groups-rings-fields

[grizzled crane]

Let me check again

[topaz solar]

Check +

[coral spindle]

There are a few issues with it, in fact

[topaz solar]

if we did Z -> Z instead, is it a ring?

[coral spindle]

I was gonna say lol

[topaz solar]

Do be careful

[grizzled crane]

The example i have is from R to R

[topaz solar]

Ah, but are there any added conditions

Do we look at all functions?

[delicate orchid]

yup

[topaz solar]

Are we doing the same multiplication?

[grizzled crane]

Nope the multiplication is different

Not composition

It’s fg(x)= f(x)* g(x)

I just started with all of them, sorry for terrible mistakes

Let me try the routine check of ring axioms here

see if i get them right

First I have to check whether the set of all functions from N to N under point wise addition is an ableian group

I think N doesn’t have inverses to undo the effect to point wise addition

@topaz solar

[topaz solar]

Yep

[grizzled crane]

Hopefully Z to Z works

as in form an abelian group under point wise addition

[coral spindle]

I'd like to hear your conclusions :)

[low wyvern]

Can someone please explain why {2,5}*{2,5} = {1,4} (why the 1 is there...). H is a normal subgroup of Z6 for context.

[coral spindle]

Remember we define Ha * Hb = H(a+b).

[slim kayak]

thats notation sure is something

[coral spindle]

Now what this means in terms of sets is the following

We know that if a is in some coset, then the coset is Ha

So in particular, we know {2, 5} = H2 = H5

So we can now calcualte any which way we like

[low wyvern]

I don't get how 2+2mod6 has become 1

[coral spindle]

It has not.

I am going to explain, please be patient.

[low wyvern]

ye np

[coral spindle]

H2 * H2 = H(2+2) = H4

Now what is H, again? H = {0, 3}

[low wyvern]

it is

[coral spindle]

I know, I am stating, not asking.

So H4 = {4, 3+4} = {7, 4} = {1, 4}

Note: we never said 2+2 mod 6 is 1

[low wyvern]

I see

[coral spindle]

So that is one way to calculate this.

Now another way is the following

You can check for yourself that if X, Y are cosets of H, then X * Y = {x+y | x in X, y in Y}

So we have {2, 5} * {2, 5} = {2+2, 2+5, 5+2, 5+5} = {4, 7, 7, 10}

Now of course, this last set is equal to {1, 4}.

This should be clear.

Happy?

[low wyvern]

Ye, I realise why I was thinking incorrectly. I was confusing it with direct product groups....

[coral spindle]

You should keep in mind that cosets are almost never groups.

[low wyvern]

Ok 👍

[celest cairn]

Not that anyone wants to check the calculations (don’t waste your time) but I think I found the minimal polynomial of $\mathbb{Q}(\zeta_3 + \zeta_6 \zeta_9) / \mathbb{Q})$.

[cloud walrus]

Sapphire

[celest cairn]

kind of.

In better terms, it’s just $\Phi_{9}(\frac{(2\alpha^2)}{5})*\Phi_{9}(\frac{2(\alpha-2)}{5})$.

[cloud walrus]

Sapphire

[rocky cloak]

,w minimal polynomial (exp(2pi i/3) + exp(2pi i/6)exp(2pi i/9))

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jagr2808

[rocky cloak]

@celest cairn

[celest cairn]

Dang I probably did something wrong, thanks.

[steel folio]

Can someone explain why point c) should be true? Both implications seem false to me

[rose prism]

why do they seem false

[steel folio]

if the 2 functions are equivalent then gx = xh^-1 for every x in G no? How can I conclude from this that g and h lie in the center of G? And viceversa, if g and h lie in the center of G how can I conclude that gx = xh^-1?

[rose prism]

yes those are the things you need to answer

but why do you think theyre false?

[formal ermine]

it's been like a year, are you still computing minimal polynomials

[steel folio]

For the second one, if g and h lie in the center of G, considering gx and xh^-1, we should have gx = xh^-1 if and only if g = h^-1, but there could be some other element in the center

For the first implication maybe because we have gx = xh^-1 for every x in G and by letting x = 1 we have g = h^-1, so gx = h^-1 x and h lies in the center, and xh^-1 = xg and g lies in the center, and we're done

[rose prism]

yes i think that proves the first implicaiton

[steel folio]

but the second seems false to me

[rose prism]

yeah

[celest furnace]

You canceled an x on the right and on the left in the same step?

[rose prism]

im not sure its true

it seems stronger than the “deduce that…”

[celest furnace]

You could plug in x = h though and get to the same thing so yeah this is weird

[steel folio]

because h and g lie in the center, we have xh^-1 = h^-1 x so i cancel only on the right (or, equivalently with g, only on the left)

[celest furnace]

Ah ur right

[rose prism]

with sigma_g = tau_h => g,h in Z(G) can you already prove the last thing?

[steel folio]

I've tried searching on internet but I get only results about character theory and I don't understand anything

[rose prism]

the conclusion is much weaker

so i think its just a mistake

you can prove the conclusion

about pi(G) n lamdba(G) = pi(ZG)

[rose prism]

from the implication here

(the implication proves pi(G) n lambda(G) \subset pi(Z(G)). the reverse inclusion is trivial)

[steel folio]

yeah

[rose prism]

whats an example

of a result

[steel folio]

like this one

https://math.stackexchange.com/questions/1613135/showing-that-the-right-and-left-regular-representations-are-equivalent

[rose prism]

this is something else

the exercise is weird

idk what it means

though tbf i know like no rep theory

but i dont think this is so closely related to the question of equivalence of reps

it is way stronger

[steel folio]

thinking about it, isnt' g = 1 and h some other nonidentity element of G that lies in the center a counterexample?

[rose prism]

counterexample to what

[steel folio]

to the reverse implication (if g and h lies in the center then the 2 functions are equivalent)

[steel folio]

I don't know, it's just what pops up if i try to google

[rose prism]

yes that implication is wrong

without a doubt

[steel folio]

a clear example is considering g = 1 and h = r^2 in D8

it can't be right

[rose prism]

yes

it is wrong

but the conclusion of the exercise is right

they clearly made a mistake

in trying to structure the exercise

[steel folio]

yeah the conclusion is right

thanks 👍

[celest cairn]

Yep, though I’ve gotten a lot more into Galois theory and proving simple things using abstract algebra.

[formal ermine]

that's awesome to hear 👍

[celest cairn]

Ty man. Also I’m having my first abstract algebra tutoring session on the 1st. Should make things a lot easier now that I don’t have to teach all of it to myself.

[frank cosmos]

Is this proof of cauchy theorem (for abelian) right? Claim: G finite abelian has element order prime p | |G|

We prove this by induction on |G|. Take arbitary g. Then <g> has an element q of prime order, if this is p we are done, if not then G/<g> has order |G|/q by Lagrange's theorem. This group has an element order p by hypothesis and then the preimage under canonical surjection has order dividing p, hence order p as desired.

[coral spindle]

... the preimage under canonical surjection has order dividing p ...
This isn't true

It is true that it has order divisible by p though, so you can still fix it.

[frank cosmos]

i see. if this preimage is m, |m|=kp, then does m^k work then

[coral spindle]

Yes that's exactly right

Perhaps unsurprisingly then, this theorem about finite Abelian groups boils down to one about finite cyclic groups. It's almost as if finite Abelian groups are made up of cyclic groups or something crazy like that...

[south patrol]

interestingly i have never sene that proof lol

[frank cosmos]

proof of cauchy theorem or the fundamental theorem

[south patrol]

cauchy

[delicate orchid]

upon

[coral spindle]

Right, because usually we would just prove Cauchy in the general case

[south patrol]

but tbf yeah that's why

[coral spindle]

And it's just

a cleaner proof

lol

[cobalt heath]

Hmm does F_{p^k} appear in the classification of finire abelian groups 🤔

[delicate orchid]

C_p^k-1...

[south patrol]

yes

well C_p^k i guess but yes

[delicate orchid]

oh the addtive group

yur

[south patrol]

i guess lol

this is unfortunate in that you mean C_(p^k -1) and i mean (C_p)^k

but yes

[delicate orchid]

C_(p^k-1) is what I me- ok yeah

[coral spindle]

A shakespearian tragedy tbh

[cobalt heath]

Ah, it has cyclic group structure for its unit, ye

(C_p)^k confuses me a bit in that I am not sure what multiplication structure to give to it >.>

[delicate orchid]

it's a direct product of groups

[coral spindle]

A group only has one operation

The classification of finite Abelian groups tells us nothing about how the 2nd operation works

[glossy crag]

There's a very neat one that has Z/pZ act on p-tuples of G such that g_1*...*g_p=1.

