finitely presented group with exponent finite? I know it is not always the case if the group is finitely generated, but maybe the condition of finitely related is enough to make it finite

[acoustic igloo]

oh yeah i missed {(0,0),(1,1)}

[toxic zephyr]

for a problem, i proved that for a list of pairwise comaximal ideals I1,...,Ik that
R=Ik+intersection_1^(k-1)Ij
but is this just like... not as much something to prove and more just like a better and more intuitive way to think about comaximal ideals?
like i was thinking of it as just each pair individually generate the ring, but like maybe it's better to think that you can complete any single ideal into the ring by just adding stuff that's in all the others? it sure as hell made proving the CRT easier at least haha
i'm not sure if i'm quite making sense...

[tough raven]

I may be giving bad advice here, but I feel like this is subsumed by CRT and is better remembered as a consequence of it (or an intermediate lemma along the way to proving it): indeed, if x ≡ 1 (mod I_k) and x ≡ 0 (mod I_1, ..., I_{k-1}), then 1 = (1-x) + x shows your comaximality.

Specifically, I feel like it shouldn't be necessary to remember any more than the following: (i) pairwise comaximal ⇒ product = intersection (ii) CRT (iii) I, J comaximal ⇒ I + J + K = R ⇒ I^k + J^l + K^m = R for any positive integers k, l, m (similarly for more ideals).

I might decide this is a terrible claim and take it back later.

If R is a local commutative ring with maximal ideal m and M an R-module such that M/mM is finitely generated, M need not be finitely generated. A counterexample is ℤ[1/p] over the localisation ℤ_{(p)}, which becomes 0 modulo (p). Can this be done without completely killing M modulo m?

More precisely, does there exist R, m, M such that M/mM is finitely generated and non-zero and M has no non-zero finitely generated quotient or at least summand? (Otherwise we can "cheat" by adding a finitely generated module to a counterexample.)

Wait, M/mM is a finitely generated quotient. No finitely generated summand then...

[south patrol]

I'm confused lol

[south patrol]

Oh okay you clarified

Well why should M have any nontrivial proper summands to begin with? That should give easy counterexamples too

[tough raven]

That's just to prevent a counterexample like ℤ[1/p] (+) ℤ/pℤ.

I'm curious if there's an example where the non-fg part is "really" responsible for the non-zero fg quotient.

[tardy hedge]

Question: what is Z[1/p]?

[tough raven]

{m/n : n is a power of p}.

(inside ℚ, say)

[south patrol]

Localisation at element p is the general thing

[tardy hedge]

So Z[1/p] is the smallest ring containing Z and 1/p?

[south patrol]

Smallest subring of Q yes

[tardy hedge]

Thx

[south patrol]

That is in general what the square brackets means ig

But ye you can do this even when you can't talk about Q

Like given a commutatve ring A and an element f you can formally invert f to get A[1/f]

And there is a natural map A -> A[1/f] which may not be injective (indeed we can have A[1/f] = 0 even if A isn't 0)

[tough raven]

This is all true, but I only really need its module structure, specifically because it has the presentation <a0, a1, ... | an = p a_{n+1}> that ensures each generator is killed modulo p.

[tardy hedge]

Is A[1/f] the ring of fractions when taking the multiplicatively closed subset to be the {f^n}

[tropic obsidian]

if u \in End(M) for a module M then if im(u) \cap \ker(u) = 0 then must im(u) + ker(u) = M?

wait nevermind its not true

[tardy hedge]

^dat shit had me fucked up

But im also high rn

[coral spindle]

peel off your skin

[tardy hedge]

WAT

[south patrol]

I guess this is false even just over fields

with like the right shift on k^N which is injective but not surjective

Similarly for multiplication by 2 on Z

For the same reason lol

[tropic obsidian]

true for fin dim vector spaces, thought it would be the same

but I guess modules suck

[coral spindle]

If you assume the module is Noetherian you might get something nice

[tough raven]

So these conditions are equivalent to ker(u^2) = ker(u) and im(u) = im(u^2).

[tough raven]

Ah, of course.

[south patrol]

Well even just oo dim vector spaces

[tough raven]

It is true for finite-length modules, because len(ker(u)) + len(im(u)) = len(ker(u^2)) + len(im(u^2)) = len(M).

[tropic obsidian]

it's true for inf dim vector spacse?

[south patrol]

Ig no cause of the multiplication by 2 on Z example

[south patrol]

No, I gave a counterexample

[tropic obsidian]

ah yeah makes sense

[coral spindle]

No that's precisely what potato is saying: the finite-dimensionality is what's making it nice

[south patrol]

And indeed this means you can do this whenever you don't have a field (or 0)

[coral spindle]

And Raghuram has identified a suitable generalisation to modules

so it's not modules that are nasty, it's infinite things

[tough raven]

It might be true for Artinian modules...

[coral spindle]

I could believe that

[tropic obsidian]

yeah that sounds right

[tardy hedge]

Woah you guys are acting like mathematicians rn

[tough raven]

Away from infinity and before!

[tardy hedge]

Are yall mathematicians r smth

[coral spindle]

Did you peel off your skin yet?

Doesn't it feel uncomfortable?

Just take it off

[tardy hedge]

No i will not do such a thing

[coral spindle]

Hope you're having a good trip

[tardy hedge]

Chat went silent

😂

[south patrol]

Giving a maths talk where my advisor is in the audience

Fear

[coral spindle]

You'll do great

[south patrol]

Thankss I hope it'll be ok

[coral spindle]

Just practise diligently

like some kind of.... clerk perhaps. some kind of diligent one.

[south patrol]

I have learnt by now to always prepare for pedantry by writing footnotes in my notes lol

Not from him but in general

[tough raven]

If A is a Dedekind domain with field of fractions A, K its field of fractions, L/K finite Galois and B the integral closure of A in L; and p is a prime of A, is it true that when we base-change to A_p, B becomes a product over the primes lying above p?

[tough raven]

Interesting; do you experience people nitpicking a lot in talks?

[south patrol]

Well this is a bit more like a formal reading group set up and people often do this with grad students ig

Like more nitpicky than they would be if it were somebody from another uni or whatever

[tough raven]

Ah, OK.

[tough raven]

I see.

[south patrol]

Idk how one is supposed to do this for maths talks aha

[coral spindle]

How do you mean?

[south patrol]

I mostly just ensure I write up smth nice beforehand so I have good notes to go from

[coral spindle]

You really just practise the talk until you're fluent

You shouldn't have to check your notes every five seconds, for example

[south patrol]

But idk if people like go over it a ton and memorise bits or whatever lol

Yeah sure

It may be too late for me now anyway lmao

[tropic obsidian]

is splitting presered by direct limits?

[tough raven]

Wait, everything is a subring of L so there is no chance of this.

[south patrol]

Splitting is preserved by any functor whatsoever

[tardy hedge]

I didnt know that^

[tropic obsidian]

that's op

[tardy hedge]

Ong

[south patrol]

Well okay for an appropriate notion of splitting

retracts are

[tough raven]

It's just that gf = 1 ⇒ F(g) F(f) = 1...

[south patrol]

That functors preserve retracts is a very useful fact

[tardy hedge]

Ohh bruh

[tough raven]

• an ses splits iff some section-retraction pair exists

/ that a direct sum is a coproduct IG

[tardy hedge]

When does one learn that kind of math?

[tropic obsidian]

could you elaborate why is this equiv to splitting?

