groups-rings-fields

[dark wave]

There happen to be 6 factorizations
Any permutation on the 6 vertices defines a permutation on the factorizations
This automorphism has no analogue for any other size except 6

[cinder hill]

should you do all the exercises from dummit and foote

[rotund aurora]

was there a characterization of Hom(A^k, B) ?

where k needn't be finite

A, B abelian groups

[wraith cargo]

Is A^k here the direct sum or direct product

[rotund aurora]

direct product

I think in general there is no easy way to deal with this, apparently Hom(Z^N, Z)=bigoplus_N Z but this is not trivial

[coral spindle]

/unjerk There’s no rule, you should do as little or as much as you feel you want to.

/rejerk If u don’t then dummit will appear at your door with a ski mask and a machete. Foote will arrive with a gun

[tribal moss]

Beware of algebraists bringing gifts, though.

[south patrol]

Yeah I don't think there is a nice characterisation in general

[rotund aurora]

they define co-induced like this (here Lambda=Z[G]), where X is taken with a trivial G-action I assume. And Hom(A, B) is given the structure of a G-module, for G-modules A and B, by the rule (g·phi)(x)=g(phi(g^{-1} x)).

But then they say that Hom(Z[G], A) is co-induced for A any G-module

so what is the actual definition

I guess what you want is H^i(G, A)=0 for i>0 for co-induced A, so why should H^i(G, Hom(Z[G], A))=0 for i>0 ?

[rotund aurora]

or idk if they suddently just changed the action?

[agile burrow]

I think the point here is that you can forget the G-module structure on A and then coinduce as above. Equivalently, you restrict A to be a module over the trivial subgroup, and then coinduce from that back to a G-module.

[agile burrow]

This is Shapiro's lemma, but it's essentially an application of tensor-hom adjunction + cohomology of the trivial group vanishes

[rotund aurora]

so here what would be the G-action on A^* ?

[rotund aurora]

the version of Shapiro's lemma I know gives this G-action on the Hom sets

I suppose if X is an abelian group, you can take G'=1 and then Shapiro's lemma tells you that H^q(1, X)=H^q(G, Hom(Lambda, X)) where the action on Hom(Lambda, X) is (g · varphi)(x)=varphi(x g^{-1}), but before the G-action was on the other "side" (inside varphi)

[rotund aurora]

namely

[agile burrow]

Yeah, so this is why cohomology of coinduced modules vanishes. But just change actions to make everything consistent, the proofs shouldn't change at all since left modules are equivalent to right modules

[agile burrow]

I think it should be (x . phi)(g) = phi(xg) or something to make the embedding a G-module homomorphism. I don't have paper on me right now, but it's either a homomorphism or an antihomomorphism. If I'm wrong, then one should just change how some of the module structures are defined so that everything is a left module for the sake of convenience

[rotund aurora]

mmh ok I'll look carefully into this

[agile burrow]

But yeah, the point of that module structure on A* is obtained from forgetting the G-module structure on A and then coinducing again

[rotund aurora]

but just to be clear. If A is an arbitrary G-module and we endow B=Hom(Z[G], A) with the structure of a G-module by (g varphi)(x)=g(varphi(g^{-1} x)) it is not necessarily true that H^q(G, B)=0 for q>0, right?

[rocky cloak]

Should be (x . Phi) (g) = phi(gx)

[agile burrow]

I agree for now. I will think a little bit more about it

[rotund aurora]

To avoid confusions: varphi in Hom(Z[G], A), x in Z[G], and g in G. So g is "multiplying" varphi, and the action is specified by what g·varphi does to each x in Z[G]

[icy lake]

what are the subgroups of the dihedral group D6 as permutations of the numbers 1-6?

[rotund aurora]

just a sanity check, here in the definition of the map N, you could also define N(a)=sum s · a, right? because if (s) is a system of representatives for G/H so is (s^{-1})

[mighty kiln]

s^{-1} is a system of representatives for H\G

[rotund aurora]

that's interesting

ty

[cloud walrus]

[redacted]

[marsh scaffold]

One observation i think is we can notice that Inn(G) behaves likes G

If Z(G) is trivial

Then I don't think this leads to any fruitful direction

Actually for this problem does knowing sth about automorphism groups specifically is important?

How does conjugation work in automorphism groups

Can we like interpret them in some way

[rocky cloak]

So if f is in the center, then in particular it commutes with inner automorphisms. Hence
f(gxg^-) = g f(x) g^-
Setting x = yg, plus some fiddling gives that f(g)g^- commutes with f(y), so f(g)g^- is in the center of g

[marsh scaffold]

Ofc it would commute with the inner automorphisms too

Lol here I was thinking that this was very difficult

You're amazing, thanks

[rotund aurora]

If you take a finite group G, you can consider the sequence G, Aut(G), Aut(Aut(G)),... must this stabilize?

[rocky cloak]

I remember looking this up before. There is some interesting results of you allow transfinite iteration

https://mathoverflow.net/a/5638/157483

[topaz solar]

See a paper by Simon Thomas, JDH, etc

I know of this one offhand because it’s nifty

I mean, I dunno if there’s any real uhhhh utility to it all, didn’t know he had a book about it, but like

It’s very silly and funny to say it ends eventually (taking that direct limit sorta thing) though, so eventually there’s no new info, but I dunno what you could glean from, say, G having a tower that terminates in exactly \omega steps (if a limit ordinal can ever occur)

Ah wait yeah, the Grigorchuck thing linked there has height omega

I’ll need to grab that book & look at that paper later

[rotund aurora]

What is the probability that |G|<|Aut G|

Maybe calculate this probability for G of order <=n and take n to infinity. Maybe the limit doesnt exist but whatever

[topaz solar]

Well, that’s not exactly well posed w/o a distribution yada yada, and we sure would have issues doing it for arbitrarily large G without tricky shenanigans

[rotund aurora]

The limsup and liminf info would also be interesting

[topaz solar]

Just do the uniform on iso classes of groups of size n (or less) or wtv?

I do have a small idea

We might be able to somehow jerryrig something from 0-1 laws

[rotund aurora]

Yeah, I am just wondering if there is any observable "tendency" to the size of Aut G compared to the size of G

[topaz solar]

If you can do something for like, subgroups of size k or less, then you might could invoke something like

[rotund aurora]

That seems the most natural, but it is not clear at all to me if the limit even exists or not

[topaz solar]

For any first order formula, probability of phi being satisfied in groups of size <= n as p_n, then lim_n p_n is 0 or 1

I forget what exact hypotheses are required etc etc, but groups work

Finite language structure things should, esp relational, because stuff works on graphs and we can do some trickery there iirc

[rotund aurora]

What is a nontrivial example of this

[topaz solar]

Ok, it’s groups as relational structures not w/ functions

This is over all structures not models though

I was just an idiot :3

[topaz solar]

But uhh there’s some fairly recent something by Kharlampovich on some similar stuff

Like, a formula of low quantifier complexity has an associated probability 1 among some distributions on groups iff it holds in F_2

For finite group things, because that density should be over some finite generation thing instead(?), I think we should have something similar by some Fraïssé-ish thing?

Yeah that’s Hall’s group

(Which, incidentally, is Sym^\omega(S_3) :3)

[topaz solar]

Anyway, my clowning aside, if you can express it through some trickery on relational languages you should have something, or something suitably simplistic, that might be usable. Otherwise it sounds like a hassle to find some other trickery to show such a limit exists or compute it

[long obsidian]

Are the Laurent polynomials Z[x,x^-1] the free commutative Z algebra generated by X or { x,x^-1}? Are they a free object at all?

[coral spindle]

They're not a free commutative Z-algebra. That would be Z[x]

You can shoehorn them into being a free object in some slightly contrived category, I think

[tardy hedge]

are they not free because x*x^-1 = 1 ?

[coral spindle]

Sure that's one way to show they're different

But they just don't satisfy the universal property

Like, you can't choose a value for x that isn't invertible, so the universal property is broken.

[tardy hedge]

wdym choose a value for x

for the map or something?

[coral spindle]

Can you picture what the universal property should be?

So let X be some set of symbols and let f : X → R be a map from this set X into a ring R. We want the free commutative ring (i.e, Z-algebra) to have an associated map f' : F(X) → R which restricts to f on X

But this cannot happen here, e.g. with the map sending x to 2 in the ring Z.

[tardy hedge]

Ok so bear with me cause this is over my head but its at the level where i can understand after some digging

Firstly, saying free commutative Z algebra is literally the same as saying free commutative ring because ring <-> Z algebra

so why say Z algebra

[tardy hedge]

true?

[coral spindle]

It is true

Idk why arclength decided to phrase it that way

[rotund aurora]

uh the proof of Hamkins result is quite neat, although no clue how the centreless result is done

[rocky cloak]

Seems somewhat complicated, but you can read it here
https://sites.math.rutgers.edu/~sthomas/tower.pdf

[dull ginkgo]

Hom(Z,Z/(n)) i proved is iso to Z/(n)

But I'm a bit unsure about the second one, Hom(Z/(n),Z)

Z/(n) is cyclic, so homomorphisms are determined by the image of 1

[coral spindle]

Yes

Go on

[dull ginkgo]

But i think it's trivial

[coral spindle]

What's trivial

[dull ginkgo]

Hom(Z/(n),Z)

[coral spindle]

Why

[dull ginkgo]

I defined \phi: V = Hom(Z/(n),Z) -> Z
\phi(\mu) = \mu(1)

it's injective because it has trivial kernel (obviously)

but the only subgroups of Z are (n)Z (infinite)

[coral spindle]

nZ you mean

aka (n)

[dull ginkgo]

yeah

(m) for some m

[coral spindle]

This is also false btw

[dull ginkgo]

aka infinite

[coral spindle]

There is a finite subgroup of Z

[dull ginkgo]

i mean

nontrivial

wait

[coral spindle]

But you are also drifting from the point

[dull ginkgo]

i realized my mistake

[coral spindle]

a map in Hom(Z/nZ, Z) is determined by where 1 goes

so what

[dull ginkgo]

gimme a sec to rewrite it on paper

I conflated Z/(n)Z and V for a hot second

[tribal moss]

Alternatively, you could argue that a homomorphism must ||take torsion elements to torsion elements||.

