groups-rings-fields

[glossy yoke]

what about diagonal matrices?

or rather, things of the form c*Id.

[cyan marten]

Maybe it's a semidirect product.

[bronze trench]

that'd suck because I'm kinda using it to describe a semidirect product lol

[bronze trench]

Yeah I'll have to check that thanks. My issue is that I'm obviously used to real vector spaces so one over a finite field is basically crashing my brain. I thought for some reason something like "oh finite field means scaling isn't like in a real vs" but then interpreted that as "no scaling" lol

ugh I suck at GT

[cyan marten]

You are trying to construct a semidirect product whose normal subgroup is Z/p x Z/p?

[bronze trench]

btw the "big" question I'm tacking is to describe all semidirect products of P by Q where |P|=p^2 and |Q|=q
I'm done with the case where P is cyclic so I'm trying to describe the other case for P by seeing it as a vector space over F_p

not construct one, actually describe them all

so I want to see what the automorphisms of the thing are

also please don't solve it for me lol
but hints are welcome 🙂

[cyan marten]

I don't think this problem is really easy, but maybe that's just me.

[bronze trench]

well either way my prof thinks it's manageable for us, this is homework 😂

actually the question is a bit harder, but reduced to that case if you do things right

[sturdy marsh]

the order of Gl(2, p) is (p^2 -1)(p^2-p) = (p-1)(p+1)(p)(p-1), p-1 and p + 1 will have prime factors q less than p (for p>=3)

[bronze trench]

he puts really hard things on our HWs sometimes but he's a bro grading it afterwards

[sturdy marsh]

so you will always have elts of order q for q < p

unless p = 2

[bronze trench]

oh I was hoping that wasn't the case, much easier to work with a trivial thing lol

ok thanks for that, I'll think about it 🙂

[sturdy marsh]

thanks!

[bronze trench]

I don't even need to describe the whole of Gl(2,p) for what I want, probably identifying the things of order q is enough to then construct the thing

the thing being the homomorphism that will give me the semidirect product

rather, construct all the things

this is the actual question if anyone cares

[bronze trench]

tbh it seems like one of the easier ones of the sheet (or I'm just dumb) so this isn't encouraging lol

[sturdy marsh]

it isn't very hard

[bronze trench]

ok yeah thanks. I think I managed the hardest part which was reducing it to the semidirect thing but that wasn't too hard at all because it's just noticing some clever lemmas and theorems to use

so now it's just "getting my hands dirty" actually looking at the things themselves

[cyan marten]

I found a note describing them and it is three pages long, and the proof wasn't written in a leisurely way.

[bronze trench]

maybe it proves some things I'm taking for granted because we proved in class or something

[sturdy marsh]

I'm pretty sure it shouldn't be hard

I'll pick up some lunch and brb

[cyan marten]

If your course is traditional, then I doubt it.

The problem is to classify all elements of order q up to conjugacy.

[sturdy marsh]

I'm back

[sturdy marsh]

hmm, if X^q-1 splits then all the matrices of that order will be diagonalizable

oh wait a second

the thing I had in mind was wrong

yeah I fucked up

the problem is not as easy as I claimed

but should still be doable using low-tech tools and a lot of casework

[cyan marten]

The proof I have seen uses mostly linear algebra (results about similarity).

Even though it's low tech, I think it's considerably involved.

Is it easy to find finite groups with arbitrarily large derived length?

[viscid pewter]

yes

just do the cyclic group of that order

then it gets harder, lol

[bronze trench]

well it's a grad course. I mean here it's different from the US but I'm in a master's degree and this is my 2nd GT course so far

[cyan marten]

Ohh, I assumed it was undergrad for some reason.

[bronze trench]

nah it's ok 😛

[cyan marten]

I don't want groups with arbitrarily large order

[viscid pewter]

oh, i can't read

[bronze trench]

actually this direct product thing was a motivation for the extension problem and cohomology but we didn't get far in cohomology

semidirect*

[cyan marten]

I know that if N and G/N have lengths a, b, then G has length (at most?) a + b, so I thought I should take semidirect products of some sort.

[cyan marten]

Yeah, a + b is just an upper bound. But I think a lower bound should be max{a, b} + 1

[cyan marten]

That kind of justifies the difficulty of the problem.

[bronze trench]

oh yeah I can see how you might have thought this was really hard if you thought it was on an undergrad context 😂

[maiden ocean]

hnn

direct product of graded rings is confusing

how does the gradiation work

like say Z[alpha]/(alpha^3) oplus Z[alpha]/(alpha^2)

whats the graded structure hhh

[maiden ocean]

nvm i got it

disregard

[viscid pewter]

is there a good reason why the order of an element and the order of a group both use the word 'order'?

[uncut girder]

@maiden ocean what did you get

@viscid pewter order of a generator of a cyclic group and order of the cyclic group are the same

[viscid pewter]

ok

but ehhh

[uncut girder]

You might say order of an element is a special case of order of a group

Because given any group

And any element in that group

It generates a cyclic subgroup

And the order of that subgroup is the order of the element

[scarlet estuary]

i'd argue the more important connection is lagrange's theorem.

[maiden ocean]

@uncut girder yea i understand it now its the obvious u just take the ith group in the first ring and ith group in the second ring and direct product them

i got confused cuz i was doing a topology thing and messed up w/ the graded structures on the original rings cuz i was overthinking it

[uncut girder]

Makes sense

[maiden ocean]

hhhhh

if A is a free Z-module can we show that Hom(A otimes Zp; Z) = A otimes Zp

[latent anvil]

no but Hom(A, Z) (×) Fp ≈ Hom(A, Fp), which is what you actually wanted

[maiden ocean]

die

[next obsidian]

Reported

[vital quail]

yep, reported sham

[latent anvil]

I'm being bullied for the simple act of mocking a child

what is this server coming to?

[next obsidian]

What no?

I was reporting Moth

lmao

[latent anvil]

yeah but arch was bullying me

[next obsidian]

Given a SES of complexes we get an LES in homology

is that a delta-functor?

or is the other condition for a delta functor, the one about the boundary maps commuting not always satisfied?

I checked Weibel, I think if you're a positive degree one then the answer is yes

[maiden ocean]

singular/simplicial/cellular homology yes for sure

well

cohomology

[latent anvil]

huh I guess homology is the derived functor of 0th homology

That's weird

Oh this only works on nonnegative degree complexes like chm said (I think)

[viscid pewter]

oh, i'm a fool, V4 isn't a subgroup of Q8, it's just that Q8/Z2 = V4

that would explain it

why is learning painful

[next obsidian]

Cuz u don’t know things. But then you know things

[chilly ocean]

Let H be a subgroup of order 6, let a be an element of order 2, b element of order 3

If c = ab is an element of order 4, then subgroup generated by a and b can't exist due to Lagrange's theorem right?

[old lava]

c itself cannot exist

in H

[next obsidian]

Are a and b in H?

[chilly ocean]

Yeah

[next obsidian]

Then it’s not possible

By Lagrange

[chilly ocean]

Alright thanks

[next obsidian]

So c itself can’t exist

Also this isn’t like

This doesn’t have anything to do with a and b

Or their orders

Something of order 4 csnt exist period

[old lava]

ya, even if a and b were order 2 or something

that c is impossible

[chilly ocean]

but ab actually gives something with order 4

[old lava]

no, that's imposible

[next obsidian]

No

Not if a and b are in H

[chilly ocean]

a = [i, j, k] (cycle length 3)

[next obsidian]

Or if their product is

[chilly ocean]

and b = [k, l] (cycle length 2)

then c = [i, j, k, l]

[next obsidian]

Those don’t exist in a subgroup of order 6

[chilly ocean]

You mean a and b?

[next obsidian]

Yes

They are not part of a subgroup of order 6

[chilly ocean]

Yeah but I'm supposed to prove that

[old lava]

can you post the entire question

[next obsidian]

Okay

[old lava]

could clear it all up

[next obsidian]

I see your point

My point here was just

You can’t have the product of two things in a group of order 6 be order 4

[chilly ocean]

yep yep

[next obsidian]

In your example this shows no subgroup of order 6 contains both

[chilly ocean]

thanks thats all I needed

[chilly ocean]

Has anyone worked through the rings chapter in Nathan Jacobson basic algebra I?

there are a couple exercises distinguishing left and right inverses

and I can't seem to tell why cause from the way things were defined in the chapter there is a two-sided identity (a1=a=1a for all a in R)
and unless there's something wrong with this proof https://proofwiki.org/wiki/Left_Inverse_for_All_is_Right_Inverse
they should be the same, no?

