groups-rings-fields

[south patrol]

or is this just one of those like lol "take characters" things

I imagine there must be a prettier argument though but idk

[wraith cargo]

Last line
gS=Sg does not imply that any s in S is in the center

[solemn dew]

oh right

since gs = sg does not have to be true

[south patrol]

Oh

[delicate orchid]

Oh?

[south patrol]

Nvm i thought i got it but i didn't lol

[delicate orchid]

Wdym by Hom_S_k btw

[south patrol]

(internal) hom of S_k-representations

[delicate orchid]

Oh right isn’t this just decomposing into irreducibles?

No not quite

[south patrol]

Hm the summands on the RHS needn't be irreducible i fink but yeah i assumed is related

But it does seem a lil sus how RHS like can be rewritten as End(V_pi) tensor E^otimes k right

[delicate orchid]

So you’re taking each irreducible and looking at the different ways of embedding it into E

Can we not tensor-Hom adjunct our way out of this one

[south patrol]

Perhaps

[delicate orchid]

Yeah maybe something like this

[south patrol]

But they are kth powers so idk

[night onyx]

it comes from End(V_pi) \isomorphic V_pi \tensor Hom_G(V_pi, V) where V is a representation of G

[south patrol]

I think thata's missing smth right

End(V_pi) tensor V i guess which would match w what i argued? hm

Okay so is like

$\bigoplus_\pi \mathrm{End}{\Sigma_k}(V\pi) \simeq \mathbb C[\Sigma_k]$ or smth

[cloud walrus]

potato

[south patrol]

I think so right? like artin wedderburn

If that's true then it follows immediately

[delicate orchid]

I’m still confused by this notation

[south patrol]

Sorry

[delicate orchid]

pi is a permutation I’m presuming

And V_pi the corresponding dude

[delicate orchid]

*partition

[south patrol]

So it's meant to be that {V_pi} is just any like enumeration of the reps

but yes pi is a partition

But this is, I assume, jujst a general thing for all G and V rather than S_k and E^otimes k

[south patrol]

So yeah I assume this is Artin-Wedderburn uhhh

[night onyx]

it's true in general for finite groups, the group algebra decomposes into the sum of End_G(V_pi) for pi irreducible, for the symmetric group irreducibles correspond to congugacy classes => correspond to partitions of n

[south patrol]

Yeah Oh yeah it is true lol

[delicate orchid]

Well since V_pi is simple any endomorphism is 0 or an automorphism so I buy it being artin wedderburn

[south patrol]

Nice

Yeah it is just a form of that

[night onyx]

yeah

[south patrol]

I feel bad about being rusty with reps already 😭

Seeing as I took an exam on them like a couple months ago lol

Thank you guys

[delicate orchid]

It’s my area of study and I still cannot cope with actual modules

Anything other than a character and I start crying

[wraith cargo]

I feel like for this problem there's some group action nonsense going on

[rocky cloak]

If M is a completely reducible kG-module and S is simple, then the dimension of Hom(S, M) is the number of times S appears as a summand in the decomposition of M.

So you're just summing up all the irreducible summands of E^k

[south patrol]

Yup yeah

[solemn dew]

@wraith cargo My book doesn't cover that 🥲

[delicate orchid]

I should've gone with my gut lol

[delicate orchid]

need more confidence in my bullshit

[rocky cloak]

I think it will be difficult to do this without breaking things into semidirect products. What tools have you been given thus far?

[delicate orchid]

this is a group of order pqr surely that can help

[wraith cargo]

It tells you one of them is normal lol

[delicate orchid]

yeah which is what we've already shown

wunderbar

[rocky cloak]

Alright, so I see automorphism, so then you probably know about inner automorphism?

[solemn dew]

yessir

[rocky cloak]

Let x generate the sylow 11 group and let y be an element of order 3

[delicate orchid]

oh this is smart

[rocky cloak]

What can you say about yxy^-

And what is the automorphism group of Z/11

[solemn dew]

since 11-sylow subgroup is normal it suggests that yxy^-1 is contained in X

[rocky cloak]

Indeed, so y induces an automorphism of the group

[delicate orchid]

now think about the order of such an automorphism

[delicate orchid]

and this

[solemn dew]

o(yxy^-1) = o(x)

[south patrol]

Oh this paper I'm reading is cool wew lol

[delicate orchid]

the order of the automorphism

not it's image

[south patrol]

Well like identifying the rep ring of S_n in a different way to what I'm used to

[delicate orchid]

send

[south patrol]

can link it if you want

https://www.maths.ed.ac.uk/~v1ranick/papers/atiyahp.pdf chap 1

Lmk if this is just standard though

[delicate orchid]

CW-complexes
I'm meant to be a break from triangle nonsense

[south patrol]

That's why i said chap 1 sir

That bit is purely algebraic

[solemn dew]

i don't know the order of the automorphism

[delicate orchid]

the uhh action of a dude on the kth tensor power reminds me a lot of how you compute representations of wreath products

[delicate orchid]

it's conjugation by an element of order 3

[south patrol]

Uhhh

[delicate orchid]

so have a guess

[south patrol]

Oh okay idk about wreath products lol

[delicate orchid]

that's fine it was just a quip for the fans in the back

[wraith cargo]

It means how many time so you have to compose the automorphism to get the identity

[solemn dew]

oh

[delicate orchid]

I mean, what else would order mean

[solemn dew]

11 since (yxy^-1)^11 = yx^11y^-1 = yey^-1=e

[delicate orchid]

no, that's not composing

that's multiplying

let $\phi$ send $x$ to $yxy^{-1}$

[cloud walrus]

wew

[delicate orchid]

then $\phi^3(x) = y^3xy^{-3} = x$

[cloud walrus]

wew

[south patrol]

One thing i findi ntersting wew is this like

S_k-trace thing

Idk if that is a common thing lol

pg 168

[solemn dew]

and x is the identity?

[delicate orchid]

no dude

x -> x is the identity map

and phi^3 maps x to x for all x

[solemn dew]

alright alright i got you

[delicate orchid]

and then we know this has to be the order cause 3 is prime but w/e

[delicate orchid]

ok, piece number dos

[delicate orchid]

ngl I absolutely hate this notation

[south patrol]

Yeah i find it a bit hard to read the notation at times lol

Is that trace thing a common construction?

[delicate orchid]

well if it's taking the trace of a rep it's a character

but I literally cannot decipher what they're doing here

I personally have never seen this before

I figured out what they were doing lol

the \pi(E) notation is so, so stupid

oh wait no no no

it's just this but in characters, I think... just spotted a tensor product which doesn't make much sense

[south patrol]

I do think it is a lil hard to think about

I just wonder where the motivation for all this comes from

P cool tho

Gives another way construct adams operations on rep ring I think

[delicate orchid]

I sat down and explicitly worked out why \pi(Trivial) was Sym^k(n) and \pi(sign) was Alt^k(n) and then it made sense to me

[pure mulch]

so i had this short exact sequence of groups, and we had to find the groups upto iso which satisfied in place of G here:
$0 \to Z\2Z \to G \to Z\2Z \to 0$
and i managed to prove that the order of G has to be 4, but now i have to find which kind of group it is, and im not sure how to proceed there

[cloud walrus]

Shyshu of the Golden Integers✓
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[pure mulch]

i know that groups of order 4 are of 2 types, Z_4 and Z_2 cross Z_2, but im not sure how to prove that either

[tawny magnet]

Use lagrange's theorem to prove that

[pure mulch]

ohhhhhhhhhhhhhhhhhhhhhhhh

holy shit ofc

[tawny magnet]

so by lagrange order has to divide 4 so either there exist order 4 or not

split the case

[delicate orchid]

$0 \rightarrow \bZ/2\bZ \rightarrow G \rightarrow \bZ/2\bZ \rightarrow 0$

[cloud walrus]

wew

[delicate orchid]

why tf are you looking at group extenstions already

[pure mulch]

it was in the pset lol, we did a bit on SESs back when we were doing lin alg, so this isnt exactly the first time i have seen it

wait group extensions?

oh it is mentioned here huh

[delicate orchid]

anyway if I'm remembering my homology correctly :trollface:

[pure mulch]

homology

[delicate orchid]

there's two extenstions here, the split one and the non-split one

Z/2Z x Z/2Z is the split one and Z/4Z is the non-split one

or you be sensible and use lagrange

[pure mulch]

i have no clue what this means

[delicate orchid]

exactly. So why are you thinking about group extenstions at all

[pure mulch]

it was an optional problem in the pset 😭

[delicate orchid]

society if all extenstions split

[tawny magnet]

what does it even mean by a sequence in this case?

