groups-rings-fields

[abstract rock]

well that's true regardless because there's usually more rhan one group for a given order (you'll see an example of this in the proof of the fundamental theorem for abelian groups) but what i mean is that there are multiple cayley tables that look different but are generated from the same group

if you've learned about what an automorphism group is, you can show this corresponds to certain automorphisms of the group

[white oxide]

will this be sufficient to show that every maximal ideal in Z[x] is generated by 2 elements?

[white oxide]

<@&286206848099549185>

[south patrol]

Hm I mean it may need a bit more work

Since the Z-basis of I_n could be large enough

All this proof really shows is that every ideal of Z[x] is finitely generated

[white oxide]

rip

ta always tryna murder us fr

[south patrol]

And you can adapt to any PID

Well

This is basically Hilbert's basis theorem if you're interested

if every ideal of R is finitely generated, same is true of R[x]

This is a special case of R = Z, where the hypothesis is immediate since we have a PID

[white oxide]

right

i don't see how we can adapt that to proving my problem tho

[south patrol]

Ye that's kinda my point lol like the proof method is too general

[white oxide]

yea lol

it's okay i'll find something online

[south patrol]

I can't actually fully remember what's the easiest way to do your problem lol

the only way i have is a bit too powerful

[white oxide]

lol do you mind dropping it?

[south patrol]

actually nah it's okay

[white oxide]

lmao

[south patrol]

Well basically

Let m be your maximal ideal and consider m intersect Z

that's a prime ideal of Z, hence either 0 or pZ for some prime p

But then that is the harder bit

[rocky cloak]

If you know how prime ideals interact with localization, you can say that a prime ideal of Z[x] that doesn't intersect Z corresponds to a prime ideal in Q[x], hence principal

[south patrol]

Yeah i think that's too fancy for okey's class though

I would say Z is Jacobson (and so is Z[x] as a result) which also shows m intersect Z is maximal

[white oxide]

wait so in that case would (p, x) = M

[south patrol]

yes

Wait sorry

[white oxide]

hmm i'm not sure what jacobson or localization is

[south patrol]

I meant pZ[x]

Lol

[white oxide]

ah okay lol nw

[south patrol]

Sorry i messed up lol

[white oxide]

ur good

[south patrol]

Yeah idk a non-fancy way to do this lol

Rip

fwiw the fancy way I know is actually like lol

m maximal, so m intersect Z is maximal and hence of the form pZ as jacobson

then we have inclusions 0 < pZ[x] < m

[white oxide]

Yeah I’m not surprised

[south patrol]

but a maximal ideal of Z[x] containing pZ[x] corresponds to a maximal ideal of Z[x]/pZ[x] = (Z/p)[x]

[white oxide]

Ta is a menance

[south patrol]

but this is a PID, so the maximal ideal is just generated by f mod p for some f in Z[x]

and hence m is generated by p and f

Oof

This is why a lot of the common examples are (2,x) and stuff

since 2 is a prime and x mod 2 is irreducible

But e.g. x^2 + 2 is irreducible over Z, but (2, x^2 + 2) isn't even prime

[white oxide]

Oh wait so would that proof work with that fancy machinery and stuff

[south patrol]

wdym

[white oxide]

Nvm lol I didn’t know what I meant

[white oxide]

Ohhh ok I see

[rocky cloak]

How about this:
So you have an ideal generated by f1, ..., fn

Consider the ideal in Q[x] generated by the same elements, then this will equal (g) for some polynomial g. (We may assume the coefficients of g are relatively prime)

Case 1: g=1, then there's some combination of the f1 that makes 1, multiplying out the denominators there's some Z[x] combinations of f that make an integer, hence the ideal intersects Z, and you reduce to the Z/p case.

Case 2: the ideal doesn't contain g. Then adding g gives a bigger ideal, hence not maximal.

Case 3: g is contained in the ideal. We know that fi = ghi for rational polynomials h. Choose p relatively prime to all the denominators of the hi coefficients and the coefficients of g. Then the ideal is contained in (p, g) hence equals (p, g)

@white oxide

[white oxide]

thank you, i'll take a look at that

[rocky cloak]

(I guess case 2 is redundant)

[rocky cloak]

This assumes you've already proven Z[x] is Noetherian though, which you more or less did here

[white oxide]

Could I also use the hilbert basis theorem for that

Z noetherian

I think

[rocky cloak]

Yes, that works

[white oxide]

Yeah

Cuz PID

[torn warren]

hello, I have a question, is my understanding correct?

the red line is translated to red line, and blue line translated to blue line

[white oxide]

i'm a little confused as to why every element of the tensor product is not of the form a tensor b, if my definition the cosets of K in F are of the form a tensor b?

[delicate orchid]

a (x) b + b (x) a is not an elementary tensor

[white oxide]

what's an elementary tensor?

[delicate orchid]

x (x) y

[white oxide]

ohh ok that makes sense i think

[delicate orchid]

as it says, the tensor product is generated by those elements

generated being the key word

[tribal moss]

A useful main example to keep in mind is the tensor product of two vector spaces of dimension n and m. That simply consists of all n×m matrices.

Elementary tensors are the matrices of rank 1.

[white oxide]

oh i understand now

oh god i forgot what rank means

[white oxide]

oh intersesting

[tribal moss]

(Matrix rank, not tensor rank here).

[long obsidian]

What does the quotient $\frac{\mathbb{Z}[x]}{<2>}$ look like?

[cloud walrus]

HausdorffT1

[delicate orchid]

F_2[x]

[long obsidian]

So it just amounts to quotienting the underlying ring?

[delicate orchid]

well see if you can show that

[tribal moss]

v otimes w is then the matrix v w^T.
(In particular e_i otimes e_j is the matrix with all zeroes except a 1 in the element pointed to by their indices, which tells us that those tensors are a basis for the tensor product).

Since v w^T has a one-dimensional column space (the column space is generated by v, unless w=0), not all matrices can have that form.

[white oxide]

hmm ok i'll definitely think about that for concreteness, thanks

why exactly do we choose K to be generated by elements of the form (i)-(iii)? can somebody explain it to me like i'm five lol

[delicate orchid]

those are the relations a bilinear function satisfies

if you put a "= 0" on the end because we're quotienting

[white oxide]

is that the same thing as middle linear?

[void cosmos]

these relations

are exactly what u need

so that u go from bilinear maps in M x N to linear maps in M tensor N

cuz they are equivalent now lmfoa

[next obsidian]

Tf is middle linear

[void cosmos]

u now quotient the free module by all of these relations

[next obsidian]

Dawg, lol

What are you reading lmfao

[void cosmos]

hungerford

[next obsidian]

But yes this is a bilinear map

Or really it’s a like

“Balanced map” or something

Never in my life have I seen this term lol

[white oxide]

i think that makes sense...

a little bit..