The chapter on rngs comes much later so we are definitely supposed to assume two-sided identity for the monoid in a ring

[thorn delta]

what do two sided identities have to do with left/right inverses?

*your confusion with left/right inverses?

[chilly ocean]

the proof I linked and its similar version for right inverse for all being left inverse relies on existence of left (right) identity

so unless there's something wrong with it what am I missing as to why the author is distinguishing between right and left inverses, maybe it's just supposed to make the exercise harder and give me more intuition for more general cases?

[thorn delta]

we're talking about question 6, right? I don't really see how this is related. question 6 is not saying that every element has a left inverse

[chilly ocean]

yeah it's not, exactly, why is it distinguishing between the two?

[thorn delta]

because left inverses are not right inverses in general.

[next obsidian]

You can have a left inverse but not a right inverse

Consider functions

[chilly ocean]

that's true but in rings according to that proof I linked they are

[next obsidian]

No

[thorn delta]

no, the proof you linked only works when every element has a left inverse

[chilly ocean]

oh true! thank you

that was it

wait hm

sure the way the proof I linked is worded is only taking about semigroups where every element has a left inverse

but doesnt the same proof work just for showing a single element with a left inverse has the same right inverse

what fails in the proof if you were to try to do that

[sturdy marsh]

try writing it down

[chilly ocean]

oh

It's the (xx_l) has a left inverse assumption

got it, thanks

[sturdy marsh]

is a noetherian local ring a regular local ring if and only if the completion at the maximal ideal is a regular local ring?

the dimension of the ring and the completion are the same, so I guess my question is "is dim m/m^2 preserved under completion?"

[stoic rose]

Yes, as $R/m^2=\widehat{R}/m^2$

[cloud walrus]

radiateur-man

[sturdy marsh]

👍

thanks

[cyan marten]

If I is a nilpotent ideal of R, and we have a homomorphism of R-modules F : M --> N, then it induces a hom. f: M/IM --> N/IN. It turns out that if f is surjective then so is F. Is there a nice "explanation" of why is this true?

[stoic rose]

@cyan marten Let $C$ be the cokernel of $M \to N$, so that $M \to N \to C \to 0$ is exact. Then $M/IM \to N/IN \to C/IC \to 0$ is exact (because tensor product is right exact), so $C/IC=0$, so $C=IC$. Hence $C=I^nC$ for all $n$, so $C=0$ as $I$ is nilpotent. Hence $M\to N$ is surjective.

[cloud walrus]

radiateur-man

[stoic rose]

Also if N is finitely generated and R is local and noetherian, then that actually works for any proper ideal by Nakayama

[cyan marten]

I am still not comfortable enough with tensor products. Do we tensor the first sequence by IR?

I also don't get why C/IC = 0

[cyan marten]

Oh, right. Thanks.

[cyan marten]

Oh wait, it's just the hypothesis. Sorry.

[stoic rose]

We tensor it by $R/I$. In general for any module $M$ $M/IM=M\otimes R/I$.

[cloud walrus]

radiateur-man

[stoic rose]

And $C/IC=0$ because $M/IM\to N/IN \to C/IC \to 0$ is exact and the first map is onto

[cloud walrus]

radiateur-man

[cyan marten]

Makes sense.

[inland bough]

i'm trying to show that every ideal in a noetherian commutative ring with unity can be finitely generated

i've got it when the ideal is countable

but i'm having trouble when it's uncountable

if i can show every ideal can be generated by a countable set i think that would do it but i can't show that either

can anyone give me some hints?

[carmine fossil]

Do you know ACC

[inland bough]

yeah that's my definitino of noetherian

[carmine fossil]

So,Take an element x_1 and generate an ideal,and if x_2 is not in that consider (x_1,x_2) and so on

[inland bough]

ohh got it

so by ACC it has to terminate which means at that point you've generated the full ideal

[carmine fossil]

Yes

[inland bough]

thanks

[fading wagon]

[crossposted from a different Discord server]
Hmm, for Iwasawa's Lemma (if G has primitive action phi (G->Sym(X)) on X, G=G', soluble normal subgroup A of Stab(x) whose conjugates in G generate G, then G/ker(phi) is simple)... is there a converse of this form:
For every simple group G, is there a set X and group H such that G=H/N, H has a primitive action on X, H=H', soluble normal subgroup A of Stab(x) whose conjugates in H generate H...
For PSL and PSp, they act on one dimensional subspaces of V...
for An, we cannot choose {1...n} and H= An, because Stab(n) is isomorphic to A(n-1), which is simple for large n.
(Should be not hard (i.e. as hard as classification itself) to show from classification if true, but idk?)

[cyan marten]

Can you analyze the soluble subgroups of A6?

Actually the stabilizer of a point is A5*

[fading wagon]

hmm, what do you mean A5*?

[fading wagon]

Not sure why we want to analyse soluble subgroups of A6?

I don't think it is likely to work for cyclic groups of prime order, since that means G acts transitively on X, and there's essentially only one transitive action from C_p. Is my intuition correct here?

[cyan marten]

The star indicates that I am correcting a mistake.

[cyan marten]

Well, my intuition was no soluble subgroup of A5 (the stabilizer of a point, if we let A6 act naturally) can generate the whole of A6, so your statement is false.

[fading wagon]

but that is if we let A6 act naturally. Is there a less natural action that would work?

[cyan marten]

I think so. If we have a transitive action of G on X, that's the same as an action of G on the coset space G/H for some subgroup H.

[fading wagon]

And there is a soluble (but not normal) subgroup of A5, just take any subgroup generated by a 3-cycle. Then conjugates of it generate A6.

[cyan marten]

If I remember correctly, a primitive action with stabilizer of a point = H allows us to write A6 = H U HgH for some g not in H.

[cyan marten]

Oh, right. The question isn't useful anyway because A5 is simple.

[fading wagon]

idk probably worthy of a post on mathematics stackexchange to see if that community has some ideas?

[cyan marten]

But wait. A6 can act transitively on at most 60 sets (number of divisors of A5).

And primitively on even less.

It could be fun to try doing it by hand.

[fading wagon]

hmm maybe, but idk how to do it by hand

that's like checking all homomorphisms from A6 to A60 !?

[cyan marten]

If G acts transitively on X, then it's primitive iff the stabilizer of every point is a maximal subgroup of G.

[cyan marten]

@fading wagon nevermind this

[cyan marten]

So A6 can act primitively only on the coset spaces A6/M where M is a maximal subgroup of A6.

Taking M = A5 yields {1, 2, ..., 6} .

According to groupprops, the maximal subgroups of A6 have orders 24, 36 and 60. Consequently, the possible sizes of sets on which A6 acts primitively are 6, 10, 15.

[fading wagon]

It's not clear we want a maximal subgroup of A6. We just need A6=H/N, and we need a maximal subgroup of H instead.

[cyan marten]

I have been assuming that the action is faithful, so always taking H = our group and N = 1. Also, if G acts on X transitively, that's the same action as that of G on G/H for some subgroup H. If G acts primitively, then H must be maximal.

We can see what the maximal subgroups of A6 are (given in groupprops), which are potential stabilizers-of-a-point, and see whether they have a normal soluble subgroup which generators A6.

I think I got it. Take the action of A6 on the cosets of (some embedding of) S4 (a set of size 15).

S4 has the normal soluble subgroup A4, whose conjugates certainly generate A6. So your conjecture actually holds in this case.

[fading wagon]

whoa cool

shall I credit you in the math stackexchange post?

hmm, not sure if I can construct the action though

[cyan marten]

I think we should try A9, say.

[cyan marten]

It's just left multiplication. The action is guaranteed to be primitive because S4 is maximal (again, according to groupprops).

[fading wagon]

Oh, coset spaces. I see

https://groupprops.subwiki.org/wiki/Subgroup_structure_of_alternating_group:A6
Is it the "standard twisted S4 in A6" or "exceptional twisted S4 in A6"?