[delicate orchid]

a SES most likely

[pure mulch]

Short exact sequence

[tawny magnet]

oh i seee

[hidden haven]

shashs all grown up

I still remember the day he asked what negative numbers are 🥹

[hidden haven]

Whenever you have a SES
0 → A → G → B → 0
G is called an extension of B by A.

[pure mulch]

wut

bro i was here when i was in 11th 😭

[pure mulch]

ic

[delicate orchid]

or if you want to go the other way, if G has a normal subgroup N, then G is an extension of G/N by N

[hidden haven]

Yeah I remember the 3' 11" shashu

🥹

[pure mulch]

i havent been 3' 11 in literally a decade

[pure mulch]

[hidden haven]

Rebellious phase 🥹 so cute

[pure mulch]

bro 😭

i dont really even remember being that small

[topaz solar]

He’s learning the deep arts

Not long till he succumbs to spectral sequences

[hidden haven]

Yeah usually you would remember the world being bigger 😊

[pure mulch]

someday™️

[pure mulch]

[tawny magnet]

this is so wholesome 🥺

[summer path]

Wholesome algechill

[pure mulch]

something wholesome for my sad life 🥹

[tawny magnet]

cheer up buddy ur doing group theory ahahah

[pure mulch]

thats whats keeping me sane

funny groups

[delicate orchid]

so that's where my sanity is going

[pure mulch]

you were ever sane?

[delicate orchid]

no

[pure mulch]

as expected

[tawny magnet]

Bloody hell why is everyone so gloom today

[pure mulch]

i have been gloom for a while

[hidden haven]

Oppenheimer release weekend

[delicate orchid]

it's not gloom it's just a medical fact

[rocky cloak]

Maybe they got attacked by phantom Ganon

[delicate orchid]

jagr gamer reference... a powerful combo...

[tawny magnet]

How about barbie

[pure mulch]

lmao

i got stabbed by them gloom swords

[hidden haven]

It wasn't Kenough

[tawny magnet]

poetry in action 😂

[sly flume]

Hi I wonder if the follow is ok for a proof that the mapping of a group is onto
G->G L(x)=ax a,x in G
If L(x)=L(y) then ax=ay so x=y 1-1
NOW ONTO
I would say a in G and x in G so ax in G so for each x there is a L(x) in G is this ok?

[chilly ocean]

i don't understand your proof

[tawny magnet]

when G is finite yh i guess but when G=Z lets say u cant say injectivity implies surjectivity

[rocky cloak]

A function is surjective if for every y in G there is an x such that L(x) = y

Can you find such an x?

[delicate orchid]

oh right we're showing group multiplication is bijective?

[sly flume]

This is from Fraleigh proof of cayley theorem so you need to show every group is isomorphic to a group of permutations so need to first give a candidate mapping and a perm is defined as a bijective function.

@rocky cloak I think is just L(a^{-1} y) = y so x=a^{-1} y

[rocky cloak]

Indeed

[sly flume]

need to say as a in G then a^{-1} has to exist right to be in a group

[rocky cloak]

Yes, a group contains inverses

[sly flume]

you need to say that in a proof as the a inverse may not exist

since we're in groups we're ok added benefit of Groups 🙂

[rocky cloak]

Yeah you need G to be a group

[sly flume]

I only bring it up as other parts of maths that gets messed up I'm thinking matrices det=0

Thanks 🙂

[sly flume]

Any

sorry anyone a big group theorist?

or studying any group theory in general?

[rocky cloak]

Is it clear why in characteristic p a group has the same representation type as its sylow subgroup? For example is there some way relationship between the representation theory of P, N(P)/P and G or something like that?

Alternatively does someone have a reference for a proof of this theorem?

[rocky cloak]

Not really a group theorist, but some group theory have appeared in my work (hence my question)

[sly flume]

yes what work are you doing?

[rocky cloak]

Representation theory of algebras / homological algebra

[sly flume]

ok cool

[rocky cloak]

You studying some finite group theory now?

[sly flume]

Yes not quite on to sylow theory yet just studying the books Fraleigh and Dummit looking to get into the computational side as well

[rocky cloak]

Like computer algebra stuff? Or what does computational mean?

[sly flume]

yes it is a whole field computational group theory the book by holt is a good reference " handbook of computational group theory " there is also a free software called GAP

https://www.gap-system.org/

[rocky cloak]

I've used GAP a bit (though not for group theory)

[sly flume]

oh alright cool how did you get on?

was it any use

[rocky cloak]

It's not an amazing programming language, it's a bit painful to debug. But it's the only one that can really do representation theory of quivers

[summer path]

iirc spamakin was working with gap but idk if he's around

[delicate orchid]

hmmm, only intuition I have is that the Sylow p-subgroups of G completely encodes the p-local data of G and the p'-parts of elements behave as expected over fields of characteristic p
Actually, modular characters are completely determined by their values on p'-elements of G so perhaps that's something? I will look for the proof now

[rocky cloak]

As far as I'm aware anyway

[sly flume]

Yeah it seems GAP is a bit of a pain. there;s no need to have a separate software for it is should be just in c++ or matlab or something with functions you can call. But I think it is the way it is coz it's so old

[rocky cloak]

Hmm, but modular characters only describe irreducible representations right? Or do they work for general representations?

[delicate orchid]

yeah, they only characterise simple modules up to isomorphism

and I cannot find this proof anywhere

it could be in here? https://link.springer.com/chapter/10.1007/978-3-7091-2814-5_37

[rocky cloak]

Yeah, but there are only a finite number of simple modules, so not sure if it will help

[delicate orchid]

yeah, I thought it was for indecomposables but no such luck

[delicate orchid]

I'm gonna give up looking now lol

[rocky cloak]

Im outside my springer access network atm, but I can check this one out tomorrow

[delicate orchid]

it's probably not in there, but that's the only thing I could find

[rocky cloak]

Also, I don't know what p-local data means. If it's not too much groups trouble could you explain in an example like G=S3, p=2.

If I'm not mistaken S3 should have 3 indecomposable representations, while C2 has 2. And S3 had 2 irreducibles while C2 has 1.

[delicate orchid]

it's a nebulous idea, it's more productive to think about the modular character bit

I don't really have it straight enough in my own head to properly describe what I mean, sorry

[rocky cloak]

I see, and p' elements mean elements of order coprime to p?

[delicate orchid]

yus

\chi_M(x) = \chi_M(x_p') for all modular characters and all x in G

so for your example, the 2 irreducibles of S_3 should only really care about (123), (132)

and this nicely shows that for a p-group P, there's only one irreducible over a field of characteristic p

[rocky cloak]

Makes sense, but is also strangely opposite of what I'm after. Since it's sort of relating everything that isn't in the sylow group

[delicate orchid]

yes it is odd isn't it

[rocky cloak]

Maybe something like a representation either "comes from" the sylow group somehow or is irreducible

That would make sense

[delicate orchid]

thinkin bout these algebras now

there's a way to move from the K[G] to the k[G] that I'm forgetting

[rocky cloak]

Hmm, understanding OG sounds hard

[delicate orchid]

yeah you tend to start with OG

[rocky cloak]

I mean you can just compose (co)induction and restriction of scalars I guess

Not sure if that's what you're thinking of

[delicate orchid]

would work but determining what splits and how would be annoying

perhaps we should think about Brauer characters? They're defined for all k[G]-modules right

[rocky cloak]

Thought that was the same as modular characters. What's the difference?

[delicate orchid]

you take your roots of unity in this O rather than k iirc

yeah that's it, you lift the eigenvalues of the action of a p'-element on a k[G]-module to those of an action on an O[G]-module

which you can do by some theorem somewhere

@coral spindle you like modular reps u got any thoughts on this mess?

[rocky cloak]

Yeah the definition of brauer character I know is that you take the group of roots of unity in (the algebraic closure of) your field, and pick an injective group homomorphism to the roots of unity in C. Then pretend all your eigenvalues are complex numbers

[delicate orchid]

yeah I think that's equivalent to doing what I did then mapping O[G] into K[G], K is required to be a splitting field of the group in question

[rocky cloak]

That would make sense then yeah

[delicate orchid]

I can do the cyclic case inductively I think

uhh how does it go

jordan block decomposition shows C_{p^k} has finite type over F_p

C_p \times C_p has a non-finite type over F_p

[rocky cloak]

That C_p^k has finite type I can do pretty easily, but if G has C_p^k as its sylow p group...

Why is G still of finite type

[delicate orchid]

I'm just trying to get a handle on why the conditions on the sylow subgroup are the way they are before thinking about how it corresponds to G

intuitively it's because being char p "kills" the p-data

but what does that actually mean

[coral spindle]

Idk Brauer stuff I'm afraid, sorry

[south patrol]

Why not?