[void cosmos]

take these for granted for like 3 minutes

he then proves that this gives out the universal property instantly

hopefully it should click

[delicate orchid]

holy shit you need to stop reading lang

this is garbage

[void cosmos]

everyone hates hungerford idk why 😦

[delicate orchid]

or hungerford whatever

if it's the same book as yesterday this is the last straw

[void cosmos]

i think he is too far to go back now

[white oxide]

NAH BRO HUNGERFORD CLEARSS

[void cosmos]

btw the motivation is ( i think ? ) the same

as with tensor products of vec spaces

which u probably saw in linear alg

[white oxide]

nah we didn't even get to jordan normal form and determinants and complexification and stuff in lin alg

😹

[void cosmos]

happens ig haha

this section has one of the coolest shit ever in math for me

the "adjoint" shit with hom and tensor

hopefully it blows ur mind too

[white oxide]

how is this easy to see :(

[void cosmos]

(8)

[white oxide]

am i supposed to see that $a \otimes 0 + a \otimes 0 = a \otimes 0$ or smt

[cloud walrus]

okeyokay

[white oxide]

o

[delicate orchid]

no because it isn't

[white oxide]

o

wait now how tf does 0 \otimes b = 0 \otimes 0

OH

wait nvm

wait yeah

wait nvm

wait where does one use the fact that x + x = x implies x = 0

cuz

[delicate orchid]

0 \otimes b = (0*0) \otimes b = 0(0 \otimes b) = 0 \otimes 0

[white oxide]

$0(0)b \otimes b = 0 \otimes 0b = 0 \otimes 0$

[cloud walrus]

okeyokay

[delicate orchid]

yeah there you go

[white oxide]

ye

but like how does 0 being the only element that satisfies x + x = x satisfy it

actually nvm

i proved it forget about it lol

hungerford my beloved

❤️

[delicate orchid]

middle fuckin linear

[white oxide]

LOL

[delicate orchid]

what would he call multilinear maps then

like the name is moronic

[white oxide]

he also calls isomorphisms in categories equivalences

W OR L

[delicate orchid]

that's based

[white oxide]

wait rlly

i saw some ppl hating on that

[delicate orchid]

nobody cares about categories up to isomorphism

[white oxide]

well Lang calls ufds factorial rings and integral domains entire rings, so imma have to go w/ hungerford sadly

[delicate orchid]

yeah that's understandable

[void cosmos]

hungerfords better

superior textbook

[white oxide]

hungerford does just the right amount of hand holding

lang kicks you to the curb and says good luck

[void cosmos]

yea man not really that was a joke

the only cool part is

we know more definitions

like quasi-regular

lmfao

[sly crescent]

I do

[next obsidian]

Why is the choice only between Hungerford and Lang????

This is like only babymode or hardcore mode

[void cosmos]

babymode is lang

ofc

[white oxide]

What other selections are there???

[next obsidian]

Aluffi, Dummit and Foote

Artin

[delicate orchid]

fartin

[white oxide]

never heard of those; ergo they must not be good

[next obsidian]

….

You must live under a rock

[delicate orchid]

how

[next obsidian]

[white oxide]

LOL

nah i'm just playin i have

[delicate orchid]

https://tenor.com/view/heath-ledger-and-here-we-cope-joker-gif-9467009

[white oxide]

do bilinear maps send (0, 0) to 0

[chilly ocean]

prove it

[delicate orchid]

yeah have a think about that one

considering we previously showed that a (x) 0 = 0 (x) 0 there's a slightly more powerful statement you can say

[white oxide]

hrmmm ok

cuz i was tryna show that $g_1\bigl((a + a', b) - (a, b) - (a', b)\bigl) = 0$, so I got $g_1\bigl((a + a', b) - (a, b) - (a', b)\bigl) = g(0, -b) = g(0, 0)$

[cloud walrus]

okeyokay

[night onyx]

the map (a, b) -> a tensor b is (by definition) linear in both variables, and linear maps send 0 to 0

[cerulean trout]

Is F(V)=K here?

[cobalt heath]

Uh

If it were, it would be an affine variety

[cerulean trout]

yeah

[cobalt heath]

Perhaps use your intuition?

[cerulean trout]

but V(xy-1) = {xy=1, x,y in K}

and every y has an inverse in K right

[cobalt heath]

perhaps?

[cerulean trout]

so then F(V) would include every y???

sorry if im misunderstanding something terribly here ;-;

[cobalt heath]

Nah actually this is a trivial one so let me apell it out

Soo what is the inverse of 0?

[cerulean trout]

anything?

[cobalt heath]

Huh

Wdym anything?

[cerulean trout]

ok nvrm nothing

0 doesn't have an inv

[cobalt heath]

Yep!

[cerulean trout]

ah ok ic I think

so V(xy-1) = K \ 0

and I(K \ 0) = 0

but V(0) = K

is that right?

[cobalt heath]

Yep

[cerulean trout]

ahhh

thank you

[south patrol]

a drawing actually helps here lol

one fun thing is that there is an analogous thing when you consider R (with its standard topology)

It's an example of a projection map which isn't closed

[abstract rock]

was lookin through that open question article someone posted from arxiv awhile ago and the following seemed like a bit deceptively simple for a 60 yr old problem. Though my question is the original author asking about abelian extensions as abelian galois extensions from galois theory or abelian by abelian extensions in the language of group theory.

[cobalt heath]

It took me a few hours to go through first 2 sections of hartshorne chapter 2. How can I go faster?

[next obsidian]

A few hours is good pace

That’s fast

[cobalt heath]

I see, thanks. I wish I could go faster to keep up with the scheme homology class.

[chilly radish]

HOMOLOGY?

But muh injectives...

[cobalt heath]

Huh

Ah right

Scheme cohomology*

Honestly it’s just index reversed

[chilly radish]

It's not tho

[cobalt heath]

Hmm

Why?

[chilly radish]

I mean there's a lot of fundamental reasons why it's more than just reversing the arrows, but there's cases that you can't really define a homology theory but you can define a cohomology theory

e.g. in the case of sheaf cohomology, you can't define a reasonable notion of sheaf homology, at least at first (you can provide alternative definitions which end up being equivalent)

So for sheaves this is because the category of sheaves might not have enough projectives, bur it always has enough injectives

To define a suitable dual notion of a homology theory you would first need to be able to generalise your definition of derived functor

Which is what sheaf cohomology is built upon

While these will end up being formally dual in some sense, one is much easier to compute in general

And some properties are best seen through the lens of one

[cerulean trout]

I'm a little confused by the notation here...y_j is a just a a variable in the polynomial, not a function, right? I suppose this is just referring to F_j?

[torn warren]

Do I understand this correctly? I translate the red line into red line, and blue line into blue line.

[tiny jolt]

y_j is the coordinate function

[cerulean trout]

what is the coordinate function?

the word coordinate does not appear in the textbook prior to this ;-;

[tiny jolt]

Ah

It just outputs the jth entry of an element of K^m

[cerulean trout]

Oh i see

[tiny jolt]

So composing it with f will give the jth entry of f

I.e f_j

[cerulean trout]

alr cool

thanks!

[cobalt heath]

I just do not know the fundamental differences between chains and cochains.

Aren’t they just differ by how their index is given?

[torn warren]

can anyone help me on this understanding..

[cobalt heath]

Seems correct ime

[rocky cloak]

Algebraically there is no difference between homology and cohomology, except the indexing convention. Both are just kernel modulo image.

But usually you don't just consider the cohomology of a random complex, but you are interested in using homology to study certain objects (topological spaces, sheaves, groups, algebras, smooth manifold, lie algebras). Whether you call something homology or cohomology then comes from how you construct a complex from your object. As a rule of thumb, if the process is closely related to Ext then it's cohomology and if it's closely related to Tor then it's homology.

[cobalt heath]

Indeed

So convention and the origin is important, I guess?

[rocky cloak]

It depends I guess. If you're just studying modules and chain complexes for themselves, and not as an invariant to study anything else then it makes no difference.

So for me, for example, it doesn't really matter whether I think of it as homology or cohomology. Just have to be consistent with indexing of course

[cobalt heath]

Is valuation just multiplicative homomorphism R -> V? Or is there more to it

[delicate bloom]

what kind of response are you fishing for here

[cobalt heath]

Sorry, I’ll google for definition when I get time.