[delicate orchid]

no no, Boyt. Absta is clearly asking what fusion system to place over (C_p)^k, which are in one to one corrispondence with subgroups of GL(k, F_p)

[cobalt heath]

Yes, but sometimes I forgor this.

[cobalt heath]

Lmao

[delicate orchid]

unfortunately the isomorphism classes are not in 1-1 corrispondence

[cobalt heath]

Hmm, GL(k, F_p) ?
Yea sad

[delicate orchid]

sad!

[keen shuttle]

This is wonderful. My appreciation.

[glossy crag]

What's a non-trivial example of a separably closed field (e.g. not an algebraically closed one)? Just found out that's a thing.

[south patrol]

Yah that's the standard one innit

[glossy crag]

I don't think it's the standard one (at least it wasn't till recently), the classical proofs are different. This is one of those nifty "math golf" proofs.

[south patrol]

Oh lol

Fair

That is the only proof I have ever seen

[delicate orchid]

I just quote Sylow 1

[south patrol]

(other than just specialising sylow 1 lol)

[glossy crag]

Well some proofs of Sylow1 build on Cauchy.

[delicate orchid]

Oh I don’t care about being circular. I care about going back to bed asap

I do think you can prove it independently iirc…??

[glossy crag]

You can, I just meant this as an example of the possible usefulness of Cauchy.

E.g. I don't think the standard proofs demonstrate that there's a chain of p-subgroups (building up to a maximal one), but this can be done via Cauchy.

[rotund aurora]

You can take the separable closure of F_p(t) which is not perfect

(separable closure means you dump in all the roots of separable irreducible polynomials)

[glossy crag]

ik

[glossy crag]

Hm, ok, that works I guess (for any non-perfect field). Thanks.

Is anyone here familiar with "higher" local fields or w/e they're called? Vostokov/Fesenko have a definition, iirc it's the local field definition, except instead of requiring the residue field be finite, you say it's a local field of degree n-1 (when defining a local field of degree n). I wonder how big of a deal they are, in books I've only meant the regular kind of local fields so far.

[south patrol]

proofs i know don't use cauchy lol

[delicate orchid]

you think I know how to prove Sylow 1? ah ah ah

[summer path]

I don't know how to prove any of the sylow theorems

(Please don't tell me, I don't want to remember)

[rotund aurora]

I think in the proof of Sylows there is half of elementary group theory

[formal ermine]

i learnt the proofs by heart once

forgot all of them again

[rustic crown]

det no remember the proofs, but remember the ideas which can be converted to proof given 15 minutes ><

[south patrol]

i wonder if i roughly remember sylow I but am not sure

😭

[summer path]

the only thing i remember is, the first time i learned them, it was ; the second time, it was ; and the third time, it was

[south patrol]

I think it's smth like

Let $G$ be of order $p^k m$ where $(p,m)=1$ and consider set $S$ of subsets of $G$ of size $p^k$. Direct computation shows $|S| \not \equiv 0 \mod p$. On the other hand, consider the action of $G$ on $S$ by left multiplication. The stabiliser of each $s \in S$ has cardinality $\le p^k$. If the cardinality of each stabiliser is $< p^k$, then each orbit has order divisible by $p$, so that $|S|$ is divisible by $p$, contradiction

[cloud walrus]

potato

[south patrol]

So there must be an element with stabilisr of cardinality exactly p^k and that's a sylow subgroup

[cobalt heath]

Idea of Sylow thm, is that something one can recall?

Like I always manage to forgor.

[south patrol]

I just remember to consider this S

and that you need a stabiliser

[rustic crown]

i like the proof that relies on cauchy

from there you slowly build a normal series of p-subgroups

if H was a p-subgroup, you can let H act on G/H by left multiplication to get that [G:H] = [N(H):H] (mod p). so if p divides [G:H], there is an element of order p in the group N(H)/H, subgroup generated by that looks like K/H giving you a larger p-subgroup and you also get H is normal in K.

[rustic crown]

idea is just build it inductively, so not really something to forget

[coral spindle]

Ya this is the proof I know (or know the idea of) too

It's really quite pretty

[rustic crown]

[coral spindle]

Plus relies on facts about p-groups you ought to know anyway

[cobalt heath]

But how to tower above the smaller subgp

[coral spindle]

Det just explained that...

[cobalt heath]

How do I recall* to tower above the subgp?

Like I will forget the normalizer.

[rustic crown]

the whole shit with sylow is act by a non-trivial p group G on a cute enough set X, and then use |X| = |X^G| (mod p)

[coral spindle]

You know you're looking for a larger p-group in which the smaller one is normal, so it is totally natural to look at the normaliser. It's the most obvious thing to do

[cobalt heath]

Ah, yep. I see

[rustic crown]

in this case the set X = G/H is the set of left cosets. and gH lies in X^H iff for each h, we have h(gH) = gH, which is equivalent to gHg^-1 = H, so gH in N(H)/H.

[rustic crown]

yep, this action stuff tells you that as long as H is not already the sylow-p-subgroup (i.e. the index [G:H] is still divisible by p) then the index [N(H):H] is also divisible by p so you can continue the argument

proof of other sylow are also pretty easy, just mix and match various actions of a p group on interesting sets

[keen shuttle]

i don't really understand the first isomorphism theorem. here is how i've thought of it:
Suppose f : G -> H is a group homomorphism. Let K = ker(f).
We are looking for a bijective homomorphism induced by these knowns.
But bijectivity requires injectivity, and we know a trivial kernel is sufficient for an injective homomorphism.
Now, the group where K is the identity is G/K, so now we want to find a suitable codomain X where f' : G/K -> X sends K to the identity in X. If we can find this, then we have an injective homomorphism because the kernel is the identity K.
Then this part is easier to see; it's going to be the original codomain of f (i.e. H), for K just represents the things sent to the identity in H.
We restrict the codomain of that homomorphism to its image (im(f)) so that our new homomorphism is surjective and we are done: f' : G/K -> im(f) is a bijective homomorphism.

but the more popular intuition is this idea of "factoring out whatever was making f not injective"

i cannot understand where this comes from

G/K is the set of cosets of K

[delicate orchid]

f is injective if and only if ker f = 0

quotienting by ker f is constructing a group where we "set" ker f = 0, that's what a quotient is

[keen shuttle]

yes

wait

i dont get how

quotienting by ker f is constructing a group where the identity is K

does it follow that under the same homomorphism, this K becomes a trivial kernel?

[delicate orchid]

alternatively what we're saying with first iso is, if a homomorphism maps a subgroup to 0, we can actually set that subgroup to be 0 via a quotient and this gives us the image of f

this isn't really "alternatively" but it's clearer

[coral spindle]

Let's talk about the first isomorphism theorem for sets, first.

Let A, B be sets. Let f : A -> B.

[cloud walrus]

Boytjie

Boytjie

Boytjie

[coral spindle]

The first isomorphism theorem for groups is precisely the same.

There is really very little adjustment to be made.

(The appropriate generalisation of this observation is massive -- the first isomorphism theorem is really a part of a topic called universal algebra, which encompasses an enormous variety of algebraic structures.)

[keen shuttle]

wdym by 'I know when f(x) = f(y)'

[coral spindle]

Keep reading for now

[keen shuttle]

like how?

[delicate orchid]

in groups you know f(x) = f(y) if and only if x = ky for some k in ker f, for instance

[keen shuttle]

yes; the cosets of the kernel are the fibres of f

that makes sense (kind of)

[coral spindle]

So... I know when f(x) = f(y) even without knowing the values of f(x) and f(y) waoh

[delicate orchid]

tbf that part reads a little weird lol

[coral spindle]

I cooked up this explanation on the fly, cut me some slack

[cobalt heath]

Ah, is universal algebra good to learn

[coral spindle]

Not really no

[cobalt heath]

I see.. sad

[delicate orchid]

from what I understand about it, it's just model theory right

but like, less general

[coral spindle]

It's very restricted model theory

[keen shuttle]

ok but i dont see how this has to do with the first iso thm, which means i prosb dont understand ur explanation that well

[coral spindle]

OK we're doing this now

Reminder!

We're working with groups now!

So $f: G \to G'$ is a group homomorphism

[cloud walrus]

Boytjie

[cobalt heath]

I thought it might give an insight on the rough structure of everything, hopefully helping with how to e.g. classify

[keen shuttle]

ur saying that the cosets are equivalence classes, so mapping cosets to image is an isomorphism

[coral spindle]

Let me speak!