I know less than you so i am allowed to joke that

[coral spindle]

Well ostensibly because I only do characteristic-zero rep theory,

but in reality it's because I suck and everyone's gonna find out and my supervisor is gonna think I'm a failure and I'm hyperventilating rn and ahhhhufdshafdsa

I mean

[delicate orchid]

oh wait really?

[coral spindle]

In reality it's because I've been focusing on some other parts of rep theory, and modular reps hasn't really come up so much. Plus I'm a bit weak in algebraic number theory

Well 'ostensibly' is the operative word...

[delicate orchid]

the word that I don't know the definition of

ok I know now

[coral spindle]

ostensibly means something along the lines of "evidentially"

as it turns out the ell-adic Schur indices are strongly related to modular reps

so it's more like

I haven't got there yet

[coral spindle]

This was the wrong comparison. It's more like "apparently" lol although that could just be seen as a synonym of the last example I gave...

[south patrol]

Aw no I'm sorry hug

[coral spindle]

I was of course joking, but I'd be lying if I said I was happy in my ignorance

[slim kayak]

Someone have some ideas how to solve this? Given some amalgamated free product $G = A *_{C} B$ and a map $\phi: G \rightarrow H$ s.t. the restrictions of phi to A (as a subset of G) and B are injective, then the kernel of phi is free? Trying to refresh on some material but I feel like I forgot something obvious for solving that problem

[cloud walrus]

kerrmode

[topaz solar]

Well, the kernel obviously has no elements from A or B

So it’ll only be things that’ve been added in because it’s A*B

[slim kayak]

Well yeah, the non-zero elements of ker phi all have length at least 2 but I dont see how that brings me closer to the existence of a basis?

[topaz solar]

Well, if you have a subgroup contained entirely in those free product objects, can you have any sorts of relations

[slim kayak]

Free product objects?

[topaz solar]

The ones that aren’t just included from A or B

clearly (a*b)(a’*b’) would have an issue with falling outside the free product objects

[slim kayak]

Whats wrong elements of the form ba and a'b with a and a' from A and b from B? They arent contained in either A or B but their product isnt reduced?

[topaz solar]

Well, if baa’b is reduced into something in A u B, then aa’ = e

Or b is e

[slim kayak]

Okay?

[topaz solar]

So you won’t have both in your kernel unless the whole thing is an inverse

Suppose you had a subgroup contained in A*B - (A u B)

And you had two words s, t such that st=e, then t is just s in reversed order and each element inverted right?

Can you write that inverse in any other form?

(Excluding introducing trivial things like (a a^-1) into it)

As in, can you have any extra relations, which means your thing wouldn’t be free

An obvious example of such a subgroup would be <ab> for some non-identity a, b in A, B resp.

Or, what any subset of those pair words (a_i b_i) for all not the identity, can add in ones like (b_j a_j) too so long as you don’t get b_i = b_j^-1 or similar for a

And these are all explicit free groups since there’s only one (nontrivial) way to write inverses

[untold turret]

F is a field, any clue why beta is algebraic over F(B')?

[next obsidian]

Definition of transcendence basis

[untold turret]

oh, since B' is a maximal algebraically independent set

[next obsidian]

What? Is that your definition?

I mean, that’s true yeah

But the definition I have is algebraically independent and K/F(B) is algebraic

I mean they’re very easily seen to be equivalent but

I guess isn’t “by definition” if that’s your definition

Oops

[untold turret]

heheh ty!

[slim kayak]

I am not quite seeing the relationship here. Right now I want to do something like this: Take random words from ker phi w_1 to w_n, their product wont be ensured to not be 1 since just letting w_n be a product of the inverse words w_n-1 to w_1 would work. To fix this I might declare my basis elements to be "minimial subwords", so for example starting with w_n I start from the left and cutoff the word when I find a substring which is also an element of the kernel, splitting the word in two. Since A u B doesnt meet the kernel I know that this process leaves me with words of length at least 2. Once I've done this the aforementioned situation can't occur, since if w_1 ... w_n = 1, then I couldve split up w_n into subwords so that the product would contain neighbouring inverses. I suppose this might work?

[topaz solar]

Well, I mean

Free groups aren’t always over a finite basis

Do you have that a subgroup of free groups is free?

[slim kayak]

Yes

[topaz solar]

So, let’s say I took (A-e) x (B-e)

And defined a map into A*B by (a, b) |-> ab

Group generated by the image is disjoint from A u B, right?

I claim these pairs also form a free basis

No wait I’m wrong in allowing all pairs

But you get the idea that if we don’t have ab and ab^-1 or a^-1 b we won’t get any A u B elements and it’ll be free if you do

[slim kayak]

Just to make sure, but the initial situation was about looking at a map going out of an amalgamated free product not into

[topaz solar]

Ye, I was suggesting finding a free basis for a subgroup of A*B

And it’ll be disjoint from A u B but contain your kernel

[slim kayak]

Wait, what would the image of (a,b)(a^-1,b^-1) look like?

If it sends (a,b) and (a^-1,b^-1) to ab and a^-1b^-1 respectively the result would be aba^-1b^-1 no?

[topaz solar]

well it’s not a group mapping lol it was just something to generate it with

So like uhh, say you had a generating set of some fashion

You could look at the first symbol in the words

I believe you should be able to find a free group containing your kernel in that way

And subgroup of free are free

[slim kayak]

It seems that what I am looking at is apparently related to the kurosh subgroup theorem, at least for a related problem it 1. shows that the kernel is a free product and 2. since the restrictions are injective it is actually a proper free group

I can't wait for combinatorial/geometric group theory to be over 💀

[noble hedge]

Can someone explain to me how the highlighted bit follows?

idk if I'm just missing something or what, but one of the comments says x^3-c for any non-cube is irreducible and has discriminant -3(3c)^2, which would be a square in the finite field if the field has a primitive 3rd root of unity, right?

but how does that then generalize to every irreducible oubic over a finite field?

All irreducible cubics generate the same (well, isomorphic) splitting field, right, Since irreducible implies separable in characteristic p? So the Galois groups should all be isomorphic, and as long as there's one isomorphic to C_3, they'd all be isomorphic to C_3, which would imply that the discriminant is square?

idk if that's the right direction or not, and if it is I have no clue how to find an irreducible cubic whose Galois group is isomorphic to C_3 for any characteristic p

[tidal sleet]

What would be a good place to ask about proofs of basic maths?
Ik a somewhat good amount of math but i hate "following blindly"
I can do subtractions of negative numbers and multiplication but i really do want to see a "why" of this, so far i only ever followed along.
Especially a proof thats simple in nature yet makes sense

[echo bluff]

#proofs-and-logic

[tidal sleet]

ty

[cloud walrus]

nrs

[untold turret]

solved, ignore above!

sanity check: if t is an element in k[x_1,...,x_n], then obviously we can't have k[t] = k[x_1,...,x_n], but what's the correct formal reason for this?

[next obsidian]

Dimension

Is one reason

[untold turret]

in the sense of having a different upper bound for the lengths of chains of prime ideals, right?

i.e. the dimension of k[x_1, ..., x_n] is exactly n, while that of k[t] is at most 1, right?

[next obsidian]

Yes

[untold turret]

nice, thank you

[prime dock]

wait guys won't the dihedral group (of say 4 sided regular polygon) have 2 identities? rotating it 4 times and flipping it twice would give you the same element on which you're doing this operation. but how is this possible? identity must be unique

[delicate bloom]

group elements don't track their past state

like for instance 1*1 and (-1)(-1) are both 1, we didn't distinguish between them back in middle school algebra class

[prime dock]

meaning?

[rocky cloak]

So irreducible doesn't imply seperable in general, but over finite fields it does. And you're right that they all have the same Galois group, namely the Galois group of a finite field is always cyclic.

And the Galois group of a cubic is cyclic iff the discriminant is a square.

[rocky cloak]

Rotating 4 times or flipping twice are both equivalent to doing nothing. Doing nothing is the identity operation.

Just like how for numbers under multiplication, both 1 and (-1)*(-1) is the identity.

[prime dock]

so basically no rotation and no flipping is the identity?

[rocky cloak]

Yes, leaving the square in it's original state

[prime dock]

oh I understand now

[void cosmos]

what are the is and what are the js

this proof looks easy but its confusing me

can anayone help

[wraith cargo]

thought this was prime avoidance for a sec lol

[wraith cargo]

they're indices?