[dull ginkgo]

Assume mu(1) = m, then mu(n*1) = n*mu(1) = nm = 0 which can only happen if m = 0 as n is nonzero

[coral spindle]

Correct

[dull ginkgo]

I got a little brain hurty for a second typing it out because our "1" is in Z/nZ and not Z, module pain

[coral spindle]

Then write 1 + nZ

[dull ginkgo]

i used an overline on paper

[coral spindle]

i.e. the actual coset

[dull ginkgo]

that also helps lmao

[dull ginkgo]

i assume the same idea is used for 2

\mu(1 + nZ) = k, then \mu(n(1 + nZ)) = \mu(nZ) = mZ = nkZ

[coral spindle]

\mu(1 + nZ) = k + mZ you mean

[dull ginkgo]

yes, typing that out is harder than the damn question

[coral spindle]

But yes

Same idea

[dull ginkgo]

\mu(1 + nZ) = k + mZ
\mu(n(1 + nZ)) = \mu(nZ) = mZ = nk + mZ
nk must be in mZ

[coral spindle]

So...

[dull ginkgo]

(writing it down brb)

isomorphic to Z/(gcf(n,m))?

wait no

[dull ginkgo]

i'm stuck here

[coral spindle]

What's the order of mu(1)

Or more precisely, what are the possible orders.

[dull ginkgo]

common divisors of n, m

ah

[cobalt heath]

O, doing number theory?

[tulip saddle]

Hello friends. Any chance I can get some help with some elementary group theory?

Alright then

My proof, at first, relied on the assumption that $\forall{a\in S} \exists{b,c \in S}$ such that $a=b/c$

[cloud walrus]

juand8

[tulip saddle]

But after some thought, I don’t think the assumption is justified

So I tried to prove it solely on the definitions in the problem but I need to show that a/1=a to do that. I tried but haven’t been able to do it

[rocky cloak]

Here's a hint: what happens if you plug in b=c=1 in (2)? Can you compute (a/1)/a ?

[tulip saddle]

You get (a/1)/(1/1) = (a/1)/1

[rocky cloak]

Well you should get (a/1)/1 = a/1

[tulip saddle]

Yes. You are right

Do I use (1) after that?

[rocky cloak]

Yeah, or you've already used (1), but you should use it again at some point yeah

[tulip saddle]

Ok ok

I think I got it now

[tulip saddle]

Is the assumption justified though?

[rocky cloak]

I mean b=a and c=1 gives a = a/1

So yeah

[tulip saddle]

How would you show associativity?

I got stuck on that one

[acoustic igloo]

hi

this isn't a group, is it?

beta shifts right and alpha toggles the first two elements

[next obsidian]

Isn’t it?

Maybe it isn’t associative

But I feel like it has to be

[acoustic igloo]

idk...

different stuff happens depending on where you start

at the top, beta is the identity but not anywhere else

at the top beta*alpha = alpha

but not anywhere else

ohhhh wait...

[acoustic igloo]

it's gonna be like this, isn't it

[acoustic igloo]

I think it's A_4

[shell pilot]

Just noticed that <4> doesn't generate Z_30 lol

So deleted my old question

However, why is <4> not a subgroup of Z_30?

[mighty kiln]

It is a subgroup

<S> is a subgroup for any subset S

[shell pilot]

It says "these are the only subgroups of Z_n"

[mighty kiln]

<4> = <2>

[shell pilot]

So they just don't list it because it's redundant?

[mighty kiln]

Yea

[shell pilot]

Thank you very much!

[tender wharf]

if you look a few pages down the line gallian tells you how to find the generators of each subgroup

[marsh scaffold]

So apparently i think this has sth to do with projections and I was trying to take the inverse image of identity in each projection

And projecting that onto the other component

But the problem is idt it works

Also what if there is nothing in the inverse image of e

What if G_1=G_2 and H is diagonal subgroup

Ok but it apparently works

Hmm but I don't see why it would work in the general case

[topaz solar]

In the diagonal one, H1=H2=G, and N1=N2=1, and psi = Id works

[marsh scaffold]

Yes I agree

So is it true that pi_i(H) and pi_2(H) are the required subgroups

And pi_1(pi_2^-1(e)) and pi_2(pi_1^-1(e))are the required normal subgroups

I am just guessing here

[rose axle]

If I have an infinite group G acting on the set X_n = {(g1,…,gn) | g1 … gn =1} by conjugation of each elt, how can I tell if X_n/G is infinite or finite?

[topaz solar]

Ok, anything in G1 x 1 \cap H will be in pi_2^-1(e)

[marsh scaffold]

Yes

[topaz solar]

So pi_1 applied to that is just pi_1(H)

Ah no I’m stupid

[marsh scaffold]

What I am trying to do here is break the components of coordinates down

pi_2^-1 (e) is just a kernel

Similarly the other

[topaz solar]

Ye, im just silly on what that kernel is

Do verify that pi_1 applied to it is normal in H1 there though

[marsh scaffold]

Well H_1 is pi_1(H) here

And it's a surjective image

[topaz solar]

ye

[marsh scaffold]

So normal maps onto normal?

[topaz solar]

now, the next bit, are the quotients of these isomorphic

[marsh scaffold]

That's what driving me mad

The projection notations are too much

💀

Maybe I can write this in some other way

[topaz solar]

That is probably a bit annoying yes

Well, h1 -> e in H1/N1 iff (h1, e) in H or something yeah?

[marsh scaffold]

Hmmm

Maybe

[topaz solar]

Well, (h1, e) are gonna be what the kernel of pi_2 looks like right?

[marsh scaffold]

I agree

[topaz solar]

And N1 is the projection of that

[marsh scaffold]

Yes

[topaz solar]

So, if h1 -> e in Q1 = H1/N1 (not to be confused with the flu), then we gotta have (h1, e) in H ye?

And, of course, the same by swapping to H2

[marsh scaffold]

okay

This is making me feel uneasy, I never knew this could happen with algebra 😭

Is there sth of a nice interpretation of this

Apart from the symbolism

[topaz solar]

If they’re isomorphic, the psi(h1 N1= N1) = eN2 kinda stuff should be correct then, yeah?

[marsh scaffold]

yeah ig

[topaz solar]

And some similar such arguing says (h1, h2) in H iff this holds because (a,1)(1, b) = (a, b)

[marsh scaffold]

Okay alright

[topaz solar]

So this is reasonable, if you can show them isomorphic

[marsh scaffold]

Fair enough

But what exactly is happening with all this

Like I am asking is there some way to summarise this in a very intuitive manner

Like we are looking at components

Inorder to make H a graph

And somehow taking maps and inverse images like this ...works??

Okay maybe I'll think about this part later , lot of assignments due lol 😭💀

Thanks for the help so far @topaz solar

[topaz solar]

Yeah

Well it’s the graph of psi H1/N1 -> H2/N2

And then you kinda lift it back up to a thing on H1 x H2

[marsh scaffold]

Ah okay but N_1 and N_2 are artificial

[topaz solar]

So (h1, h2) is in this relation (H, that is) iff h1N1 and h2N2 are corroborated by the graph

[marsh scaffold]

or are they...hmmm

[topaz solar]

Which is saying you’re in the h2 coset of G1 x 1 in H

Ig?

There’s probably a more coherent way of writing it down

[marsh scaffold]

Maybe

But it cleared more of my doubts

[topaz solar]

But it’s kinda flattening the (x, e) and (e, y) coordinate things

[marsh scaffold]

flattening in the sense?

[topaz solar]

In a way which is multiplication coherent

[marsh scaffold]

okay

[topaz solar]

In the sense that they correspond to e -> e in the iso

[marsh scaffold]

ah

Okay

Thanks again man

I appreciate it a lot

[marsh scaffold]

Oh I looked up this in Wikipedia there's a very nice summarisation of the ideas you wrote about,
It says that "that the subdirect product of two groups can be described as a fiber product and vice versa."

[rotund aurora]

is the definition of solvable and nilpotent for infinite groups the same as for finite groups?

[wraith cargo]

yep

[rotund aurora]

is G(bar Q_p/Q_p) solvable?

[long obsidian]

If I have a commutative algebra like C[x] then it is generated by the monomials 1,x,..,x^n,...

Is the generating set of the quotient algebra just the image of the generators? That is does \pi(1),\pi(x),\pi(x^2),...generate C[x]/I as a commutative algebra where pi is the quotient map?

[topaz solar]

That is how linearity works indeed

[chilly radish]

a generating set. There are many

[tardy hedge]

Why is it called “exact” in exact sequences

How should i be thinking of the significance of image = kernel

What is this saying

[weary frost]

I think it comes from calling the homology (ker modulo image) a "defect", so no defect being exact

[rocky cloak]

From looking around it appears the term "exact differential form" was around since 1899. And the DeRham complex is exact if all closed forms are exact, so it would make sense if the terminology was taken from there. Can also be a combination of that and what sonihi said.