[south patrol]

Tbh this is mostly just smth you will learn at some point doing algebra or whatever lol

[tough raven]

Sure, as soon as I know which kind you're talking about.

[south patrol]

Though not in undergrad prolly lol

[tardy hedge]

I think ive been kind of bored with galois theory and number theory rn

[tough raven]

can't relate in my (5th? 15th?) number theory arc rn

[tardy hedge]

Cuz for an ses to split u need fg = 1

[thorn jay]

you're just constantly travelling along a tiiiny arc segment of that arc

[south patrol]

Well more specifically ig look at the splitting lemma

[tropic obsidian]

ohh

[tardy hedge]

Yeah number theory is a bit interesting i guess. I think we are only now finally getting to the more interesting bits with fractional ideals or class number things i guess

Its an intro class so idk much

[tough raven]

Ah a class

[tardy hedge]

Yea lol

[tropic obsidian]

sections get sent to sections and you use one of th equivalent def. for splitting?

[thorn jay]

which of the like 4 definitions is this referring to?

[tough raven]

The one with lectures and maybe assignments and/or exams.

[thorn jay]

oh yeah

[tropic obsidian]

sorry, but what is the other way kind of splitting in which what I said is not true? I am prob missing something very basic

[thorn jay]

groups and modules have different notions of splitting, in a sense

[tardy hedge]

Why?

[tough raven]

It's true for all of them, but you wanted me to explain what "splitting" has to do with pairs (g, f) satisfying gf = 1, and that depends on which kind you're talking about.

[rotund aurora]

are you familiar with this?

not sure if that's what you are asking

[tough raven]

Yep. I was wondering about just localising.

[thorn jay]

If A, B, C are just regular groups, then r having a section need not imply B = AxC or q having a retraction like for modules

[rotund aurora]

ah, ic ic

[thorn jay]

but then the sequence is still called split

[tough raven]

OK actually I've only seen this for the fields, let me think for a minute if the ring version follows obviously.

[thorn jay]

WHY does it use both the big and small \mathcal{O}

[tough raven]

Because one is small and one is bigger.

🥱

[thorn jay]

I

[rotund aurora]

Neukirch's taste is incomprehensible to mere mortals

[thorn jay]

I thought my notation was bad /j

[tardy hedge]

r having a section means there exist a p map back into B with ..

[thorn jay]

with rp = id_C

[tardy hedge]

Oh so they differ by something in the kernel

[thorn jay]

it means that r fixes the image of p in B

i.e. every element can be represented as a pair (a, c) where a in A, c in C, due to the sequence also being exact

[tardy hedge]

Ok im tryna think about it

[thorn jay]

multiplication then is of the form
(a, c) * (a', c') = (a \phi_c(a') , c c')
for some automorphism \phi_c of A

[tardy hedge]

Grad school is too quick for me, i wish i could spend more time thinking about earlier concepts

[tough raven]

OK, so O is free, so we can view the LHS as sitting inside the tensor product with \hat{K} = K (⨯)_K \hat{K}, and since O (⨯)_o K = L, that's L (⨯) \hat{K}. Of course the RHS sits inside \prod \hat{L}_w. Using a basis of O, the LHS is elements whose components lie in \hat{o}; similarly for the RHS upon choosing bases of \hat{O}_w over \hat{o}. Now if we can just see that the change of basis is defined over \hat{o}...

Maybe one just has to redo the proof with fields.

[rotund aurora]

did you do undergrad in math?

[tardy hedge]

Yeah but i didnt get into theory much

Wouldve been cool if i did

I did a lot of applied stuff back then and also didnt know how much i liked algebra and other further math topics

All i knew was calculus / analysis type of math and so i thought i just didnt like math too much

[rotund aurora]

ah, interesting

[tardy hedge]

Because i didnt take any algebra besides some theory lin alg course that , looking back now, i COMPLETELY didnt understand yet somehow got an A+ 😂

Makes no sense

[tawdry venture]

I already proved part a. Is the proof for part b just stating that all ideals are left ideals?

[thorn jay]

No, because you've proven that Ann M is an ideal

But (b) asks about Ann K for any subset K of R

[tawdry venture]

Ah ok thank you!

[tardy hedge]

were your guys's math programs in undergrad really fking hard? Because the school i went to for undergrad made their pure math courses so damn hard it scared me away from ever doing it

averages of like 50-60% when all the students there are really good students

[tough raven]

Is it true that any non-zero prime ideal of an integral domain contains a minimal non-zero prime ideal?

[tough raven]

Probably higher than 50-60 on average (maybe 75? 70?), but it was pretty tough.

[ivory ore]

for (i) may i get any hint

[ivory ore]

never went to school 🦊

[tardy hedge]

Thats like trivially true for pids right

[tough raven]

Yes, or more generally any 1-dimensional integral domain.

[tardy hedge]

1-dimensional?

Ohh

I see

[tough raven]

All non-zero prime ideals are minimal prime ideals.

[tardy hedge]

Krull dimension

?

Whats an example of a ring like that thats not a pid?

[tough raven]

Dedekind domains

Probably some UFDs that I can't come up with

This is a bit esoteric, but valuation rings with value group dense in ℝ

[tardy hedge]

That is over my head atm lol

[tough raven]

Roughly speaking, anything like k[X_1, ..., X_n]/(f_1, ..., f_{n-1}) (more intuitively, the solution set to the polynomial equations should be 1-dimensional).

Hmm, I really don't have any simple example that isn't basically "Dedekind domain" though.

[tough raven]

Even more generally, if every prime ideal p has a finite height (maximum length n of a chain of prime ideals p_0 ⊊ ... ⊊ p_n = p) such as if the ring is Noetherian, or more generally if there are no infinite descending chains of prime ideals. (Because you just keep shrinking the prime ideal until you can't any more.)

Also, any UFD: any prime ideal is generated by irreducible elements, so P ≠ 0 ⇒ P contains an irreducible x ⇒ (x) ⊆ P is a minimal non-zero prime ideal.

[tardy hedge]

Nice 👍

[tardy hedge]

Why are prime ideals generated by irreducible elements?

Oh i think i got it

[tough raven]

If x ∈ P, one of the irreducible factors p of x has to lie in P, and x ∈ (p) ⊆ P.

[tardy hedge]

P has a finite generating set of irreducible elements?

Oh is it prime ideals only contain irreducible elements

That makes sense

[tough raven]

Not necessarily finite (e.g. the ideal generated by all the variables in a polynomial ring in infinitely many variables).

[tough raven]

Well, no: (2) contains 4, for example. Rather that if it's prime, I can find some generating set which is just irreducible elements.

[tardy hedge]

Oh

And (x) is a prime ideal because x is irreducible? Irreducible element always implies prime element?

[tough raven]

For a UFD specifically.

Indeed, it's false in a Dedekind domain that is not a UFD. For example, in ℤ[sqrt(-5)], 6 = 2⨯3 = (1-sqrt(-5))⨯(1+sqrt(-5)) and you can show that all factors are irreducible, but none of them divides the other. So eg (1+sqrt(-5)) divides 2⨯3 but neither 2 nor 3.

Is it true that if R ⊆ S is an inclusion of integral domains such that S is finitely generated over R, and R is a Jacobson ring, then R and S have the same Krull dimension?

[next obsidian]

No

[next obsidian]

Do you mean finitely generated as an algebra?

If so just look at like Z < Z[x]

[tough raven]

Right, that was dumb.