[round hull]

isn't it basically the same thing

[void cosmos]

yea but i dont get how K is not in U P_j where j !=i

isn tthis the assumption

isnt*

[wraith cargo]

no

the assumption is that

K \subseteq U_i P_i

but now choose some P_i in this union

now the assumption they make is that if you remove any of these P_i's, then K is no longer in the union of what's left

[void cosmos]

ok

so they removed P_i

from the union

so that K is not contained in the rest of them

as in U P_j , i not equal to j

right

[wraith cargo]

yes

you can imagine this as like

there's an element of K in every P_i

so K is contained in the union

but if you remove any of the P_i's then an element of K is missing

so K can't be contained in the union of what's left

[void cosmos]

yea

i get it

tysm

[cloud walrus]

hausdorff

[median pawn]

the star is the free product

[coral spindle]

Yes

[delicate orchid]

Free groups are determined up to isomorphism by their rank

[median pawn]

how should i think of this? i'm trying to produce an isomorphism using universal properties

[delicate orchid]

Use the inclusions of S into F(S) and your star product

[warm wyvern]

they kinda explained their meaning right after that message

[prime dock]

ok but like you're late now and the doubt is already resolved

[warm wyvern]

[rocky cloak]

It was refferenced in there https://projecteuclid.org/journals/duke-mathematical-journal/volume-21/issue-2/Indecomposable-representations-at-characteristic-p/10.1215/S0012-7094-54-02138-9.full
Turns out that when P is a sylow subgroup of G, then every module of G appears as a summand of a module induced from P. So they are representation finite/infinite at the same time. Still not quite sure how that translates to wildness, but it's probably pretty simple.

[delicate orchid]

nice find

[median pawn]

hmm that'd do it

although consider the following; let's say we define the free product of two groups as the categorical pushout (and not in terms of words, etc.)

and we only want to show that the free group on two generators is isomorphic to Z * Z

(i.e., we just know that the pushout exists, and nothing about that the underlying set of the free product is the disjoint union)

how'd u go about the arrow stuff?

[delicate orchid]

the general pushout is free product with amalgamation, and here the amalgamation is trivial so that collapses a lot of stuff already

[median pawn]

yup it's the trivial one

[delicate orchid]

so the "left corner" in your pushout diagram is the trivial group which is initial, so the maps into Z are uniquely determined immediately - what's left over should be reminisent of a coproduct

take the "tip" of your pushout diagram to then be F(S), you can construct an equivalent diagram coming out of it

I'll draw it, one second

[median pawn]

ooo i can imagine it already

[cloud walrus]

wew
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

compile error yet it looks correct, most interesting

where S = {s_1, s_2}

I think this works

[delicate orchid]

oh right it doesn't like the 2nd level lines, oh well

[rocky cloak]

Guess the really short proof is that the free functor is left adjoint to the forgetful, hence preserves colimits. Pushout is a colimit, and 1=F(Ø), Z=F({*})

[glacial moss]

Can the product of two non-invertible elements in a ring be invertible?

[coral spindle]

Let a, b be non-units and suppose x is the inverse of ab. Then xa is the inverse of b, which is contradictory.

[noble hedge]

So then how would I go about proving the galois group of an irreducible cubic is cyclic if I wanted to conclude its discriminant is a square (still over a finite field)?

[rocky cloak]

Over a finite field, every Galois group is cyclic, generated by the Frobenius automorphism

[noble hedge]

Ah. Okay that’s a proof I gotta attempt then. Don’t think I’ve seen that result quite yet.

[rocky cloak]

If you're really only interested in cubic equations, then I guess just proving that special case is a little easier.

Let F be a field with q = p^n elements, and let y be the root of an irreducible cubic. In any finite field of characteristic p, The map f: x |-> x^q is an automorphism (check this). And f fixes F. Thus y^q is again a root. Since every element of F is a root of x^q - x, and a polynomial of degree q has at most q roots we must have that y^q is different from y.

Thus F(y) is the splitting field of the cubic. F(y) has degree 3, so the Galois group has order 3, hence equals C3

[solemn dew]

Any hints?

[wooden ember]

Proceed by induction using the fact that sylow subgroups are normal and that PQ is iso to P x Q for distinct sylow subgroups (since they’re both normal and have trivial intersection)

[wooden ember]

Haven’t really been able to figure this out but I did manage to show this: the factorisation for invertible ideals is unique since factors of an invertible ideal are invertible. I’m thinking maybe I could go for another strategy: if I can show every prime ideal contributes to the factorisation of some invertible ideal (trying with principal), by the above I’ll get every prime ideal is invertible and I can conclude from there and the work I did before

[rocky cloak]

Might be a viable strategy

[white oxide]

Here, if $F$ contained the primitive roots of unity, would $L_1 = F(\alpha_1)$? And if it didn't, would $L_1 = F(\alpha_1, \xi)$ where $\xi$ is a primitive nth root of unity?

[cloud walrus]

okeyokay

[white oxide]

Would that also explain why $\alpha_2^{n_2} \in L_1$? because either $L_1 = F(\alpha_1)$ or $F(\alpha_1, \xi)$ and the condition that $\alpha_2^{n_2} \in F(\alpha_1)$ implies that?

[cloud walrus]

okeyokay

[white oxide]

<@&286206848099549185>

[delicate orchid]

insanity

[white oxide]

wew

save me

you're a smart mann

[rocky cloak]

Yes, everything you said here is correct

[white oxide]

okay thanks a lot

ur the goat

[rocky cloak]

I'm curious, what do they go on to use this f2 polynomial for. I don't really see what they need it for.

[white oxide]

oh this is the rest of the proof

it references a previous proof so lemma post that real quick

[rocky cloak]

I don't see why they don't just take the splitting field of x^n - a2^n over L1. Are they afraid to say that the extension of solvable groups is solvable or something

So weird

[white oxide]

yeah i was confused about that too, tbh i have no clue

[untold turret]

let $t_1, \dots, t_n \in k[x_1, \dots, x_n]$ and $\mathfrak a = \langle t_1 \rangle$ be an ideal of $k[x_1,\dots, x_n]$. then, $x \in \mathfrak a \cap k[t_1,\dots,t_n]$ implies $x = t_1x'$ with $x' \in k[x_1,\dots,x_n] \cap \text{Frac}(k[t_1, \dots, t_n])$. why is the field of fractions suddenly appearing here?

[cloud walrus]

nrs

[rocky cloak]

I guess x' = x/t1. But I would think that intersection just ends up being k[t1, ..., tn]...

[untold turret]

ohhhhhhhhhhhhhhhhhhhhhhhhhhh yeah that makes a lot of sense

yeah the author specifies that this ends up just being k[t1,...,tn]

but i was very confused at why that field of fractions appeared

[delicate orchid]

Like… what is a group…. I don’t get it….

Ok I get it now

[rocky cloak]

Please enlighten us

[hidden haven]

He is being a shameless farmer

[rocky cloak]

Who can blame them

[hidden haven]

Me

[delicate orchid]

It’s just a lil dude

[glossy crag]

If K is a non-archimedean field, is it the quotient field of the valuation ring {|x|<=1}? This has got me stumped.

[hearty meteor]

hi

I need an opinion

and also some resources

what is the best resource that can be recommended to learn group theory?

in A.A

Also can it be learnt over the span of a few weeks?

[glossy crag]

Depends on your knowledge, but the chapter on groups of pretty much any "intro to algebra" books would do. Personally, I'm partial to Herstein's "Topics in Algebra".

[untold turret]

i personally like to recommend Gallian's Contemporary Abstract Algebra because there's a solution manual

[hearty meteor]

thanks 🙏

[wraith cargo]

Yes

(I think)

[glossy crag]

Your reasoning?

[wraith cargo]

note that K-A (here A is the valutation ring) consists of the inverses (in K) of the non-unit elements of A

so taking Frac(A) we just get back the entire field since all the stuff in A gets inverses

I feel like this isn't the most rigorous treatment of the problem but it sounds reasonable

[glossy crag]

Hm, you're right

[south patrol]

I've found stuff

Wait

1 sec just paranoia

OK ye it is

[wraith cargo]

what were you scared of

[glossy crag]

I was trying to show K={ab^{-1}:a,b\in A}, but was hung up on writing elements with |x|>1 as "proper" fractions, i.e. was trying to find non-identity elements with ab^{-1}=x when the entire time i could've just taken a=1

[south patrol]

That i was wrong in saying it was standard

[glossy crag]

It was stupid

Thanks.

[deft zinc]

what is stupid?

[coral steeple]

Given groups G, G' and a homomorphism \phi: G -> G', if an element of G is the only element in its conjugacy class, is

... \phi automatically an injection?