(is the exceptional twisted S4 obtained from the standard twisted S4 by one of the outer automorphisms?)

[cyan marten]

@fading wagon The smallest index of a proper subgroup of A6 is 6, so there are no subgroups of index 3 or 5. Now a subgroup strictly containing a subgroup of order 24 must have one of these indices, which is impossible. So we don't even need a particular isomorphism class (much less a particular embedding).

[fading wagon]

Ah, okay.

[cyan marten]

It seems so

I think we lucked out in this particular case because S4 has a normal sol..etc..

[fading wagon]

yeah, possibly... A7 might not look so good lol

https://math.stackexchange.com/questions/3959002/possible-converse-to-iwasawa-lemma

[cyan marten]

I had no idea the maximal subgroups of A_n were so complicated. Honestly I thought they were just the smaller alternating groups.

@fading wagon I might have made a mistake. I assumed that A4 was a subgroup of S4, but actually this was the twisted A4. We still need to check that its conjugates generate A6, which is no longer evident.

Actually I think the twisted A4 has no 3-cycles at all!

[fading wagon]

oh no

[fading wagon]

Twisted A4 as I written there has (1,2,3) in it

[cyan marten]

It has (123)(456)

[cyan marten]

Exactly.

[fading wagon]

I included the inclusion map?

that shouldn't be an issue since (1,2,3) seems to be an element of twisted A4 with the inclusion "σ↦σ if σ is even and σ(5,6) otherwise"

A4 in twisted S4 is just A4

[cyan marten]

But there's a mysterious subgroup of order 36 which does have a 3-cycle, so maybe we wil have more luck with that.

[cyan marten]

Ohh, I see

[fading wagon]

do you think it's worth it to include the proof of iwasawa?

[cyan marten]

No.

[cyan marten]

That makes perfect sense but I can't convince myself that (123) is indeed in that weird looking thing.

[fading wagon]

Well, (123) in S4 maps to (123) in A6 via that map

[cyan marten]

I see

The maximal subgroups of A7 have orders 72, 120, 168, or 360.

The groups themselves aren't given in groupprops, unfortunately.

I suspect the last three occur as S5, PSL(2, 7) and A6, respectively.

But I have no idea why a group of order 72 (index 35) has to be a maximal subgroup of A7.

[fading wagon]

hmm, but just as well the group of order 72 is solvable. Hmm, idk what that group is

[cyan marten]

There are 50 groups.

Also a 3-Sylow subgroup isn't obviously normal.

[fading wagon]

is it something along (even of S4)(even of S3) with (odd of S4)(odd of S3)?

that would give a subgroup of 72 elements

and (123) is in it.

but idk if this subgroup is maximal

[cyan marten]

I don't follow. Do you mean it's generated by something like (12)(34)(567) and (1234)(12)?

[fading wagon]

Yeah, both of them are in the subgroup

but idk if those generate it

it's generated by (123), (567), (234), (12)(56) I think...

[cyan marten]

I am not sure why they generate a subgroup of order 72 (honestly I suspect they don't).

That's it, I will run GAP

[fading wagon]

Consider in S7. Then all permutations that do not cross 1234 and 567, there are 144 of them.

Then take those even permutations, there are 72 of them

[cyan marten]

You're right!

I checked it using GAP

[fading wagon]

cool, the question is "is it maximal"?

[leaden finch]

anyone know whats a good book to learn advanced calculus

[fading wagon]

this is #abstract algebra lol

not calculus

[leaden finch]

ohh wheres the other channel lol

[cyan marten]

Again, using GAP,< (1, 2, 3)> is normal.

[cyan marten]

Well, it can only be contained in a subgroup of index 5 or 7.

[fading wagon]

hmm, so subgroup of index 5 cannot exist, and subgroup of index 7 is isomorphic to A6??

[cyan marten]

But A7 has no subgroup of index 5, and since this group doesn't fix 1, 2,.., 7 it isn't a subgroup of any of the conjugates of A6 contained in A7. However, these are precisely the subgroups of index 7.

[fading wagon]

alright so we are done with A7 lol

[cyan marten]

Yes, I think.

[fading wagon]

time for A8?

[cyan marten]

Let me consult GAP

I am not good at it though

[fading wagon]

ConjugacyClassesMaximalSubgroups(AlternatingGroup( [ 1 .. 8 ] ));?

  AlternatingGroup( [ 1 .. 7 ] )^G, Group( [ (1,3)(2,4), (2,3,4), (5,7)(6,8), (1,2,3,4)(5,8,7,6), (1,2,3,4)(5,6),
      (1,5)(2,6)(3,7)(4,8) ] )^G, Group( [ (5,7)(6,8), (2,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ] )^G,
  Group( [ (5,7)(6,8), (1,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ] )^G ]```
hmm... not sure what we are looking at

maybe I should get their orders

[cyan marten]

I just used MaximalSubgroups

[cyan marten]

There's the "StructureDescription" function.

[cyan marten]

And got 157

[fading wagon]

eh, should mod out by conjugacy

[cyan marten]

Yeah

[fading wagon]

there seems to be just 6

[cyan marten]

Yeah but for some reason I can't use StructureDescription

[fading wagon]

Let's analyse by hand...
Group( [ (1,2,3,4,5), (3,4,5), (6,7,8), (1,2)(7,8) ] )^G, (seems to be even on a 5-3 split)
Group( [ (1,2,3,4,5), (4,5,6), (1,2)(7,8) ] )^G, (even on 6-2 split, i.e. S6)
A7,
Group( [ (1,3)(2,4), (2,3,4), (5,7)(6,8), (1,2,3,4)(5,8,7,6), (1,2,3,4)(5,6), (1,5)(2,6)(3,7)(4,8) ] )^G, (order 576)
Group( [ (5,7)(6,8), (2,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ] )^G, (order 1344)
Group( [ (5,7)(6,8), (1,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ] )^G (order 1344)

hmm maybe I can get their orders

[cyan marten]

I thinj it's because it's a conjugacy class, not a group per se.

[fading wagon]

okay I found the groups

at least their orders

[cyan marten]

The first one has a normal subgroup of order 3.

In fact it's C3 : S5

[fading wagon]

C3:S5??

[cyan marten]

Of order 360 (!)

[fading wagon]

Direct product of C3 and S5?

[cyan marten]

No, semidirect.

The only nontrivial one.

[fading wagon]

ah I see

[cyan marten]

Precisely what we need.

[fading wagon]

never seen this notation before

[cyan marten]

So we're done with A8.

[fading wagon]

eh, there should be an easier way to knock out cases

[cyan marten]

I checked with GAP and it turned out it doesn't have a 3-cycle.

[fading wagon]

maybe we haven't spotted the pattern yet

[fading wagon]

what doesn't have a 3-cycle?

[cyan marten]

The twisted A4

[fading wagon]

generated by?

in A6?

[cyan marten]

Yes.

The one we looked at

[fading wagon]

eh, the A4 of twisted S4 in A6 is untwisted

what generators were you using?

[cyan marten]

Ohhh, my bad. Sorry.

[cyan marten]

(123)(456),....

[fading wagon]

looks like the exceptional twisted?

[cyan marten]

That makes sense, but I don't think the maximal subgroups of An are easy to find.

[fading wagon]

A9 time, if you are up for it

ConjugacyClassesMaximalSubgroups(AlternatingGroup( [ 1 .. 9 ] ));

Group( [ (1,2,3,4,5), (3,4,5), (6,7,8), (7,8,9), (1,2)(8,9) ] ) [5-4 split]
Group( [ (1,2,3,4,5), (4,5,6), (7,8,9), (1,2)(8,9) ] ) [6-3 split]
Group( [ (1,2,3,4,5,6,7), (5,6,7), (1,2)(8,9) ] ) [7-2 split]
A8
Group( [ (1,2,3), (4,5,6), (1,2)(4,5), (7,8,9), (1,2)(7,8), (1,4,7)(2,5,8)(3,6,9), (1,5,2,4)(3,6) ] )
Group( [ (2,7,6)(3,4,8), (1,2,3)(4,5,6)(7,8,9), (2,4,9)(3,7,5) ] ) PGammaL(2, 8) ???
Group( [ (1,3,4,5,6,7,2), (1,3,5)(4,7,6), (1,4)(2,8)(3,7)(5,6), (1,7)(3,6)(4,5)(8,9) ] )```

so we get the expected 5-4 6-3 and 7-2 splits

[cyan marten]

It has only nine maximal subgroups.