I would indeed appreciate some intuition tho

[delicate orchid]

It’s the p-adic valuation but more general

[plush tree]

In chapter 7 of Hydon's book on Symmetry methods, he describes a method to find first integrals of ODEs using characteristics of symmetries. The method very roughly takes various permutations of the Wronskian of n+1 (or more) symmetry characteristics for an nth order ODE, and their ratios under some conditions are first integrals. The system o first integrals is then purely algebraic, allowing you to solve the ODE. Since every integral is a first integral of some first order ODE, could this method be used to evaluate arbitrary/undocumented integrals?

[dense raven]

When can the multiplication operation of a finitely generated group be extended to its asymptotic cone? I know it works for finitely generated abelian groups.

[tardy hedge]

Does problem #25 basically reduce to “odd + odd = even”?

[next obsidian]

ya

[tardy hedge]

Nice

[white oxide]

what's the importance of this theorem? it seems as if it's saying that right modules preserve exactness under tensoring with certain conditions... is there anything more to this?

[next obsidian]

It says tensoring is right exact

[white oxide]

ohh hm ok

right i guess i've seen that term before

[delicate orchid]

have you seen that Hom is left exact

[rocky cloak]

Fun fact:
Any right exact functor F: mod R -> mod S that preserves direct sums is given by tensoring with a bimodule

[delicate orchid]

I buy it

[white oxide]

uhh i went over it like once over the summer i think

[delicate orchid]

yeah so this is just that but going the other way

[white oxide]

is it this

[delicate orchid]

yur

[white oxide]

oh so when hungerford says "hom is a sort of dual notion to tensoring" or whatever

[delicate orchid]

note how the 0 for the tensored sequence is on the other side

[white oxide]

is this kinda what he means

[delicate orchid]

they're adjoint functors

[white oxide]

i haven't gone over what adjoint functors are yet

but i'll look it up

[white oxide]

oh ye

wait so

i'm assuming the functors would be you fix a module D, then A \mapsto D \otimes A

and then for the hom functor it's just A maps to Hom_R(A, D)

[delicate orchid]

yur

you'd write this as Hom$(-, D)$ and $- \otimes D$

[cloud walrus]

WewGhostTbh

[south patrol]

I've always wondered if this admits any nicely categorical proof lol

[delicate orchid]

what the fact that uhhh right/left adjoints between module categories are left/right exact?

[south patrol]

No lol

[delicate orchid]

oh

[south patrol]

I mean the thing okeyokay posted

It always seems a little bit of a mess to prove

[delicate orchid]

that's the same thing

[south patrol]

Not quite

The "if" part

[delicate orchid]

ah

[south patrol]

That bit always feels dodgy

[delicate orchid]

yeah I see what you mean now

[south patrol]

But perhaps we can just pick appropriate D and use Ext to make it work

¯_(ツ)_/¯

[delicate orchid]

apply Ext to the whole thing, and then since they are inverses you get back to where you started :thecosmoshumswithatunemostsweet:

[coral spindle]

show us the emoji, wew

[delicate orchid]

the velvet curtain remains drawn

[coral spindle]

NO

is that also an emoji

[delicate orchid]

no that's just a thing I had in my head

[coral spindle]

isn't that what all emojis are

[delicate orchid]

perhaps...

[rocky cloak]

I mean A -f-> B -> C -> 0 is exact iff C is the cokernel of f. And the definition of cokernel is just that the sequence remains exact when you apply Hom(-, D) for every D

[unkempt mountain]

what's the point of learning abstract algebra, can it lead to proving other concepts in math that can do something within the real world, does it just stay in mathematics, or can it directly be applied to reality

[delicate orchid]

heard of "quantum mechanics"?

[rocky cloak]

Abstract algebra does indeed have many applications, both to other fields of math and the sciences

[unkempt mountain]

interesting

[rocky cloak]

For example cryptography and coding theory is something you interact with in the real world every day, and heavily uses abstract algebra.

Representation theory of groups apears in solid state physics and chemistry.

Noether's theorem is a theorem in Lagrangian mechanics, relating symmetry groups to preserves quantitees in physical systems.

[unkempt mountain]

that's cool I always presumed that abstract algebra was just purely mathematical as the abstract name suggests lol

[rocky cloak]

I mean most mathematicians simply call it algebra, the name "abstract algebra" is just to differentiate it from the primary school subject people call "algebra"

[alpine island]

Really I think they should rename the primary school one.

[karmic moat]

is this 'extending scalars' just saying you can take an S-module and turn it into an R-module

[tardy hedge]

For Z4 x Z2 are all subgroups {e} x Z2, Z2 x Z2, Z4 x {e}, {e} x {e}?

[rocky cloak]

Yeah, or rather it's giving you one way of turning S modules into R modules

[tardy hedge]

And Z4 x Z2

[delicate orchid]

looks right

[tardy hedge]

The question itself asked for all cyclic subgrps of that . So i first listed all subgrps and now i will see which ones are cyclic

[delicate orchid]

don't forget Z2 x {e}

[rocky cloak]

Think you're missing at least the one generated by (2, 1) for example

[tardy hedge]

Oh yeah

[rocky cloak]

A better way to find all cyclic subgroups is probably to consider the subgroup each element generates

[tardy hedge]

Yea, prt of the reason im talking about this is cuz i feel there are better ways

[delicate orchid]

tensor the subgroup graphs ;3

but yeah

[tardy hedge]

Oh yeah cause a cyclic group must have some element generating it

[delicate orchid]

exactly

[tardy hedge]

So for this low order group its easy to check them all

[karmic moat]

i like rotman

maybe i'll buy a physical copy

[delicate orchid]

the rot

[summer path]

you can always print a physical copy

(don't do this)

[karmic moat]

oh so true

if i print physical copy

i can put 3 inch margins

so i have room to write annotations

you're a genius!

[white oxide]

wait bro

i'm literally going over the same theorem now

or at least i think so

my question was how does alpha being a homomorphism imply that it's well-defined

[karmic moat]

i think i learned most of this tensor stuff in a linear algebra class

just generalizing to modules now

[white oxide]

oh damn we didn't cover tensors in lin alg

[delicate orchid]

tensors in a lin alg is cracked

[karmic moat]

it was kind of something we shoved at the end because we had time

[summer path]

why are all these tensors so large

[karmic moat]

we didn't go in depth

kind of just defining it and giving some properties

[delicate orchid]

that's all of mathematics

[karmic moat]

real

we defined tensor products, using quotient spaces to define bilinear relations, proved existence of tensor products, and then did extension of scalars

and then i guess he didnt feel like doing more so we moved onto the group of matrices and invariant subspaces

non commutative algebra is weird

im not used to thinking about left and right modules

[summer path]

who actually cares about noncommutative algebra

(i don't know)

[karmic moat]

non-commutative algebraists

[dense raven]

geometric group theorists

[white oxide]

what am I doing wrong here? $\alpha(r_1 + r_2 \otimes b_1 + b_2) = r_1(b_1 + b_2) + r_2(b_1 + b_2) \neq r_1b_1 + r_2b_2 = \alpha(r_1 \otimes b_1) + \alpha(r_2 \otimes b_2)$

[cloud walrus]

okeyokay

[rocky cloak]

Expecting the two sides to be equal maybe.

I don't see that you're doing anything wrong.