Since there can't be anything in the kernel, all the left cosets are trivial

[coral steeple]

Or have I completely misunderstood the meaning of these words lol

[dim bane]

if G is abelian then every element is the only element in its conjugacy class, yet there are certainly homomorphisms on abelian groups with nontrivial kernel

[coral steeple]

Wait but if $a$ is our element in a singleton fibre of $\varphi$, and there is an $n\in \operatorname{ker}\varphi,;n\neq 1$, then we have $\varphi(an)=\varphi(a)1=\varphi(a)$. This means $a\equiv an$, but since $n\neq 1$ we have $a\neq an$, a contradiction

[cloud walrus]

person2709505

[dim bane]

i'm confused what this has to do with conjugacy classes, and i'm also confused why $\varphi(an)=\varphi(n)\implies a=an$. isn't that condition violated precisely when $\varphi$ is not injective?

[cloud walrus]

suremark

[coral steeple]

My book defines the conjugacy class of a as the inverse image of $\varphi(a)$. Maybe that's nonstandard?

[cloud walrus]

person2709505

[next obsidian]

[dim bane]

oh.

[coral steeple]

Uh oh

[dim bane]

uhhh let me think lol

I was using the definition on wikipedia

[next obsidian]

No way that’s their definition

Please show me a photo of this, cuz that’s crazy

[ivory trail]

i'm trying to see why anyone would ever write that

[next obsidian]

I don’t even see there being any map at all which could possibly have this be the conjugacy class of a

I mean maybe this is the conjugacy class of a under ker phi?

[coral steeple]

?

[ivory trail]

a conjugacy class is an orbit under the action of G on itself by conjugation

[next obsidian]

Is this talking about quotient groups?

[ivory trail]

but

that doesn't help

[next obsidian]

Or is N specifically the kernel here?

[coral steeple]

Nope not yet

[hidden haven]

Congruence classes not conjugacy

[coral steeple]

Oh wait

Oops that explains it

[dim bane]

and im assuming N is the kernel of that map?

[coral steeple]

Yeah I confused "congruence" and "conjugacy"...

[coral steeple]

Yes

Ok, so if a is the only element in its congruence class then is phi an injection?

[dim bane]

well, just to be extra precise, I don't think there is one choice of congruence classes for a group, like clearly in this case the congruence classes are determined by phi

but yeah

not saying u implied that but i wanted to be precise about that

[coral steeple]

Well assuming phi is already chosen yes

[dim bane]

conjugacy class on the other hand is independent of any particular mapping

[ember field]

Guys Can we construct every finite extension as some finite number of towers of simple extensions.

[rocky cloak]

Sure, pick a basis and adjoin one element after another.

[celest cairn]

Is there any way I can do this without having to raise a trinomial to the 4th power?

[rocky cloak]

If you can find the Galois group of your extension. Then the minimal polynomial has the Galois conjugates as roots. Then you could just expand the polynomial

(of course expanding a 4th degree polynomial is also a bit of work)

[celest cairn]

Thanks. Is there any way I can find the minimal polynomial from my last few steps without having to redo everything?

[rocky cloak]

I mean you are pretty much done I guess. you have
$\alpha^2 + \alpha -1 = -(2\alpha + 1)i$
So just square both sides. It does involve squaring a trinomial though, but I'm not sure you can avoid that.

[cloud walrus]

jagr2808

[celest cairn]

Oh whoops forgot about that zeta_{4}^6, lol.

Thanks.

[rocky cloak]

Also, you seem to have dropped a 1 at some point, so that -1 should actually cancel

So I guess you dont have to square a trinomial after all

[celest cairn]

Sweet, thanks man

[formal ermine]

brooo

are you still finding minimal polynomials

[rocky cloak]

,w minimal polynomial exp(2pi*i/3)-i

[formal ermine]

just w

[cloud walrus]

Results provided by WolframAlpha

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

Requested by jagr2808#0

[uncut girder]

Its been 2 years sapphire still doing min polys

[formal ermine]

ong

[rocky cloak]

Thx, just wanted to test that

[formal ermine]

he started doing min polys before I even got into math

[uncut girder]

Sapphire at this point you should just write a computer program to compute min polys for you

Automate the process, you don't need to do it by hand

[rocky cloak]

Don't even need to write the program since WA exists

[celest cairn]

Yep. Just trying to learn anything and everything I can about them. Quite a fascinating subject.

[celest cairn]

Yea but it’s fun to compute them manually.

[lethal dune]

what's the mini poly of π

(no field specified)

[celest cairn]

x - pi over R

[rocky cloak]

x^2 + x - pi^2 - pi over Q(pi^2 + pi)

[raven sonnet]

Where K would fit in {|x|<=1} ?

[static yew]

So if I'm grasping this stuff correctly

Q[a,b,c,...] is isomorphic to Q[x]/h(x) where h(x) is an irreducible polynomial with roots equal to a,b,c,...

Theres a subtlety here that I'm not sure about - h(x) obviously can't have an unrelated root, like h(x) for Q[sqrt2] can't have a sqrt3 as a root or it would'nt be isomorphic to Q[sqrt2], but h(x) = x^2 - 2 has -sqrt2 as a root but we dont say Q[sqrt2, -sqrt2]

Not quite sure how that line is drawn

[wraith cargo]

-sqrt 2 is contained in Q[sqrt 2]

so why add an object that's already in the field?

[static yew]

Right. I know that. But I'm focused on the isomorphisms between Q[a,b,c,...] and Q[c]/h(x) where roots of h(x) are a,b,c,...

[raven sonnet]

Anyone willing to help me?

[static yew]

It seems like it would get real crazy with cube roots

[wraith cargo]

I don't understand your question

[wraith cargo]

Generally the rule is, if like b is contained in Q[a] then you don't need to write b as being adjoint

so if all the roots of h(x) are contained in let's say Q[a] you don't need to write Q[a,b,c,...] you can just write Q[a]

in your example

sqrt 3 isn't contained in Q[sqrt 2]

so you need to write out that you're adjoining that as well

[static yew]

Anyway the thing I'm ultimately trying to understand is a white paper on elliptic curves over finite fields Fp

It uses weierstrass form so y^2 = x^3 + ax + b

[raven sonnet]

He said that, I was disturbed by how. Where K would fit in {|x|<=1} ?

[raven sonnet]

.

[wraith cargo]

K isn't in the valuation ring

[raven sonnet]

Why ?

[wraith cargo]

K is the field containing the valuation ring

[raven sonnet]

Got it, thanks

[static yew]

And the paper goes into Fp[x,y] / (x^3 - y^2 + ax + b)

But wouldn't all the points on the curve be in the same congruence class (0) then?

[wraith cargo]

all the rational points sure

[static yew]

Wdym

[wraith cargo]

okay wdym by conguence class here lol

I think I might've misunderstood you

[static yew]

Okay so F2[x] / x^2 + 1: x^2 + 1 = x^3 + x = 0

We say that 0, x^2+1, and all multiples of x^2+1 are in the same congruence class

They are all congruent to 0

[glossy crag]

Got a question about local fields on MSE https://math.stackexchange.com/questions/4742230/number-of-extensions-of-a-local-field-of-fixed-degree, please upvote if you can so it gets some exposure.

[rocky cloak]

So Fp[x, y] are not points on a curve, but functions. Fp[x, y]/(x^3 - y^2 + ax + b) is the ring of polynomials on the curve. The equivalence class of (0) are the polynomials that vanish on the curve.

(this is a truth with modification as you really need to consider the curve in the algebraic closure, but whatever)

[static yew]

Hrm, ok.

The goal of this white paper is to explain a certain algorithm for counting points on an elliptic curve over a finite field. So there's a bridge somewhere there. Guess I gotta keep reading

[rocky cloak]

Probably goes through the spectrum of the ring

[static yew]

!! One of the steps along the way involves considering the frobenius transform over the algebraic closure

I haven't grasped what exactly that does though, cuz frob(x) = x for prime fields

[static yew]

https://tenor.com/view/the-ring-crawl-ghost-ring-ring-movie-gif-12040507

Spectrum was mostly B&W

(No idea what a spectrum is in this ctx)

[delicate orchid]

the set of prime ideals

[static yew]

Dont really grasp ideals either yet

The order of this book is so weird to me (stillwell elements of algebra)

Section 6.4 is automorphisms and groups. Words make sense but it seems to focus on the "identity automorphism" which, I get it, that obviously exists but how is that knowledge useful to me?

The existence of an identity transform is useful if you're trying to solve a puzzle where you have to use exactly n steps but the real solution only requires e.g. n-1. Add on as many identity steps as needed and bam! Solved

But where in mathematics is the identity automorphism useful?