So it's only closed points which pull back to closed points...

[tough raven]

What about dim S ≥ dim R?

[tough raven]

Or rather: is it possible to reduce the dimension of a Jacobson integral domain by a localisation inverting a single (or finitely many) elements?

[crystal vale]

i mean there is always 0 in h(R/p) \cap f(M1)

[toxic zephyr]

let s be in ann(m) (which is nontrivial) for m nonzero in an R module M. let S={1,s,s^2,...} which is a multiplicative system. then we can define the localization S^-1M right? and in the localization, and for any x in <m>, we will have x/s'=0/s' for any s'? meaning that we can have a nonzero/s'=0/s' even if 0 is not in S?
so in general, if we have something in S that lies in a nontrivial annihilator of some of M, then that element will be... equal(?)/congruent to 0 in the localization?

am I right?

[tough raven]

Yep.

To be precise, define (0 : S)_M := {m : sm = 0 for some s in S} (this is a submodule iff S is multiplicative). Then the kernel of the natural map M → S^{-1}M: m ↦ m/1 is precisely (0 : S)_M.

This is the reason why equality in a localisation is defined by m/s = n/t ⇔ u(mt-ns) = 0 for some u in S --- you need to kill anything if it annihilates/is annihilated by u, or u can't be invertible.

[sturdy spear]

My exams are over, now i will soon start AA (one of the subjects I am excited to start)

[tough raven]

As the special case of this where R is an integral domain and S = R \ {0}, the kernel of M → S^{-1}M is precisely the torsion submodule of M. S^{-1}M kills the torsion submodule of M and turns the quotient into a vector space over S^{-1}R (the field of fractions of R).

[crystal vale]

I don't get how N' and N has SFR?

I got it

[violet spade]

Hi, I've been trying this problem for a while, but I can't seem to find where to start. Could someone help?😃

[coral spindle]

Hint: I write p(x) = x^2 + x + 1. Note that (x-1)p(x) = x^3 - 1, so the roots of p(x) are primitive 3rd roots of unity.

[violet spade]

I know that the roots are equal to $e^{2i\pi/3}$ and $e^{4i\pi/3}$. But what do i do with this information? @coral spindle

[cloud walrus]

joel

[coral spindle]

No, this is not true

We are not working in C.

There is no meaning to those symbols in F_p.

I am saying that to be a root of p(x), x must be a primitive 3rd root of unity

Hopefully you know when the quotient F[x]/( p(x) ) is a field based on a commonly understood property of polynomials. So use this property and the observation above.

[violet spade]

rightt so for example $X^2+X+1$ is reducible in $F_3[X]$.

[cloud walrus]

joel

[violet spade]

So then the quotient is not a field

by chinese remainder theorem right

But what about for example $F_{11}[X]$? How do we prove that this is irreducible in there?

[cloud walrus]

joel

[coral spindle]

Why

[violet spade]

Because $X^2+X+1$ has a zero for $\alpha=1$

[cloud walrus]

joel

[coral spindle]

What is alpha?

You mean X = 1

[violet spade]

yeah sure

just evaluate

[coral spindle]

So when is X^2 + X + 1 irreducible

[violet spade]

when it has no zeros

in F_p

[coral spindle]

And what did I say the zeroes of the polynomial were? I did implicitly assume that p =/= 3, which is a case you have dealt with now.

[violet spade]

well the roots are primitive 3rd roots of unity

that's what you siad

said

So $X^3=1$

[cloud walrus]

joel

[coral spindle]

No, that's merely a 3rd root of unity, not a primitive one

The good news is there's only one 3rd root of unity that isn't primitive in any field

But I digress

So when is the polynomial irreducible in F_p?

[violet spade]

When X^3 is not 1?

[coral spindle]

You don't seem to be connecting the dots here so I will just say it

p(x) is irreducible in F_p (where p =/= 3) iff there does not exist a primitive 3rd root of unity in F_p.

Using your knowledge about the group of units of F_p will allow you to solve this now

[delicate orchid]

holy hecking chonkers \sqrt{-3} \not \in F_p^\times le quadradic reciprocity has arrived

[violet spade]

?!

[violet spade]

So no element of order 3?

[violet spade]

this was for the other person

[coral spindle]

That is what a primitive 3rd root of unity is, yes

[delicate orchid]

Boytjie's method is far better but this is immediately what came into my mind:
X^2+X+1 factorises if and only if the discriminant, -3, is a square mod p. Apply quadratic reciprocity (and multiplicativity of Legendere symbols) to get (-3 | p) = (p | 3). The only non-zero quadratic residue mod 3 is 1, hence X^2+X+1 is reducible only when p = 1 mod 3.

[coral spindle]

This truly is a wholesome chungus quadratic reciprocity moment

[delicate orchid]

as was claimed

[coral spindle]

Did you know it has 99999billion proofs/?!?!?!

[delicate orchid]

I once had to work out precisely when X^2+Y^2, X^2+XY+Y^2, X^2+3XY+Y^2 were all irreducible mod p there was so much fuckin reciprocity bullshit

[coral spindle]

This doesn't sound fun

[delicate orchid]

p = 13 is the smallest case there's your hint

[elfin wraith]

Never work out when things are irreducible, the only acceptable proof is “tried to find factors, couldn’t”

[delicate orchid]

it was essentially the same proof I just gave but for -4, -3, and 5 respectively and then you chinese remainder theorem'd it all together. It wasn't too bad just a bit tedious

[coral spindle]

Tedium repels me

[valid night]

Can someone give me a hint on what I should do to prove that $X^4 +2$ is irreducible in $\mathbb{F}_{5^3}[X]$ please

[cloud walrus]

Kroros

[steep bramble]

heyo

is there like a rule to create these commuative diagram? like why is there a dashed aroow?

just curious

[thorn jay]

Dashed arrow means "there exists"

And often also "there exists a unique"

[steep bramble]

where can i learn more

[thorn jay]

By learning category theory

[steep bramble]

without getting into catregory

theory

alr

i was about to

[thorn jay]

Well it's literally category theory

[steep bramble]

i saw it in like some grp theory note so

;-;

[thorn jay]

Category theory is applied everywhere

For a non abstact-nonsense algebra focused book I recommend Aluffi's Algebra Chapter zero

[steep bramble]

for category theory?

[thorn jay]

It doesn't do pure category theory, rather how categorical thinking and is applied in abstract algebra

[thorn jay]

In a sense, I assume you're more interested in how it's used in algebra

This book is great for that

[steep bramble]

any books for category theory?

recommandation

i know like 0 stuff ;-;

[thorn jay]

For pure category theory I have been recommended "category theory in context"

[steep bramble]

alright

thanks for the help

[thorn jay]

👍

[crystal vale]

anyone know shorter proof of 4.13?

[sturdy spear]

what book is that

I think i should make a thread for D&F

[crystal vale]

you should

[crystal vale]

it is not book it is just online stuff

[sturdy spear]

i see, seems interesting

is it commutative alg?