[white oxide]

oh sorry i meant as in i'm trying to show that alpha is a group homomorphism

and that's what i need to show

that $\alpha(r_1 + r_2 \otimes b_1 + b_2) = \alpha(r_1 \otimes b_1) + \alpha(r_2 \otimes b_2)$ right

[cloud walrus]

okeyokay

[white oxide]

but i expanded them and they're not equal

[rocky cloak]

No, that's not something you want to show

[white oxide]

whatever i'll come backc i've seen enough $\otimes$ for today

[cloud walrus]

okeyokay

[rocky cloak]

Being a group homomorphism just means that f(x + y) = f(x) + f(y), no complicated cross addition of the two factors in the tensor like this

[south patrol]

Okay nice

[elder wave]

It’s a little big, very unhandy

i have one

[spice whale]

rep theorists
also ShiN

[rocky cloak]

If your algebra isn't noncommutative then you're just a geometer in disguise

[karmic moat]

NOOOOOOOOOOO

i was hoping it would be a small-ish size

[elder wave]

How would that work with >700 pages

[karmic moat]

NOOOOO I DIDNT KNOW IT WAS >700 PAGES

day literally ruined

i give up. im not studying hom alg anymore

[torn warren]

For this theorem, I don't want to use the way as the textbook (see the circled part), after the step irr(beta, F) divides irr(alpha, F), can I prove it in this way?

[torn warren]

hello, is anyone here?

@south patrol are you still there?

[south patrol]

Helo

[torn warren]

🥹

[south patrol]

Okay this is unfamliar notation so i shall perservere

[torn warren]

which notation?

[south patrol]

irr(alpha,F) but dw it's clear what it means

Yes I think yours works just fine

If an irreducible polynomial divides another, then they're the same up to units

and since both are monic they agree

I do think symmetry arguments are nice too but yes

[torn warren]

thank you!

[south patrol]

np

[torn warren]

I have another question, can you help me too?

[ivory trail]

you should just post the question and people will help if they want, without making someone commit first

[torn warren]

okay, let me type it

If I take an example, say let $f=(x-\alpha)^m(x-\beta)^n$, after I apply $\tau_x$ to $f$, do I get $\tau_x(f)=(x-\beta)^m(x-\beta)^n$? My question is it seems undefined for $\tau_x(\beta)=?$

[cloud walrus]

WT

[kind pier]

what is (c) asking me to do if not to prove (a)?

[torn warren]

just redo part(a) for a,b in Z

[kind pier]

oh lmfao i didn't notice it said Z+ until now thank you

[torn warren]

what is lmfao?

[sly crescent]

Laughing My Fucking Ass Off

[rocky cloak]

This feels like a lucky 10000 moment

https://xkcd.com/1053/

xkcd: Ten Thousand

Saying 'what kind of an idiot doesn't know about the Yellowstone supervolcano' is so much more boring than telling someone about the Yellowstone supervolcano for the first time.

[torn warren]

If I take an example, say let $f=(x-\alpha)^m(x-\beta)^n$, after I apply $\tau_x$ to $f$, do I get $\tau_x(f)=(x-\beta)^m(x-\beta)^n$? My question is it seems undefined for $\tau_x(\beta)=?$

[cloud walrus]

WT

[torn warren]

I mean this proof..

[rocky cloak]

That's not what you get from tau(f) though. Should be

(x - beta)^m (x - tau(beta))^n

[torn warren]

but what is tau(beta)? it is undefined, since beta is not in F(alpha)

[rocky cloak]

tau is defined on the algebraic closure of F, so it's not undefined. Exactly what tau(beta) is is irrelevant to the proof

It will just be some conjugate of beta

[torn warren]

oh, right, I get wrong, I thought $\psi$ is undefined on beta, right? $\psi(beta)$ is undefined

[cloud walrus]

WT

[rocky cloak]

Yeah, psi(beta) would be undefined
(Unless beta happens to be contained in F(alpha) I guess)

[torn warren]

thank you, yes, and is there a theorem to say about when the conjugate can be in the same extension field? for example, $Q(\sqrt2)=Q(-\sqrt2), R(i)=R(-i)$, but $Q(2^{1/3})\neq Q((2^{1/3}e^{i2\pi/3})$, while $2^{1/3}$ and $2^{1/3}e^{i2\pi/3}$ are conjugate.

[cloud walrus]

WT

[rocky cloak]

I'm not sure whether there is any helpful theorem, but it's equivalent to the Galois group having same order as the degree. And in particular the Galois group will be abelian.

So any degree 2 polynomial have the property. Don't know if you can say much more than that.

[torn warren]

the degree is the highest order of the irr poly, say $n$, and the order is $p^n$ for some prime p as the characteristic. If they are equal then $p^n=n$? do I understand correctly?

[cloud walrus]

WT

[rocky cloak]

I'm not sure what you're saying...

[plush quartz]

This is not true. Let S= {-2,-3,-4,0,2,3,4}

[rocky cloak]

2 + (-3) = -1

[delicate orchid]

4+4 = 8

[plush quartz]

nvm thanks for correcting my incorrect claim lol

i didnt list a subgroup

[torn warren]

oh, sorry I thought Galois field, I haven't learn Galois group..

they are very different?

🥹

[rocky cloak]

The Galois group is just the group of automorphisms of a field extension. And you talk about the Galois group of a polynomial by considering the Galois group of the splitting field

Galois fields are just finite fields, so they're not really directly related.

[torn warren]

is the number of automorphisms of a field extension equals the number of distinct roots in the spliting field?

[uncut girder]

Not necessarily

I encourage you to think of an example

[rocky cloak]

Almost. The order of the Galois group is the dimension of the splitting field.

So tieing back to what I said before, if f is an irreducible polynomial of degree n, with root alpha then F(alpha) is the splitting field of F iff the Galois group has order n

[torn warren]

Thank you, I will keep in mind this after I learn that chapter, currenlty I am reading spliting field and separable extension

@rocky cloak @uncut girder

[tardy hedge]

If G has a unique element a of order 2, show that a is in the center of G

[uncut girder]

Nice

[tardy hedge]

How we do this

[uncut girder]

Ate u reading dummit and foote @torn warren

[torn warren]

I am reading Fraleigh

[uncut girder]

Whats the order of the conjugate of an order 2 element by an arbitrary group element?

[uncut girder]

Cool

[tardy hedge]

Idk what conjugate means

aga^-1 or somethingv

?

[uncut girder]

Yes

Say g is the unique order 2 element and a is arbitrary group element. Whats the order of aga^-1?

[tardy hedge]

Ah yeah

Thanks

[delicate orchid]

can we view this as a degree 2 extension of K(y)? like K(V) = K(y)/(y^5-x^2)? If so then this is obvious

if we can't, then #algebraic-geometry will be more help

[cobalt heath]

Yea I think that's what's happening here

Btw, is any sheaf on noetherian scheme a coherent sheaf? And is quasicoherent subsheaf of coherent sheaf a coherent sheaf

[delicate orchid]

it's a degree 2 extension so {1, x} forms a basis of K(V) as a K(y)-vector space

alternatively, any higher terms x^n are either equivalent to y^m or xy^m for some m

this is of course assuming it is actually a degree 2 extenstion, I kind of got that completely from vibes

[delicate orchid]

I highly, highly suggest going to #algebraic-geometry

I can’t help you here lol

[charred crescent]

can i ask a linear algebra question in here or nah

[summer path]

#linear-algebra

[open sluice]

#nonlinear-algebra

[cobalt heath]

Let me ask again.

Is any sheaf on noetherian scheme a coherent sheaf? And is quasicoherent subsheaf of coherent sheaf a coherent sheaf?

I thought so after learning their definition, but professor talked about quasicoherent subsheaf of the trivial bundle, which seems to imply that this wohld not hold.

[next obsidian]

No, and then yes

[cobalt heath]

Well so much for me saying "any sheaf on noetherian sheaf"

I meant to say the trivial line bundle.