[fading wagon]

is the 6-3 split a semidirect product of S6 and C3?

[cyan marten]

Eight*

[cyan marten]

No idea

We can just use StructureDescription.

[fading wagon]

Looks like it, hmm...

Maybe the n-3 3 splits work in general

for the S5-C3 semidirect product, which was the normal subgroup?

[cyan marten]

The one that appears first, I think.

[fading wagon]

hmm, what command did you use in GAP to find that?

[cyan marten]

I am not sure what that means, but if we have a normal group of order 3 then we're done.

[cyan marten]

StructureDescription.

[fading wagon]

ah okay

[cyan marten]

Well, not exactly. We need both factors to be solvable.

[fading wagon]

yeah it's the first one

oh yeah whoops

but S5 isn't exactly solvable??

hmm looks like we don't need that?

yeah it's C3:S6

[cyan marten]

Yeah!

We don't need it.

We only need C3 to be solvable, and it always is.

[fading wagon]

Ah looks like a pattern we can try to prove

[cyan marten]

Yeah. So A9 acts primitively on 84 points in a way so that the hypothesis to Iwasawa's theorem are satisfied. This is so cool.

[cyan marten]

Yeah

Take G = <(123), (45...n)>

[fading wagon]

???

[cyan marten]

But it isn't always in An..

Ohh and it is usually not maximal either, being so small.

Maybe we should try embedding C3 : S_(n - 3)..

[fading wagon]

An can act on left cosets of (permutations of n-3)(permutations of 3) by left multiplication.
It suffices to show

1. An=An' (trivial)
2. This subgroup is maximal
3. This subgroup is isomorphic to C3:S(n-3), having C3 normal subgroup
4. The normal subgroup is conjugate to (1,2,3)

[cyan marten]

4 is also trivial

[fading wagon]

well, if you come up with the isomorphism then 4 is trivial.

[cyan marten]

Wait! What's the centralizer of (123)?

[fading wagon]

basically everything that commutes with (123)? in An?

[cyan marten]

Nevermind. It's isomorphic to C3 x S_(n - 3)

[cyan marten]

That's in Sn, though

So the centralizer is C3 x A_(n - 3).. Not what we're looking for.

But the normalizer of <(123)> in A8 is isomorphic to A6.

Which is maximal.

[fading wagon]

hmm, let me see if we can prove this, and what we need

[cyan marten]

Well, it seems like the normalizer is an alternating group usually

I think we should try embedding C3 : S_(n - 3). Let's call this group G_n.

Its index in A_n is n(n - 1)(n - 2)/6, which is not promising at all.

I mean we can't apply the index trick.

[fading wagon]

Okay, it is trivial that (n-2, n-1, n) is normal in that subgroup.

[cyan marten]

Yes.

[cyan marten]

I think we should investigate Gn first then see how we embed it. We haven't shown (although it's intuitively obvious) that we can embed it in An.

Gn = {(s, f) : s = 0, 1, 2, f in S_(n - 3)}

[fading wagon]

G_n?

[cyan marten]

.

[fading wagon]

hmm... I see

but I defined it as the subgroup preserving {1...n-3}

I managed to intuit <(n-2,n-1,n)> is a normal subgroup

I'm left with showing that it is maximal...

[cyan marten]

That's much nicer.

Wait, you defined Gn as the subgroup preserving these?

[fading wagon]

yeah

preserving that set

[cyan marten]

But that's S_(n - 3)

[fading wagon]

so {sigma(1)....sigma(n-3)}={1...n-3}

no it's not

remember it can mess around with n-2, n-1 and n

[cyan marten]

Ohhh, right.

I think I get it now.

[fading wagon]

Yeah, it doesn't work for A_6 for some reason

[cyan marten]

So Gn contains the normalizer of a 3-cycle.

[fading wagon]

yeah looks like with works with A_7

[cyan marten]

It's contained in one*

[cyan marten]

The normalizer if (123) in A7 is isomorphic to (C3 x A4) : C2

And G7 has order 36

[fading wagon]

hmm, what's with the normaliser?

[cyan marten]

!

[cyan marten]

Oh nevermind

[cyan marten]

Let me see

[fading wagon]

ConjugacyClassesMaximalSubgroups(AlternatingGroup( [ 1 .. 6 ] ));

looks like that subgroup is not maximal

oh wait

we can swap over 1,2,3 and 4,5,6, no wonder it's not maximal

[cyan marten]

Probably the quirks of the outer automorphism

[fading wagon]

nah, just 6=3+3

[cyan marten]

Fair.

How do you define Gn in GAP?

[fading wagon]

https://math.stackexchange.com/questions/3959002/possible-converse-to-iwasawa-lemma
omg we have an answer

[fading wagon]

Gn in GAP? idk I don't really define it, but I just type it out with the generators

[cyan marten]

Great!!

That the converse holds.

[fading wagon]

lol shall we finish up this argument with alternating groups? idk

[cyan marten]

I just checked that the normalizer of (123) is maximal for n > 6

Actually I assumed that n < 10 as well.

I think we should try to show that this normalizer is always maximal, and we will be done.

[cyan marten]

What do you think?

[fading wagon]

would be cool

I posted what we have so far, but we just need to show that subgroup is maximal I guess?

[cyan marten]

Yeah. I think it's better to work with the normalizer, rather than Gn.

[fading wagon]

Well, the groups are the same for n>6 I think...?

[cyan marten]

No. In the cases I checked, Gn seems to be of index 2 in the normalizer.

It's 5:23 AM here.. I must go. Good day/night!

[fading wagon]

gd/n!

[dire bramble]

pardon noob question but I am doing algebra and reading knapp and there is this statement

what the fuck is the inductive argument being made here for showing that the set is spanning

it's unclear to me what he is trying to say

[carmine fossil]

If you know x^n=g(x) for some g(made up of monomials of degree <n) ,you can say x^(n+1)=ax^n +h(x) where h(x) is made up of monomials of degree less than n

So x^n+1=a g(x) + h(x) and the rhs is made of monomials of degree <n

[dire bramble]

okay that makes significantly more sense

I'm a bit confused why the fuck he writes theta^n+m instead of just doing the case of m=1 to make the induction obvious

[carmine fossil]

I think you are supposed to figure that out on your own

[dire bramble]

whatever, thank you anyway lol

[latent anvil]

hello I am stuck

Let $A$ be a valuation ring of $K$ with valuation group $\Gamma$, and let $v : K \to \Gamma \cup {\infty}$ be the valuation. Suppose $\Delta$ is a subgroup of $\Gamma$ such that whenever $g\geq h \geq 0$ and $g\in \Delta$ then $h\in \Delta$. Then the set $\mathfrak{p} = {a\in A : v(a) \notin \Delta}$ is a prime ideal of $A$

[cloud walrus]

shamifold

[latent anvil]

So I'm good with all of that so far

But I'm supposed to show the valuation group of $A/\mathfrak{p}$ is $\Delta$

[cloud walrus]

shamifold

[latent anvil]

And I am completely out of ideas

So one thing I know is that $v^{-1}(\Delta) = A_p^\times$

[cloud walrus]

shamifold

[latent anvil]

But I don't see how to turn that into like a map $(A_p/pA_p)^\times \to \Delta$

[cloud walrus]

shamifold

[golden pasture]

the rough idea is that Δ is a isolated subgroup and we have the bijection Δ = <v(A-p)>

oh wait we dont need to theres a more direct argument

We have $\mfk p=A-v^{-1}(\Delta)$ then go by contradiction, suppose $xy\in\mfk p$ which implies $v(xy)\notin\Delta$ and wlog suppose $v(x)\geq v(y)$ then using the fact $\Delta$ is isolated show if $v(y)\in\Delta$ then $v(x)\in\Delta$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[latent anvil]

Um

What?

Is this proving p is prime?