Exact differential form appears to be coined by Poincare, while exact sequence seemed to be coined by Eilenberg and Steenrod, and first used in print by Kelley and Pitcher. No one gives a reason for choosing the term as far as I can tell.

[dull ginkgo]

a_L(x) = ax

Assume a_L is in End_R(M)

then for each r in R, x in M

a_L(rx) = a(rx) = ra_L(x) = r(ax)
(ar)x = (ra)x

(ar - ra)x = 0 for each x implying [a,r] is in Ann(M) for each r

Does this mean a_L is in End_R(M) if and only if:
a + Ann(M) is in Z(R/Ann(M))

[dull ginkgo]

Yeah this works

I want to see if the following holds

Assume $M$ is a $R$-module. Assume $J$ is a subideal of $\mathrm{Ann}_R(M)$

Then is it true that $\mathrm{Ann}R(M) / J = \mathrm{Ann}{R/J}(M)$

[cloud walrus]

Arzela-Açaí Theorem

[dull ginkgo]

Actually I think this is a consequence of the morphism theorems for rings if I consider Ann_R(M) to be the kernel of the left-scaling morphism from R to End(M)

yep that works suprisingly

Also i keep forgetting that $\ker(b \circ a) = a^{-1}(\ker(b))$

[cloud walrus]

Arzela-Açaí Theorem

[dull ginkgo]

S be a subset of ring R, then define the submodule:
Tar(S) = {x in M : rx = 0 for each r in S} [target submodule]

Assume M is simple

Assume proper two-sided ideal J contains Ann(M):

• Tar(J) is a submodule of M, thus Tar(J) is trivial or M
• If Tar(J) is trivial, then for each r in R: rx = 0 iff x = 0.
• Then for each r in Ann(M): rx = 0 iff x = 0. This contradicts M being nontrivial
• Thus Tar(J) = M. Thus for each r in J: rx = 0 for all x in R. By definition, this means J = Ann(M).
• Thus, Ann(M) is a maximal two-sided ideal

Let K = R/Ann(M), then M is a faithful K-module

• M is simple, so for each x in M*: Kx = M
• Define the K-module morphism f:K -> M, f(r) = rx
• im(f) = Kx = M
• Ann(M) is maximal, so K is simple
• Ker(f) is a two-sided ideal of K, so Ker(f) = (0)
• Thus f is an isomorphism by FIT

I need to do the other way

[long obsidian]

Consider the ideal of even integers J in Z and the inclusion hom into the polynomial ring Z->Z[x]

What is the extension of J? How do we find it

[dull ginkgo]

Z[x] is generated by Z and x, so (2Z,x) = 2Z[x]

[dull ginkgo]

Observe R/I as an R-module. Assume V is a submodule of R/I. Then rV = (r + I)V = V. Thus, V is a left ideal of R/I. By Lattice Isomorphism Theorem, V would correspond with a left ideal of R containing I, but I is maximal so either V = R/I = M or {0}/I = {0 + I}. This means R/I is simple

[dull ginkgo]

and for this last one it's real short. M_1 and M_2 are irreducible. Let f be a nonzero morphism, then im(f) cannot be {0_2}, and ker(f) cannot be M_1, then by irreducibility of both, im(f) = M_2, and ker(f) = {0_1}, thus f is an isomorphism.

Then each nonzero endomorphism of irreducible M is an automorphism, and thus End_R(M) is a division ring as every nonzero element is a unit

[rocky cloak]

In general if you have a ring homomorphism f:R -> S and an ideal I in R, then the extension is Sf(I), i.e. all S-multiples of things in f(I)

[marsh scaffold]

Hey

So I have been trying this problem and wasn't able to make any progress without the hint

So I read the hint and I don't understand what is that naturally defined map here

One thing (quite trivial) is G"/G''' is an abelian group right

So it has a cyclic subgroups decomposition

But other than that what naturally defined map are they talking about here

[marsh scaffold]

Given that the order is squarefree we can see how the automorphism would look like, generators have to map in same components only

[cobalt heath]

The map would be like, G-action on G"/G"'.

There is a canonical one, isn't it?

[marsh scaffold]

And what exactly is that action

Huh

[cobalt heath]

Like, start from what would be the action of G onto itself.

[marsh scaffold]

There are many ways G can act on itself

Uh maybe i don't understand what you're aiming at here

[cobalt heath]

But there are the actions you would usually go for

[marsh scaffold]

Like conjugating and left action etc

[marsh scaffold]

Like what 😭I'm sorry I am not very well versed with this

[cobalt heath]

Yeah, the two actions

And only one would translate onto G"/G"'.

[cinder hill]

When a question says that Sn acts on a set, can I assume it’s the normal action

[marsh scaffold]

So? Left action does not work? So I would take conjugation?

So we want to map into inner automorphism of Aut(G''/G''')

Oh

Okay

Now I see 👍

No wait

G"/G''' is very different

An inner automorphism of it would be like uh

The conjugation on G'' induces an automorphism in both G" and G'''

[cobalt heath]

I'd rather say conjugation does not work

[marsh scaffold]

Fck

How do they call this natural

Does that mean I'm unnatural 😭

[cobalt heath]

gxg^(-1) = [g, x] x, so I think it only gives trivial action onto G"/G"'.

[marsh scaffold]

Now but I meant the inner automorphism on G induces an automorphism on G'

G"

And so on..?

Cause they are normal?

[cobalt heath]

Ah wait, it may work. Hmm

Anyway, usually "naturally defined map" like here means you get from common choice of actions.

I guess choosing the correct one is part of the problem.

[marsh scaffold]

[marsh scaffold]

What is the common choice of action

It's def not common to me

[cobalt heath]

Uh, I mean

Left action and conjugation are usual ones, you know

[marsh scaffold]

Left action doesn't work ig

It totally dosent seem like it leads to sth

Also action is a very general terms imo actions map onto permutations of cosets

And not automorphisms specifically

Why exactly would finding a g action lead to a map to the automorphism?

[cobalt heath]

My bad with left action, it does not yield G -> Aut(G) even

But usually, you can think of G -> Aut(H) as a G-action for obvious reasons.

It's a little bit specific one, but not that much.

After all, a G-action on set X is basically G -> Aut_Set(X).

[marsh scaffold]

I agree

I am saying that this action like you said above is very specific

[cobalt heath]

Not really

Like, in which way is it "specific"?

[marsh scaffold]

Obv just any action wouldnt work 💀

And you said conjugation dosent either

Also apart from all that how does this hint motivate the problem

[cobalt heath]

Yeah, that was my mistake that conjugation might not work

[marsh scaffold]

I mean we are taking of a stabilisation thing, why would one want to look at such a map

[cobalt heath]

Conjugation action by arbitrary G does not give trivial one, but I mistook it.

[marsh scaffold]

So every inner automorphism on G is the required natural map

Caused by induced automorphisms

[cobalt heath]

Conjugation is always closely related with these stuffs.

[marsh scaffold]

I see, is there like a intrinsic reason why

Like the hint to me seems very "out of the blue"

Why would one think of looking at such a map in the first place

[cobalt heath]

Yeah going for G" is a little out of the blue, but considering conjugation action is one of the main things you could try.

Ah the problem asks for G" = G"'. Then, indeed you want to 1. look at G"/G"'

Then, 2. How do you investigate G"/G"' ? By actions!

[marsh scaffold]

Hmmm okay seems reasonable

Okay so Aut(G"/G''') isn't that isomorphic to so U_p1xUp_2...xUpn as the order is square free

So somehow they want me to show that it's trivial

So that would make G"/G''' of order 1 or 2?

Ok anyways thanks for the help so far @cobalt heath

[cobalt heath]

Ah, no problem!

Tbh I dunno how square-free order part would interact, so I cannot give you further help. Still, I hope it vaguely helped!

[cinder hill]

I’m having a lot of trouble with this question. I managed to prove that if the stabiliser is a maximal subgroup then G is primitive, but I can’t do the other direction. I really don’t see why if the stabiliser is contained within a larger proper subgroup it would make G not be primitive.

The stabilisers of blocks contain the stabilisers of each element of the block, but I don't see a reason why a larger subgroup containing some Ga has to be a stabiliser of a block (or even contain Gb for some other b in B)

Sorry if I'm not completely coherent I'm just really confused

[rotund aurora]

what is the Galois group of Q(zeta_n, p^(1/n)), where p is a prime number

I think p could be replaced by any integer different from +-1,0 and the result would be the same?

[rocky cloak]

It's the semidirect product of Z/n with Aut(Z/n).

You can see this because it's the composition of Q(zera_n) and Q(p^1/n), and Q(zeta_n) has Galois group Aut(Z/n) and Q(p^1/n)/Q(zeta_n) has Galois group Z/n

[rotund aurora]

so, Aut(Z/nZ) acts on Z/nZ, and you take the semidirect product with respect to that?

[rocky cloak]

Yes

[rotund aurora]

ahh good point, I see

[rocky cloak]

I don't know if you figured this out in the below discussion, but the natural action is indeed conjugation.

Some further hints:
||G''/G''' is square free abelian, hence cyclic.||
Therefore ||Aut(G''/G''') is abelian||
Hence ||G' acts trivially on G''/G''' under conjugation||
||What does this say about the group G'/G''', and what can you then conclude about G''?||

[rotund aurora]

what if you take G=Z_p and H=pZ_p?

isn't pZ_p closed? but I don't think there is a non-trivial mapping Z/pZ-->Z_p

[rustic crown]

ig they don't require sigma to be a group hom?