[next obsidian]

This also isn’t really the right place for such a question, belongs more in #algebraic-geometry

[cobalt heath]

I see, thank you.

[mighty spade]

Anybody know if there are any free modules (not vector spaces) such that the determinant is still an Euler-Poincaré map?

[sonic coral]

Let G be a finite group of order n and H a nontrivial subgroup of G. Let k = [G:H]. If n does not divide k! then G is not simple.

For groups that aren’t simple, does this imply the existence of a subgroup that is “big enough” to show the group is not simple

or is that what the sylow theorems are for

[errant shadow]

Are centres always finite abelian groups?

[topaz solar]

Consider the center of an infinite abelian group

[errant shadow]

Ah right, my bad.

[woeful sage]

vector spaces are just modules over fields

[teal vessel]

just making sure: this is the multiplicative group of the integers mod 2^n, right? i.e. Z (mod 2^n) - {0}

[coral spindle]

Yes

[teal vessel]

ok, that'll be interesting to work with.

[abstract rock]

Compute the euler totient, should give you what you need

[night onyx]

specically those which are coprime to 2^n

[teal vessel]

yeah, got that pretty quick. I was thinking too smoll

all coprime elements at n=3 square to 1, that's interesting.

[lusty marlin]

Yes, the multiplicative group of integers mod 8 is isomorphic to ℤ₂×ℤ₂

[noble hedge]

Is there a nice way to prove that for a field ext K/F that no homomorphisms from K to F exist when [K:F]>1?

It seems intuitive but I'm a bit concerned about a counterexample involving some weird cardinality.

[coral spindle]

Well unfortunately this isn't true: the algebraic closure of C(x) is isomorphic to C

[noble hedge]

Oh shit really?

[coral spindle]

Yes

The degree of this extension is infinite of course

[noble hedge]

C not as in complex numbers, right?

[coral spindle]

C is the complex numbers.

[crystal turtle]

Weird cardinality argument comes to the rescue again

[delicate orchid]

this is true for finite fields by an easy cardinality argument

[noble hedge]

Ah okay

[noble hedge]

Yeah mb I misattributed algebraic closure to C and not C(x)

Okay wait what about extensions of the form [F(x):F]? That'd still be infinite, but F(x) wouldn't necessarily be algebraically closed right?

For infinite fields F

[crystal turtle]

https://math.stackexchange.com/questions/2756388/there-is-exactly-one-algebraically-closed-field-with-prescribed-characteristic-a

I assume this is what one uses to prove this: Just check the alg closure of C(x) has same cardinality as C

[coral spindle]

Well a homomorphism there is determined by where you send x, which is only well-defined if you choose some element in F which is not a root of a polynomial in F... I'll leave you to see the issue there

[noble hedge]

All elements a of F are roots of the polynomial x-a right?

And the not-well-defined part comes from the fact that you can construct multiple polynomials that would be mapped to the same element, should you have picked an element that is a root of some polynomial in F?

[delicate orchid]

multiple polynomials that would be mapped to the same element
immediate contradiction as field homomorphisms must be injective

[noble hedge]

Right, that's what makes it not well-defined.

[teal vessel]

double check to make sure I'm not stupid and misremembering things:
for finite, abelian groups, given the orders of x and y are a, and b, respectively, the order of xy should be the LCM(a,b), yes?

[viscid pewter]

yes

[delicate orchid]

G = Z/6Z. o(1) = 6, o(2) = 3, o(1+2) = 2 \neq LCM(6, 3) = 6

[teal vessel]

so yes, I am being dumb and misremembering things

[rocky cloak]

Let F be the the field of rational functions in infinitely many variables. Then F is isomorphic to F(x).

I guess an interesting question is whether F can be isomorphic to K if [K:F] is a finite extension

Meh, not that interesting either, as Q(x) is isomorphic to Q(x^2) for example

[noble hedge]

That was actually the example that made me fear for a weird example involving uncountable cardinalities.

[rocky cloak]

Well Q(x) is countable

[teal vessel]

oh, actually, for a finite abelian group, given |x|=a and |y|=b, |xy| must divide lcm(a,b), so I still have what I need, cool.

[noble hedge]

Uncountable cardinalities seem even less intuitive to me in these situations and allow for some unfunnyness to occur sometimes.

[delicate orchid]

yeah cause (ab)^n = a^nb^n = e implies o(a)|n, o(b)|n

[rocky cloak]

Sure, but this time you can't blame uncountability since Q(x^2) < Q(x) is a finite extension of countable fields

[noble hedge]

I meant as a consideration for the occurrence of a counterexample to my original question.

[rocky cloak]

Yeah, so there are counterexamples that are countable, so you don't have to do any weird uncountable chenanigans

Maybe I just don't understand what you're trying to say

[noble hedge]

I wasn't looking specifically for a counterexample, it was more of just an intuition that a counterexample could exist, and that if it were to exist, where it might arise from.

[teal vessel]

therefore, |-5| divides |5| in all multiplicative groups Z mod 2^n, now I just need to show that it always contains at least one distinct element for n>=3, and I'm good.

(the proof of the statement is a previous exercise I don't feel like typing out for context, but is easily proven)

[delicate orchid]

this was uhh (Z/2^nZ)* not cyclic right

[teal vessel]

yeh

[delicate orchid]

there's gotta be a funny way we can use the fact that there are 3 square roots of unity mod 2^n to do it
and the fact that it's a 2-group

yeah, cyclic 2-groups have a unique involution, but (Z/2^nZ)* has three

[teal vessel]

if there are three unique square roots of unity mod 2^n then that's sufficient on its own, since every cyclic group has a unique subgroup of order [divisor of the group's order]

[delicate orchid]

very good observation

[teal vessel]

but I then have to find the form of that third root outside of 1 and -1. time for a scavenger hunt ig. will be back with a form in a couple minutes.

[delicate orchid]

oh I didn't even count 1

that makes 4 roots

[teal vessel]

sanouth oesndi'b3pai

I love how you can tell what kind of keyboard I use by my keysmashes

lol

[delicate orchid]

I was using "square root of unity" as a synonym for "involution" but that's not true

cause 1^2 = 1

[rocky cloak]

Better upgrade to "primitive square root of unity"

[delicate orchid]

I would if I knew what primitive meant

assuming it's just "root of unity of maximal order"

[south patrol]

Ye

[mossy lintel]

How do I prove the following two definitions for splitting field are equivalent?

It's clear if E is a minimal extension then it splits f(x) over F, but how can I get the other direction?

[south patrol]

well both say it's minimal such that the polynomial splits

maybe the missing thing is that if n= deg f, having n roots is equivalent to splitting into n linear factors - is that what confused you?

[mossy lintel]

I'm confused about why two splitting fields must have same dimension?

[south patrol]

hm that isn't in the definitions

But yeah that is a non-trivial theorem

The key idea is: if you add a single root of an irreducible polynomial of degree k, then the extension is of degree k

Because that polynomial is the minimal polynomial

Then that polynomial factorises in that new extension, and if you continue inductively you get the same thing (up to isomorphism) no matter how you add roots

But that is just a sketch

[mossy lintel]

I'm sorry but what is "the minimal polynomial"?