[crystal vale]

no

[sturdy spear]

oh

[crystal vale]

yes

[sturdy spear]

cool

[crystal vale]

i think #1272080242004983819 is not active

[sturdy spear]

Afzal's Abstract Algebra thread

[sturdy spear]

yeah, prob bcz neam will be busy

[rocky cloak]

Because 4 and 3 are relatively prime it suffices to check that it's irreducible over F5. This you can do quite easily by brute force or by noticing that 2 is a primitive root modulo 5

[valid night]

I see, thank you

[midnight wolf]

what was a field again I forgor

[pearl fog]

A ring but everything except 0 gets an inverse

[midnight wolf]

I see

so a groupe with multiple binary operations and every element except the identity gets an inverse

[coral spindle]

We usually require commutativity too

[tardy hedge]

where do division rings come up? i havent seen it before

i mean tbf i barely seen any noncommutative ring theory so

[coral spindle]

Endomorphism rings of simple modules

[tardy hedge]

simple modules have no proper submodules right

[coral spindle]

Yes

[pearl fog]

(d) My approach(perhaps anyone's approach) is breaking down LHS into $[F(\alpha_1, ..., \alpha_r):F(\alpha_1, ..., \alpha_{r-1})]....[F(\alpha_1, \alpha_2):F(\alpha_1)]$, I played with a few examples like $Q(2^{1/4}, 2^{1/6})$ and $Q(2^{1/4}, 3^{1/4}, 6^{1/6})$ and I think the statement is true. So $F(\alpha_i) \cap F(\alpha_j)=F$ for all $i \neq j$ should imply (I found counterexamples for intersection!=F) that minimal polynomial of $\alpha_i$ over F should still be irred over F(1, ..., i), but where do I start?

[cloud walrus]

Cro

[rocky cloak]

I'm not sure I have a good hint for this other than trying to get a feel for it through examples.

Q(2^1/3) and Q(2^1/3, third root of unity) could be good examples to consider

[pearl fog]

Is it an counterexample to that equality? cuz I think 9=3*3 over F=Q is fine. I mean like is it some linear algebra property? What property?(I didn't take department of math linear algebra last year not familiar enough, guess is something like linear dependent over larger field is still linear dependent over smaller field according to stackexchange)

https://math.stackexchange.com/questions/578459/minimal-polynomial-is-invariant-under-field-extensions

[rocky cloak]

3*3 = 9 which is bigger than 6

[pearl fog]

This doesn't use intersection but still gives min poly invariant

[pearl fog]

only 6?

oh right 2^1/3 * 1^(1/3) is next root

And indeed 2^1/3 in three directions, 2^2/3 in three directions spans Q(2^1/3, 3rd root 1)

[acoustic igloo]

if a group has at most m solutions to x^m=e then it has at most phi(m) elements of order m (phi is euler's totient function)
does this sound right?

thanks 🙏

[tardy hedge]

when talking about irreducible elements in a domain we ignore 0 right

[crystal vale]

yes non-zero non-unit element

[tough raven]

Is there a relatively elementary proof from the classical Hilbert's Theorem 90 that if L/K is a finite abelian extension of fields and N(y) = 1 then y ∈ L^⨯ is a product of finitely many g(x)/x's, where each g ∈ Gal(L/K), x ∈ K^⨯?

Serre uses this in Local Fields for K a maximal unramified extension of a local field (i.e., a complete DVF with algebraically closed residue field, I think) using a Galois cohomology computation, but it seems like it should be provable directly from the cyclic case by induction (unless it's not true without the hypothesis on K).

https://doi.org/10.4153/CMB-1965-055-2

[tardy hedge]

Elements here have inverses because the modules are simple, and is multiplication in endomorphism rings just usually not commutative or?

[coral spindle]

Yes and yes

multiplication in endomorphism rings is composition

[thorn jay]

I cannot imagine there being many commutative endomorphism rings

Is there a known criterion?

(Nontrivial criterion i should add. The module being trivial is not a useful criterion)

[tardy hedge]

Man idk why im being stupid, because the modules are simple the maps are all injections but are they surjective as well?

[thorn jay]

im f < M
So im f = 0 or im f = M

[tardy hedge]

Ah shit

Yea

[thorn jay]

I.e. f is either 0 or a bijection

[tardy hedge]

Thanks

[thorn jay]

I.e. division ring

[tardy hedge]

Nice

[thorn jay]

Pesky 0

[next obsidian]

Binging true

Big if true*

[frigid epoch]

What are some good example classes for simple modules

[thorn jay]

cyclic abelian groups

[frigid epoch]

F over F a field ofc

[thorn jay]

prime*

[frigid epoch]

Ah right

In these cases the endomorphism rings are commutative though

Annoyingly

[thorn jay]

any simply module must be cyclic, clearly

so isomorphic to a quotient R/I

submodules of R/I are the ideals in the interval [I, R] by the correspondence theorem

so if R/I is simple then [I, R] is the 2-element lattice

i.e. I must be a maximal ideal

there you go, classification of simple modules using UA

[frigid epoch]

Why is this clear

Oh cause otherwise

[thorn jay]

suppose M is not generated by a single element
let m =/= 0 in M
=> <m> =/= M and <m> < M
contradiction

[frigid epoch]

You'd have two different generators a, b spanning two different submodules

Yeah that's the better way to phrase it

I meant "linearly independent", I suppose!

[tardy hedge]

Clearly? That was a whole exercise in dummit and foote! Hehe

But ya i remember that one was easy

lol

[thorn jay]

it was clear to me qwq

[frigid epoch]

And why does this hold

[thorn jay]

because R is the free R-module on one generator

[frigid epoch]

Oh

[thorn jay]

and if a module M is generated by the set X, then M is isomorphic to a quotient of the direct sum of R indexed by X

this is a general UA fact, too!

although usually phrased in terms of free algebras but who caares it's all the same

[coral spindle]

When you look at finite groups' reps over char 0 fields that aren't algebraically closed, it happens much less often than one would perhaps expect. And ofc this just doesn't happen in positive characterstic since all finite division rings are fields

[frigid epoch]

So all simple modules over commutative rings are fields,,,

[coral spindle]

Uh no?

M_2(R) is a simple M_2(R)-module

[thorn jay]

a simple module over a simple ring R is either R or 0

[frigid epoch]

Dammit typo

I just want the interesting endomirphism division ring already

[coral spindle]

You want an example?

[tardy hedge]

Is I the kernel of R->M by r -> r*m with m the generator of M

[coral spindle]

There is a unique simple 4-dimensional $\bR Q_8$-module, and its endomorphism ring is $\mathbb H$

[cloud walrus]

$\mathbf{Boytjie}$

[thorn jay]

exactly

that map must be surjective, hence we get R/I \cong M

[thorn jay]

that's awesome

like

of course

why wouldn't it

[coral spindle]

It is quite poetic

[frigid epoch]

While I appreciate the example, the quaternions were the precise division ring I did NOT need to see lol

[thorn jay]

😭

[tardy hedge]

Hahahaha

[thorn jay]

trauma response

[tardy hedge]

I’m have a hw question about them i still havent done

[coral spindle]

Well you'll have to look away from real division algebras, naturally.

[tardy hedge]

Q5

Still dont know what the dimension of separations is referring to

[frigid epoch]

I assume you can get stuff like R<x,y> in two noncommuting invertible variables x,y

[thorn jay]

invert 1+x then

[coral spindle]

Anyway, fin-dim central division algebras can be constructed via homological methods using the Brauer group. So if you really want to just pump out a bunch of examples there are methods for doing this, but they're tedious

[thorn jay]

what's the Brauer group?

[coral spindle]

Tl;dr it classifies central simple k-algebras

And it's a group

[tardy hedge]

Whats a central algebra

[coral spindle]

A central k-algebra is a k-algebra whose centre is k

[thorn jay]

I love it when a classification of stuff has a group structure

[tardy hedge]

Ok coll

Cool

[thorn jay]

coll

[coral spindle]

E.g. Z(M_n(k)) = k

so M_n(k) is a central simple k-algebra

[tardy hedge]

n x n matrices over k?