Z/pZ is discrete here, so any such section would be automatically continuous

[rotund aurora]

yeah, I guess

ty

[marsh scaffold]

Wait what kind of conjugation are we talking about here

[rocky cloak]

Conjugation by elements in G.

x is mapped to gxg^-1

[marsh scaffold]

Uh

Oh okay wait

It's well defined cause of normality

Why is that action trivial again

G' on G"/G'''

[rocky cloak]

If you have group homomorphism from G to an abelian group then G' is in the kernel

(this is like the main deal with G')

[marsh scaffold]

Oh okay sure

So that's what we mean by trivial action

Oh now we can use isomorphism theorems ?

G/G' =~ G"/G'''

Then uh

I have actually no clue how we relate all this to G'/G'''

[rocky cloak]

No, you have no such isomorphism

[marsh scaffold]

Oh my bad there can be more things in the kernel

[marsh scaffold]

So how do I find out more about G'/G'''

[rocky cloak]

That's not the problem. There's no map G -> G''

[marsh scaffold]

Wdym there's no map

[rocky cloak]

Well, I don't know which isomorphism theorem you're thinking of. But you don't have any map G -> G'' to use first iso

[rocky cloak]

Think about how G'/G''' acts on itself by conjugation

[marsh scaffold]

Like how any group does

You're telling me to look at it's centre...uh...I don't get it 😭

[rocky cloak]

Allright, so let's breath slowly.

You figured out conjugation by G' on G''/G''' is trivial. So now we want to say something about conjugation of G' on all of G'/G'''.

So think about how G'/G''' looks like, maybe think about sequence

1 -> G''/G''' -> G'/G''' -> G'/G'' -> 1

[marsh scaffold]

How is the map G'/G''' -> G'/G" defined here

[coral spindle]

If you write down the first thing that comes to mind, that's probably the map.

It's the simplest possible thing

[marsh scaffold]

uh aG"' mapsto aG"?

[coral spindle]

There you go

[marsh scaffold]

Sorry for being so dense but I am unfamiliar with how we can get an idea of how G'/G''' looks like from this sequence, i understand that it's a series of maps and can infer nothing more than that

[rocky cloak]

Well, you know that G'/G'' is cyclic, so you can lift the generator to an element in G'/G'''. Then think about which elements commute

[marsh scaffold]

Wait aG" maps to aG''' is not well defined map is it

Lifting the generator means it's image wrt some homomorphism in G'/G''

?

[coral spindle]

No, it depends on the representative. You can just choose one, though, and get a section.

[marsh scaffold]

I think I may not be prepared enough for this problem , I just picked up some notes and was going through it's excercises (turns out that it's a bad idea)

Thanks for your help @rocky cloak and @coral spindle

[tribal moss]

The help you're getting does seem to be vey obscure, too.

[coral spindle]

Oy very obscure

[rocky cloak]

Is there an easier solution?

[tribal moss]

I don't know; I don't even know one solution -- but it seems to be pretty unhelpful just to keep saying "think about [blah blah blah]" without actually giving any information.

[rocky cloak]

Fair. It's a little hard to thread the line between hints and just straight up giving the solution

Like you want to prove that G'/G''' is abelian, and I don't really have a clever way except, just think carefully about what the group looks like

[tribal moss]

That's the "think about" again ...

After this long a conversation my hypothesis would be that there's some fact or property that the helpee needs to use, which is obvious to you but which he is not aware of -- and no amount of "think of this" or "think of that" is going to make that knowledge appear in his mind.

[rocky cloak]

Well, I feel like I'm giving hints one fact at a time. But there might have been a better order to give hints in idk

I guess the next hint could be that if a group is generated by two elements that commute with each other it's abelian.

It also fits into a more general theorem that if a group modulo it's center is cyclic, then the group is abelian

[tribal moss]

I'm not sure if it was ever stated in the conversation that you wanted to prove that G'/G''' is abelian?

At least I can't find such a statement.

[rocky cloak]

Well, I said that G' conjugates trivially on G'/G'''

[tribal moss]

You did? I must be blind; I can't find that.

[cobalt heath]

I am sorry that I am bad at this.

[tribal moss]

My apologies, I wasn't even reading far enough back to your attempt.

[cobalt heath]

Ah. I mean judging myself, I failed to be helpful with this problem.
Maybe my general incompetence regarding this problem (To be honest, I am not sure I understood the problem myself)

[rocky cloak]

We will strive for better hints in the future

[tribal moss]

I hope my criticism did not sound too personal. It just seemed clear that there was a failure of communication: they were unsure which intermediate conclusion they were supposed to reach, while you assumed they already knew which way they were going and gave hints based on that assumption.
Saying "think about X" doesn't help unless one already has some way of recognizing which X-adjacent facts are relevant to the problem.

[long obsidian]

For an ideal J of a commutative ring R and the inclusion map I:R->R[x] is the extension of J via I always J[x]?

I'm pretty sure this ideal contains the extension of J

[coral spindle]

Yes

[tardy hedge]

whats the difference between isomorphic extensions and equivalent extensions?

in a short exact sequence

[chilly ocean]

Isomorphic extensions are equivalent but not all equivalent extensions are isomorphic

[tardy hedge]

Isnt it the other way around

I think its the other way around, at least in D&F

[chilly ocean]

Hey guyz. I am not really sure where to ask this. But can someone recommend me a good source (book/youtube) for self studying groups, rings and fields? I am an undergraduate in my final year of studies.

[boreal inlet]

Richard Borcherds maybe?

[chilly ocean]

are they on youtube?

[boreal inlet]

yes

It's not like detailed to the point it's a substitute to a video lecture though, but he explains nicely

[chilly ocean]

Thanks 🙂

Are there any good books?

[boreal inlet]

An unrelated question, are you by any chance, Indian?

[chilly ocean]

yes I am

[boreal inlet]

There are plenty yes

[boreal inlet]

Try following NPTEL IITM if you want detailed lectures

[chilly ocean]

how did you figure that out? XD

[boreal inlet]

or IITB

[boreal inlet]

The way you said "Groups, Rings and Fields"

that's the UGC name convention for the course

[chilly ocean]

ohh

are you indian by any chance?

[boreal inlet]

Yes

I passed 3rd Year, currently Int M.Sc. 4th Year

[chilly ocean]

Well thats cool.

Will IITM NPTEL be enough for a good understanding? I have never followed them

[boreal inlet]

DnF is always a start if you're new, I prefer Micheal Artin. But those common books aside, Jacobson's Basic Algebra Book 1, and while a bit terse and you need to fill out the details yourself as exercises, Lang. Though I don't recommend lang if you're a first timer

[boreal inlet]

it should be enough help, rest you gotta do it yourself because nobody will teach you how to solve problems

it's practice only

But do read Lang letter on, many courses on Galois theory follow Lang

You can also use Hungerford but that's basically just lang anyways

[chilly ocean]

Alright

thanks a lot

[boreal inlet]

you're welcome

[chilly ocean]

i will start with NPTEL

[boreal inlet]

anyone else can also suggest other better books also, so look out for those also

Do read the books in your own time, I understand how packed the courses are so you can even read them after your masters, there's plenty of time before your PhD

or anything else

[surreal sluice]

Could someone explain the highlighted statement to me? I don't see how E is the directed union of Ms.

I understand the directed union here. But I don't see Ms satsify the property.

[rocky cloak]

You haven't defined E, so it's a little hard to say

[surreal sluice]

Ah, I was assuming it would be all the Ms. Perhaps, that is a problem. Maybe I need to enlarge it 🤔

[rocky cloak]

No I'm saying, I don't know what E is

It's not stated in your screenshot

[surreal sluice]

Ah, right. Sorry, my mistake. This is the proposition that I am trying to understand the proof of

[rocky cloak]

Alright, and you're just asking if you have two such Ms: M1 and M2 why they are contained in a common larger M?

You just put the xs that give you M1 together with the xs that give you M2 together into a larger finite set, and then construct M from that.

[surreal sluice]

Yes. I was thinking each M is generated by the orbit of a single element.

[rocky cloak]

M is generated by the orbits of all the xs

[surreal sluice]

Ok, I see. Thank you so much for taking the time to answer my novice questions 🙂

[rapid tinsel]

Is there a good criterion to tell when the G-invariants of a tensor product of (char 0) G-representations vanishes?

In general (A \otimes B)^G is different from A^G \otimes B^G (unless G acts trivially on one of them), e.g: https://math.stackexchange.com/questions/364471/g-invariants-of-tensor-products-of-g-modules

But I wonder if the multiplicity of the trivial subrep is an easier thing to determine...

[coral spindle]

Do you care about infinite groups?

[rapid tinsel]

No, only finite (non-abelian) groups

[coral spindle]

Are you working over an ACF?

[rapid tinsel]

Yup

Let's say the reps are over C

[coral spindle]

OK so you are indeed talking about the multiplicity of the trivial rep

Hm

[rapid tinsel]

I guess the group can be abelian too, but I cannot even think of anything concrete atm

[coral spindle]

I almost assumed that A (x) B was a GxG-module which would've made things easier but this is ofc not the question

[rapid tinsel]

(side note: I know almost zero representation theory)

[coral spindle]

OK so using character theory we can at least make this a straightforward calculation in terms of the characters of A and B

Let me write this out really quickly and see if there's a nice shortcut

[rapid tinsel]

What's the character of A^G in terms of the character of A?

[coral spindle]

Well, A^G is the isotypic component of the trivial representation in A

So in particular it's going to have character <1_G, chi>1_G where chi is the character of A

[rapid tinsel]

Ooo yah of course!