[latent anvil]

I know how to do that

[golden pasture]

oops busy completing my 40 hours
ye this is proving p is prime

[frank haven]

hey, any recommendations for a good introductory alegbra textbook for self-study

my prof next sem is supposed to be bad so im prepping early

[thorn delta]

artin
jacobson
dummit and foote

[frank haven]

ok, ty ill check those out

[thorn delta]

np!

[old lava]

although I'm not the biggest fan of it at times

gallian is also good for self study

so you could check it out

[frank haven]

ooh gallian is free to

at least the 8th edition

[old lava]

ya, I think so

it's a very easy textbook

my issue with it is that it doesn't cover all the content I'd like it to

[thorn delta]

i dislike that group homomorphisms are on page 200.

[old lava]

yes

that is another issue

[frank haven]

ok, well the course is only up to ring theory i believe

[old lava]

but you could use it as an easy intro, and learn the rest in your course

[frank haven]

yeah thanks for the advice

[latent anvil]

So does anyone have any thoughts on the valuation rings thing I posted earlier?

[latent anvil]

this

[hot lake]

so like, you have to show that for every nonzero class, the valuation on it is constant and in Delta ?

[latent anvil]

I don't think so, that seems false

I'm supposed to show $\kappa(\mathfrak{p})^/(A/\mathfrak{p})^ \cong \Delta$

[cloud walrus]

shamifold

[latent anvil]

Or equivalently find a surjective valuation $w : \kappa(\mathfrak{p}) \to \Delta\cup{\infty}$ such that $A/\mathfrak{p} = { x \in \kappa(\mathfrak{p}) : w(x) \geq 0}$

[cloud walrus]

shamifold

[latent anvil]

I tried defining w in terms of v but I couldn't figure out how

[viscid pewter]

tmw you're not sure if it's that you don't understand the concepts well enough or if the proof is written a little obtusely

[stoic rose]

@latent anvil Actually for $x,y \in A_p^\times$ such that $x - y \in pA_p$ we have $v(x)=v(y)$. Hence $v$ is well-defined on $(A_p/pA_p)\times$

[cloud walrus]

radiateur-man

[stoic rose]

The reason for this is that if $v(x)\ne v(y)$ then $v(x-y)=\text{min}(v(x),v(y))\in \Delta$ but this is a contradiction as $x-y \in pA_p$

[cloud walrus]

radiateur-man

[stoic rose]

(I use $p$ for $\mathfrak{p}$)

[cloud walrus]

radiateur-man

[latent anvil]

ahh I think I see my confusion

So I got that but I thought there could be units of the fraction field which weren't units in A_p

But if xy = 1 mod pA_p then xy = 1 + z for z in pA_p, and as pA_p is the Jacobson radical of A_p, 1+z is a unit, so x is too

Omg I feel so dumb for not spotting that, ty!

Okay yeah an element of a local ring is either not in the maximal ideal (and so a unit in the quotient) or is in there, and so zero (not a unit) in the residue field

Blergh

[stoic rose]

Lol yes

All non-units are in the maximal ideal so are killed when passing to the quotient

[latent anvil]

Thanks again

[stoic rose]

Pleasure, that was a cool problem to do

[latent anvil]

The whole problem it's taken from is cool

From goertz & wedhorn's AG book

[stoic rose]

Nice

[latent anvil]

Say you have a sequence $X_1 \to X_2 \to \ldots$ of closed immersions of schemes. Does this have a colimit?

[cloud walrus]

shamifold

[latent anvil]

I would ask in #point-set-topology but it's busy lol

[fierce perch]

No take a colimit that is a formal scheme

It’s true for affine schemes

And more generally it’s done in a paper by bhargav bhatt

[latent anvil]

I don't know anything about formal schemes and can't find that paper

Could you tell me how to construct the colimit in the affine case?

[stoic rose]

In the affine case I guess you just take the inverse limit of the corresponding rings

(the spectrum of this inverse limit)

[latent anvil]

Oh duh

[fierce perch]

Yeah

[oblique river]

well be a little careful, the colimit of Spec (R_n) is the same as Spec(lim R_n)

the thing on the left is the "formal spectrum" of Spec(lim R_n)

for example, Spec(Z_p) has 2 points but colim Spec(Z/p^nZ) has one point

[latent anvil]

Wait buncho I'm confused

[latent anvil]

Do you mean not the same?

because it seems like you said they're the same and then gave a counterexample

oh yeah okay my concern was while Spec of the limit is the colimit in Aff it might not be the colimit in Sch

[hoary lagoon]

thinking of doing some self study on this course during my semester off. Anyone having any book reccomendations?

[sturdy marsh]

what do you want to learn about

[hoary lagoon]

just like an introduction to the course would be nice

[sturdy marsh]

course?

[pallid ember]

elements of modern algebra is a good intro book

[sturdy marsh]

did they say a course on algebra?

[pallid ember]

well it's the algebra channel

so I'm assuming

[hoary lagoon]

oh yea sry

abstract algebra

[sturdy marsh]

dummit and foote + aluffi is also good

[hoary lagoon]

rn im looking at Beachy and Blair

[sturdy marsh]

I have not heard of it, so I can't comment

[viscid pewter]

moderne algebra

[hoary lagoon]

would you say abstract alg is more proof heavy than analysis?

[sturdy marsh]

they're the same

[hoary lagoon]

cool

[sturdy marsh]

almost everything goes: definition, proposition, proof

[oblique river]

@latent anvil sorry!! yes, I meant not the same

if R is a complete local ring wrt some ideal I then there's Spec(R) which is what you expect and there's also Spf(R) (Spf = "formal spectrum") which is defined as the colimit of Spec(R/I^n)

Spf(Z_p) is like the infinite order tangent space of the point (p) in Spec Z

[latent anvil]

hmm okay

So I'm trying to construct a k-scheme X which is covered by closed subsets X_n (for each integer n), each isomorphic to A^1, and such that X_n and X_{n+1} each intersect at one point

For finitely many n in [-N, N] I can construct a space like this as a closed subset of A^2N

And each will embed into the next as a closed subscheme of the next

So if I can take colimits of a sequence of closed immersions of (affine) subschemes I thought that would give me the X I wanted

(i first tried constructing X as a subset of A^infty but it wasn't locally closed and so I didn't know how to show it was a scheme)

[stoic rose]

@latent anvil Can't you just take infinitely many copies of Spec k[x,y]/(xy) and glue the y-axis of the i'th copy with the x-axis of the (i+1)^th copy?

[latent vector]

hey

can i get some intuition on the first isomorphism theorem?

i understand how modding out by the kernel makes it injective, but where does the surjectivity come from

also merry christmas

[carmine fossil]

G/ker(T) is iso to im(T)

Because, You can think of each element of G/ker(T) as a preimage of an element in im(T)

[latent vector]

wait slow down

if G is homomorphic to H

then the first isomorphism theorem says

G/ker(T) is iso to H

[carmine fossil]

(I am using T to denote a map)

[latent vector]

oh fuck

mb

i'm sorry it's 6am

but yeah

wait hold on

G/ker(T) is isomorphic to im(T)

right

[carmine fossil]

If T is surjective,im(T) is the codomain

[latent vector]

hold on i had it for a second

why is G/ker(T) iso to im(T)

T maps G->H

and then modding out in one

yeah

doesn't that assume

oh fuck

nah that's dumb

why the fuck is T surjective

[carmine fossil]

Because im(T) is image of map T?

[latent vector]

no i mean

i guess i'm still stuck on

why G/ker(T) is surjectively mapped to im(T)

fuck it i'll do it tmrw

thanks for your help

good night man

[carmine fossil]

You know how the map T from G to im(T) is surjective?

G/ker(T) is G,but you are grouping together all elements which give the same image

[latent anvil]

What do you mean by gluing here? The only theorem I know about gluing schemes in general is about gluing along open subschemes, and the ones you're talking about are closed

[latent vector]

@carmine fossil it makes perfect sense

i thought the theorem said somehting different

not that G/ker(T) is iso to im(T)

but that like

G/ker(T) is iso to H

[carmine fossil]

If T is surjective,im(T) will be H

[latent vector]

t isn't necessarily surjective

which fucked me up for a sec

on the website i was reading it assumed it was surjective ig and said im(T) = H

[stoic rose]

Well actually take the axes minus a point

That's open

Hum wait no that won't work immediatly

It should work by removing two points though I think

Say we glue the y-axis of the first copy minus (0,0) and (0,1) with the x-axis of the second copy minus (0,-1) and (0,0), with the identification (0,y)<--->(y-1,0)

Something like that, glue the pink part with the green part

Just do that infinitely many times

[rigid cape]

anyone done quaternions?