[teal vessel]

I've narrowed it down to a form:
(2m+1)^2 is congruent to 1 (mod 2^n) iff m(m+1)=k2^(n-2) for some integers m,n,k. Since these integers have equivalence classes in this modular arithmetic environment, I can probably limit m and k to both be only unique up to congruence mod 2^n, no, smaller congruence, tbd.

m mod 2^(n-1)

[delicate orchid]

have u tried thinking about some obvious choices for other 2nd roots of unity?

like you'll probably want a 2^(n-1) in there somewhere, as it squares to 0

[teal vessel]

I have some guesses like the numbers halfway around the cycle or so

[teal vessel]

The numbers to either side of k=2^(n-1) (the halfway point) are k±1, and when squared they become k²±2k+1=2^n(2^(n-2)±1)+1. Therefore, they both square to 1, thus we have two distinct subgroups of order 2 (as long as n≥3, else it would be 1²=(-1)²=1). This directly implies that (Z/2^nZ)* cannot by cyclic. QED.

[delicate orchid]

yup!

[teal vessel]

I am still really curious if I could go about it with the notion of showing that |5|=|-5| while <5>≠<-5>, because the fact that |5| is consistent in this group family was an interesting thing to show.

[delicate orchid]

yeah that would work

I'm gonna spill the beans now

$(\bZ/2^n\bZ)^{\ast} \cong C_{2^{n-2}} \times C_2$ with the annoying $C_2$ being $\langle -1 \rangle$

[cloud walrus]

WewGhostTbh

[delicate orchid]

so <5> and <-5> would indeed be different, and as you say because |5| = |-5| this implies the group isn't cyclic

[alpine island]

Oh so that's why 4 has a primitive root

[delicate orchid]

sure, whatever that means!

[alpine island]

A number $n$ has a primitive root iff $(\mathbb Z/n)^\times$ is cyclic

[cloud walrus]

nHail

[alpine island]

and the primitive root is the generator of that cyclic group

So 2 is a primitive root of 5

this is really useful for solving some modular equations

Especially equations of the form $a^x\overset{p}{\equiv}b$ (solving for x)

[cloud walrus]

nHail

[delicate orchid]

yeah the primitive root theorem

[old spire]

Question: Given an alternative algebra A, is the subalgebra generated by two commutative, associative subalgebras B and C always associative?

In the special case when B and C are both generated by one element, the answer is yes

That's "Zorn's Theorem" or something like that

[tardy hedge]

I did my group theory midterm and it was pretty easy

Surprised he made it easy like that

It was just like “give example of non abelian order 8 group” and other basic questions

I get disappointed at myself though cause it feels like anytime im met with a question that forces me to think a bit , i fumble

Idk it is what it is

[teal vessel]

usually because the human mind is really good at pulling up similar problems and solving classes of problems at a time, so when you hit something you have to genuinely think about, it's because you haven't seen something like it before. If you had a basis to approach it from, it would be easy, so difficult (not to be confused with tedious) problems are also novel to you.

[open sluice]

if you’re thinking then that should be celebrated

[tribal moss]

There are problems that are good to learn from, and there are problems that are good to evaluate students on, and they're often not the same. For midterms it's the second property that's relevant.

[teal vessel]

the best problems to learn from are usually the ones that twist your brain up in knots for a couple days and refuse to let go until you have fundamentally changed how you perceive the properties you're working with (at least a small amount)

[torn warren]

when we say splitting field of some poly f(x), it means the minimal field to contain both F and the zeros of f(x), right? So we can say x^2-2 splits on R, but R is not the splitting field of x^2-2 ?

[south patrol]

Yes

Though note that the splitting field of a poly is always relative to the field the poly is in and is not unique

so one often says "a" splitting field

[torn warren]

I try to understand the word "relative", could you give an example?

let f=(x^-2)(x^2-3), where f is in F[x], if we set F=Q, then the splitting field for f is Q(sqrt2, sqrt3). If we set F=Q(sqrt2), the splitting field is still Q(sqrt2, sqrt3)?

[tribal moss]

But if you set F=R, then the splitting field is R itself.

[torn warren]

yes, is there a non-trivial example?

[tribal moss]

For example, the splitting fields of x²+1 over Q and R are different?

I'm not quite sure what Potato means by "not unique", but my best guess is: Suppose we know about quaternions, and we ask for splitting fields of x²+1 over R. Alice says "the" splitting field consists of the quaternions of the form a+bj with a,b in R; Bob says it consists of the quaternions of the form a+bk with a,b in R. Those two fields (they are fields) are not the same from the point of view of H, but they are equally good splitting fields.
They are isomorphic, though.

[torn warren]

For your example, the splitting field is Q(i) and R(i), right?

[tribal moss]

Yes, R(i) also known as C. 🙂

[torn warren]

If let $p=x^3-2, a=2^{1/3}, b=2^{1/3}e^{i2\pi/3}, c=2^{1/3}e^{i4\pi/3}$, and $F=Q$, let $S$ be the splitting field of p on Q, then the degree of $S$ is 6. However, the degree of $Q(a,b,c)$ is 27, hence $Q(a,b,c)$ is not the splitting field of p. Is this correct?

[cloud walrus]

WT

[torn warren]

Is the degree of $Q(a,b,c)$ to be 27?

[cloud walrus]

WT

[south patrol]

It's unique only if you fix an algebraic closure

[tardy hedge]

Bruh the fact that all that homogenous and particular solution shit from ODEs and matrix equations are just consequences of group homomorphisms 🤯

I love that kind of stuff

So fun

First time learning this so its prob not cool anymore to others lol

[delicate orchid]

Everything is a consequence of everything else

Nah it’s still cool

[tardy hedge]

Yeah

The connections between everything

[abstract rock]

Nah b its still somewhat of a hot topic further in you go. although groups are insufficient for depicting general solutions of PDEs they can give a solution with special properties

there's also ergodic groups and groups that describe dynamical systems

[alpine island]

a hot topic you say? I suppose they do have rings there.

[abstract rock]

bahah

[tardy hedge]

Thats awesoke

Awesome

[torn warren]

anyone can help me🥹

If let $p=x^3-2, a=2^{1/3}, b=2^{1/3}e^{i2\pi/3}, c=2^{1/3}e^{i4\pi/3}$, and $F=Q$, let $S$ be the splitting field of p on Q, then the degree of $S$ is 6. However, the degree of $Q(a,b,c)$ is 27, hence $Q(a,b,c)$ is not the splitting field of p. Is this correct? The spliting field is not simply put the roots into Q( . )?

[cloud walrus]

WT

[white oxide]

yea i understand that, but don't we want to show that $\alpha(r_1 \otimes b_1 + r_2 \otimes b_2) = \alpha(r_1 \otimes b_1) + \alpha(r_2 \otimes b_2)$

[cloud walrus]

okeyokay

[chilly radish]

How did you conclude the degree of Q(a,b,c) is 27?

[white oxide]

or are we fixing r

[delicate orchid]

yeah that's right, if I remember ur question

it was many a moon ago...

[white oxide]

lol

so we can't reduce the left-hand side to $\alpha(r_1 + r_2 \otimes b_1 + b_2)$?