[coral spindle]

So letting $\chi$ and $\vartheta$ be the characters of $A$ and $B$ we have that the multiplicity of $\mathbb 1_G$ in $A \otimes B$ (which has character $\chi\vartheta$) is:
[ \langle\chi\vartheta, \mathbb 1_G\rangle = \frac1{|G|}\sum_{g\in G} \chi(g)\vartheta(g) ]

[cloud walrus]

Boytjie

[coral spindle]

And this is just mildly annoying isn't it? We can't really get this in terms of the similar sums for chi and theta

Yeah I don't see any nice way to transform this

The mathoverflow question you link has some nice 'counterexamples' which is for the sign representation of Z/2Z

So certainly we need more conditions than simply the multiplicities of the trivial characters in each of the other modules

[rapid tinsel]

Yup!

[coral spindle]

Yeah, I just don't think there's gonna be a nice condition here. Even more annoyingly, the homological condition they state in the question won't generalise. Woof.

So yeah I think this is about as good as we can get. I guess at least it is easy to calculate?

Aside: I didn't need to ask whether or not we were working over an ACF, now that I think of it

[rapid tinsel]

I thought all this theory just works as long my field characteristic doesn't divide #G

[coral spindle]

You have to be careful when using the inner product, as over non ACFs we don't get that the characters form an orthonormal basis.

[rapid tinsel]

Ooo oof

[coral spindle]

(And you also get questions regarding what happens if you're not working in a subfield of C – what's conjugation then?)

But in this case it works fine because of something called the Schur index.

[rapid tinsel]

My fields come with an embedding to C, so I can treat them as reps over C

[coral spindle]

It works so long as you look just at the trivial character

[coral spindle]

Yes & in general since we only actually care about the algebraic values, we can always view char 0 reps as complex

[rapid tinsel]

Why is tensor product so messed up lmao

[coral spindle]

Ig it's kinda more natural to consider the tensor product of two modules as being over a different ring

So I mean like, A (x) B is more naturally a GxG-rep

For example the tensor product of any two simple G-modules is a simple GxG-module

And this would even allow us to calculate this above quantity more nicely since it really is just the product of the multiplicities

[dark wave]

Can someone help explain double covers of Sn?

[rocky cloak]

Explain like classify them?

[dark wave]

Just intuitively what are they, what are the elements, how do I multiply them

So far my understanding is: Assign each permutation either + or - and a permutation corresponds to its + version and its - version. There are two covers depending on whether an element squares to 1 or -1. Odd permutations flip sign.

[tardy hedge]

So an equivalence of exact sequences is saying something like .. the sequences are exact in the same way ?

But merely an isomorphism of exact sequences mean they are exact in some sort of potentially different way ????

Kind of weird concept

Also im just taking about short exact sequences

But within both cases, we are stil talking about B an extension of C through A, B/A = C

But these are distinguishing between different ways of coming up with maps that give u the right quotients ..?

[rocky cloak]

Well I'm not super familiar with them, but from the description on
https://groupprops.subwiki.org/wiki/Double_cover_of_symmetric_group
You basically introduce a new element of order 2, and then in the first case instead of transpositions squaring to the identity, they square to this new element.
And in the other case, the commutator of disjoint transpositions is the new element, instead of the identity

[dark wave]

Makes a lot more sense now

[rocky cloak]

It's not really about different ways of being exact.

It's more like a short exact sequence
0 -> A -> B -> C -> 0
is a description of how B is built out of A and C. And then a sequence
0 -> A -> B' -> C -> 0
is equivalent if B' is built in pretty much the same way.

While an isomorphism is a little weaker, were you can also replace A and C with isomorphic objects.

Basically the point of the equivalence is exactly what you need to make the pullback of a short exact sequence into a well defined short exact sequence.

While isomorphism is just an isomorphism of complexes, so is not specific to short exact sequences.

[coral spindle]

A subgroup is not uniquely described up to isomorphism of either the subgroup or the supergroup. You have to specify how the subgroup embeds. In a similar way, an extension is equivalent to another if the subgroup and quotient are embedded and quotiented(?) in the same way. This is the idea.

So idk, they're isomorphic and the way they relate is preserved by the isomorphism.

[noble lynx]

I dont get why the left and right exactness conditions are actually different, if given 0 -> A -> B -> C -> 0 and say your functor was left exact then could you not extend to the right so that the infinite chain 0 -> A -> B -> C -> 0 -> 0 -> .... is exact and then apply your functor to get 0 -> F(A) -> F(B) -> F(C) -> 0 -> 0 -> 0 ... ?

[hollow imp]

well left-exactness and right-exactness only preserve the exactness at the left + middle/middle + right terms in short exact sequences I'm pretty sure

[noble lynx]

oh they dont preserve infinite chains like that?

if 0 -> A_1 -> A_2 -> A_3 -> ... then you dont get exactness of 0 -> F(A_1) -> F(A_2) -> F(A_3) -> ... if the functor is left exact?

[hollow imp]

nope

[still dew]

Okay so I am confused if this question should go to foundations or here but here goes:

A commutative ring with unity of not the zero ring always has a maximal ideal

The proof ig uses zorn's lemma and the set of proper ideals being non empty due to {0} and ordered under inclusion and every chain is bounded by it's union which is also an ideal

But apparently if we consider a commutative rng i.e just remove the identity then this fails

Like there are non zero commutative rngs with no maximal ideals ?

What exactly goes wrong if I try to write the same proof as rings here

[coral spindle]

The proof works for unital rings because the presence of 1 indicates equality with the ring. You just show that there is a maximal element in the poset of ideals without 1, and the upper bounds of chains of these ideals -- namely the unions -- preserve this property. For rungs you cannot characterise maximal ideals like this and so you have no such guarantee.

There are simply examples of rungs where this fails. See this example.

[still dew]

Ah okay i see

This underlying assumption was actually quite difficult to spot for me

[still dew]

Ah also sorry to bother but how did u text that link like that

[coral spindle]

[text](link)

https://support.discord.com/hc/en-us/articles/210298617-Markdown-Text-101-Chat-Formatting-Bold-Italic-Underline

[still dew]

Thank you , you're awesome 😎

[novel loom]

Never mind.

I figured it out. xD

[coral spindle]

I hope you inferred that x has no inverse in k[[x]]

[novel loom]

Yes.

Because the powers of -1s accumulate.

1 + (1-x) + (1-x)^2 + ...

doesn't have a well-defined constant term.

Ergo, it isn't an element of k[[x]]

[coral spindle]

Yes, infinite sums are generally not defined

[south patrol]

Hm but you have to make that expression make sense in the first place to make that argument rigorous right

[novel loom]

On the other hand, if R = k[[a,1/b]]...

[cloud walrus]

ComplexVariable

[south patrol]

Note also that the map k[[x]] -> k sending x -> 0 is a ring hom

were x invertible that could not be the case

this also shows (x) is maximal

in fact it is the unique maximal ideal

[novel loom]

Why?

[south patrol]

it induces an iso k[[x]]/x -> k

I'm assuming k is a field

[novel loom]

Ah.

[shell pilot]

What about (1 2 3 4) = (1 4)(1 3)(1 2)? Isn't that an odd count of \beta's

espilon is the identity permutation

[barren sierra]

What is epsilon

I feel like we're missing the full context

[shell pilot]

Edited now, sorry about that

[south patrol]

But then neither side is the identity

[shell pilot]

Isn't (1 2 3 4) the identity?

[south patrol]

No, it sends 1 to 2 etc

[shell pilot]

Wow, thank you. I was mixing up the matrix notation with the cycle notation

So could I have written {1, 2, 3, 4} as S_4 = (1 4)(1 3)(1 2)?

Or is that epsilon equation looking for the matrix notation ((1 2 3 4), (1 2 3 4))

[placid heart]

Let $M$ be a finitely generated projective $R$-module. Prove that $M*$ is a finitely generated projective $R$-module.

[cloud walrus]

Casiel368

[placid heart]

I tried to use that M can be put into direct sum with some T that is isomorphic to some R^(S)

But is Hom_R(R^(S), R) finitely generated and projective?

If T is too big I can't guarantee that even if M is finitely generated

[chilly radish]

You can guarantee that S is finite since M is finite. Now, Hom preserves (finite) direct sums, and you can conclude what you want from there

[placid heart]

Oh right, S being finite was the only step missing. Thank you

[shell pilot]

Why are we trying to "move the a element further to the left in the cycle?

Is it because if a goes to the very first 2-cycle then the full cycle no longer creates the identity since a gets mapped to something other than a?

[still dew]

They're doing that so as to obtain a in the action of that 2 2 cycle product

It's basically all done to use inductive hyp

[south patrol]

Fun argument which I share only because it generalises: fg projectives are precisely retracts of R^n for some n. (-)* is a functor and preserves any retracts. So M* is a retract of (R^n)* = R^n and hence fg projective

Bit silly but the idea of thinking of projectives as retracts is nice since those are preserved by basically anything

[shell pilot]

Why does r > 2 mean that r - 2 is even by induction? What if r = 3 or 5 or 7 etc.?

[rocky cloak]

The proof shows that if you have the identity written as a product of r transpositions for r>2, then it can also be written as the product of r-2 transportations.

So if it was possible to write it as the product of 3 transpositions it would also be possible to write it as a product of just one transposition, but that's impossible.

[shell pilot]

That makes much more sense. Thank you!

[crystal vale]

an element x in A is nilpotent if x^n = 0 for some n>0. A nilpotent element is a zero divisor(unless A = 0).

0 is a nilpotent element but it is not zero divisor, yes every non-zero nilpotent element is zero divisor.

[rocky cloak]

This just depends on whether you include non-zero as part of your definition for zero-divisor or not.

If you include non-zero in your definition of zero-divisor, then it would make sense to do the same for nilpotent.