[coral spindle]

Yes

[tardy hedge]

This is a k-algebra

?

[coral spindle]

You are aware of what I mean when I say "k-algebra" right?

[frigid epoch]

Wdym

[thorn jay]

I think it is probably one of the more famous k-algebras out there

[coral spindle]

I mean a ring with a central embedding of a given field k

[thorn jay]

besides

well

[tardy hedge]

Yeah idk how i havent seen it

[thorn jay]

k

[tardy hedge]

Yea

[coral spindle]

So a ring R with a map f : k -> Z(R)

[tardy hedge]

K embeds into the 1x1 matrices i guess right lol

[coral spindle]

So a central k-algebra is one where f is surjective

[coral spindle]

it embeds into nxn matrices

No matter what n

[tardy hedge]

Oh ok nvm yea

[coral spindle]

If you meant to say diagonal, then yes

[tardy hedge]

Yea

[frigid epoch]

When I think k-algebras I just think polynomial rings over R, and other function algebras

Holomorphic, continuous and so on

[coral spindle]

Ok

[frigid epoch]

And the silly C and H ofc

[tardy hedge]

Not silly not silly

[frigid epoch]

Well C is a quotient of a polynomial ring over R

[tardy hedge]

So what?

[coral spindle]

Every fin-dim k-algebra is

commutative ones at least

and C is indeed a field extension of R.

[frigid epoch]

Indeed

[thorn jay]

me when
k[X] is the free commutative k-algebra on generators X

[coral spindle]

Yeah

[thorn jay]

absolutely crazy

[coral spindle]

Frankly blows my tiny brains out

kaboom! you know

pink mist everywhere

[frigid epoch]

The quaternions slightly annoy me because they overshoot being algebraically closed

More square roots of -1 than necessary

[thorn jay]

what polynomial objects in the variety of noncommutative rings does to a mf

[tardy hedge]

Wat

Omg det

[rustic crown]

det went when me first saw proof of existence of alg closure.

[tardy hedge]

Like how every field has an algebraic closure)

?

[rustic crown]

you take X = {x_f : f in k[t] \ k}, i.e. a new variable for each non-constant poly

[thorn jay]

I wonder if there is such a theorem for polynomial algebras in arbitrary varieties

probably not..

[delicate orchid]

oh. my. science. it's the hecking chonkers schur index

[thorn jay]

that would be too nice

[coral spindle]

🥺 you're too good for this world

[delicate orchid]

I don't know how that interacts with the endomorphism algebra though.

I also don't particuarly care about endomorphism algebras

[coral spindle]

Well for a start the schur index is going to be the sqrt of the dim

[delicate orchid]

yeah

[frigid epoch]

It's so crazy how we call objects in algebra "algebras"

[coral spindle]

As long as I choose the right field, but let's move past that for now

[thorn jay]

wh

what

[frigid epoch]

It's like calling differential manifolds "analyses"

[thorn jay]

I know right

[frigid epoch]

Oh but topology has topologies

[thorn jay]

at least we universal algebraists have actual good reasons to call them algebras

[coral spindle]

But yeah fundamentally it doesn't tell us very mcuh about the algebra. You are right.

[delicate orchid]

very sad

[coral spindle]

Idk

We don't really care about the algebra that much

[thorn jay]

that's discrimination

[coral spindle]

If we just want to get a handle on the representations this works very nicely

Like, I can plod along and calculate all the character values

That's not hard at all

[delicate orchid]

why do we care about endomorphism algebras again

[coral spindle]

Well I mean

I don't?

[delicate orchid]

are they the blocks

[coral spindle]

Oh yes

yeah they are

[delicate orchid]

yeah that's it then

I knew I gave a shit for a reason

[thorn jay]

wdym by that

[coral spindle]

There's like, M_n(E) sitting in the group

[delicate orchid]

yurrrr

[coral spindle]

where E is the endo of some module

yeah

Many such cases!

Wew is this a wholesome chungus artin-wedderburn?

please confirm

[delicate orchid]

caring about block iso classes in char 0 is woke

[coral spindle]

It is woke

I am awakened

[delicate orchid]

maybe. Give it a solid 70%

[coral spindle]

Praise me

[thorn jay]

what are blocks here?

[delicate orchid]

blocks of the group algebra

[coral spindle]

You know how matrices can have blocks on the diagonal? Imagine a ring of matrices having blocks too

[tardy hedge]

Ong 😂

[coral spindle]

This is honestly a helpful description I swear lmao

[tardy hedge]

We forget the simple things in life sometime

Sometimes

[thorn jay]

a bit like cosets?

are they disjoint?

[frigid epoch]

And like

[coral spindle]

Uh no

[frigid epoch]

Algebra is a weird variant on the name al-khwarizmi

[coral spindle]

It's like direct product decompositions

[thorn jay]

ah okay

[delicate orchid]

a block is a primitive central idempotent e of a ring R and is also eR

[frigid epoch]

It's like calling sets and set theory

[delicate orchid]

the phrasology is goofy

[tardy hedge]

Once upon a time i was going crazy when i first learned quotient groups. I thought it was so cool

[coral spindle]

Wew Wew Wewwwwwww did you know that Galois conjugate characters have the same Schur index but they may not have the same Brauer class of endomorphism algebra?????

[tardy hedge]

I think wew was here when i was going off

In 2023

[frigid epoch]

Or no

[delicate orchid]

I'm gonna be sick

[coral spindle]

So fucked up right? I love it <3

My little Schur indices are menaces

[delicate orchid]

all fields are algebraically closed UNFORTUNATELY for you so

or are Q... you can have Q

[coral spindle]

Q_p

bitch

[delicate orchid]

no

[coral spindle]

Yes

[delicate orchid]

no

[coral spindle]

Navarro says Yes oxoxoxoxoxoxox

[delicate orchid]

who?

[coral spindle]

[tardy hedge]

U guys are both british pls dont argue ❤️

[coral spindle]

He had his 60th bday conference recently

[tardy hedge]

One Love ❤️

[delicate orchid]

fuck I was going to type "cares" but I can't now

[thorn jay]

do they have their british license?

[delicate orchid]

I did hear about this conference

[coral spindle]

You presumably didn't go

[delicate orchid]

I don't go places

[coral spindle]

I may have........

[thorn jay]

that's not how you fucking spell license idiot

[delicate orchid]

thanks for the invite

[thorn jay]

what are you gonna do, travel back in time?

[delicate orchid]

I would've probably been out of my depth anyway

[coral spindle]

Oh Wew it was actually really good. And now I feel bad

[delicate orchid]

don't feel bad I actually don't care

[coral spindle]

Phew

[delicate orchid]

where was it hosted

[coral spindle]

In Valencia

[delicate orchid]

oh YEAH that's why I didn't go

[coral spindle]

Moreto and Rizo presented a really cool McKay-esque conjecture based on these "picky" elements

they're very cool

[delicate orchid]

definitely cannot afford to go abroad for conferences

[delicate orchid]

I've had enough of McKay-esque conjectuers for one lifetime

[coral spindle]

OK fine >:(

I don't think they've put the paper on the arxiv yet

But it's very cool

[delicate orchid]

it does sound kind of cool if you would tell me what a picky element is

[coral spindle]

I was typing it out!