[latent anvil]

Oh that sounds totally plausible radiateur

[rigid cape]

k, if anyone has done quaternions, i made this simple program in python for using quaternions to rotate 3d coordinates. It is pretty simple:

import math

class Quaternion:
    def __init__(self, vector, angle):
        self.rad_angle = math.radians(angle)
        
        self.w = round(math.cos(self.rad_angle / 2), 10)

        self.x = vector[0] * math.sin(self.rad_angle / 2)
        self.y = vector[1] * math.sin(self.rad_angle / 2)
        self.z = vector[2] * math.sin(self.rad_angle / 2)
        
        self.quaternion = ["Rotation: " + str(self.w), "Vector: " + str(self.x) + ", " + str(self.y) + ", " + str(self.z)]

    def log(self):
        print(self.quaternion)

quaternion_one = Quaternion((1, 0, 0), 90)
quaternion_one.log()

quaternion_two = Quaternion((0, 1, 0), 90)
quaternion_two.log()

Any suggestions

?

[latent anvil]

What do you want suggestions for?

[rigid cape]

any improvements, or if im doing something wrong which there is a high chance of.. would you say that this quaternion rotation is correct?

[golden pasture]

suggestion: dont use floats

floats are bad

[rigid cape]

are you sure?

[golden pasture]

yes

[rigid cape]

since they give you high accuracy tho

[golden pasture]

no

floats bad

[rigid cape]

from ur suggestion i can tell they gave you a hard time didn;t they

[golden pasture]

quaternions are useful for rotations cuz you can in some cases represent as rationals

and screw storing sin and cos values

ouch

[rigid cape]

ur suggestion seems to have messed up my program tho

@golden pasture sorry but imma have to say no to your suggestion

[golden pasture]

T_T

[rigid cape]

cus this suppposed to be the correct output

but instead i was getting 0, 0, 0

as the output for the vector part

[latent anvil]

did you just just replace the floats with ints?

Also this isn't really relevant to this channel imo

[rigid cape]

yes

[rigid cape]

oh sorry

[fierce perch]

Lol

[sturdy marsh]

tolaria what's the thing in your pfp

[chilly ocean]

[fierce perch]

Dessin d'enfants

[pallid ember]

typo in the second line

shouldve been -/-/-__ instead

[molten silo]

Help

[mint gulch]

Think about the subgroups of that group

There are exactly two cases for that

[chilly ocean]

for non-isomorphism of the two given groups, look at cyclicity

for order 4 you could probably even brute force multiplication tables

[old lava]

probably, it's only 32 entries in total

[mint gulch]

Hint: Since the group has order 4 only can have subgroups of order 1, 2 and 4

[old lava]

lol this reminds me

of this d&f problem

[mint gulch]

See Problem 11a, my reasoning was to take the least common multiple and since is Z, the ideal has to be of the form aZ, then, the Annihilator is 600Z, right?

I mean, since is an Ideal and 600 belongs to the annihilator, then 600 Z is a subset of the Annihilator. And if x belong to the annihilator, then, in particular x(1, 1, 1) = (0,0,0), that means 24 divides x , 15 divides x and 50 divides x, hence 600 divides x. Is my reasoning ok?

🤔

[molten silo]

https://cdn.discordapp.com/attachments/496784958430380033/792197003269636127/unknown.png

So i used the fact that it can only be isomorphic to z4 if and only if it cyclic

Otherwise its multiplication table is the same as z * z or the Klein - Four Group

Thus making it isomorphic

[mint gulch]

Let me see the oroof

Proof

@molten silo

But, the idea is ok, but I want to know how you get that

[molten silo]

Didnt write it up, just trying to get an idea

How about this one

[short geode]

Chinese remainder theorem is very useful here.

[carmine fossil]

You don't need crt for this

[fading wagon]

identity obvious, need to show operation is closed, inverse exists and operation is associative

[molten silo]

but what is the inverse

[fading wagon]

use the bezout's lemma

[molten silo]

But i cant put the pieces together

[short geode]

for closure I think you will need crt

[carmine fossil]

You don't

You need gcd(m,a)=1 and gcd(n,a)=1 implies gcd(mn,a)=1

[molten silo]

ok

but what about the inverse?

[carmine fossil]

(Hint:||Look at the Hint given||)

[molten silo]

i hate my life

[short geode]

we are talking about mod not gcd. While gcd=1, mod is varying

[carmine fossil]

If gcd(mn,a)=1 then gcd(mn (mod a),a)=1

[molten silo]

yes

[carmine fossil]

So, That's sufficient

[short geode]

Identity = 1

[molten silo]

iverse?

inverse

[short geode]

Can we say if a belongs to G then n-a is it's inverse

[molten silo]

isnt that for addition

this is for multiplication

[short geode]

yes, if a is coprime to n then n-a is still a coprime.

[molten silo]

i see

[short geode]

nvm

[molten silo]

ok

[glossy yoke]

what? no.

[short geode]

why?

[molten silo]

ok

[viscid pewter]

multiplicative??

[carmine fossil]

Multiplicative inverse of a(mod n) is b(mod n) such that ab+nm=1

[glossy yoke]

the multiplicative inverse of 3 mod 7 is not 4.

[viscid pewter]

the multiplicative inverse of 1 mod 3 is 1 not 2

[short geode]

okay dammit

I am sorry

OOF

[molten silo]

thank you

[glossy yoke]

multiplicative inverses can be found using bezout's identity.

[molten silo]

i see

its simple

[short geode]

smh, I am an idiot.

[carmine fossil]

I don't see how crt can be useful here

[viscid pewter]

crt?

[carmine fossil]

This

[viscid pewter]

oh that

[molten silo]

thank you boys

[viscid pewter]

yeah so inverses are covered by the word 'coprime'

[glossy yoke]

rings are annoying. i like groups. don't say rings are groups, ok?

[viscid pewter]

rings are groups

.>

[molten silo]

Haven't done rings yet

Next semester

[viscid pewter]

they're just groups with more stuff on top

[latent anvil]

Rings are groups is a bad take

It's like saying groups are sets

[inland bough]

if S is a subset of affine n-space over an algebraicly closed field k, then if I is a prime ideal then Z(I) is irreducible

i'm having trouble finding a counterexample to I is an ideal, if Z(I) is irreducible then I is a prime ideal

[stoic rose]

Take for example n=1, I=(x^2)

Then Z(I) is just a point so is irreducible

More generally as Z(I)=Z(Rad(I)) take any non-prime ideal whose radical is prime, such as P^2 for any prime P

After taking the radical though I think you get an iff, for I radical Z(I) is irreducible iff I is prime

[vestal snow]

Show that if X is a closed algebraic set of A^n and Y of A^m then X x Y is a closed set of A^{n+m}
I think that X x Y = Z(I' + J') where X = Z(I) and Y = Z(J)
I' is the ideal generated by I in A[x1,...,x{n+m}] and J' is defined similarly
Is this correct?

[latent anvil]

Yup

[vestal snow]

Thanks

[meager sonnet]

I'm new to groups and fields. Why are f(x)=x and f(a+sqrt(2)b)=a-sqrt(2)b the only field automorphisms of Q[sqrt(2)]? How about f(x)=-x?

[delicate bloom]

try to show it's not a field automorphism

it doesn't even fix Q

it's better if you try to play around on your own with these kinds of things to see, like does f(ab)=f(a)f(b) and f(a+b)=f(a)+f(b)

[devout crow]

if $a,b$ are elements of a group, is $a^n = b^n = e \implies (ab)^n = e$ always true?