[crystal vale]

They don't defined 0 as zero divisor

[rocky cloak]

If they explicitly define zero-divisors to be nonzero, then yeah you'll get that as an exception to nilpotent elements being zero-divisors

[crystal vale]

Okay, thank you

[tawdry plover]

What is the significance of defining equivalences like this , or is the exercise supposed to just be an exercise or does this procedure lead to something

[coral spindle]

This is something called localisation which is a bit like adding inverses of every element in S to your ring R, hence the notation.

See for example what happens when R = Z and S = Z\{0}

[tawdry plover]

Oh we get Q?

Sth isomorphic i mean

[tawdry plover]

Where can I read more about this

[coral spindle]

Any book on commutative algebra will cover this

[tawdry plover]

Thanks!

[tardy hedge]

If you have two SES between the same three A,B,C, but the maps between them may be different, are they necessarily isomorphic SES?

Can u always find an isomorphism between the ses

[dull ginkgo]

For an associative algebra K over commutative ring R, say an element v is integral iff p(v) = 0 for some R-polynomial p.

Then Cayley-Hamilton says for a finite-dimensional R-module M that every element of End_R(M) is integral.

Does the same hold if we weaken it to finitely-generated?

[agile burrow]

What is a finite-dimensional R-module? I've only seen the term dimension used for modules in the context of vector spaces

[mighty kiln]

Free modules?

[agile burrow]

Oh maybe

[tardy hedge]

I do not understand the last paragraph

I cant see whats going on

[agile burrow]

But yes, the more general statement for finitely generated R-modules is that any endomorphism is integral (and one can even say something about the structure of the coefficients)

You can find a proof in like chapter 2 of atiyah-macdonald or something using this standard determinant trick

[soft tiger]

For exercise D part 2, how exactly would you state this

I’m trying to say that if any composition of any n elements under the binary operation G is equipped with is equal to the identity element of G, then the product of any cyclic permutation of k places to the right of these elements is also a equal to the identity, where k and n are natural numbers and 1≤︎k≤︎n.

Does this make sense?

[dull ginkgo]

Generated by a finite linearly independent subset

[mighty kiln]

What's a linearly independent subset

If it's linearly independent in the sense that ∑rivi = 0 → ri = 0 then it's a free module

[dull ginkgo]

Linearly independent set is a set such that finite scaled sums from the set are 0 iff all the scalars is 0 ye

That’s the word yeah

Free

I don’t really think there is an issue with extrapolating the adjugate though so

[mighty kiln]

Well if it holds for finitely generated modules it certainly holds for finite dimensional free modules

[dull ginkgo]

The main problem I think with finitely generated modules is with the generating sets and the whole idea of matrices probably not being unique

I view generating sets generating ofc, but basises generate uniquely

[mighty kiln]

Well the result they cited works for finitely generated modules over a cring in general

[dull ginkgo]

(Differences are 0 iff coeffs are equal)

[dull ginkgo]

Actually I think you can define the determinant for End(M) for finitely generated M

[mighty kiln]

I'm not sure what you're looking for

[tardy hedge]

Why are we restricting to identity on those sets, im quite lost on thus

Not sure what to be thinking of

[tardy hedge]

Is this purely for this example, considering the fact that for both sequences , the first map from A to B is the same?

Im confused on if this is saying something more general or if this is still related more specifically to this particular example

For an automorphism B->B, what does restrict to identity on a quotient of B even mean?

[barren sierra]

The map you get from first isomorphism theorem would be identity

[tardy hedge]

So its not just any quotient

Specifically kernel of that original map

[chilly radish]

Which is the image of A, which is what it says

[barren sierra]

Well in that paragraph it specifically says B/psi(A) = B/ker(phi)

You can't just quotient by your favorite random subgroup and expect to get identity

[tardy hedge]

Yea for that example because its an exact sequence?

[shell pilot]

Why are the only orders of elements of A_6 1, 2, 3, 4, 5 and not 6?

[coral spindle]

Well this is exactly to say that there is no element of order 6 in A_6

So what would such an element look like, in terms of cycle type? Hint: there are only two ways it can be.

Then simply inspect it an note that such an element of S_6 cannot be an element of A_6

[shell pilot]

I'm using this as a reference. So wouldn't there be a 6-cycle which would have order 6 and then a 3-2-1 cycle which would have order 6?

Oh, neither of those would be able to be written with an even number of transpositions

6-cycle would have 5 transpositions (odd) and a 3-2-1 cycle would have 5 transpositions as well (odd)

[hollow imp]

3-2-1 would have 3 I'm p sure

[shell pilot]

So the 6-cycle would be (1 2 3 4 5 6) and the 3-2-1 cycle could be (1 2 3)(45)(6) which could be written as (1 3)(1 2)(4 5)(6) or do you not include the (6) because you should only include transpositions?

[coral spindle]

I don't really care about that distinction, either is fine. But in any case you have your answer don't you?

[tardy hedge]

But why does the kernel of B to B need to be psi(A)

Ok honestly i think im genuinely completely clueless on this example 😂

[shell pilot]

Yes, correct, I was counting the 1-cycle as a transposition but obviously it is not. Just the 3-2 which can be written as 2-2-2

[chilly radish]

You said it yourself

It's an exact sequence

It's actually doesn't have to be the kernel for the map B->B, but exactness + the defn of equivalence of extensions (commutativity of a certain diagram) gives you that the kernel of phi_i, which is the image of psi contains the kernel of your map, so you can mod out by psi(A)

It's not exactly the first isomorphism theorem since you might not get an injective map

Ok let.me rephrase

You have an automorphism B->B, and you wanna get an induced map on the quotient

The classic thing to do is consider the composite
B->B->B/psi(A)
Where the second map is the quotient map

Now, the kernel of this composite map will be exactly the preimage of psi(A)

Equivalence says that this should be psi(A)

[tardy hedge]

“You wanna get an induced map on the images”, sorry can you explain a bit more about this

[chilly radish]

Quotients*, sorry

Quotient by the image

[tardy hedge]

The image of B->B should be what set?

Psi(A)?

[chilly radish]

You can also just show that if f:B->B is your automorphism then the map
f'(b+psi(A))=f(b)
On B/psi(A) is well-defined

[chilly radish]

B, since it's an automorphism

[tardy hedge]

Oh yeah oops

[chilly radish]

Once you do this you can check that it's the identity when they're equivalent by diagram chasing. You basically want to show that if b\in B then f(b)-b\in psi(A)

[tardy hedge]

Sorry im just trying to remember what psi(A) is for and why we are doing something with quotients

And still just trying to think of what the overall picture is of what we’re trying to do

[chilly radish]

Understand a criterion for the equicalence of extensions

Maybe reread the relevant section

[placid heart]

Let $R$ be a commutative ring, $N, M, M', M''$ be $R$-modules such that $M'\to M \to M'' \to 0$ is exact. Prove that there exists an exact sequence $0 \to Hom_R(M'',N) \to Hom_R(M,N) \to Hom_R(M',N)$.

[cloud walrus]

Casiel368

[tardy hedge]

Why do we need to restrict to identity on psi(A)?

This one seems easier to understand at first

[placid heart]

I have my morphisms already defined and everything works fine up to the last step

If the initial e.s. has morphisms f, g, I define f*(phi)(m')=phi(f(m')) and the same for g

They are morphisms, $g*$ is monomorphism and $Im(g*)$ is contained in $Ker(f*)$

I need the last part

[tardy hedge]

In this example, is the fact that in both sequences, the map from A to B is the same (psi) relevant?

[cloud walrus]

Casiel368

[placid heart]

How do I see the converse?

i.e., how do I show that $Ker(f^) \subseteq Im(g^)$?

[cloud walrus]

Casiel368

[tardy hedge]

I think so, right? Thats why we are saying we need the B->B map to be identity on psi(A)?

So that first part of the diagram commutes

Ok that MUST matter right? Or else i feel like this is more complicated

[chilly radish]

Very

[tardy hedge]

Phew!

Ok i am a lot closer to understanding this lol thank you

Not 100% there yet but

And from the exactness is how ker(phi1) = ker(phi2) = Psi(A)

[lone niche]

Start with an element x in Ker(f star), and show that there exists some element y in the domain of g star, such that g star (y) = x. Expanding the definitions you have an easy candidate.

[placid heart]

Thank you!

Here's another one I'm struggling with

Let $R$ be a PID. Prove that any finitely generated submodule of a free $R$-module is free.

[cloud walrus]

Casiel368

[placid heart]

Okay so I have some basis for some free module L

And m_i generators of a submodule M

I can write the generators of M with the basis of L and try to reduce that by using that A is a PID

[next obsidian]

@placid heart the best proof of this is via induction IMO

I guess for that though you need to specify the over-module is finitely generated

But you can WLOG to that case

[hidden wind]

i love wlogging

[tawdry plover]

So apparently when we take the set of all idempotent elements in a commutative unital ring and define addition e plus k as e + k - 2ek and e*k is the usual then it itself becomes a ring

I would like to know how did one come up with this choice for defining addition like this

[placid heart]

Umm why can you do that? Also I don't see how the induction would work

[next obsidian]

Write down a finite generating set. Look at which basis vectors they reference, then look at the free module those basis vectors span. Now your module is a submodule of a finite free

[hollow imp]

if you have access to the classification of f.g. modules over a PID, it suffices to show that there are no torsion elements

[placid heart]

Okay I think I can give it another try

Meanwhile here's another one I got stuck with

Prove that a module $M$ is projective iff it has a dual basis.
A dual basis of $M$ is a set ${(m_i, \phi_i)}{i\in I} \subseteq M\times Hom_R(M,R)$ such that for all $m\in M$ the following hold:\
${i/\phi_i(m)\neq 0}$ is finite\
$m=\sum{i\in I} \phi_i(m)m_i$

[cloud walrus]

Casiel368

[hollow imp]

think about how to construct the map using the dual basis

[placid heart]

And the converse?