An element is picky if it lies in exactly one Sylow p-subgroup

So p-picky I suppose

And I'm not stuttering

[delicate orchid]

https://tenor.com/view/toletole-cat-lockin-gif-12000667884327844099

you have my attention

[coral spindle]

I would love to tell you more but I can't quite remember the details...

[thorn jay]

"It's p-picky, b-baka"
looks away embarrassed

[delicate orchid]

this is equivalent to the fusion class being fused by... something

automorphisms of S I think

yes

[coral spindle]

Malle gave a talk which classified the picky elements in groups of Lie type. They have a lot bc of things called regular unipotent elements, amongst other things

But yeah you should expect picky elements to have a lot of interest

[delicate orchid]

I suppose if Sylow subgroups are your local data these elements could tell you exactly WHERE locally you are

[tardy hedge]

Do people collaborate on papers here?

[delicate orchid]

it has happened

[tardy hedge]

Nice

[delicate orchid]

I think? like once

[thorn jay]

I mean, this is a handy place to meet people

[tardy hedge]

Yea^

[coral spindle]

Yes exactly!

[tardy hedge]

It would make sense to happen

[coral spindle]

This is huge in groups of Lie type

[delicate orchid]

scrumptious simply scrumptious

[coral spindle]

you get such great info

Like, induction from the Sylow is just really strong

[thorn jay]

although the server is fairly heavy on non-researchers I suppose, and given the amount of branches of math..

[delicate orchid]

AH IT IS THIS! If every element satisfies this then your Sylow must be normal, your fusion system essential rank 0 AHHHHHHHHHHHHHHH mmmmmmmmm GOOD

[thorn jay]

same

[delicate orchid]

oh my hecking chonkers it's the stable element theorem

but for the wrong mackey functor

[coral spindle]

Guys help, he's speaking Martian

[delicate orchid]

that's got me thinking now. About something unrelated I will now google

[thorn jay]

you both are speaking Martian

[coral spindle]

If/when the paper comes out I will let you know

[coral spindle]

NOOOOOoooooooooo he's being weird !!!! not me !!!!!!!!!!!

[thorn jay]

we all tell ourselves that

smh

[delicate orchid]

my goofy ahh forgetting about the Segal-Carlsson completion theorem

[delicate orchid]

I'm locked in buddy

ok I did just lock out but I've locked back in

[thorn jay]

we lock ouin

[delicate orchid]

ok I've locked back out again

[thorn jay]

😭

neurodivergent ahh

[delicate orchid]

if you coun't locking out as reading about G-spectra that is

[thorn jay]

holy shit

reminds me of a paper I didn't have the time to read

[thorn jay]

are G-spectra perhaps the sets of normal subgroups P such that for normal subgroups N, M, we have
[N, M] < P => N < P or M < P
?

[glad osprey]

Quick q: the splitting field of x^3 - 1 over Q has degree 3 right? The solution manual says the splitting field is Q(sqrt(3)i), ie. degree 2

[coral spindle]

It has degree 2

Because X^3 - 1 = (X-1)(X^2 + X + 1)

And the latter factor is irreducible

And -- tbh this is the most important bit -- a single root of the latter factor genuinely does generate the whole extension

[glad osprey]

aah, I was thinking x^3 - 1 was irreducible

[coral spindle]

You may want to have a quick look at the cyclotomic polynomials

These will explain the situation

[glad osprey]

I will, thanks

[coral spindle]

In general X^n - 1 will have a splitting field of degree phi(n) over Q, for example

but yeah

That's the search term

[thorn jay]

qwq they are not

[tardy hedge]

It does so here because its quadratic right

[thorn jay]

another time UA terminology is bound to be inferior because of obscurity

[tardy hedge]

Or it generates the whole splitting field because thats the only irreducible factor of the polynomial

[delicate orchid]

I'm basically trying to understand why this is true

[glad osprey]

I guess you could observe that 1 is already a root of x^3 - 1, so you just need to adjoin an element of degree 2 to get the splitting field?

[coral spindle]

Sure that's true. It works because it's quadratic. But the thing is that it works in general

For any prime p, x^p - 1 factors in a similar way

[tardy hedge]

Ya thats assuming the remaining factor is irreducible over base field

[thorn jay]

haha I understand nothing about this besides

• functor
• finite
• group
• G-Set
• category

[tardy hedge]

Ya the point u were making is that for any x^n-1, adjoining a root to x^(n-1) + … + 1 factor generates all roots ?

[delicate orchid]

surely you understand "equivalent"

[coral spindle]

Definitely not for all n!

But for primes

[thorn jay]

I was too lazy to write that down

[tardy hedge]

Ah ok yea

In my number theory class we were looking at cyclotomic fields but we only kept it to primes

[coral spindle]

But yeah, the roots of X^{p-1} + X^{p-2} + ... + 1 are the primitive pth roots of unity, and it's just not terribly hard to see that one such thing generates all of them

[tardy hedge]

Ya its cyclic group type shi

[delicate orchid]

I'm so glad all of the coefficients of cyclotomic polynomails are \pm 1. I've checked the first 104 and it seems to hold!

[thorn jay]

the devilish large number:

[delicate orchid]

the devilish 105:

[tardy hedge]

Whats \pm 1

[delicate orchid]

plus minus 1

[tardy hedge]

Oh ok

[thorn jay]

ah okay I presumed it was 105

but out of fear of being wrong I just said large number

[delicate orchid]

I don't remember why it's 105

[tardy hedge]

For 105 its not ? Lol

[coral spindle]

$\pm$

[delicate orchid]

let me google

[cloud walrus]

$\mathbf{Boytjie}$

[thorn jay]

lovely 3 * 5 * 7

[delicate orchid]

yeah it says it's because it's the product of 3 distinct odd primes

ok? whatever

[thorn jay]

😭 such an arbitrary fucking reason

"why don't your congruences 3-permute?"
"yeah cuz these random ahh fucky wucky maps aren't surjective"

[coral spindle]

Well that is the smallest such thing

It's not a coincidence either that the smallest non Abelian group is of order the smallest squarefree composite

[delicate orchid]

well yeah I think it's clear any multiple of 105 also fucks up

[delicate orchid]

https://tenor.com/view/huh-cat-pissi-twitch-7tv-gif-7223094372649290081

[coral spindle]

I love the number 6. It is my favourite number. You already know why.

[delicate orchid]

6? Group orders have to be equal to the power of a prime buddy

[coral spindle]

Quiet.