[cobalt heath]

Ah gotta just learn model theory then

[delicate orchid]

ebepbpepbpebpepbepbpebpepbep

[cobalt heath]

Lmao

[coral spindle]

Now $H$ we'll set to the kernel of $f$.

When is $f(x) = f(y)$? When $f(x)f(y)^{-1} = e_{G'}$, right?

[delicate orchid]

classify... what...

[cloud walrus]

Boytjie

[coral spindle]

So, after simplifying...

when $xy^{-1} \in H$.

[cloud walrus]

Boytjie

[coral spindle]

I.e....

[cobalt heath]

Groups

[coral spindle]

when they lie in the same coset of H

[keen shuttle]

also (sorry to interject, i wanted to say this earlier) this makes sense but it doesnt justify why every1 says 'factor' in my mind, like i dont get the notion of factoring the in traditional sense (i.e. ab + ac = a(b+c)) is used as a heuristic here but this might be an overfixation on my part and i shud rlly just kep my initial idea of what its about)

[coral spindle]

I cannot even begin to explain how useless universal algebra is at trying to classify groups

[delicate orchid]

oh I'm with you on this. I don't like calling them factor groups either,
in a way you can think of writing any g in G as g = xh for some h in H, then the "factorisation (quotient)" sends g -> x

[delicate orchid]

ALL of them?

[coral spindle]

So anyway as I was SAYING

This relation ~ I described up above for sets — it's exactly the relation whose classes are cosets of the kernel

So again — it's exactly the same thing!

[cobalt heath]

Dunning Kruger is getting me, like perhaps all group could be somewhat classified

[coral spindle]

The extra structure that comes from the group you essentially get for free

[delicate orchid]

finite groups are. Infinite groups are hopeless

[keen shuttle]

Why do we choose the kernel?

[coral spindle]

Classifying just finite groups is extraordiarily hard too tbh

[cobalt heath]

You mean finitely generated groups can be classified, right

[keen shuttle]

this isn’t true for any normal subgroup of G, what’s so special about the kernel?

[delicate orchid]

and the classification theory behind finite simple groups is one of the most incredible achievments in the history of mathematics, with a proof spanning over 11,000 pages

[cobalt heath]

Right? /j

[delicate orchid]

all normal subgroups appear as a kernel of a homomorphism

[coral spindle]

It cannot be any other subgroup. There is no choice involved.

The kernel describes exactly when things in the image are equal, in other words, when things in the original group differ by something that's gonna go to the identity.

Remember how I explained cosets? The 'power' of the kernel of f is gonna get killed off once we apply the homomorphism f. It's just gonna be the identity. So things that differ by something in the kernel are gonna be the same thing once we apply f.

Clear now?

[cobalt heath]

1 -> K -> G -> G/K -> 1 moment?

[delicate orchid]

perhaps I misunderstood the question

[coral spindle]

Finitely generated groups cannot be meaningfully classified, no

[delicate orchid]

watch this

[coral spindle]

I'm watching bish

[cobalt heath]

Why tho

[coral spindle]

Because

they're really complicated

and there's lots of em

[cobalt heath]

I mean, is there fundamental obstructions doing this

[delicate orchid]

the word problem isn't computable? maybe?

[cobalt heath]

Ah. Hmm

[coral spindle]

The isomorphism problem isn't computable either

[delicate orchid]

lol

same diff

[cobalt heath]

Gödel incompleteness might get us here then, I guess

[keen shuttle]

okay that makes sense. I’ll think about it more. Thanks

[coral spindle]

Don't sink into cranky territory man...

Like just looking in computable land for a second, at purely finite groups

We have a list of the finite simple groups, which is wonderful, and this has been extremely hard

But we currently have no hope of classifying all finite groups. Those are just finite ones!

This is because as you seemed to allude to, the group extension problem is really hard.

[cobalt heath]

Ahh

[coral spindle]

There doesn't need to be some deep reason for it, it's just really fuckin' hard!

[cobalt heath]

I mean, word problem being uncomputable simply means you cannot classify right

Or is it more complex than that

[coral spindle]

No that doesn't follow

[cobalt heath]

Ah

Now ig why you said 'do not be crank'

[cobalt heath]

What

You guys said earlier that mathematicians beautifully classified all finite groups

[coral spindle]

No, all finite SIMPLE groups

[delicate orchid]

I kind of disagree with this morally. Sure we can't write them all down but we basically know all of them with the classification of the simple ones

[coral spindle]

Nah, knowing a composition series is not enough

[cobalt heath]

Hmm.

[delicate orchid]

hmmm

[coral spindle]

If that were true we would understand p-groups perfectly

[delicate orchid]

I do

[coral spindle]

I believe you

[sand cradle]

every finite group is isomorphic to a finite group 😎

[cobalt heath]

Lmao

[cobalt heath]

We don't?

[coral spindle]

Speak to wew about that one apparently

[delicate orchid]

we understand all of the interesting examples of finite groups, then

[coral spindle]

And like, there's still mountains of work with reduction theorems. There was a big one iirc recently regarding the McKay conjecture – they'd reduced it to the FSGs? I can't remember now

[delicate orchid]

McKay conjecture mentioned
[neuron activation]

[coral spindle]

But my point is that like, there's still a lot of work involved in this stuff. We don't get a free understanding of groups from the simple constituents at all, we really have to work hard to use it

And ofc there's nothing close to a list, like for FSGs

[delicate orchid]

FSG standing for?

[tribal moss]

Finite simple group.

[coral spindle]

The acceptable acronym for groups. I once tried the same with Abelian ones and it didn't go down well

[delicate orchid]

right

[delicate orchid]

upon

anyway McKay conjecture is trivially true because I said so

[coral spindle]

You should write a paper on that

[delicate orchid]

consider the fusion system of F_S(G) and F_S(N_G(S)) etc. etc.

[coral spindle]

A final comment: I don't think we're ever going to get a classification of finite groups beyond composition series; I truly think that's the best we're going to get.

Not justified well, just an opinion

[delicate orchid]

https://webspace.maths.qmul.ac.uk/r.a.wilson/pubs_files/mckayweb.pdf the title of this paper goes hard

thought I'd share

the paper does not go hard. It's case checking on steroids

[keen shuttle]

Oh ok I see

This is super slay

Thank you

[keen shuttle]

Could you elaborate on the ‘power’ bit please tho?

Oh

Nvm I get that

Ok cool

Tu

[cobalt heath]

Huh.

Hmm, you mean like there could be infinitely many ways to compose groups?

[delicate orchid]

You can’t get all of the group’s properties by seeing how it decomposes

[grizzled crane]

what is this set Z[i]?

[summer path]

{a+bi | a,b \in \Z}

[grizzled crane]

is this anyway related to field extension?

The reason it popped up is I have seen motivation of complex numbers from a+ bsqrt(2), a,b in Q

sort of they incorporated just sqrt(2) on top of rationals

as in starting with Rationals and we want the to include sqrt(2)

further working its way to a+ib, to include sqrt(-1)

[prisma ibex]

I mean it's the ring of integers in Q(i)

[sly crescent]

Example?

[tribal moss]

I'm thinking of a group which has a normal subgroup isomorphic to C_p, and the quotient is C_2.
C_p and C_2 are both simple, but am I thinking of C_{2p} or D_{2p}?

[hidden haven]

You're thinking of D_p 😌

[sly crescent]

It’s the latter because C_{2p} has a second decomposition with the factors swapped

[tribal moss]

Then how about a C2 as a normal subgroup with C2 as a quotient. Is that C4 or V4?

[sly crescent]

The former because V4 has three of them

[celest furnace]

This doesnt make sense

[rotund aurora]

Could you classify groups of composition series length bounded by some integer k? When k=1 these are the finite simple groups

[coral spindle]

Well perhaps you can for specific values of k, but this is already a monumental task for just k=2.

[rocky cloak]

To be fair, it's already a monumental task for k=1

[coral spindle]

Lol yes, but at least that task is complete

[rocky cloak]

I can do k=0

[coral spindle]

I may just be ignorant, but I don't believe there is a complete understanding of extensions of the FSGs by FSGs, let alone a classification of these extensions up to isomorphism

Groump extemtiom promblem hard uwu

[rotund aurora]

sure. But in principle, it smells possible

[coral spindle]

It smells more possible than k=3 lol

[rotund aurora]

I mean for any given k. I have no proof, just an opinion lol. It would be pretty cool

but you would not be able to put all of them together

[slim kayak]

Maybe familiarize yourself with the k=1 case?