[hollow imp]

well that's vague

[placid heart]

I think I get what you said tho. I can try to use the basis so that the map from the definition of projective exists

[placid heart]

Yes thank you

How can I prove the converse?

[hollow imp]

for the converse, I think you can consider the diagram [\begin{tikzcd} M \ar[d, dashed] \ar[dr] \ F(M) \ar[r] & M \end{tikzcd}]

shit that's the wrong diagra

[cloud walrus]

Alphyte

[placid heart]

F is field of fractions right?

No wait M is not a ring

What is F?

[south patrol]

It should be the free module on the underlying set of M

The point is that ||M is a direct summand of a free module, which you can take to be F(M)||

[placid heart]

The generators of M?

Oh

[south patrol]

Idk what you mean by "the generators"

Just the underlying set

[placid heart]

So then F(M) happens to be isomorphic to M

No

No I don't get it

How does that help

Where does the dual basis come out from

[south patrol]

This is basically never the case

Well the point is that having a dual basis is equivalent to being a direct summand of a free module, if you stare enough

[tardy hedge]

You can have two different homomorphisms A->B with the same kernel, right?

I.e, distinct ways you can embed A\ker into B

oh i mean for sure for sure

heh . . .

😩

😳

[topaz solar]

phi and 2phi

:3

(Depending on what your structure is but like, you sure can for vector spaces eg)

[tardy hedge]

I was thinking of like Z2 x Z2

Some direct product thing

[dull ginkgo]

Hey guys, quick question here

Lets say we have a finitely generated module M over a commutative ring R

Of course, if $v_n$ is our generating set, then every element takes the form of $\sum_{n = 1}^{N}{r_n v_n}$ but not necessarily uniquely. Can we still define a multilinear determinant function based off this generating set, or do we need linear independence for it to be well defined?

[cloud walrus]

Arzela-Açaí Theorem

[topaz solar]

Well, those are isomorphic kernels, are they equal tho. But yeah, (1,0) -> a, (1,0) -> b are two different maps

Which can have the same kernel if (0,1) -> 0

[tardy hedge]

Yeah i didnt work out the details but you could have two different maps into Z2xZ2 with same kernel im sure

Ok maybe i should actually think harder

[tardy hedge]

Why were you saying “well..”

[topaz solar]

Ah, into that as B

[tardy hedge]

Yeah so it works out with what i was trying to do im pretty sure right

[topaz solar]

I thought you meant something with that as A, and the kernels being copies of Z2

[tardy hedge]

Oh

[topaz solar]

I mean, you have two different injections Z2 -> Z2^2

[tardy hedge]

Yeah

[topaz solar]

Which has the same kernel, 0

[tardy hedge]

So kind of trivial example but

[dull ginkgo]

Like I know I can define it for linearly independent generating sets as every element has a unique expansion for it

[placid heart]

Let $A=\mathbb{Z}_5^9 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}2^3$. Describe al abelian groups $G$ such that $End{\mathbb{Z}}(G)$ is isomorphic to $A$.

[cloud walrus]

Casiel368

[dull ginkgo]

As a ring or as a module (Abelian Group)

[placid heart]

Module

[dull ginkgo]

ah i see

i was about to use the existence of idempotents

[placid heart]

Okay so I know that if $d_1|d_2| \cdots | d_r$ are the invariant factors of $G$, then the $i$-th factor of $End(G)$ is $(\prod_{j=1}^{i-1}d_j) \times d_i^{r-i}$ (is this right?)

[cloud walrus]

Casiel368

[placid heart]

And the factors of $A$ are $(5,5,5,5,5,10,10,10,20)$

[cloud walrus]

Casiel368

[placid heart]

Since no integer to the 9th power will be 5 then there are no such abelian groups

I suspect this is horribly wrong but why

[dull ginkgo]

https://proofwiki.org/wiki/Cayley-Hamilton_Theorem/Finitely_Generated_Module

Is this incorrect?

It says that it's defined for a finitely-generated module but this feels extremely incorrect because I don't feel like you can define a determinant on a finitely generated module

for instance, Z/2Z is a finitely-generated Z-module. However 2(1 + 2Z) = 2Z = 0 in this module

but the determinant in the 1d case is just the scalar, but then 0 = 0det(1 + 2Z) = det(2Z) = det(2(1 + 2Z)) = 2

which is factually inaccurate

[dull ginkgo]

oh, it's just because of projectiveness

cool

[acoustic igloo]

what's an upside-down large pi symbol?

https://en.wikipedia.org/wiki/Burnside's_lemma
before the examples

[dull ginkgo]

Usually for coproduct, in the case of sets, a disjoint union

[kind temple]

just in case, its typed in latex as \coprod, and if you want a smaller version of it, use \amalg

[still axle]

categorical sum

[acoustic igloo]

oh, thanks

[mighty kiln]

Amalg is a binary operator while coprod is a large operator

[tawdry plover]

If R is a commutative unital ring and let B(R) denote its set of idempotents then
There is a bijection B(R) to B(R/√0) where √0 is the nilradical of R

---

What I've got so far is injectivity when considering the projection map

But I am not able to show that they're surjective?

I only have that r^2-r belongs in the nilradical

Or better yet r^n-r belongs in nilradical for all n >1

Oh wait

If r is idempotents in that quotient then so is 1-r

So their product is also idempotent

But if but if r-r^2 in nilradical then it would become zero in the quotient ring under a large enough power

[rocky cloak]

Yeah, so it's a little tricky. But a key fact is that
(1-r) + r = 1
and then this of course remains true when you raise both sides to some power.

Then there's still a bit of messing around to construct a lift, but that's a place to start

[tawdry plover]

Wait didn't we get that r(1-r) is idempotent in R/√0?

That dosent do anything does it 💀

I totally forgot 0 is idempotent

Wdym by constructing a lift

[rocky cloak]

Like constructing an idempotent in R that maps onto r in R/nilradical

[tawdry plover]

Oh okay

What but the project is supposed to be an isomorphism right

So the lift that would be constructed would have the form r+nilpotent?

[rocky cloak]

Yeah, that's right

[tawdry plover]

Ok so is it some algebraic manipulations

Cause I am not able to see any ;-;

[naive lance]

Thought coproduct?

[coral spindle]

Yes

A common name for many coproducts is a sum

[rocky cloak]

Here's a little bit more hint

So let's rename 1-r as s for convenience of notation

You have that
r + s = 1
and that sr is nilpotent. Say (sr)^n = 0. Now raise both sides to 2n. Then you get

r^2n + ... + s^2n = 1

Notice the first n terms are multiples of r^n and the last n are multiples of s^n, and that all the middle terms are nilpotent.

[languid trellis]

I don't quite see why the (S,R)-bimodule structure is important here, or why this can be viewed as a "change of base"?

I imagine it's to make sr ⊗ m = s ⊗ rm, but I dont' see why these should be equal given that sr = sf(r)

My question isn't very precise because my understanding isn't very good so I would appreciate any pointer in the right direction (:

[south patrol]

The point is basically that

1. the right R module structure let's you define the abelian group S (x)_R N
2. the left S module structure lets you put a left S module structure on the above abelian group

As for changing base, eh, I mean the term doesn't already have a meaning at this point right

It's just you are changing the ring of "scalars"

Honestly people will still call it extension of scalars too even if R -> S isn't injective anyway

[languid trellis]

The right R module structure lets me define the abelian group S ⊗_R N because it guarantees relations such as (sr ⊗ n) = (s ⊗ rn), right? point 2 is clear.

The sticking point for me is why (sr ⊗ n) = (s ⊗ rn) should hold, when the right R-module structure on S and the left R-module structure on N seem "different" in some sense

[coral spindle]

(sr ⊗ n) = (s ⊗ rn) literally comes from the definition of the tensor product over R; we simply impose that relation

I really think of it as 'zipping up' S and N along the R-action. It doesn't really matter if they're not compatible, we'll just get some fucked up module that way

Like Z/2Z (x) Z/3Z being trivial

[languid trellis]

So, we just need right R-module structure on S so that sr is defined when we're in the free Z-module on S x N?

then we quotient out by the module relations

[coral spindle]

I'm not sure what you mean by the first bit

It's defined for free

[languid trellis]

I mean, "sr" makes no sense if S is not a right R-module

[coral spindle]

But it is, via f

We are defining a right R-module structure on S via f

sr = sf(r)

That's the structure

Is it arbitrary in some sense? Yes

[languid trellis]

Yeah, in that message I was just trying to explain to myself what is happening mostly

I wasn't disagreeing

Okay, that's cleared things up

the relations agree by definition of the tensor product not due to the multiplication in the respective modules

[coral spindle]

Yes

[languid trellis]

but the multiplication in the respective modules means that the tenor product may be funky on occasion

like Q/Z ⊗_Z Q/Z = 0

or Z/mZ ⊗_Z Z/nZ = 0, m n coprime

Okie doke

[coral spindle]

Yes

[languid trellis]

Thank you both (:

[coral spindle]

Like this is the example I was talking about

Things can get killed off by that zipping

[languid trellis]

Yeah

I can't wait to get to the exercises because this has been way too much content to digest

[coral spindle]

Yh tensor products are a lot

[prisma umbra]

What next?

[mighty kiln]

What power of this is the identity?

[junior badge]

Why can we immediately conclude that f zero mod (x_1 - a_1, ..., x_n - a_n)?