[thorn jay]

Latin squares of order up to 6 but not including 6 are nice

[delicate orchid]

are latin squares just cayley tables

[coral spindle]

Uh almost

[thorn jay]

of quasigroups, yes

[delicate orchid]

yeah kind of

[coral spindle]

They're a bit more free

[thorn jay]

I love Latin squares

[delicate orchid]

quasigroupoids

quasigroup object in a quasigroupoid

n-quasigroups

there you go. Now go write a paper about them

[thorn jay]

did you know that a reduced Latin square is the cayley table of a group iff the stabiliser of the group action of permuting the rows and columns has the same order as the latin square, and the Latin square is then a Cayley table of it's stabiliser?

that also immediately proves Cayley's theorem

[delicate orchid]

wait hold my chungus lemme lock in

what does reduced mean

[thorn jay]

first row and column the same

[delicate orchid]

I see

[thorn jay]

usually
1, 2, ..., n

[delicate orchid]

order is size?

like, side length

[thorn jay]

the n there, yes

[delicate orchid]

oh yeah

that's nifty

[thorn jay]

it's so cool

[delicate orchid]

regular representation moment

[thorn jay]

you can go even further by looking at the isotopy class of such a Latin square

and then you have a nice formula for the size of that class

anyhow, that was my talk on Latin squares

oh yeah, if a Latin square is symmetric along its main diagonal, then its stabiliser of the abovementioned group action must be abelian

this has to do with said stabiliser being one of the nuclei of the underlying quasigroup (if L is reduced), which must be a subsquare, hence abelian if L is symmetric

then you can do a symbol transformation and yada yada I actually have a way cooler proof

involves giving a closed formula for the elements in the stabiliser

[delicate orchid]

no idea what a nuclei of a quasigroup is but I'll take your word for it

but I believe the symmetric => abelian thing

[thorn jay]

set of elements that associate with the rest of the elements in some way

[delicate orchid]

like a jacobi relation or

[thorn jay]

no

left nucleus:
elements x such that for all a, b
x * (a * b) = (x * a) * b

middle nucleus and right nuclus similarly

[delicate orchid]

I see

[thorn jay]

okay so you know how that the underlying group of that group action (I'll call it the transformation group action) is S_n x S_n?
cuz you permute the rows and columns and that's two independent permutations and such

if L is symmetric, then the stabiliser must be elements of the form
$$(\sigma, \sigma^{-1})$$

[cloud walrus]

.enpeace_music

[thorn jay]

and then of course it must be abelian

[delicate orchid]

so it's a subgroup of the twisted diagonal embedding of S_n into S_n ^2 but how does that mean it's abelian

am I le stupid

[thorn jay]

cuz stabiliser is closed under composition

so

$$(\sigma_1, \sigma_1^{-1}) \circ (\sigma_2, \sigma_2^{-1}) = (\sigma_3, \sigma_3^{-1})$$
for some permutation $\sigma_3$

[delicate orchid]

oh ok yeah

[thorn jay]

which is like, genuinely one of the coolest proofs ever imo

[delicate orchid]

if you write it like that sure but like, (1,1^-1)(2, 2^-1) = (12, 1^-12^-1) = (12, (21)^-1)

[thorn jay]

but maybe I'm silly

[delicate orchid]

this notation is confusing me

[thorn jay]

what's confusing about it

[cloud walrus]

.enpeace_music

[delicate orchid]

that just isn't true

that's not the definition of a direct product of groups?

[thorn jay]

no but it must be true in the stabiliser

as every element of the stabiliser is of that form

[delicate orchid]

sure

[thorn jay]

hold on, let me restate this

[delicate orchid]

I got that

[thorn jay]

Let $T^L$ denote the stabiliser of a symmetric square $L$.
Then let $(\sigma_1, \sigma_1^{-1}), (\sigma_2, \sigma_2^{-1}) \in T^L$.
As every element in the stabiliser must be skew-symmetric like that, there must be a permutation $\sigma_3$ such that

[cloud walrus]

.enpeace_music

[thorn jay]

$$(\sigma_1, \sigma_1^{-1}) \circ (\sigma_2, \sigma_2^{-1}) = (\sigma_3, \sigma_3^{-1})$$

[cloud walrus]

.enpeace_music

[delicate orchid]

so we've shown that 1^-12^-1 = (12)^-1 which is of course equivalent to it being abelian

[thorn jay]

ye

[delicate orchid]

my complaint was about a minor confusion in the notation more so than a conceptual misunderstanding

this proof is nice

[thorn jay]

I'm sowwy I promise the actual proof I wrote down is actually good

[delicate orchid]

BUUUUUURRRRRRRPPPPPPPPPPPPPPP

no need to apologise to me I AM a moron after all

[thorn jay]

same ngl

[thorn jay]

thank you! probably one of the cooler proofs I've written

although perhaps not that substantial of a result

[thorn jay]

except maybe the proof of some Noetherianness condition related stuff for congruence relations

[tough raven]

Well, the endomorphisms of the R-module R are R, which can be commutative. More generally, the endomorphism ring of a simple representation of a finite-dimensional algebra over a field F is a finite-dimensional division algebra over F. This could be commutative and if F is algebraically closed (or e.g. a finite field) it has to be commutative.

If you have a direct sum M1 (+) ... (+) Mn of simple modules which are not isomorphic to each other, then the endomorphism ring is the product of endomorphism rings of the individual Mi's, so it has a chance to be commutative.

Otherwise I can't think of a lot of examples.

[tough raven]

But examples where F is an algebraically closed field (usually taken to be ℂ) and the endomorphism rings of simple modules are F (either because they're finite-dimensional or for other reaons) are quite common.

[tough raven]

Oh, that sounds interesting. Has this been studied non-trivially outside of ℝ (or real-closed fields)? I know there is real/complex/quaternionic type in that case, but never found any similar structural results if you wanted to look at e.g. representations over ℚ(1^1/257) or something.

[coral spindle]

Yes it has been studied a lot!

Well

There's a lot that is still mysterious

But the key term is the Schur index of a character

[tough raven]

Alright, that looks interesting. It seems to have lots of surprising properties.

Thanks!

[coral spindle]

Do you know Isaacs' book? Chapter 10 is what you want. That and chapter 9 if you're not familiar with changing the field from C

Isaacs' Character Theory

Christ now that I say that I realise it doesn't scratch the surface

Unger's paper here has a good survey of various things you need to know about the Schur index

[noble adder]

i'm trying to show that if F extends E and E extends K, then [F:K] = [F:E][E:K]
i've tried using the bases of these spaces, but i can't seem to figure anything out from that

is there another approach i should take

[tough raven]

proceeds to spend the next 10 years working with 2000 groups where this is just all elements

[tough raven]

I do not at all.

[tough raven]

Wow, Discord has Markdown support for links?

[sly crescent]

This is all elements in all groups

Because, by definition, groups are associative

[tough raven]

Yes, that was my point.

[thorn jay]

Quasigroup theory is very silly

Indeed, we cope by mainly studying axioms and consequences of those axioms

[crystal vale]

anyone can help me to go through this

i can share my doubt

I think lang can help me here

[wraith cargo]

What is a FR

[glad osprey]

Using the bases is the correct approach; try to write a basis for F over K in terms of the basis of F over E and E over K

[next obsidian]

Finite resolution I think k

This is about finite free resolutions I think

[wraith cargo]

Ah interesting

[tough raven]

Hmmmm

[crystal vale]

Yes p\subset Q'_i \cap R

[crystal vale]

I understand up to that d lies in every associated prime of N. After that I don't get it

i don't get it how every one of the factor module in filtration of N admits a FR?

[crystal vale]

i got it

[crystal vale]

i don't get it how (f) is isomorphic to R_0[x] as R[x] module?

f maps to 1 + p[x], right?

[arctic trail]

Does SL(n,Z) surject into SL(n,Z/p) ?

for n = 2 this is simple to prove

[tough raven]

Qi' is the preimage of an ideal of R0[X], so it contains the preimage of P0 ie P. Hence Qi' ∩ R contains p. As they have shown, it also contains d ∉ p, so the inclusion is strict.

p was chosen to be maximal among intersections P ∩ R for P such that R[x]/P does not have finite FR. So Qi' ∩ R ⊋ p means R[x]/Qi' = R0[x]/Qi has finite FR.