[rotund aurora]

I wonder if there is a rigorous notion for what it means to classify stuff. I guess it's something like a finite number of constructions parametrized by tuples of integers (for example A_n, PSL_n(q), etc) and rules (again, a finite number of "schemas" or something) that say precisely when two instances of such constructions are isomorphic

[long obsidian]

Hey is there notation for the ring of fractions of an arbitrary integral domain R?

[rotund aurora]

Frac(R)

[long obsidian]

Oh thank you!

[rocky cloak]

https://en.m.wikipedia.org/wiki/Field_of_fractions

[delicate orchid]

I love the giant Z Wikipedia adds to every ring theory embed. It’s like a mark of shame

[coral spindle]

Algebra is just eepy

Z z z z z

[rotund aurora]

Given a rigorous definition of what it means to classify structures, you should be able to prove that groups are unclassifiable. I wonder if things like this have been attempted, or if they are treated in more generality in the context of model theory

[slim kayak]

Usually something like "Any X with property P is isomorphic to one of these families:"

[rotund aurora]

mmh what gives hope is that if you look at groups of order p^n then their composition series length is n

[coral spindle]

Hope? This is one of the simplest possible cases, and we have nothing even close to a classification of p-groups!

[delicate orchid]

C_p, C_p^2. Uhh Heis(F_p). C_p^2 \ltimes C_p. SD16. Uhhhhh yeah I think that’s all of them

[rotund aurora]

is it? Most groups are p-groups, right? When you look at bounded composition length, there are not so many

[slim kayak]

Can't think of another one either, you should publish

[delicate orchid]

Classification of p-groups would go so hard though. We’d probably get the isomorphism classes of all possible block algebras from it 😋

[rotund aurora]

I'm trying to see how many groups of order p^n with n fixed are there as p varies, I think it's at most polynomial in p

[coral spindle]

Restricting to literally exactly one finite simple group — and an Abelian one too — in your composition series is about as simplified as we could get, yes!

[delicate orchid]

Can I put a lil A_5 on that thang? As a treat

[coral spindle]

Wew can have little a A_5 as a treat

[rotund aurora]

well yes but the fact that there are not so many p-groups of bounded composition series length at least gives hope, that is what I meant

[coral spindle]

Well that's not true either

[rotund aurora]

it gives hope to me at least

[coral spindle]

Most groups of order <= 1024 are of order 1024

I need to get the numbers here but I remember this tidbit

The point is, there are lots of p-groups

[delicate orchid]

98% maybe. That might be 2048 though

Just SmallGroups it

[coral spindle]

https://oeis.org/wiki/Number_of_groups_of_order_n

Here's a list of # of isoclasses of groups, per order. Note how the numbers jump when you reach a prime power

[rotund aurora]

but what is the exponent? This is where I'm getting at

[delicate orchid]

Ah nope. Over 99% of all groups of order less than 2000 are order 1024

[coral spindle]

Even worse lol

[delicate orchid]

A ten term composition series…. Think about what that does to a man…

[rotund aurora]

1024 is 2^10. If you consider p^10, then the proportion of groups of order exactly p^10 in the set of groups of order <=p^10 is not as high as p gets larger, I think

[rotund aurora]

but I could be wrong

[slim kayak]

1024 tidbit was right

By a lot

[delicate orchid]

It isn’t, purely because of 2-groups existing. If you exclude q-groups with q<p I imagine the proportions are still yucky gross ew

[coral spindle]

I have no idea how to count that but yeah I doubt

[rotund aurora]

doubt what?

[coral spindle]

Doubt that it gets easier, as you predicted

[rotund aurora]

I mean, it does

[coral spindle]

Sure

[rotund aurora]

I rather attempt to classify all groups of composition series length 10 than all groups

[coral spindle]

Lmao

[rotund aurora]

and as Wews said, if 2^k<p^10<2^(k+1) then you would expect most groups of order <=p^10 to be of order 2^k, not of order p^10

[delicate orchid]

I feel like it actually only gets harder. C_2 has no automorphisms so if you even look at split extensions at the first factor there’s no choice to be made - this is not the case for p > 2.

although perhaps not. Good luck fitting C_p into C_p-1 non-trivially

[rotund aurora]

Anyway, you have this. The problem is the O(m^(5/2)), it could depend on p (probably does), so this does not imply polynomial in p. Although it could still be, I don't know the details

[delicate orchid]

I’d love to learn how people come up with these bounds

[rotund aurora]

This is from Enumerating finite groups by Neumann, chapter 5

to give lower bounds, you just construct a shit ton of p-groups lol

[delicate orchid]

My favourite!

Just fyi we don’t have a classification for p^8 yet iirc and p^7 was barely doable

[rotund aurora]

wait good news

so its at most polynomial

mmh the proof is not super long

[naive whale]

https://inquirerslab.me/index.php/2023/12/21/dihedral-group-visualizer/
So I made this toy recently, wanted to share with my friends here as maybe someone will find it helpful as an educational aid. You can visualize dihedral groups of different orders and their various actions, and it will read the current state back to you so can also be used as a sort of calculator

Was pretty fun to write, it keeps track of the state as a matrix and uses some linear algebra tricks to decode whether it's currently a rotation or reflection etc

[delicate orchid]

Conductor we have a problem

[naive whale]

yeah once in a while it bugs out

[naive whale]

Do you remember how you got that? Which group and which actions?

[delicate orchid]

Uhh D_16 and no, sorry

I wasn’t really paying attention

Are you using a look up table to map the matrices back to the group elements

[formal ermine]

s7 then r8 on n = 10

@naive whale

[naive whale]

Hmm, seems like it's s4 that it has trouble with on D16

[coral spindle]

Gonna bet it's some tan(90) business

[naive whale]

Nope, it computes eigenvalues and eigenvectors of the matrix, if one of the eigenvalues is -1 then it assumes it's a reflection and the other eigenvector gives the axis of reflection, then it uses that to find the angle. Otherwise it assumes it's a rotation and computes the angle etc

Then it figures out which element based on the angle

[coral spindle]

You really ought to represent the group, internally, purely symbolically. Once you've done so you can convert the abstract group element to a matrix — this is called a representation btw. This would be a much more computationally efficient and stable way to do this!

[naive whale]

Yeah, it would probably be cleaner to do it that way, it's handy to have a matrix anyway though because it uses the state matrix when rendering and applies to the polygon and all the vertices

when you click an action it "interpolates" between the current state matrix and the new one

Yeah, I see the issue, should be an easy fix

[naive whale]

Fixed now at least

[naive whale]

Epic that someone found an issue 10 seconds after I posted, thanks for the beta testing 😎

[fossil hound]

thats cool ngl

[rotund aurora]

Is this true? (From the Feit-Thompson theorem wikipedia article)

it sounds very ironic. I don't know why the proofs in pre 1960 group theory would be particularly short

[delicate orchid]

17 page proofs... gross...

[rustic crown]

17 pages of a single dense proof or 17 pages of more theory about new definitions

latter doesn't sound bad at all

[crystal turtle]

Read the next section in the article. It points out that afterwards several longer proofs were published. But ofc 17 pages for a single proof is still pretty long

[delicate orchid]

reformatting the classification of finite simple groups so it doesn't use any lemmas and it's literally just 11000 pages of dense bullshit

[icy bear]

I was trying to solve this problem and decided to start from SL2 since from it to GL2 is easy
from the matrix
[a, c]
[b,d]
the determinant is ad - cb, which I want to be equal to 1
so
ad = 1 + bc
since both a and d can't be 0 at the same time I have p choices for a and p-1 choices for d
now I'd think that I'd have p choices for b and c would be uniquelly determined, so it should be p²(p-1), why is it p(p-1)(p+1)?

[rotund aurora]

lol imagine it was

\begin{theorem}
blah blah
\end{theorem}
\begin{proof}
...several hundreds of thousands of lines...
\end{proof}```

[coral spindle]

since both a and d can't be 0 at the same time
sure about that?

[delicate orchid]

is this for a or b

[icy bear]

just b

[delicate orchid]

I like a. So I'm going to give an answer for that one irregardless

[icy bear]

a is easy

oh sure

[icy bear]

yeah cus otherwise the matrix wouldn't be inverteble

OHH

[delicate orchid]

Put ur vectors into a matrix... matrix is of full rank if and only if those vectors are linearaly indpeendant... full rank + square => invertible.. .bish bash bosh...