[mighty kiln]

Quotienting R[x] by (x-a) sends P(x) to P(a)

[junior badge]

Aha, nice. Thanks!

[prisma umbra]

What will be the formula when ad-bc=1 what will be the formula then?

[south patrol]

@rocky cloak btw do yoou reckon the idempotent problem you were talking about can be done like AG-y

I mean like having non-trivial idempotents is equivalent to being disconnected (besides 0 ring) and R -> R/nilradical induces a homeomorphism

But idk if it is like possible to get the bijection from stuff

[bold vector]

I am working with an elliptic curve defined over a finite field $\mathbb{F}_p$ and have a basis set of points ${g_0, g_1, \ldots, g_n}$. When I perform the FFT on these points, I obtain a new basis ${h_0, h_1, \ldots, h_n}$.
Additionally, I have a set of scalar field elements ${a_0, a_1, \ldots, a_n}$ whose FFT gives me ${v_0, v_1, \ldots, v_n}$. Here, both $a_i$ and $v_i$ are elements of $\mathbb{F}_p$.
I am trying to understand the relationship between these two bases when the scalars are exponentiated. Specifically, I want to know if the following equality holds:
$g_0^{v_0} \cdot g_1^{v_1} \cdot \ldots \cdot g_n^{v_n} = h_0^{a_0} \cdot h_1^{a_1} \cdot \ldots \cdot h_n^{a_n}$
In other words, can I say that the product of the elliptic curve points raised to the powers of the FFT-transformed scalars $v_i$ is equal to the product of the FFT-transformed elliptic curve points $h_i$ raised to the powers of the original scalars $a_i$?
To provide more context:

The elliptic curve is defined over the finite field $\mathbb{F}_p$.
The elements $a_i$ and $v_i$ are assumed to be in the multiplicative group of the field.

Any insights into this relationship or any pointers to relevant literature would be greatly appreciated.

[cloud walrus]

Nerses | Layerswap

[rocky cloak]

Well, I don't know exactly what counts as AGy, nor if I'm the right person to ask.

I guess maybe you want to think about an element in R, as a function. Then it being idempotent modulo the nilradical means it's 1+nilpotent or nilpotent at each point.

Then some kinda compactness argument means r^n is 0 on all those places it was nilpotent. Then you can multiply by a unit that is 1 modulo nilradical to get an idempotent, or something

[agile burrow]

I guess another way is that idempotents are in bijection with the connected components of Spec A. Then you're just using that Spec A and Spec A/nilradical have the same underlying topology

But idk if that bijection is implicitly using this fact

[rocky cloak]

Yeah, I guess you would need to prove that if specA is a disjoint union, then A is the product of two rings.

Feels like you would have to involve nilpotents there somehow

[south patrol]

Yes

[rocky cloak]

Also, I feel it's a misguided approach, since it's true for noncommutative rings as well. Which you would miss out on with an AG aproach

[south patrol]

Yeah fair

Tbh I'm unsure how to show uniqueness of lifts withuot commutativity

[viral halo]

I is an ideal. What is Z[x]/I isomorphic to? I orginally thought that it is isomorphic to Z/4Z, by considering the map phi: Z[x]->Z/4Z that maps a_0+a_1x+...+a_nx^n to a_0-a_1+..+(-1)^na_n (mod 4). This satisfies phi(f(x)+g(x))=phi(f(x))+phi(g(x)), but I'm not sure if it satisfies phi(f(x)g(x))=phi(f(x))phi(g(x)).

[coral spindle]

Maybe you can write down a few elements of this ideal to get a better idea

So for example if $f$ is in $I$ then you know that $\bZ[x]/I$ is a quotient of $\bZ[x]/(f)$, which gives you a good first idea of what might be going on

[cloud walrus]

Boytjie

[viral halo]

I will try do some examples

[coral spindle]

If you can find a particularly restrictive f then you may have a good idea.

Hint: maybe look at degree 1 polynomials

[south patrol]

Tbh I like the original approach

[coral spindle]

The original approach cuts straight to the answer but this might be an approach that gives you a better picture of what's going on :)

[south patrol]

Just rephrase your map in a nicer way, namely ||snds f(x) |-> f(-1)||

[coral spindle]

At least I find it easier to cut the problem up this way

[viral halo]

Idk if this is correct but is this a good start? Let f(x)=2-2x, Z[x]/I is isomorphic to (Z[x]/(2-2x))/(I/(2-2x)), and then perhaps finding an ideal J such that: J subset of (2-2x) subset of Z[x] and then applying the same isomorphism theorem again (and maybe getting something useful). Or did you have another approach in mind?

[coral spindle]

Well this is all correct but there's an easier f!

In fact I'll just show you

f(x) = x + 1 is an example

(Typo sorry)

So the quotient map Z[x] → Z[x]/I factors through Z[x] → Z[x]/(x+1) → Z[x]/I. Maybe you can phrase Z[x]/(x+1) in a nicer way (I assure you, you can) and perhaps even figure out what ideal you have to quotient by to get Z[x]/I

[viral halo]

Is it correct to say that Z[x]/(x+1) is isomorphic to Z by considering evaluation at -1? So (if we draw a commutative triangle) Z[x] → Z[x]/I is the same as Z[x]->Z (by evaluation at -1) and then from Z->Z[x]/I by taking the evaluation mod 4?

[coral spindle]

Yes Z[x]/(x+1) is indeed Z

So yeah you're very close indeed now

You can probably go back to your original approach & describe the map in a nicer way, as potat put it

[viral halo]

yes, I understand potatos solution but I think that this one is perhaps more enlightening

[coral spindle]

Yes but ofc all roads lead to rome :) I think this way just shows you how you'd get to potato's solution

rome being the 1st iso

[viral halo]

Is the mapping in potatos solution essentially the composition of evaluation at -1 and then taking the modulo?

[coral spindle]

That's right

[coral spindle]

which is what you're describing here

In explicit coset notation it's f |-> f(-1) + 4Z

[viral halo]

Very nice, ty for the help

[tawdry plover]

Sorry for the delay, this is genius so after what you've done the required lift is taking the n^th power?

[tawdry plover]

No wait that wouldn't work in the original ring

[rocky cloak]

Well, almost. You take this nth power of (r + s), and then you have to group the terms a little cleverly

[tardy hedge]

Why do we want an induced map on the quotient?

In both sequences, the B->C map have the same kernel, so for each coset of B/ker, the elements in the coset get mapped to the same element in C (but which element of C it gets mapped to is different because B->C is a different map in both sequences)

True so far?

What we are trying to do now is find conditions on the B->B automorphism so that the diagram all commutes properly right?

We know B->B must be identity on psi(A), so that that first part of the diagram commutes (true?)

But now I am trying to understand why the B->B map has to fix cosets of B/psi(A) (psi(A) = ker(phi))

[tawdry plover]

Guess I am just not clever cause it's not clicking ;-;

[chilly radish]

Because C is the quotient of B by exactness. What is your definition of equivalence? What do the maps on A and C have to be?

[tardy hedge]

“Induced map on quotient” vocabulary is still getting me. This means, The map B->B needs to fix cosets of B/psi(A)? This is the induced map we are talking about?

Im sorry its taking me a long time to understand this, it feels like a lot to understand and digest for me atm

[tardy hedge]

“Any such automorphism of B must fix the coset (0,1)+psi(A)” wht?

Why?

Does any of that have to do with making the diagram commute, or is this simply a consequence of something to do with quotients in general?

[coral spindle]

So they give a justification right after saying that – because that's exactly the coset containing the elements of order 2. Is it unclear to you why that works?

[tardy hedge]

It is unclear, im not sure why we are talking about exactly the coset containing elements of order 2 and why we are considering that etc and where it fits in with the bigger picture

[coral spindle]

Well I mean automorphisms must preserve order of elements, yes?

[tardy hedge]

Yeah

[coral spindle]

So if we have a coset that contains exactly those elements of order 2, then it must be fixed under the automorphism, because the automorphism must send those elements to those elements

So the coset is fixed

[tardy hedge]

I mean like, so this has nothing to do with the diagram commuting?

[coral spindle]

Sure it does. You're looking, in particular, for an automorphism of B satisfying certain properties. Those properties listed here are exactly what it means for the diagram to commute

Let's view it in the abstract like so.

\alpha is the inclusion A → B

So \phi \circ \alpha is the restriction of \phi to A

And the first square on the left here says exactly that this restriction is the identity on the image of A in B

Does that make sense?

\alpha' is also just another inclusion

[tardy hedge]

Phi(alpha) is the restriction of phi to A

Ok

[coral spindle]

funky notation but sure

[tardy hedge]

Yes it makes sense to me why B->B needs to be identity on alpha(A)

But also in my example alpha and alpha’ are same map

[coral spindle]

That's fine

Makes it easier tbh

[tardy hedge]

Yea i CANNOT make this any harder for myself 😅

Last night in bed i was thinking and i thought i understood why it needs to fix B/psi(A) but now im confused again

[coral spindle]

Well this comes from the other commuting square, maybe you can give it a shot

[rocky cloak]

Yeah, I don't know about more hints to give, but I can do an example:

Say that (rs)^2 = 0, then consider
1 = (r + s)^4 = r^4 + 4r^3s + 0 + 4rs^3 + s^4

Define e = r^4 + 4r^3s. Then 1-e = 4rs^3 + s^4, so (1-e)e is a multiple of (rs)^2, hence is 0. So e is idempotent.

Also e = r^4 = r modulo nilpotents.

[tardy hedge]

I know B->B needs to send ker(Beta) to itself, so that B->B->C sends the right elements to 0