[delicate orchid]

it makes the automorphisms of a finite group a group

and groups are nice

and is also kind of completely required for you to define inverses

[static yew]

If I have a finite field F with size q(p^k)

Then x'=x^(number coprime to q-1) is an automorphism, right?

x' = x^q is frobenius which is the identity transform here

[delicate orchid]

q = 1 if F is a field

all finite fields are of size p^k

[static yew]

Sorry that was bad typing blame mobile

When I said q(p^k) that was supposed to be a clarification that q is a prime power p^k
Not a multiplication operation

[delicate orchid]

oh ok

[static yew]

Ironically I figured you would know what I meant because as you said, the multiplication interpretation fixes q=1 in which case why have a q

[delicate orchid]

then it's characteristic p and any map like x = x^{np} is an automorphism

[static yew]

Hm

[delicate orchid]

what are we actually trying to do here

[static yew]

Trying to construct scenarios to verify my understanding against reality

[delicate orchid]

down here there is no reality

[static yew]

The problem with self study with no answers is that I can't easily check my understanding. Which is difficult with a learning disability.

[static yew]

I refer you to my earlier comment: I'm not even 100% sure R actually exists

[delicate orchid]

that's fair enough but you were asking about the coordinate ring of an elliptic curve? My apologies but I don't see how this is relevant?

[delicate orchid]

irrelevantgardless it is a useful construction

[south patrol]

wdym by exist

[static yew]

Okay so

I am a software engineer. My job frequently touches on cryptography cuz I have a knack for it, so I'm the go-to guy

Historically we have used RSA

Z/nZ for prime n - a field - is REAL easy. Just use high school algebra/arithmetic, divide by n and take remainder, and everything Just Works

For composite n, it's now a ring. The only change is you cant always divide so canceling isn't always valid. But if you're always only multiplying, then lots of stuff still work.

Then there are galois fields. GF(p^k) is not isomorphic to Z/(p^k)Z so it's not as straightforward, but those drive CRCs, Reed Solomon error correction, LFSRs, and a zillion other things. Because they are fields, the addition and multiplication operations are messy but I can work it out with high school algebra and then implement it with those operations

E.g. AES-GCM computes a secure verification code of a message that can be done very efficiently because of addition and multiplication in a field are both commutative, associative, and invertible

[delicate orchid]

right

my question was asking why you're asking about automorphisms of finite fields in relation to coordinate rings of elliptic curves

I am well aware of the applications to cryptography

[static yew]

I have worked with all of these things and I understand them fairly well (except Reed Solomon)

For example, I can see that one of the constants in AES-GCM was obviously chosen because it is it's own multiplicative inverse, meaning that using Montgomery multiplication requires one less cpu register

[delicate orchid]

clever speed up

[static yew]

But now I need to expand my knowledge to cover ECC - Elliptic curves

I can do that math itself easily because its just Z/pZ - you could have a prime power field and the math still works but nobody does that

But meta-analysis gets more complex

And right now the thing I'm trying to understand is Schoof's algorithm for computing the number of points on a certain curve in a certain field

And that gets into advanced concepts like frobenius transforms and algebraic closures and multivariate polynomial rings

Except I'm completely self taught. Last actual class I took was calc 2. And since I only studied the stuff I needed to understand for my day job, I have large gaps in what I know, because i didnt take classes with prerequisites. I read pages on the internet. Lots of wikipedia

[delicate orchid]

wikipedia is a dreadful resource

[static yew]

So when I come here everyone loses patience with me cuz e.g I never had a linear algebra course. I don't really know what a vector space is. (The writeup on the wikipedia page looks a lot like how ECC point arithmetic works though.)

So I'm trying to build up the base knowledge I need to deal with these more advanced concepts

[delicate orchid]

stop reading wikipedia and pick up an algebra book like Artin

[static yew]

I picked up stillwell

Elements of algebra

But just cuz I read it doesnt mean I understand it

[delicate orchid]

do the exercises

[static yew]

Of course there are exercises but there dont seem to be any answers.

So how do I know if I got it right?

Anyway I gotta get back to work

Never mind it is suddenly pouring out

[delicate orchid]

right, my philosophy is if a book claims to be an introduction to a topic but does not provide answers it is a dog shit introduction

hence my recommendation of Artin

which I think has answers?

[static yew]

But yeah im asking weird questions because this is how im verifying that my understanding is correct

You said that that x^(np) = x is an automorphism

[delicate orchid]

I was wrong, it's x^(p^n) = x

you just compose the frobenius automorphism with itself

[static yew]

So hold on I'm processing this

This book says an automorphism of a ring maps elements of R -> R such that f(a)+f(b) = f(a+b) and f(a)f(b) = f(ab)

[delicate orchid]

yup

[static yew]

So x^p=x in any finite field

[delicate orchid]

since when

[static yew]

Or no

[delicate orchid]

it is in Z/pZ

[static yew]

X^p=1

[delicate orchid]

if you're thinking about fermat's little theorem its x^{p-1} = 1

and only applies for Z/pZ

[static yew]

Which one is fermat's lil theorem

[delicate orchid]

for a general finite field we have x^q = x

[static yew]

Ahhh okay

[delicate orchid]

as the group of units is cyclic of order q-1

[static yew]

Okay so I was communicating poorly and we were talking past each other

[delicate orchid]

but since F_q is still characteristic p, (x+y)^p = x^p+p(...)+y^p = x^p+y^p

so it's still an automorphism

[static yew]

Wait so x^(p^n) is an automorphism in F_(p^k)?

[delicate orchid]

yes

[static yew]

Even if n!=k

[delicate orchid]

well if n = k then it would be the identity

so why would I bring it up

the frobenius automorphism is not dependant on the order of the field, only the characteristic
x -> x^p is always an automorphism in a characteristic p field

[delicate orchid]

you can prove this by expanding out the brackets in (x+y)^p, like I did up here (and (xy)^p = x^py^p obviously)

[static yew]

This is obvious for prime fields but lemme check something

[static yew]

Hrm. I gotta spend more time processing that. Is there a proof for it somewhere?

[delicate orchid]

I can write it up now if you recall the formula for binomial expansion

[delicate orchid]

Recall $(a+b)^n = \sum_{k=1}^n (n \text{ choose } k)x^ky^{n-k}$

[cloud walrus]

wew

jagr2808

[delicate orchid]

Now note that if n = p, p choose k = p!/k!(p-k)!

Which has a factor of p if k isn’t equal to 0 or p

[delicate orchid]

This sum should go from k = 0, sorry

Anyway

[delicate orchid]

So, in a field of characteristic $p$, [p \text{ choose} k= \begin{cases} 1 & k = 0, k = p\ 0 & \text{ otherwise}\end{cases}]

What now

Right fuck off then

[barren sierra]

/]

at end

[delicate orchid]

P choose k is 0 if k isn’t 0 or p

So all of the terms are 0 except for the x^p and y^p one

So (x+y)^p = x^p+y^p

[cloud walrus]

wew

[delicate orchid]

Good enough I’m tired of texing on mobile

[hidden haven]

Too much theoriposting, we should build an atomic bomb and test this out

[delicate orchid]

bitch I am the bomb

[glossy crag]

The bomb... exploded?

[hidden haven]

[wraith cargo]

Could anybody give me a hint for this 🥺
Idk how to set up the SES

[white oxide]

could somebody help me understand why hbar is a element of the set curly S? I can see that it lives in B, but i guess i'm confused about seeing how Im g \subset H + Rb

[glossy crag]

A very neat intuitive proof of Hilbert 90 via H^1 https://math.uchicago.edu/~may/REU2013/REUPapers/Lingle.pdf, to anyone who's interested.

[rocky cloak]

I'm sure you can just brute force calculate this, but a clever solution is to notice that such a complex is just a chain map and the total complex is the cone. Then you can use the long exact sequence in homology

[wraith cargo]

okay wow that's very clever

thanks!

[rocky cloak]

By assumption the image of g is a subset of H, so then it's also a subset of H + Rb.

[white oxide]

ahh right thanks!!

[white oxide]

Why is it that $[\mathbb{Q}(\omega)_{\langle \phi^3 \rangle} : \mathbb{Q}] = 3$?

[cloud walrus]

okeyokay

[white oxide]

Sorry should've specified, here $\omega$ is a primitive 7th root of unity and $\phi$ is determined by sending $\omega \mapsto \omega^3$

[cloud walrus]

okeyokay

[white oxide]

Is it because deg$(\omega + \omega^{-1}, \mathbb{Q}) = 3$?