[coral spindle]

is [0 1; 1 0] invertible?

[delicate orchid]

find me an inverse.. you can't//../.

[coral spindle]

despite having a = d = 0?

[icy bear]

I messed d and b up

or d and c

[delicate orchid]

I highly recommend doing GL first

just fyi

[icy bear]

there is a bijection between invertable matrices and bases

[icy bear]

starting to think a is trying to tell me to do that

[delicate orchid]

it's purely combinatorial and you don't need to worry about determinants... that comes later (dark night rises ref)

it might be!

[icy bear]

well translating from GL2 to SL2 is as easy as

unless I have a terrible misunderstanding I'm not in the mood of solving

[delicate orchid]

yes going both ways is easy

but working out |GL_2(F_p)| is far easier

just my opinion! no hate!

[icy bear]

hate!

yes hate!

[delicate orchid]

MODS

[icy bear]

you're right tho!

[delicate orchid]

anyway yeah. Just think about what choices you can make for an element in GL_2(p)

[icy bear]

since (c,d) just can't be a multiple of (a,b)

I'm stupid

that's the only limitation that's way easier

[delicate orchid]

see if you can work out the formulas for GL_n(p) from the same idea

[icy bear]

hmhmuhmumh

[delicate orchid]

it's slightly more tricky, as "a multiple of" becomes "a linear combination of" but they're still easy to count. Worry about n = 2 first though

[icy bear]

still am not sure tho

(a, b) (c,d)

p choices for a and p-1 choices for b since if one is 0 the other can't

[delicate orchid]

not true

[icy bear]

(0,0) basis vector?

[delicate orchid]

you can pick any pair (a,b) as long as you don't pick (0,0)

[icy bear]

if one is 0

the other can't

if a is 0, b isn't

[delicate orchid]

yes, but you've assumed a is non-zero

[icy bear]

no

[delicate orchid]

so you've left out columns/rows that look like (0, b)

[icy bear]

but like, it's symmetric

ohhh

I think I understood

[delicate orchid]

there are p^2-1 choices of the first row

of which p^2-p are when a \neq 0

[delicate orchid]

so you need the p-1 of this form

[icy bear]

noo don't say it

[delicate orchid]

sorry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

my hamster geraldine bartholomew VI ran across my keyboard

I was powerless to stop her

[icy bear]

oh that sucks

now it's just factoring smh

tho I would have come to that conclusion probably

[delicate orchid]

fret not, there is swag on the horizon

[icy bear]

https://media.discordapp.net/attachments/547549702909067274/1158032577576321196/LET_THEM_CONSUME_YOU_WHOLE.gif?ex=651ac561&is=651973e1&hm=247542eb651d4df138f939ba5cab34fa085bd63da226003456e72fbd156d8667&

the sad thing about being me is that I can't just ignore that you made the challenge and move on with my life

tho ig I could do some induction

some

btw I really liked this problem love combinatorics specially in random stuff like algebra

[delicate orchid]

induction irrelevant. Just keep applying the logic you came up with here

you have basically free choice for the first row right? it just has to be anything non-zero

[icy bear]

oh so now we talking about zfc?

[delicate orchid]

for the second row I'm sure you can tell me what condition you need

[icy bear]

not be a multiple of the last one

[delicate orchid]

yur

so what about the third row

[icy bear]

not be a linear combination of the last two

which I already chose

oh so

[delicate orchid]

so how many different third rows can you pick?

[icy bear]

I think I have the knowledge

wait wait

don't say anything

[delicate orchid]

anything ah ah ah

[icy bear]

mods

[rotund aurora]

Is there something like LMFDB but for groups? There is groupprops but is more of a wiki

[delicate orchid]

what's lmfdb

some database?

[rotund aurora]

Yeshttps://www.lmfdb.org/

[delicate orchid]

ok I googled it

https://en.wikipedia.org/wiki/ATLAS_of_Finite_Groups

[rotund aurora]

right, that is probably helpful. Thanks

[delicate orchid]

this is so cute I didn't know about this

[coral spindle]

It's a beeeg book too. Like not very thick but it's beeeg

like a real old atlas

ugly as sin typesetting, oh well

[icy bear]

oh actually it's easier than what I was trying to do

[delicate orchid]

I've never seen a physical copy unfortunately

[icy bear]

it's p³ - all linear combinations of 2 vectors

which is p

so p³-p?

[icy bear]

or well no wait

no no it's p²

since to linearly combine 2 vectors I need to chose 2 coefficients

and since they are linearly independent I don't need to care about repetition

(that may seem obvious but I was very concerned about that)

[rotund aurora]

In his lectures on yt, Benedict Gross said he would bring a copy to class one day for his students or something. I think he didn't, what a scam

wait the typesetting is fine

nvm I got fooled, it's just the introduction

[delicate orchid]

Good luck

[cloud solar]

Let R be a finite ring with order n. Show that R is boolean iff the order of R[X] is n^2

How can I show that

[delicate orchid]

Yes, each linear combination of two vectors can be associated to a pair (a, b) in F_p^2 so there are p^2 of them

[icy bear]

so GL3:
(p³-1)(p³-p)(p³-p²)?

[delicate orchid]

Yusssss

[icy bear]

SO

[delicate orchid]

Now just follow the pattern hahaha

[icy bear]

thank god I didn't distribute

(p^n-1)(p^n-p)(p^n-p^2)... until yk where

[delicate orchid]

I dunno if I buy this. R[X] is infinite order even if R is finite

[icy bear]

thank you that was a good challenge I wouldn't have thought of doing myself genuinelly thank you

always apreciate a good math problem

[cloud solar]

But if the problem says iff there are exaclty n^2 polynomial functions f:R->R ?

[dire siren]

@cloud solar are you talking about polynomials or polynomial functions?

[south patrol]

Lol that explains it

there being only n^2 polynomial functions I believe

R[x] refers to polynomials in the sense of sequences of coefficients, and there are infinitely many (take 1, x, x^2, ... for example)

[delicate orchid]

Right yeah. Obviously if everything is idempotent these would coincide

[south patrol]

Anyway, if you want a hint: how can you use the fact a ring is boolean to simplify polynomial functions?

oh yeah computing # GL_n(F_p) is a good problem lol (above)

[dire siren]

and a hint for the other direction: count polynomial functions of degree <=1

[south patrol]

Well that hint sort of gives away what i wanted them to think about aha but sure

Not too big a hint anyway so it's fine

[dire siren]

I modified it

I just wanted to kinda make sure they don't mix polynomials and polynomial functions

[south patrol]

I'm now curious about something for this but yes

Yeah me too

:)

[slim kayak]

you can press shift to make things faster

[south patrol]

wdym like deleting stuff?

[slim kayak]

yeah

[south patrol]

thank

[cloud solar]

Yes sorry

[cloud solar]

Yeah now it is easy:))

[delicate orchid]

;)

[chilly radish]

This encompasses Z/6 by CRT

[cloud solar]

If A is a finite ring, U(A) has order p prime, every element a in A\U(A) is nilpotent, and the number of nilpotent elements is >= the number of invertible elements then the ring A has 4 elements

I called N(A) the set of nilpotent elements

I took f:N(A)->U(A), f(x)=1+x

f injective and because A finite, f bijective so ord(N(A))=ord(U(A))=p

So the ring A has 2p elements

And I am trying to find a contradiction with p>=3

The group (A,+) cyclic isomoprhic with Z/2pZ

[frank cosmos]

what does this mean?

[coral spindle]

They're asking you to describe the function in words. Calculate its values and see what you get.

You will see that x^2 - x is not its simplest description.

[frank cosmos]

f sends [0] and [1] both to [0]

[formal ermine]

yeah

[frank cosmos]

"note f(x) !=0"

we are treating f(x) as a formal element of R[x] is my understanding

so i dont get how to get a "simpler form" in this scnario

[coral spindle]

Yes, a polynomial is not a function!

[frank cosmos]

yes

but they do induce a ring homomorphism

the "function"

[coral spindle]

No there's no ring homomorphism here

It's just a function

But yes, you can take a polynomial and get a function

(This is actually a ring homomorphism R[x] -> R^R now — the process of converting polynomials to functions)

[frank cosmos]

the polynomial function is 0, is that right?

[coral spindle]

That's right.