[cloud walrus]

xdres

[carmine fossil]

No,Take a=(1 2 3),b=(2 3 4), ab=(1 2) (3 4) in group S_5

[devout crow]

ty 😃

[viscid pewter]

'rings are groups' isn't as bad as 'groups are sets' because the first operation is a massive gamechanger; the second thingy is more of the same

that said they're both correct, more or less

[golden pasture]

thonk?

rings are very very very different from groups

it is as bad as groups are sets

the things you care in rings is different from the things you care in groups

also different from what you care in sets

[viscid pewter]

meh

groups and rings both have structure, at least

both have operations and symbol-shuffling

[stoic rose]

But having 2 operations is an important difference

[viscid pewter]

yes ofc

[stoic rose]

Also the additive group of a ring must be abelian while general groups are usually not abelian, abelian group theory is actually pretty boring

[viscid pewter]

but not as different as groups and sets imo

[golden pasture]

it is very very ver yve ry very veryver yvery different

pls

[cyan marten]

It's a theorem that for any distinct m, n, r there is a group with an element x of order m, y of order n, and |xy| = r

[viscid pewter]

in general, it's very false

i feel like

ah, finally, found a counterexample

wait

yeah

consider GL2(R), i think that's the notation

[0, 1; 1, 0] has order 2

[1, 0; 0, -1] has order 2

their product is [0 -1; 1, 0] i believe, which has order 4

in general i think it's only true for abelian groups

[cyan marten]

Sometimes a weaker law holds, like $(ab)^n = a^n b^n [a, b]^{\frac{n(n+1)}{2}}.$

[cloud walrus]

benedictusnoctis

[viscid pewter]

ah, i've never heard of that one

[cyan marten]

It should be (n-1)

Rings are essentially monoids acting on abelian groups, right?

Is there a way to formalize this notion?

If we have an abelian group A, we can also regard it as a monoid, as we can End A. Then if f : A -> End A is a monoid homomorphism, we can define an operation * that distributes over addition by a * b = f(a)(b).

But A is equipped with a different monoid structure...

[inland bough]

@stoic rose thanks, so basically if I is an ideal, we have rad(I) is prime iff Z(I) is irreducible

also i was not aware Z(I)=Z(rad(I)) but yeah i get why it's true

i prefer this because given any subset of polynomials T we can figure out whether Z(T) is irreducible by just taking the ideal generated by it, and then it's radical

[stoic rose]

Yep exactly

And the Nullstellensatz states that there is actually a perfect one-to-one correspondence between varieties in A^n and radical ideals, so for example two ideals with different radicals will yield different varieties

[inland bough]

i can't believe i missed this tysm

should be algebraic sets and radical ideals but yeah

or actually there is one between varieties and prime ideals

[stoic rose]

If your definition of varieties requires irreducible then yes

There are varying definitions for varieties

[sturdy marsh]

A British mathematician was giving a talk in Grothendieck's seminar in Paris. He started "Let X be a variety...". This caused some talking among the students sitting in the back, who were asking each other "What's a variety?". J.-P. Serre, sitting in the front row, turns around a bit annoyed and says "Integral scheme of finite type over a field".

I think we also assumed separated in my class

[stoic rose]

Right looking at wikipedia it seems like most definitions of variety include irreducible

[sturdy marsh]

yeah I've never seen the integral assumption dropped

[stoic rose]

The book I'm currently reading doesn't though

I think

[sturdy marsh]

which book?

[stoic rose]

Milne's Algebraic groups

Probably because you can often have nonconnected algebraic groups

[sturdy marsh]

ig you only need reduced to make all the algebraic groups theorems work

[stoic rose]

More precisely he asks for geometrically reduced

[sturdy marsh]

hmm

also by "all the algebraic groups theorems" I meant Cartier's thm on smoothness

that's the only one I really know

[stoic rose]

Hum Liu also doesn't require integrality

Actually he doesn't even require reduceness

[sturdy marsh]

I think stacks project, hartshorne and mumford do

[stoic rose]

I don't know much either, I just started to read that

[sturdy marsh]

cool

I will probably learn Lie stuff before I try to learn algebraic groups

I made a mistake not taking the lie groups course last year 😦

[stoic rose]

I am personnaly more of an algebraist than an analyst so I prefer to dive directly into algebraic groups

[sturdy marsh]

that's fair

it's just that I've been told a lot of the ideas are motivated by Lie theory 🤷

[stoic rose]

Yeah, they are pretty similar I think

Most Lie groups are also algebraic groups

And you can associate a Lie algebra to an algebraic group in a similar fashion as for a Lie group

And I think you can use this to classify semisimple algebraic groups from the classification of semisimple Lie algebras, similarly to what is done with Lie groups

[sturdy marsh]

hmm

interesting, wikipedia says that this works over char p as well

[stoic rose]

There are some complications in char p but yeah you can do some stuff there too

Which is pretty cool

[broken vale]

this is a dumb question but can someone please explain this concept to me:

for example, in the polynomial ring F_3[x] where F_3 is the finite field containing 3 elements 0, 1, and 2, we have that the polynomial x(x-1)(x-2) is = 0 regardless of what element in F_3 you plug in for x. but it is not the 0 polynomial

how is that even possible? i don’t know why that’s messing with my head so much

[vestal snow]

@broken vale They induce the same polynomial function from F_3 to F_3, but they are not the same polynomials. Remember that the definition of a polynomial is a finite formal sum of the form a_nx_n + ... + a_0

And the notion of equality in the set of formal sums of this form is that the coefficients are equal in for each power of x

[broken vale]

that makes sense but it’s still a mind fuck

[vestal snow]

I actually had a related question which I was about to ask

Isn't F_3 a counterexample to this

[broken vale]

what is k^n? direct sum of n copies of a field k?

[vestal snow]

Yeah

This question is from introductory AG though

[broken vale]

what does it mean for a subset to be irreducible

[golden pasture]

wait wdym

[vestal snow]

I(Z(0)) contains the polynomial x(x-1)(x-2), but this would contradict the fact that I(Z(0)) would have to be prime using the theorem

[broken vale]

i have a lot of questions before i can try to help

[vestal snow]

It's okay. There are others here

[broken vale]

i want to try though.

are we considering these questions in polynomial ring k^n[x]?

[golden pasture]

wait why does I(Z(0)) contain x(x-1)(x-2)

[vestal snow]

Because Z(0) = F_3 and that polynomial disappears on all elements of F_3

[golden pasture]

this is confusing but polynomials are not defined by their values at each point

[golden pasture]

uhhh

so here we only look at F3[x]

you can have a nonzero polynomial whose value at every point are all 0

but still be nonzero

[broken vale]

bizarre but yeah i understand

[golden pasture]

think of evaluation maps as modulo the ideal x-a

in basic nt we have 0=2 mod 2

[golden pasture]

oops this analogy doesnt quite work lol

but the idea is you cant just look at the values of polynomial at every point to figure what it is

kinda not what you are used to

[vestal snow]

wait is this related to my question or his?

[broken vale]

mine i think

[vestal snow]

Okay cool

[golden pasture]

this also I(Z(0)) is just 0

[broken vale]

does this apply to Z[x]?

[vestal snow]

Why is I(Z(0)) = 0?

Isn't it defined as all polynomials disappearing on the set Z(0) = F_3?

[golden pasture]

it doesnt cuz you can figure out what the polynomial is from the values on integers with classical methods

[broken vale]

hmm. weird

so follow up question then

[golden pasture]

oh hm

lemme uh

open up sage for a moment

[broken vale]

if two polynomial in Z_n[x] are equal (meaning equivalent mod n), then are all of their coefficients equivalent mod n?

[golden pasture]

nope

[broken vale]

uh

what if n is prime

[vestal snow]

Why do you need sage for it lol?