[cloud walrus]

okeyokay

[delicate orchid]

alas, I don't think so

[white oxide]

huh, why is that so

[delicate orchid]

expand that back out and you'll get like, r_1(x)b_1+r_1(x)b_2+r_2(x)b_1+r_2(x)b_2

[white oxide]

doesn't addition of tensors just sum up to addition of cosets or something

[white oxide]

yea

that's why it didn't work

i'm confused as to how to show it's a homomorphism then lol

i'll repost the theorem just for you wew ❤️

[delicate orchid]

thanks boss

[delicate orchid]

oh wow this seems annoying

lets call our map out of R x B -> R "f"

then a(r (x) m) = ra(1 (x) m) = rf(1, m) = f(r, m) = rm ok ok just getting some things straight in my head

hopefully you can see that commuting with the scalar product is ok

a(r'(r(x)m)) = a((r'r)(x)m) = r'rm = r'a(r(x)m)

wait yeah I don't buy this. At all

now a(r(x)m)+a(s(x)n) = f(r, m)+f(s, n) = rm+sn, there's no more simplification to be done here

a(r(x)m)+a(s(x)n) = ra(1(x)m)+sa(1(x)n) = rf(1, m)+sf(1, n) = f(1, rm)+f(1,sn) = f(1, rm+sn) = a(1 (x) (rm+sn)) = a(1 (x) rm + 1 (x) sn) = a(r(x)m+s(x) n)

hold on hold on

[white oxide]

nw take ur time

[delicate orchid]

there we go

ok

here's the key trick

$\alpha(r \otimes m)+\alpha(s \otimes m) = rm+sn = 1(rm+sn) = \alpha(1 \otimes (rm+sn))$

very devious

[white oxide]

bruh usually hungerford verifications are very easy/not hard or tedious

[tardy hedge]

Thanks for the insights @open sluice @tribal moss

[white oxide]

this one was a curveball damn

thanks man

[cloud walrus]

WewGhostTbh

[delicate orchid]

no worries, it was fun to think about

very happy I got it

what clued me in is the fact that they specify the ring had a unit

[tardy hedge]

you guys are all awesome here

[white oxide]

oh yeaa

[white oxide]

wait so here $\alpha(1 \otimes (rm + sn)) = \alpha(r \otimes m + s \otimes n)$?

[cloud walrus]

okeyokay

[delicate orchid]

yeah so the explicit transformations would be

$1 \otimes (rm+sn) = 1 \otimes rm + 1 \otimes sn = r \otimes m + s \otimes n$

[cloud walrus]

WewGhostTbh

[white oxide]

right right got it

muchos gracias

sanity check: does this expression equal $\bigl(a_{11} \otimes b_{11} + a_{12} \otimes b_{12} + \dots + a_{1m_1} \otimes b_{1m_1}) \otimes c_1\bigl) + \bigl((a_{21} \otimes b_{21} + a_{22} \otimes b_{22} + \dots + a_{2m_2} \otimes b_{2m_2}) \otimes c_2)\bigl) + \dots$?

I HATE SUMMATIONS

[cloud walrus]

okeyokay

[delicate orchid]

yes

[white oxide]

lol i should know this by now but whenever you see a double summation as in the expression farthest to the right is it safe to think of it as a permutation of the indices

[delicate orchid]

it's just the associativity of addition

no, no it isn't

it's linearity

we have $(a_1 \otimes b_1+a_2 \otimes b_2) \otimes c = (a_1 \otimes b_1 \otimes c + a_2 \otimes b_2 \otimes c)$

[cloud walrus]

WewGhostTbh

[white oxide]

i see i see

[torn warren]

is it $333=27$?

[cloud walrus]

WT

[torn warren]

each root has degree of 3

[delicate orchid]

||if I'm not mistaken c is already in Q(a,b)||

[chilly radish]

Spoilers!!

[delicate orchid]

the most minor of spoilers

[chilly radish]

Give me your reasoning for why [Q(a,b,c):Q(a,b)] is 3 for example

[chilly radish]

It's kind of the whole point

[torn warren]

c=a^-1*b^2, so c is in Q(a,b), but is there a systemtic way to determine the degree of a extension field, instead of counting them?

[delicate orchid]

well if c is in Q(a, b) it's pretty safe to see that the degree of Q(a, b, c) as an extenstion of Q(a, b) is 1

[torn warren]

so c is a redundant zero, but it counts as conjugate

[delicate orchid]

I don't know what that means, my point was that Q(a,b,c) is literally equal to Q(a,b) so as a Q(a,b) vector space it has dimension 1, and is thus a degree 1 extenstion

[torn warren]

Does that mean everytime I have to try all combinations to see a parameter can be expressed by other parameters?

[torn warren]

yes, my question is there a systemtic way to determine if a parameter is redundent, rather than try all possible combinations?

[delicate orchid]

dunno! If you're quoitenting out by a minimal polynomial then you can read off the degree as the degree of the polynomial

[white oxide]

are all tensor product proofs tedious

[torn warren]

the minimal poly is p=x^3-2, so [Q(a):Q]=3, [Q(b):Q]=3, [Q(c):Q]=3

[white oxide]

it seems like most of them use the universal property

[delicate orchid]

the tensor product is the universal property ;3

[delicate orchid]

yeah. but by the tower law [Q(a,b,c):Q] = [Q(a,b,c):Q(a,b)][Q(a,b):Q(a)][Q(a):Q]

so you can't just multiply these together

[torn warren]

but what about [Q(a, b):Q(a)]? on Q(a), p=(x-a)(x^2+ax+a^2)

[delicate orchid]

well if you quotient out by that you won't get a field

[torn warren]

so the irreducible part is quadratic, so the degree is 2?

[delicate orchid]

yup!

[torn warren]

why can't we stop here, since p can be factorized on Q(a)

why we proceed to factor the quadratic part?

[delicate orchid]

can you factor it over Q(a)?

[torn warren]

p=(x-a)(x^2+ax+a^2)

[delicate orchid]

no, the quadratic part

[torn warren]

no the quadratic part can't, but my question is why we need to factor the quadratic part?

[delicate orchid]

ok I'm just going to ignore the fact that you're claiming the quadratic is both irreducible and then asking why we need to factor it

We want to extend Q(a) to Q(a, b) so we need to quotient Q(a)[x] by the minimal polynomial of b over Q(a)

[torn warren]

I see, thank you!!

[summer path]

i mean you kinda define tensor product via universal property

unless you really want to list out all the properties i guess

[white oxide]

if we're considering the space of all maps of $V \times W$ into $X$ to be a module, would we consider $\mathscr{L}(V, W; X)$ to be a submodule of $\text{Hom}_R(V \times W, X)$?

[cloud walrus]

okeyokay

[white oxide]

so that the induced operation would be given by $f * g \mapsto fg$ instead of composition

[cloud walrus]

okeyokay

[white oxide]

wait yeah

because composition wouldn't even make sense lmfao

[prisma ibex]

categories with more than one notion of tensor product

[south patrol]

No

Bilinear maps are not linear

They seem to be viewing it as a subspace of all maps V x W -> Z of sets

[delicate orchid]

How did you do it?

It’s a bijection in even characteristic

Odd order <=> odd characteristic

(-1)^2 = 1^2

I’d help but I really have no idea how to approach affine varieties over finite fields

[rocky cloak]

Hmm, so at least if q-1 is divisible by 4:

There is the 0 solution.

If exactly one variable is 0, this is the same as solving x^2 + y^2 = 0, which is equivalent to (x/y)^2 = -1. Since q is 1 mod 4, there are exactly 2 solutions. And since there are q-1 choices for y, that gives 2(q-1) solutions. Adding in that any of the 3 variables could have been 0, that's 6(q-1) solutions.

Lastly if all are nonzero. Pick i to be a square root of -1 and rescale the variables by ix3. Then it's equivalent to x^2 + y^2 = 1.

Let f(z) be the legendre symbol, so f(0) = 0, f(z) = 1 when z is a square and f(z) = -1 otherwise.

Then the number of solutions to x^2 = z is 1+f(z).

Now to find the number of non-zero solutions to x^2 + y^2 = 1 pick x arbitrarily. Notice x cannot be 0, 1 or -1. For any given x the number of choices for y is 1 + f(1-x^2). So the number of solutions is [sum 1+f(1-x^2)] where we sum over all x not equal to 0, 1 or -1.