[frank cosmos]

cool, thanks

im also struggling on this one, 2.7 just says you have an injective ring homo from R to End_ab(R)

not sure how the hint solves the problem

if c is in the center of R, then I dont know how to prove that the left action of c on R is in the center of E

[dire siren]

so charA={2,p,2p}
-if charA=2, then |A| is a power of 2, contradiction
-if charA=p, then |A| is a power of p, contradiction
-if charA=2p, then A is isomorphic to Z/2pZ (as a ring), but then |U(A)|=phi(2p)=p-1, contradiction

[prime dock]

Is there any trick to memorizing upto isomorphisms in groups
Like any group of order 6 is isomorphic to either Z6 or S3

[coral spindle]

Experience

[prime dock]

Another doubt
Isn't Z2 ⊕ Z2 ⊕ Z3 an upto isomophism for order 12

[coral spindle]

What does that mean

That is indeed a group of order 12, if that's what you're asking

[slim kayak]

whats an "upto isomorphism"

If you the usual "up to isomorphism problem" problem statements you are meant to list all isomorphism classes, and for abelian groups there is the nice friendly structure theorem finitely generated mo- I mean abelian groups

[prime dock]

like how many different type of structures a group of certain order can have
like if you have a group of order 4 then it is either isomorphic to Z4 or Z2 ⊕ Z2

[coral spindle]

So you mean a list of groups up to isomorphism

[prime dock]

yeah

[coral spindle]

but you said something very confusing

Isn't Z2 ⊕ Z2 ⊕ Z3 an upto isomophism for order 12

[prime dock]

what was I supposed to say instead

[coral spindle]

This doesn't look like a list of all groups of order 12, up to isomorphism, to me.

[coral spindle]

That's what I'm asking you

[prime dock]

I meant, isn't it one of the possible structures

[coral spindle]

Yes, as I said, it is a group of order 12

[prime dock]

Alr

[coral spindle]

But this is confusing because you could have named any group of order 12

So this doesn't mean anything

[prime dock]

Is there a pattern to it or do I just have to memorise

[coral spindle]

You just have to memorise, if you really want to.

I doubt you are expected to.

As kerr said, there is a nice way to do it for Abelian groups, but groups in general?

Nah

[summer path]

you should know some of the really low order ones, but past that, some are nice to know; but past that, no one's probably going to expect you to remember that

[prime dock]

it helps with some problems like this

[summer path]

well you are given that G is cyclic, so you're just looking for subgroups of a cyclic group

[prime dock]

Subgroups of Z12 basically

[summer path]

you aren't being asked for the number of non-isomorphic groups of order 12

yeah

[prime dock]

But this I figured because of upto isomorphism (i really don't know how to use that term exactly)
So I guess there is no trick to it I just have to memorise them all

[summer path]

well you know every subgroup of a cyclic group is also cyclic, so you can just go look at what each element generates

||but t he number of them is also just the number of divisors of 12||

[coral spindle]

When people talk about "Xs up to Y", they are talking about things in (let's say) a set X, and an equivalence relation Y on this set.

People talk about, for example, the integers up to equivalence modulo 2. If we want a list of all integers up to equivalence modulo 2, one list that works is 0 and 1. Another list is 3 and 16. This is because any integer is equivalent mod 2 to exactly one of the things in those lists.

When people ask you to count the groups of order 12 up to isomorphism, they are asking for a list of groups of order 12 such that any group of order 12 is isomorphic to exactly one of the groups you list.

Clear?

Another example is that the Jordan normal form allows you to classify matrices up to similarity.

[long obsidian]

Hey when someone talks about laurant polynomials with the notation Z[x,x^-1] does this include elements like 1/(x+x^2)?

x^-1+x^-2=(x+1)/x^2 right? Is it the case that laurant polynomials always have a monomial denominator?

[coral spindle]

Unless 1/(x^2+x) can be written as a sum of things of the form ax^n for a, n in Z, then no. I'm not motivated to check if it can be

Laurent polynomials look like $a_{-n}x^{-n} + \cdots a_0 + \cdots + a_mx^m$.

[cloud walrus]

Boytjie

[long obsidian]

Thanks

[south patrol]

Note it literally is just polynomials in x and its inverse

Rather than just being more ad hoc notation

[stark helm]

I am kind of confused, p(x) has a zero, why it also can be a irreducible factor of f(x)? if p(x) has zero, then p(x) has a root, existence of roots imply reduciblity. Do I misunderstand something?

[slim kayak]

irreducible factor in F[X]

[stark helm]

is it correct to say that p(x) is reducible in f(x)?

[slim kayak]

no

[south patrol]

it's an irreducible factor in F[x], not in the extension

[slim kayak]

f(x) is some random polynomial, but eventually if you keep dividing you get to an irreducible factor. Its either a linear polynomial or something more complex

[stark helm]

so p(x) is in F[x], and it is irreducible in F[X], but it might be reducible in extension field like F[x]/<p(x)>, and that's why p(x) can have a zero, is that true?

to make p(x) become reducible at extension field?

[slim kayak]

to make it have a zero

[south patrol]

Yup. It may be handy to think of examples.

For example, the polynomial t^2 + 3 is irreducible over R. But in the extension R[t]/(t^2 + 1) - which we may identify with the complex numbers - our polynomial splits as (t + sqrt(3)x)(t - sqrt(3)x)

[timid latch]

how do i find tbe order of elements in ℤ/Nℤ

like for rxample the equivalence class of -1 in Z/36Z

[cloud solar]

Happy new year, guys!

[stark helm]

You can just count how many -1 can be added to get 0 in Z36, it should be -1*36=-36=-36+36=0 in Z36, so order is 36

[stark helm]

Do you know mod, if you know mod, then just count how many times an element can be added in order to get 0. choose an element k in Z/nZ, and a*k= multiple of 36, where a is the smallest number you need in order to achieve multiple of 36(a means order)

[oblique matrix]

Do we actually need to include -1 in the set of generators for Q8? it's already produced by the other generators right?

[slim kayak]

its standard for group presentations

altho -1 as a generator is cursed

[south patrol]

Hm well you would also have to add in that (-1)^2 = 1 right

Or smth

Idk this looks a lil odd to me

maybe im weird

[slim kayak]

wdym, ijk-1=-1-1

i dont see the issue

[south patrol]

Wdym

[slim kayak]

xxxx with x replaced with "-1" seems really cursed to me

[south patrol]

What I mean is that if we are just viewing -1 as a symbol there doesn't seem to be enough

[slim kayak]

thats also what i mean, i just find it funny

[south patrol]

Oh okay

[slim kayak]

I guess its fine with (-1)^2=1 added

just still cursed

[south patrol]

Hm do you automatically have that -1 is central then though

[cobalt heath]

How does one prove that -1 is in the center here?

Ah wait, nvm. Quite obvious

[cobalt heath]

You just need
< i, j, k | i^2 = j^2 = k^2 = ijk >
and define -1 = i^2.

[south patrol]

yeah exactly

ok fair

[slim kayak]

how would you reduce i^4 to 1 with that presentation?

1 is the empty word, -1 really isnt a thing on its own

[cobalt heath]

Kerr: The same way with this presentation

[slim kayak]

that would be?

all relations either keep the length constant or even increase it. There is none that could take a word of length n>0 to 0 as far as I can see

I dont want to be pedantic just genuinely asking. Worst case we just need one final relation

[cobalt heath]

Yea I think this one needs some thoughts

Hmm

So ijk^(-1) = 1 here
similarly
jki^(-1) = 1
kij^(-1) = 1

So ki k jk^(-1) = 1

Which means ikj = 1.

[slim kayak]

I see, yeah

ij=k

[cobalt heath]

Now k = ij, so you get the desired one

Ye

Representation theory would have saved much hassle but >.>

[slim kayak]

nope combinatorics land here

[celest furnace]

How so?

[grizzled crane]

why do rings require ableian group under addition? maybe a motivation to the axiom?

[celest furnace]

Like why should it necessarily be abelian?

Or why it should be a group?

[grizzled crane]

abelian

[tribal moss]

If you omit the axiom x+y = y+x, you get something called a "two-sided near-ring".

[celest furnace]

I think a lot of it comes from wanting to generalize a lot about the integers

[tribal moss]

There are ways to make examples of such structures, but I'm not aware of a situation where one arises naturally such that we'd want theory about how they behave.

[grizzled crane]

even if I didn’t ask for abelian group under addition, integers would still fit in right?

it’s there but you are not asking it

[tribal moss]

Yes, the integers would still be an example, but there's less you can prove about the general class of such structures.

(The main point of the constructions I could quickly google up seems to have been just to show that x+y=y+x is not a deductive consequence of the other ring axioms).