[arctic trail]

wait my argument was bad nvm

No nevermind it wasn't

Take a matrix of the form in SL(3,p):
[a b c]
[d e f]
[0 0 g]
Take g not congruent to +-1 (mod p).

[tough raven]

I think the keywords to look up for this kind of thing are weak/strong approximation for algebraic groups (SLn is one such algebraic group).

[arctic trail]

this cannot be the image of a matrix in SL(3,Z) by the reduction homomorphism

[tough raven]

E.g. by the third sentence here https://en.m.wikipedia.org/wiki/Approximation_in_algebraic_groups, and in https://mathoverflow.net/questions/123009/simply-connected-simple-algebraic-groups someone seems to think it's well-known that SL_n is "simply connected" (whatever that means).

So the answer ought to be yes. Of course that's not a proof, but at least you know what to expect.

[tough raven]

Hmm. Why?

Are you assuming the lift also has those 2 zero entries?

[arctic trail]

oh wait, nevermind

sorry

I guess I'm too tired haha

I don't know enough algebraic number theory to understand that

[tough raven]

https://library.slmath.org/books/Book61/files/70rapi.pdf I'm pretty sure this implies it's true for SLn

[tough raven]

Oh there are no proofs here just stringing together results from the literature

[rocky cloak]

https://surya-teja.com/2010/11/09/surjectivity-of-sl_nz-to-sl_n-z-nz/

appears to be a short proof using elementary matrices

[tough raven]

(f) is a principal ideal of R0[x], which is an integral domain, so it is isomorphic to R0[x] (as R0[x] and hence as R[x] modules) by multiplication by f.

[arctic trail]

thanks!

[crystal vale]

Yes

[crystal vale]

Is there any result which is used here?

[tough raven]

Nothing that you can't prove in two lines.

[crystal vale]

I don't get how integral property works here

[tough raven]

R0[x] is an integral domain, so multiplication by f is injective on R0[x].

[crystal vale]

Multiplication by f is injective on R0[x], you mean q(x) -> q(x)f(x) is injective map

Yes it is

And it is surjective

So R0[x] isomorphic to (f) as R_0[x] module

Correct?

[fickle trail]

I just started studying quantum mechanics for research and i have a few questions, is there anyone who can help me?

[rocky cloak]

https://dontasktoask.com/

[crystal vale]

So now I have to show it is also isomorphic as R[x] module, I think we can just extend the map

[tardy hedge]

bro never asked

[coral spindle]

You can lead a horse to water

[crystal vale]

Is it general true that if A and B are isomorphic as R-module then it is also isomorphic to Q - module if there is surjective ring homomorphism Q to R ?

[tardy hedge]

are you saying to make A and B into Q-modules by setting a*x = f(a)x sort of thing?

[crystal vale]

Yes

[south patrol]

Yes. This is "restriction of scalars along Q -> R" and that is a functor

So in particular there is this case

But this is true for any map of rings

[crystal vale]

So not need to be surjective?

[south patrol]

Yeah

[tardy hedge]

the "extension" of scalars case happens when you have injection R->S and you want to make an R-module into an S-module right?

and then you need tensor products for that?

[south patrol]

Not necessarily an injection but yes

[crystal vale]

Yes

[crystal vale]

Yes

First we make S \otimes_R M

Then we define S \otimes_R M as S - module

Right?

[tardy hedge]

yeah pre sure

im tryna remember the details of this thouh

stuff like, when can you actually embed the R-module M into an S-module

with R->S ring hom

I kind of forget these details

[south patrol]

Exactly, via the S-module structure on S

And then the "compatibility" with the original action on M is encoded in the tensor product

[south patrol]

Not rly sure what you mean by this

As in, when can you literally extend scalars without changing the underlying abelian group?

[tardy hedge]

Yea! I suppose thats the question

this was discussed in dummit and foote and i remember a lemma was like, the quotient with kernel of some inclusion map is the largest quotient that can be embedded or smth

[south patrol]

Well there is a notion of descent which you can look at

[tardy hedge]

so if M -> S(x)R M is injective R-module hom or something

[south patrol]

Oh I mean that is different

[tardy hedge]

Oh because the abelian group in S (x)R M is different?

If this m -> 1(x)m map is injective, then you can say we have embedded the Rmod M into an S-module, right? But, we havent necessarily kept the underlying abelian group the same ?

[south patrol]

Like you will basically never keep it the same

[tardy hedge]

Yeah

[tough raven]

Sure. There isn't any dedicated quantum mechanics channel though, so you might end up having to post different questions in different channels (e.g. perhaps: questions about groups or Lie algebras here or #advanced-algebra , about Hilbert spaces and operators might be #advanced-analysis , solving Schrodinger's equation for specific Hamiltonians in an ODE or PDE channel, etc.).

[tough raven]

If S is a quotient of a ring R, then for any S-modules, S-linear maps are the same as R-linear maps, S-submodules are the same as R-submodules, two S-modules are isomorphic iff they are isomorphic as R-modules, etc.

[fickle trail]

Yea ok thnx

[crystal vale]

Thanks @tough raven ❤️

Can I DM you?

[tough raven]

If it helps, it is true that you can embed M into some S-module iff this map is injective, and one sufficient but not necessary condition for it to be injective is that M is flat (in particular, free or more generally projective) and R → S is injective.

[tough raven]

OK.

[tardy hedge]

yea that helps, thanks

[sturdy spear]

wouldn't it be H is closed under *|_H?

[kind temple]

binary operations on a set are always closed on that set since they are functions X x X —> X, so having a term for that would be a bit redundant.

but substructures aren’t always invariant under binary operations. so we say that H is closed or invariant under the group operation

[thorn jay]

You mean substructures with respect to the structure of G if we disregard the magma structure?

Like a topological space / manifold etc

[kind temple]

i didn’t mean substructure in a technical sense

[thorn jay]

Right

[kind temple]

do you think it holds in a more technical sense? i am interested in what you meant by your comment on top spaces and manifolds

[thorn jay]

Well, a set can have many different structures attached to it

So, for example a lie group is a set which has a smooth manifold structure and a group structure, interacting nicely (namely the group operations are smooth maps)

A substructure of a lie group is then a subset which is both a submanifold and a subgroup

[kind temple]

ah sure

[thorn jay]

In that sense it is a substructure of the underlying manifold, which is closed under the group operations

[kind temple]

right

[thorn jay]

Can't say I'm too well-versed on Lie groups though, i know the definition but that's basically it

In (universal) algebra there's a related concept called a reduct, where you forget the some of the structure (in the form of operations)
So a module, for example, has an underlying abelian group, and that abelian group would then be a reduct of the module

But that is beside the point, kind of

[kind temple]

by respecting the substructure on G, you meant, a subset of G which has some other structure but is also a subgroup of H?

i guess i’m a bit confused there

[thorn jay]

Did i say respecting the substructure on G?

I meant that youre a substructure of G if you're a subset respecting (being closed under) the structure of G

Structure being a very handwavey concept here

[kind temple]

got it

[frigid epoch]

What are class fields?

[tough raven]

#advanced-number-theory is probably a better channel to ask this question in.

[frigid epoch]

Oh ok

[glad osprey]

Does anyone have an example of a normal extension that is not a splitting field? I know such an extension must have infinite degree

wait, just Q-bar over Q is an example I guess, since it's not a splitting field

[tardy hedge]

whats a normal extension?

[glad osprey]

an extension F <= E is normal if every irreducible polynomial in F that has a root in E splits into linear factors in E