[cloud walrus]

okeyokay

[white oxide]

Oh wait it's just because of this lattice right

and the fundamental theorem of galois theory

[hearty meteor]

what background knowledge in A.A is required to fully understand group theory?

im trying to speedrun it

like in a few weeks

[open sluice]

a lot of abstract algebra courses start with group theory

whether that’s pedagogically preferable or not is debatable but it’s definitely doable

(learning group theory first, that is)

[hollow mica]

what would you start with

[open sluice]

no idea

when I opened my first algebra book I’d just needed to learn group theory for a project I was doing

but I felt it was too hard when I tried it so I decided to do rings first

maybe it depends on how much group phenomena you happen to already be familiar with

we start with rings because people usually understand +*- pretty well but maybe groups first would be more tolerable for a Rubik’s cube nerd

[hearty meteor]

thats what i was aiming for

rubiks cube

but im not a very big rubiks cube nerd

so ill try to start with ring

s

thanks though!!!

[open sluice]

I mean you can try

[formal ermine]

lol

[sacred monolith]

Hello, I am looking to self-study abstract algebra, I’ve done some research what book to use, I’ve heard of Herstein, Fraleigh's, and Dummit and Foote's. I was wondering what your guy’s thoughts on which one to use, or other recommendations. Thank you.

[summer path]

#book-recommendations and also #book-recommendations message

[sacred monolith]

oh my bad, Thank you

[white oxide]

i find it so interesting that some people find rings more intuitive

I mean yea I guess it makes more sense but you would think less operations is easier to comprehend or smt

[formal ermine]

I recommend artin

[formal ermine]

fields are the most intuitive ones imo

[white oxide]

yea that's also true

[rocky cloak]

Like Bladewood said many (most?) abstract algebra courses start with group theory. I think the reason for this is that groups are very simple (I don't mean easy, I mean has few moving parts). So if you want to deal with abstraction this view is more focused. The downside is that groups can feel more abstract.

Conversely, you already have a lot of experience working with rings, like the integers, the real numbers, polynomials, matrices. So it's often easier to reach for a specific example.

[summer path]

tangentially, i think where you start learning intro aa is kind of dependent on where you want to end up; it would be pretty awkward to start with rings/fields (stuff people are generally more familiar with) and want to cover galois theory, since you have a bit of a group theory knowledge gap in the middle

[rocky cloak]

Wouldn't you just go rings -> groups -> Galois theory as opposed to groups -> rings -> Galois theory? Or what would be the awkward gap?

[summer path]

i guess rings -> groups -> fields -> galois theory also works

but i was thinking that rings -> fields -> groups -> galois theory would be rather awkward

[delicate orchid]

groups not being first is odd to me

[rocky cloak]

Right, you're saying that fields is a nice segway into Galois theory. That's fair enough, though things don't have to be completely compartmentalized either

[rocky cloak]

How so?

[devout dirge]

Why every lie algebra homomorphism from the Heisenberg lie algebra to Mn(C) sends the center into nilpotent element?

An element of the center is a cumottator so its image has trace zero. How can i proceed from here?

[vast hawk]

Hello, I would like to receive help with a group theory exercise.
The question is to find an example of group G such that its 2nd derived subgroup is different from its 3rd derived subgroup. I know the answer but I can't understand why it's the correct answer (I can give it to you if you wish)

(I already posted a message in the channel "help-9")

[rocky cloak]

My first guess would probably be to try to find a solvable group whose 3rd derived group is trivial (but 2nd derived is not).

[vast hawk]

I think that's the right idea

But I have difficulty in finding the number n derived subgroup

[rocky cloak]

So something like a nested semidirect product of C7, C3 and C2 maybe

[vast hawk]

I thought to C2

[rocky cloak]

Instead of thinking about the derived series you can think about maps to abelian groups.

[vast hawk]

Because, if I'm right, C2xC2 give a group of order 4 that's abelian and so its derived subgroup is trivial

[rocky cloak]

That's right

[vast hawk]

(I hope I let me understand because English is not my mother tongue ^^)

[rocky cloak]

So what's the example you have?

[vast hawk]

I think if we take G = S_(4), it would work

I know the derived subgroup of S_4 is Alt(4)

[rocky cloak]

Alright, if you know that A_5 is the smallest non-abelian simple group, you know that S_4 is solvable. Then it's enough to show that the 2nd derived subgroup is nontrivial

and since A_4 isnt abelian youre done

[vast hawk]

I understand what you said but what are the 2nd and third derived subgroup of S_4 ?

[rocky cloak]

Idk

But you could calculate it

[vast hawk]

But how do you conclude that they are different then ?

[rocky cloak]

becuase A4 is solvable

so the derived series gets smaller and smaller until it reaches the trivial group

[vast hawk]

Ok, so A4 is the derived subgroup of S4 and when we calculate its derived subgroup, we get the trivial group because it is abelian

[rocky cloak]

No, A4 is not abelian, so the derived subgroup is not the trivial group

So the 2nd derived subgroup is nontrivial, and hence the third must be even smaller

(and in fact the 3rd is trivial)

[vast hawk]

Sorry but how do you know the third one is trivial

[rocky cloak]

I checked

[delicate orchid]

[A_4, A_4] = C_2 x C_2

[rocky cloak]

The derived series is {e} < C2xC2 < A4 < S4

[vast hawk]

That's the 2nd derived subgroup of S_4 ?

[delicate orchid]

yes

and the third is trivial

[vast hawk]

The third one is given by : [C_2 x C_2, C_2 x C_2], is it true ?

[rocky cloak]

yes

[cloud walrus]

wew

[vast hawk]

because C_2 x C_2 is of order 4 so it's abelian and then its derived subgroup is trivial ?

[delicate orchid]

yus

[vast hawk]

Okeeee, thanks a lot !
I think I understood

[vast hawk]

I'm back with a new question

Is it correct to say that x^2-t, the minimal polynomial of \sqrt(t) on F_2(t) is not separable?
I need to show that the extension F_2(\sqrt(t))/F_2(t) is not a Galois extension.

And I know a polynomial is separable is its roots are simple

[glossy crag]

It is, the polynomial x^p-t over F_p(t) is the prototypical example of an irreducible inseparable polynomial. Once you adjoin a single root z of it, it splits into (x-z)^p.

[vast hawk]

Ok and it's inseparable followinf the definition ?

[glossy crag]

It has just one root repeated p times, so it's inseparable.

[vast hawk]

ok ok, I was right

Thank you 😉

Would it have changed if we had taken F_2(sqrt(t))/F_2 ?

Is it possible to take it ? It has the same minimal polynomial ?

[glossy crag]

sqrt(t) is transcendental over F_2, because t is, so it no longer makes sense to speak of minimal polynomials, because there aren't any.

[vast hawk]

yep

[chilly ocean]

Hello I have a quick question: If G is a finite group of order n, f is an endomorphism of G and I={x in G with f(x)=x^-1}. Show that if |I|>3/4 × n then G is abelian

It is ok if I say: f(x)=x^-1 then f(f(x))=f(x^-1)=x. But f(f(x))=f(f(x)f(e))=f(x)f(e)=f(x) so f(x)=x but f(x)=x^-1 so x=x^-1 so x^2=e and from here the problem is simple?

[delicate orchid]

one sec

$f(f(x))=f(f(x)f(e))=f(x)f(e)=f(x)$

[cloud walrus]

wew

[chilly ocean]

Oops

Yes I see where

[delicate orchid]

middle inequality doesn't hold, $f(f(x)f(e)) = f(f(x))f(f(e))$

yeah

[cloud walrus]

wew

[chilly ocean]

Yes

[delicate orchid]

I do think it could have something to do with the number of order 2 elements though

[rocky cloak]

So if x, y and xy are in I, then they commute.

For x in I, the intersection of I and x^-I is more than half the group, so x commutes with at least half the group.

[chilly ocean]

Yes I did it

Then the centraliser is the group

We get that I is subgroup

And by lagrange I=G

[south patrol]

Hm if we have an exterior algebra (in a graded context) what does the notation $\langle x_1,\dots,x_n \rangle$ mean? Is it just notation for $x_1 \wedge \dots \wedge x_n$ lol - i'm seeing this in the context of divided algebras and these

[cloud walrus]

potato

[south patrol]

But i've yeah only seen this notation in specific contexts hm

[solar vessel]

uh

span?

or generated subalgebra?

[south patrol]

No, they refer to elements

[white oxide]

$\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$ is a finite Galois extension over $\mathbb{Q}$ since it's the splitting field of $f_1(x) = x^2 - 2, f_2(x) = x^2 - 3$, and $f_3(x) = x^2 - 5$ right

[cloud walrus]

okeyokay

[white oxide]

And they all have zeros of multiplicity one?

[south patrol]

Ye

I mean the multiplicity one bit is generally sort of irrelevant over Q since every poly over Q is separable

Now compute Galois group

[white oxide]

Also why is $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}): \mathbb{Q}(\sqrt{2}\sqrt{3})]$ = 4? Isn't $x^2 - 5$ a minimal polynomial of degree 2 with coefficients in $\mathbb{Q}(\sqrt{6})$?