[grizzled crane]

is there properties of ring, in which we specifically tweak the abelian axiom?

[crystal turtle]

Commutativity is really really nice.

[grizzled crane]

Makes sense

[crystal turtle]

Almost any interesting example of a ring would have addition being commutative.

[tribal moss]

In "near-rings" you drop x+y=y+x and also drop one of the distributive laws so you have x(y+z) = xy+zy but not necessarily (x+y)z = xz+yz. The are enough of a thing that they have a name, but I have no idea what they're good for.

[grizzled crane]

makes sense

[tribal moss]

Hmm, one example of the latter would be the set of all maps from a group to itself, with pointwise application of the group operation as "addition" and function composition as "multiplication".

[ivory trail]

rings are also monoid objects in Ab

[latent crane]

How does that work

[hidden haven]

A map A ⊗ A → A is precisely a distributive multiplication. The monoid axioms are equivalent to associativity and unitality.

Or if your question was a more general "what's a monoid object" then that's a category theory thing that you can look up. It generalizes monoids and rings and a bunch of other things.

[vast fjord]

i'm strugling with 5

more specifically with finding a x^n power that is the inverse of another y^k

and both k and n are positive integers

[tribal moss]

If x^n is in H, then (x^-1)^n is in H too.

[vast fjord]

yes but is that going to have a positive n?

likely i would have to take into account the different cases I gather?

[tribal moss]

It's the same n in (x^-1)^n as in x^n. If was positive when you raised x to it, it will also be positive when you raise x^-1 to it.

[vast fjord]

ohh ok

[naive whale]

How exactly am I to interpret the $x_1^{\phi(h_1) x_2}$ business in this composition law?

[cloud walrus]

Nomad

[naive whale]

Just not sure what the notation means. Guessing that psi(h_1) x_2 means image of x_2 under the automorphism, but as a superscript of x_1?

[formal ermine]

remove the superscript

its just x one times the psi thing

[naive whale]

So superscript is a typo?

[formal ermine]

yea

[naive whale]

Yeah I guess that would make sense

[formal ermine]

arithmetic theory of quadratic forms

[frank cosmos]

is this proof that (2,x) is not a principal ideal in Z[x] right?

(2,x) is all polynomials with even constant term. If it is generated by say r, then r must be constant since 2 \in (2,x). If r is even then (r) dosent include x+2 and if r is odd then (r) includes the constant polynomial r, both contradictyions

[south patrol]

Yes good

Essentially this is just saying that 2 and x have no common factor

Challenge question: show that all maximal ideals of Z[x] are of the form (p,x) for p a prime

[frank cosmos]

oh lol

i see that now

is that an iff though?

like if (f,g)=(d) then d is a common factor of f,g but is the converse true?

[frank cosmos]

(p,x) are all polynomials with constant term multiple of p. Suppose I is an ideal containing (p,x). Then I contains a polynomial with constant term pq+d where 0<d<p. Since I is abelian wrt +, d is in I. But then we have (d) \in I and (d)+(p)=Z since gcd(d,p)=1 thus I=Z[x]

[south patrol]

That shows (p,x) is maximal - yup. My question was showing every maximal ideal is of that form

[frank cosmos]

ohh

[south patrol]

Also, btw it's easiest to think in terms of quotients. There's an isomorphism $\mathbb Z[x]/(p,x) \simeq \mathbb F_p$ which shows $(p,x)$ is maximal - try filling in details

[cloud walrus]

potato

[frank cosmos]

where does this isomorphism come from

[south patrol]

It is a good short exercise

[crystal turtle]

(hint: anytime you have an isomorphism involving s quotient, consider using the first isomorphism theorem)

[frank cosmos]

send to constant term mod p?

[delicate orchid]

you're quotienting out by (p, x)

so you're setting p = 0 and x = 0 in Z[x]

[frank cosmos]

does this not do that

[delicate orchid]

well yes but that's not all it does

[frank cosmos]

right so then by homomorphism properties this forces the image of polynomials to be the same as all other polynomails with same constant term mod p

which gives the nessecary equivalence classes

[tardy hedge]

Homomorphism has the word homo in it

[dawn elk]

a group G acting on C^2 is generated by https://i.ibb.co/Lrn1YX7/image.png, what's the order of the group G?

[delicate orchid]

ah my apologises, I thought you meant the quotient was sending a_0+a_1x+...+a_nx^n to (a_0 mod p)+a_1x+...+a_nx^n as opposed to just a_0 mod p, which is correct

[frank cosmos]

why does this show (p,x) is maxcimal?

[south patrol]

If R is a ring and I and ideal, then I is maximal (resp. prime) iff R/I is a field (resp. Integral domain)

Important fact

[frank cosmos]

i see

is it something to try proving or to look up

[delicate orchid]

you can prove this yourself

[frank cosmos]

nice

[delicate orchid]

it's not too bad

[formal ermine]

hint: ||correspondence theorem||

[frank cosmos]

ye i got it

[frank cosmos]

also u can get this via the 3rd iso theorem instead

Z[x]/(p,x) = (Z[x]/(x)) / ((p,x)/(x)) = Z/((p,x)/(x)) = Z/pZ

[south patrol]

Sure, but you prove 3rd iso basically by doing this sort of 1st iso argument

[rocky cloak]

Try computing the order of I and J, try computing IJ and JI. Maybe you're able to compute all the elements of G.

[dawn elk]

Thanks!

[frank cosmos]

still thinking abt this

should i try and show Z[x] is noetherian?

so i can be more concrete

[south patrol]

I don't think that's relevant

wdym about more concrete?

[frank cosmos]

like i can talk about a max ideal in terms of its generators

[south patrol]

i don't think it really helps cause it can be a mess

like knowing an ideal is fg doesn't seem to help you know it's generated by 2 elements hm

Oh sorry I messed up, I mean (p,f(x)) where f(x) is irreducible mod p

[frank cosmos]

o

[south patrol]

Though I now realise the way I would prove it is overkill

[slim kayak]

This is giving me a very strong sense of dejavu

oh right

[south patrol]

Yeah I now realise the proof I know uses the fact Z, Z[x] are Jacobson lol but you can also write it elementarily

[frank cosmos]

i found a proof online

[south patrol]

lol

[frank cosmos]

yeah not something i can do lol

[cobalt heath]

Huh

[slim kayak]

I thought you'd use F_p[X] being a PID

[south patrol]

how did you prove it kerr?

[frank cosmos]

eh the idea isnt bad actually

[south patrol]

/ would you

[slim kayak]

I didnt, I hallucinate it as showig that Z[X]/(p,f) = F_p[X]/f. Since F_p[X] is a PID and f is irreducible mod p, it is maximal

[south patrol]

that is (again lol) the converse of what i'm asking aha

i'm asking that every maximal can be written in that form

[slim kayak]

oh right

[cobalt heath]

Idk how I could compute prime ideals of a UFD

[slim kayak]

I kinda assumed we had the shape of prime ideals in the bag

[cobalt heath]

Like do we have generators which are irreducible

[slim kayak]

Given a prime ideal just go through the cases. If p \cap Z the zero ideal? Then localize and so on

[cobalt heath]

Ah

Hmm, Principal prime ideals of Z[x] are precisely (p) and (x-n)

[south patrol]

Oh lol i just tried doing it and my proof ends up being more or less theirs but yeah with localisation

[cobalt heath]

Wait

[slim kayak]

localization my beloved

[cobalt heath]

So (3, x-1) is not a maximal ideal?

[south patrol]

kerr if you're interested it was like

[slim kayak]

that is a prime ideal

[cobalt heath]

I mean

[south patrol]

That is a maximal ideal.