[golden pasture]

im just running sanity checks haha

ok so

right

F_3 isnt algebraically closed

[broken vale]

x^2+1 has no roots

[golden pasture]

construct the polynomial product (x-i) over all i in the ring

you found a polynomial that is everywhere 0 and nonzero itself

[broken vale]

yes but

[vestal snow]

I don't think we need to assume that

Here is the proof of this theorem

[golden pasture]

i was thinking F3 is algebraicly closed for some reason lol

ok anyways

[vestal snow]

and it doesn't use algebraically closed

[golden pasture]

yea it should be prime

hm

[broken vale]

i’m saying that if the polynomials p(x) and q(x) are “equivalent mod n” does that just mean that “all their coefficients are equivalent mod n”? or does it mean something else

^ that’s last question then i stop interrupting i promise

[golden pasture]

i feel it needs algebraic closure🤔

[golden pasture]

coefficients equivalent

[broken vale]

so the answer to my question is yes

okay thank you

[vestal snow]

I figured it out

Z(0) = F_3 is not irreducible

[golden pasture]

oh lol

right we have a direct construction from the proof even XD

[vestal snow]

This confused me

Because it treats A_{F_3} as an affine space

Which it isn't because F_3 is reducible

[chilly ocean]

Related to the fact the centralizer of a product of group subsets is the intersection of the respective centralizers, looking at those products which are the entire group-- as well as the fact that the centralizer of a subset is the centralizer of the subgroup generated by that subset-- we can reduce the problem of computing the center of any given group to taking the intersection of centralizers of convenient choices of subgroups on the "lower end" of the lattice of subgroups
Why does this concept smell very much like that of algebraic sets? Like why are the properties all basically the same except we work with zero-locii and ideals?

[bronze trench]

Hello my people. I have a particular issue I can't seem to find a lot of resources for, since this seems to be pretty specific. If anyone can help me that's be greatly appreciated!
So this is about the approach used by Rotman's book regarding the extension problem on groups. My question is the following:

Given data $(Q,K,\theta)$ and a group $G$ which is a semidirect product, if $G$ satisfies the data, is it necessarily a semidirect product of $K$ by $Q$?

[cloud walrus]

bennycunha97

[bronze trench]

I am pretty confident it is (since there's a homework question I'm trying to solve now which I think implies that)

If I can prove that I have a proof for the HW question but that step is eluding me. If possible I'd prefer hints or observations instead of a straight proof because I don't think it'd be fair if anyone does my homework 😂

[stoic rose]

@bronze trench What do you mean by "G is a semidirect product" and "G satisifies the data"

[bronze trench]

ok I think this is quite specific which is also giving me trouble. The first part means that there are a normal subgroup A of G and a subgroup B of G such that

$G= A \rtimes B$

[cloud walrus]

bennycunha97

[carmine fossil]

I think he means Q is normal in G and Q inter K={e},G=QK and thetha is a homomorphism of K into Aut(Q)

[stoic rose]

Every group is a semidirect product of the trivial group by itself

So this condition is vacuous

[bronze trench]

hm, something's going on because I'm pretty sure the last part is false in general without some assumption then

either he means G is a non trivial semidirect product or he means that it's already a semidirect product of K by Q and I don't need to prove that part. It's a bit ambiguous, I think I'll ask the prof

[stoic rose]

Can you show me the exact question?

[bronze trench]

if I can assume that the thing just feels wrong, that makes the rest of the problem pretty easy and my prof is not one to make easy HW questions lol

yep just a sec

I'll get the needed definitions if you want me to. I'm still in (i)

[stoic rose]

Yeah what does it mean to realize the data?

[bronze trench]

but point is, any 2 semidirect products of K by Q realising the theta are equivalent and that's no hard so I don't think I can assume already that the thing is indeed a semidirect product of K by Q right from the start. I'll get those definitions, 1 sec

[stoic rose]

I would guess it just means that G is a semidirect product of Q by K with conjugation given by theta

[bronze trench]

well not quite, then group cohomology comes in and the thing being a semidirect product means it's in the 0 class of the cohomology group

satisfying data is a bit more general and in general this is used to somehow classify group extensions

one of which is the semidirect product and is the "simplest" one

[stoic rose]

Hum I got to go I'll look at that later if I have time

[bronze trench]

ok thanks 🙂

[bronze trench]

update: my question is pointless, I'm dumb at interpreting things. We are in the particular case I mentioned, where the group is a semidirect product of K by Q

[stoic rose]

I just read the def and yeah that's what I was thinking

[bronze trench]

we had a definition for a semidirect product satisfying theta and this is still that, but saying it satisfies the data is the same

implicit in that "a semidirect product" is indeed K by Q 😛

yeah my prof just cleared that up for me

also I'm happy because his homework is usually really hard but he gave us a xmas gift this time and like half of it is kinda strightforward

I mean the other half is probably still near impossible but at least it's not all of it lol

he is really cool in giving marks tho, he makes things really hard but then is very lenient while giving the grades so it kinda evens out

and tbh I prefer that to easy assignments where he profs are nitpicking every detail

[viscid pewter]

does every infinite group have to have a nontrivial subgroup?

it should, right?

because in order for an element not to generate a subgroup it has to generate the entire group

so every element is a generator? which just feels wrong, idk why

[sturdy marsh]

If an element has infinite order, you have a 'Z' in there

a more interesting problem is show that a group with only finitely many subgroups must be finite

[viscid pewter]

ah, yes, i see

ok very simple then

[next obsidian]

What do you mean brofibration, those two things you said are the same thing :^)

[sturdy marsh]

well yeah, because it's true?

they're equivalent

perhaps I should've worded it better

If G has only finitely many subgroups, then G is a finite group

Oh wait were you making an ‘all groups are finite’ joke

[carmine fossil]

Ok,Field Theory fun fact: Every field has a copy of Q(if infinite)or Z/pZ,(where p is the char of Field,if char is finite) as a subfield

[uncut girder]

no

there are infinite fields of char p

Ex: F_p(t)

[next obsidian]

Nah, I was just joking that the key in proving finitely many subgroups => finite is to realize that infinite order element means copy of Z

[sturdy marsh]

pretty much

you also need to look at the case where every element has finite order

just cover the group using cyclic subgroups

[cyan marten]

Nice. Is there a "topological" way to think about this result?

[next obsidian]

Topology lol

[latent anvil]

?

I mean it says that an infinite degree cover has an intermediate (in sufficiently nice circumstances) right?

[cyan marten]

Unfortunately I am unfamiliar with covering spaces.

If we partition the group G into equivalence classes, where two elements are equivalent iff they generate the same cyclic subgroup, then we get a set G/~ which we can naturally preorder by saying a <= b iff the cyclic group generated by a is contained in that gen. by b.

We can then endow G/~ with Alexandrov topology (open sets are upper sets) to see what happens..

Since G has finitely many subgroups, G/~ must be finite, and as Chmonkey said every element must then have a finite order, so we get nothing from making G/~ into a topological space at all.

[latent anvil]

Isn't G/~ just the set of cyclic subgroups?

[cyan marten]

Yeah!

[cyan marten]

A polynomial in Z[x] can be reducible mod p for every prime p, but not in Z. The minimal degree for a counterexample is 4. Any idea why?

I think it should be obvious (for example because reducibility = having a root for deg < 4)

For a quadratic to be counterexample, it must have no root, so the discriminant D is not a square, and we can find a prime p for which D is not a quadratic residue.

[latent anvil]

Let f be a nonconstant integer polynomial. Then lim n to infinity f(n) = ±infinity, so for some sufficiently large N we have |f(N)| > 1. In particular f(N) has a prime factor p. Then f(x) has a root mod p, namely N. This proves that any nonconstant integer polynomial has a root mod some prime

polynomials of degree 2 or 3 over a field are reducible iff they have a root

(you don't really need nonconstant here, f just needs to not be constantly ±1)

[cyan marten]

It really is obvious!

[latent anvil]

Err wait I think I flipped around your question in my head

You wanted to see why there's no irreducible polynomial over Z which is reducible mod all primes

I should know better than to do number theory 😓

[cyan marten]

Same here.

[oblique river]

This is a galois theory explanation but you can consider an irreducible polynomial f(x) in Z[x] which has galois group Z/2Z x Z/2Z over Q

the galois group of f(x) mod p must be 1) a subgroup of Z/2Z x Z/2Z and 2) cyclic

so therefore for every prime p it must have size smaller than 4 and therefore the polynomial is reducible mod p

@cyan marten

also @latent anvil if you're curious

sorry if I misinterpreted the problem

[cyan marten]

We want to show if a polynomial f is at most cubic, and is reducible mod every prime, then it is reducible in Z. I can't follow your example.

[oblique river]

oh sorry my example was giving an infinite class of examples of polynomials of degree 4 which are irreducible over Z but reducible mod every prime

which isn't what you wanted

[cyan marten]

Interesting!