[white oxide]

ye

[cloud walrus]

okeyokay

[white oxide]

imma try to draw a lattice in a sec

[white oxide]

o i didnt know that

[white oxide]

o it prolly has to do something with the tower law or something right

[white oxide]

Is it because $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) : \mathbb{Q}(\sqrt{2}, \sqrt{3})][\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{6})] = 2 \cdot 2 = 4$

[cloud walrus]

okeyokay

[rocky cloak]

Yeah, that's right

[south patrol]

True for any field of char 0

[white oxide]

oh right they're perfect

oops

a basis for Q(\sqrt{2}, \sqrt{3}, \sqrt{5}) over Q would be these guys right

[delicate orchid]

considering you just said the index is 4

and there are more than 4 things there

[white oxide]

nah boss that was Q(\sqrt(2), \sqrt(3), \sqrt(5)) over Q(\sqrt(6)) and not over Q

[delicate orchid]

oh then yeah double it

now I buy it

kind of interesting how this demonstrates that adding up binomial coefficients gives you a power of 2

[void cosmos]

yo

primary decomposition is literally magic

[delicate orchid]

yeah it's like prime decomp but better and swagger

[void cosmos]

yea

thats the point

its like number theory

is now in like polynomial theory xD

like you can have fundamental theorem of arithmetic but for polynomial rings for example

coola f

[delicate orchid]

I presume you mean a general polynomial ring

[void cosmos]

yea?

wait

what

whats a general polynomial ring

[delicate orchid]

just a general one u know

cause there are definitely many polynomial rings that have prime factorisation

[void cosmos]

you mean like

F[x,y]

instead of F[x,y]/something ?

[delicate orchid]

R[x]

[molten viper]

Would someone mind providing a hint for this:
I wish to prove that the jacobson radical is equal to the nilradical in A[x] (A some commutative ring).
So far I have: suppose a \in jacobson, then we know 1-ay is a unit for all y \in A. Specifically, f = 1 - a is a unit. I know from a previous problem that f_0 = 1 - a_0 is a unit and f_n = -a_n are nilpotent for n \ge 1. How could I show that a_0 is nilpotent?

[chilly radish]

Who are the f_n exactly

And the a_n

[molten viper]

the coefficients of f and a respectively

[chilly radish]

Ah ok

[molten viper]

I know that if x is nilpotent then 1 + x is a unit, but I have no guarantee that works in reverse (in fact I'd assume it doesn't)

[formal ermine]

-2 in Z

[chilly radish]

I mean for a=a_0+a_1x+...+a_nx^n to be nilpotent of course a_0 needs to be nilpotent

For a^2=0 to be true

[wraith cargo]

Okay so do you already know that Nil(A[x]) = Nil(A)[x]?

[chilly radish]

Compare coefficients

[molten viper]

Well I'm trying to prove that a_0 is nilpotent in order to show a is nilpotent

[chilly radish]

But the Jacobson radical is contained in the nilradical

[molten viper]

I don't know that

[chilly radish]

Oh

[wraith cargo]

wait do you not know what shi said or what I said lol

[formal ermine]

it follows immediately from the definition tho?

[molten viper]

both actually

No it doesn't

[formal ermine]

yes it does

[molten viper]

Least not the definitions I've been presented

[wraith cargo]

what shi said is easy to show
Just write the definitions of the jacobson radical and the nilradical

[formal ermine]

how do you define the jacobson and nilradical radical

[wraith cargo]

nilradical radical

[formal ermine]

I'm too tired

[wraith cargo]

What Amerika does to someone

[molten viper]

nil: set of all nilpotent elements or the intersection of each prime ideal
jacobson: intersection of every maximal ideal (and I know that a is in jacobson iff 1-ay is a unit for every y in the ring)

[wraith cargo]

use proposition 1.8 of Atiyah

[formal ermine]

wait

[molten viper]

I'm waiting :)

[wraith cargo]

every maximal ideal is prime

that's the magic bullet here

[formal ermine]

yeah lmao

what I said

I got confoooosed

[wraith cargo]

so it follows that Nil inside Jacobson

[molten viper]

which is not what i'm trying to prove right now :)

[rocky cloak]

Maybe consider y=x

[wraith cargo]

well you'd have to prove it at some point anyways :P

[elder wave]

Aren’t you trying to prove that they’re equal

[molten viper]

Yeah

[rocky cloak]

Then write out the coefficients of f in f(1-ax)=1

[elder wave]

Well

[molten viper]

But I didn't ask for a hint on that part

Thank you @rocky cloak

[molten viper]

I definitely didn't consider x because I got mixed up by notation

[rocky cloak]

x is pretty important, since the statement isn't true in arbitrary commutative rings

[molten viper]

That makes sense

I guess when it comes to textbook problems you often have to use facts about every part of the setup

if that makes sense

[south patrol]

Yes

[elder wave]

If you’re not using all given assumptions you’re usually wrong

[molten viper]

I remember when I took our rings course, we proved some fact about rings and my professor said "it's not true about vector spaces, so you should try multiplying ring elements" or something like that

[south patrol]

Which causes problems when they give too many assumptions xd

[south patrol]

Yeah this is good to think about

[molten viper]

I also once got advice that when you finish a problem, you should try playing with the assumptions and seeing if removing assumptions breaks things

[elder wave]

This is a good approach for finding counterexamples

[molten viper]

or alternatively if you can't figure one out, try using stricter assumptions and see if you actually need that set of assumptions

[elder wave]

Remove one assumption, see where it’s used in the proof, think about which kind of object it could fail for

[molten viper]

(obviously if it's a textbook problem, you don't need stricter assumptions)

[molten viper]

it was actually even simpler than that

[rocky cloak]

Oh yeah?

[molten viper]

Yeah so

because 1 - ax is a unit, if we look at the coefficients we get 1 - a0x - a1x^2 etc

so every a_n is nilpotent

[rocky cloak]

Does that follow from something you've proven already?

[molten viper]

It follows from the previous problems

[rocky cloak]

Then it's pretty simple yeah

[molten viper]

I have suffered before with skipping textbook problems and then those proofs being used in the following problems

the worst time being when a professor assigned some problems and like the first one assigned used something proved in a problem he didn't assign but it's whateverrrrrrrr

[formal ermine]

or simply a = 0 because x has no inverse

[rocky cloak]

Guess a=0 would be very nilpotent

[elder wave]

Prove it!

[formal ermine]

so true

timo why are you awake

[elder wave]

Waiting for my gf to finish reading a paper and explaining the math occurring in there

[rocky cloak]

What she reading about?

[elder wave]

Social science

[formal ermine]

bro got the study rizz

[coral steeple]

Is this proof that a normal subgroup H of a finite group G satisfies aH=Ha for a fixed element a in G valid?

Let $ah\in aH$ be arbitrary. We want to find $h' \in H$ so that $ah=h' a$, or $h=a^{-1}h' a$. Since $H$ is normal, conjugating by $a$ is an injection from $H$ to itself. Because $H$ is finite, this injection must be surjective, and $h'$ exists. The other inclusion follows by symmetry.

[cloud walrus]

person2709505

[chilly radish]

You don't need finiteness, it's always a surjection because you have an inverse

Try and think what that is

[coral steeple]

Is $h'$ just $aha^{-1}$?

[cloud walrus]

person2709505

[coral steeple]

I think it is; not a very big brain moment for me

[delicate orchid]

aha!

[molten viper]

Here's a rather specific question, if I have ideals I and J, with I \subset J, what can I say about R/I and R/J?

Is there some sort of nice homomorphism between the two?

[chilly radish]

Yes

[delicate orchid]

correspondence theorem for one

[chilly radish]

Third isomorphism theorem

[molten viper]

Let me give those a quick lookup

Ok so, I'm seeing that if I subset J, then J/I is an ideal of R/I
and
If I subset J, (R/I)/(J/I) iso (R/J)

I have to remind myself how we define J/I

[void cosmos]

what does derived category mean

i know what a derived functor is

i think 😄

[wraith cargo]

that is a very non trivial definition

[void cosmos]

yea fuck it then

[wraith cargo]

iirc it's the quotient category of the homotopy category of chains of some abelian category

[void cosmos]

quotient category 😦

yea fuck it

ty tho

[wraith cargo]

https://www.math.utah.edu/~milicic/Eprints/dercat.pdf
If you wanna learn more I recommend this PDF

[molten viper]

So just to check, J/I is the set {j + I | j \in J}?

[wraith cargo]

Here's the exact definition

[chilly radish]

Yes