[cobalt heath]

Yea

[slim kayak]

idk what you would fit into it. (p,f(x)) is as large as it gets

[cobalt heath]

If so, you need some adjustments

[south patrol]

Indeed

[cobalt heath]

Ah yep, now I recall this proof in my undergrad algebra. Elementarily, you can use content

[slim kayak]

The prime ideals should look like:
(p) with p prime number
(f) with f irreducible in Q, hence Z.
(p,f) with f mod p not equal to zero

[south patrol]

Let $\mathfrak m \subseteq Z[x]$ be a maximal ideal. $\mathfrak m \cap \mathbb Z$ is a prime of $\mathbb Z[x]$. If non-zero, then image in $\mathbb F_p[x]$ is prime and non-zero, so a poly irred mod $p$ etc. If $\mathfrak m \cap \mathbb Z = 0$, then $\mathfrak m$ corresponds to a non-zero prime ideal of $\mathbb Q[x]$ and you get a contradiction

[cloud walrus]

potato

[south patrol]

Eh that is the same but i thought about Q in a different way lol

The nice proof is that Z[x] -> Z is a surjection of Jacobson rings so it sends maximals to maximals

so in particular m intersect Z is pZ for non-zero p

[slim kayak]

uh, do you need that to see that m \cap Z must be prime?

like, you have graded structure here to see that it must be prime

[south patrol]

No, that's easy

[slim kayak]

so its either zero or maximal

[cobalt heath]

Maximal part does need Jacobson ring

[south patrol]

Yeah

Prime is easy and just assumed as a known fact in my original thing

But Jacobson is needed to send maximals to maximals which cuts out one of the cases (the harder one)

[slim kayak]

But the only non-zero prime ideals are maximal is what I mean

[south patrol]

Yeah

I don't get your point

Like sure that's what I assume in my original proof

Okay actually I omitted that line lmao sorry

If it's non-zero [then it's pZ]

[slim kayak]

I just dont see the necessity for jacobson rings

[south patrol]

bracketed is what i meant to add

Jacobson is needded for sending maximals to maximals

like, it means we know m intersect Z is non-zero

[slim kayak]

to rule out the 0 case?

[south patrol]

(because it is maximal)

Yeah

[slim kayak]

Why cant you argue that if the intersection was zero that you could add any p?

[south patrol]

I think otherwise you have to apply Gauss' lemma and stuff, which is sort of related to Jacobson property

[south patrol]

Yeah that's how it'd go without the Jacobson thing

[slim kayak]

Oh

[south patrol]

Like image in Q[x] is an irred poly over Q with coeffs in Z i.e. irred poly of Z

so that you get (f) and then Z[x]/f isn't maximal

[slim kayak]

Yeah for a generalized proof for some either type of ring jacobson would be nice

[south patrol]

But yeah that uses Gauss' lemma and then you still have to do that last step ye

I forget stuff about Jacobson rings lol

[slim kayak]

Okay.

Only difference is that I personally prefer to minimally modify the proof to just outright show how all prime ideals look like

Just because I needed that not too long ago

[slim kayak]

Most of the time these type of questions come down to doing some clever manipulations/constructions or neither and then using localizations and quotients to extract insight

they can still end up hard though, especially if you for example forget that maximal ideals have the property that either x is in M or 1-xy is for some y.

[cobalt heath]

Woah

Where can I learn this

[slim kayak]

Should all be found in a decent comm alg text

[cobalt heath]

Did I not read e.g. lang algebra properly

[south patrol]

what do you want to learn?

[cobalt heath]

The decent intuitions about connections like this.

[slim kayak]

ah, the book famous for its pedagogy

I mean if you seen all the proofs then it comes down to practice

[south patrol]

Maybe they aren't actually connected lol

I'm pretty sure they are

Like

[cobalt heath]

,ask Gauss lemma

[south patrol]

wrong gauss lemma

[cloud walrus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by abstia

[slim kayak]

Polynomial in Z[X] is irreducible iff irreducible in Q[X], here the dollar store statement

[cobalt heath]

I mean, 2x is reducible in Z[x], but I recalled what it was. Thank you!

[delicate orchid]

"a finite set of integers has a gcd" wow.

[slim kayak]

"have you tried clearing the denominator?"

[cobalt heath]

Sorry, but I cannot see this

[steel pulsar]

Question : Let n be an integer. How many fixed points on {1, . . . , n} a does permutation belonging to S_n have on average? Check it for n = 4 by explicit calculations

I did this:

4 = 4 (0 fixed point) there is a 3! cycles product which gives this partition
4 = 3 + 1 (1 fixed point) there is a 2!0! cycles product which gives this partition
4 = 2 +2 (0 fixed point) there is a 1!1! cycles product which gives this partition
4 = 2 + 1 +1 (2 fixed point) there is a 1!0!0! cycles product which gives this partition
4 = 1 +1 + 1 +1 (4 fixed point) there is a 0!0!0!0! cycles product which gives this partition

But I don't know what to do next. Could someone explain it to me in detail and step by step, please?

[rotund aurora]

what did you get for n=4?

I got 1 if I didn't mess up

[delicate orchid]

There’s a phenomenally cool character theoretic way to do this

Yes it is 1

[rotund aurora]

If you know linearity of average/expectation it is easy

[rotund aurora]

show

[steel pulsar]

Please can someone explain to me step by step and gently

[rotund aurora]

• How many permutations of S_n are there that fix exactly k elements?
• say the number above is f(k), then you want to find (f(0)+f(1)+....+f(n))/n!

wait this is annoying

cuz you would need derangements lol

[delicate orchid]

Let $\chi$ be the permutation character of the action, then it’s a commonly known result that $\chi = 1+$ the standard rep (the action on n vectors fixes the subspace spanned by the sum of each basis vector) hence $\langle \chi, 1 \rangle = \frac{1}{|S_n|}\sum_{g \in S_n} \chi(g) =1$, but the sun is exactly the average number of fixed points and we are done

[cloud walrus]

W;3w Lads Tbh

[delicate orchid]

*sum not sun

Doesn’t burnsides lemma tell you that the average number of fixed points is the number of orbits

And clearly S_n acts transitively on your set

So one orbit => 1 fixed point on average

[rotund aurora]

Calculate the number of permutations that fix a given element. Then add all of them and divide by n!

[steel pulsar]

You're skipping a lot of steps if you're already at Burnside's lemma to answer this question

I have already the correction

[delicate orchid]

What

I’m skipping no steps unless you want me to prove burnsides lemma each time I use it

[steel pulsar]

My teacher uses the lemma only at the end to average. There is some reasoning to be had before that that I need to understand

[delicate orchid]

The reasoning is the proof of burnsides lemma lol

[stark helm]

I am wondering if F is a field, and f(x) is in F[x]. When Char F=0, and f(x)=a0 where a0 is a constant, why we say f(x) is reducible here? I am considering a0 has a unit, so it should be irreducible here.

[steel pulsar]

I need to understand the reasoning for this question because it's been 3 weeks I haven't understood

[delicate orchid]

Go understand the proof of burnsides. This is a basic application of it

[rotund aurora]

@steel pulsar

[delicate orchid]

Irreducible elements generate maximal ideals. Units do not

*in PIDs

[steel pulsar]

The number of permutations in S_n is n! that's it ?

so in S_4 is 4!

[rotund aurora]

No, calculate the number of permutations that fix one element

Like you want to calculate $#{(x, \sigma)\in [n]\times S_n\c \sigma(x)=x}/n!$

[cloud walrus]

Croqueta

[rotund aurora]

when you fix x, calculating the number of pairs (x,sigma) as above is easy, so you can just sum over x

[steel pulsar]

do you have an example ?

[rotund aurora]

why do you want an example? Isn't this concrete enough?

[steel pulsar]

for me it's abstract

[rotund aurora]

It is not

[steel pulsar]

In fact I don't know how to calculate the permutations of a fixed element

[rotund aurora]

In fact, this is a situation where specializing to some n will fool you, because you may be led to think that cycle decompositions will help (as you did above), when in reality makes things more difficult

[rotund aurora]

ok, so start there

If you know how to calculate the number of permutations on n elements you should be able to do that

[slim kayak]

Doesn't your definition of irreducible state a number not being a unit?

[rotund aurora]

btw [n]={1,2,3,...,n}, in case it's notation you are not familiar with

[steel pulsar]

ok

[stark helm]

nonzero, non-uinit element is reducible

[steel pulsar]

?

[stark helm]

a0 is unit in F, so f is irreducible, so I don't know why it is reducible

[slim kayak]

Doesn't that involve counting a lot of permutations multiple times

[stark helm]

I am wondering that constant is a unit in a field

[rotund aurora]

yes? but so what

[steel pulsar]

Can someone explain to me how we calculate the permutation for a given element, please?

[rotund aurora]

Calculate the number of permutations of S_n that fix 1

[steel pulsar]

I say that I don't know how to calculate the number of permutations of a given element, you tell me to calculate the number of permutations of a given element x)

[rotund aurora]

You were gifted a brain, like all of us here. Use it

[steel pulsar]

Yes, there are things that I understand and others that I don't understand

For that, it's been 3 weeks I haven't understood

[paper flint]

Take some more time to work out what you've been suggested above

Otherwise you might as well read a solution online

[steel pulsar]

you don't understand, I already have the solution

here is the solution