Change variables to x = z-1. Then this is the same as

sum 1 + f(2z - z^2) = q-3 + sum f(2z - z^2)

Where we sum over all z not equal to 0, 1 or 2. Dividing a number by z^2 doesn’t change whether it is a square, so

sum f(2z - z^2) = sum f(-1 + 2/z)

As z ranges over all elements except 0, 1, 2, we have that -1+2/z ranges over all elements except 0, 1, and -1.

Since exactly half of all nonzero elements are squares the sum of f(z) over non-zero z is 0. Removing 1 and -1 from the sum, then gives -f(1)-f(-1) = -2.

So there are q-3-2 = q-5 nonzero solutions to x^2 + y^2 = 1. Counting for the possible values of x_3, that’s (q-1)(q-5) = q^2 - 6q + 5 solutions. Adding in the ones we found earlier there are

q^2 -6q + 5 + 6(q-1) + 1 = q^2

solutions.

you can probably do something similar if q is 3 mod 4, havent thought about it

[rocky cloak]

I mean, that's basically what I'm doing when I'm dividing everything by one of the nonzero terms

[elfin prairie]

Give an example of a subset of a ring that is a subgroup under addition but not a subring

Now how do I go about picking a ring

to work with

[tardy hedge]

Oh yea the fact that any coset of a subgroup H has the same elements as H is almost obvious cause there are at most size of H products and theyre all unique

cuz if ah1 = ah2 then h1 = h2

[south patrol]

I think basically any ring will work

[elfin prairie]

Integers under addition and multiplication won't I think

[south patrol]

It does

Well as long as you require rings to have identity

Which is standard

[elfin prairie]

yeah but bZ is also a subring

[south patrol]

It isn't normally

What is the identity in that ring

[elfin prairie]

1?

[south patrol]

Is 1 in bZ

[elfin prairie]

no

[south patrol]

So there you are hehe

[elfin prairie]

there is no mention of having or not having identity

[south patrol]

Uhh hm

Well what is your definition of a ring

[elfin prairie]

a set with 2 operations under one of whom it is an abelian group and the other operation is associative and distributes over the 1st operation

so we want the other operation to be either not associative or not distributive

[mossy lintel]

Or just not closed

[alpine island]

prof just proved 2nd iso. theorem

some guy asks "so what's the intuition for this? like if I had to prove it what should I think?"

and the prof responds "you'll get intuition from the 4th isomorphism theorem"

[teal vessel]

what's the statement of the second isomorphism theorem?

[tribal moss]

They're numbered differently in different texts.

[teal vessel]

then I suppose the better question is: what is the statement of the isomorphism theorem which your prof has called the second?

[alpine island]

A/(A intersect B) isomorphic to (AB)/B

[teal vessel]

looks like that'll be fun when I get to it.

[alpine island]

Yeah it's not hard to prove it from 1st. iso theorem

aka discussion's greatest moment

consider a homomorphism from A -> (AB)/B

then show the kernel is A intersect B

[summer path]

Everyone can count to 1, and then the rest is some arbitrary permutation

But you only really need to know the first one since the others are kinda just exercises that follow from it

[delicate orchid]

yeah but having the specific forms is occasionally handy

[summer path]

Just prove them as exercises and then hope you remember them 3 years later

(I didn't remember them 3 years later)

[amber plank]

Very simple question:
Classifying all groups of order 98.

Turns out, not so simple and I am in pain

The way I approached it is first via Sylow's theorem --> unique (therefore normal) Sylow 7-subgroup of order 49, so either C_7 x C_7 or C_49, and the Sylow 2-subgroups are all copies of C_2.

The case of the Sylow 7-subgroup being C_49 is very easy, as using semidirect products gives you all possible subgroups here.
However in the case of it being C_7 x C_7, the automorphism group explodes into being isomorphic to the massive GL_2(F_7), so I don't know how to continue from here if I were to use the same method.

I calculated all subgroups of order 2 of GL_2(F_7), but that's still 49 cases to consider, any way to simplify this further?

[delicate orchid]

,w prime factors of 98

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by wewladstbh

[delicate orchid]

uh oh!

so yeah lets consider split extenstions

ok so C_7 x C_7 is order 48 so you'd want to extend it by an order 2 subgroup of GL_2(F_7)

there aren't too many of those I don't think

[amber plank]

there are 49 of them sadly

[delicate orchid]

:uponthewitnessing:

[cobalt heath]

How do you know 49?

[amber plank]

subgroup of order 2 contains an element of order 2 and the identity, so it's just finding all elements of order 2

So matrix
(a, b)
(c, d)

when squared giving I_2

[delicate orchid]

most of these will produce isomorphic extenstions

[elfin prairie]

right

[cobalt heath]

Are there 49 of these

Huh.

[delicate orchid]

there seems to be 57 of them

[amber plank]

Ah fuck

Even worse

[delicate orchid]

now lemme think

[cobalt heath]

Ah. Conjugacy class..

[delicate orchid]

their action on C_7^2 is given by the action on F_7^2, if it's the -I_2 case then this should give us <a,b,c | a^7 = b^7 = [a, b] = c^2 = 1, cac = a^-1, cbc = b^-1>

now staring at this it really looks like some copies of (C_7 \ltimes C_2)

[cobalt heath]

Perhaps elements of the same conjugate class gives the same group?

[delicate orchid]

and since the other matrices are all conjugate in GL(2, 7) they should give isomorphic extenstions

yeah exactly

so there should be two non-abelian groups here + C_49 \ltimes C_2 \cong D_98 for 3 non-abelians in total, plus the two abelian ones for 5 in total

[amber plank]

If I understand things correctly, for context I have not done split extensions
Via semi-direct products, for any homomorphism π: C_2 -> Aut(C_7 x C_7) we can form a semidirect product [C_7 x C_7] °π [C_2] here
What I'm wondering and whether I am correct on this, two semidirect products like that are isomorphic as long as im π is in the same conjugacy class of subgroups order 2 of Aut(C_7 x C_7)

[delicate orchid]

by "split extenstion" I just mean a group formed by a semidirect product

[amber plank]

Right okay

[delicate orchid]

this is true but I can't remember why...

gimme a min

[amber plank]

Yeah that's what I was about to ask about

I'd guess it's this theorem extended slightly somehow

That would make sense

[delicate orchid]

Let $H, H' \leq \text{Aut}(G)$ and consider $X = G \rtimes H, X' = G \rtimes H'$, furthermore let $H = \phi H' \phi^{-1}$ for some $\phi \in \text{Aut}(G)$. Then I believe the map $\theta \colon X \rightarrow X'$ sending $(g, h)$ to $(g, \phi h\phi^{-1})$ is an isomorphism.

[cloud walrus]

WewGhostTbh

[cobalt heath]

I think this is analogous to some linear algebra fact

[amber plank]

Hmmm

[delicate orchid]

it will be in our case cause we're viewing our group as F_7^2

[cobalt heath]

Ah yep

[delicate orchid]

I was playing around with the idea of conjugate matrices having equal eigenvalues

[delicate orchid]

but it just seems easier to view it like this

[amber plank]

So now the question becomes, how do we show all non-identity, non- -I_2 matrices of order 2 in GL_2(F_7) are conjugate

[delicate orchid]

proof by throwing it into magma

[amber plank]

classic

[delicate orchid]

a matrix is order 2 if and only if all of it's eigenvalues are square roots of unity

[cobalt heath]

Huh, magma

[delicate orchid]

so we have 3 possiblities, (1, 1), (1, -1), (-1, -1)
first is the identity, obvs

[amber plank]

h u h

[cobalt heath]

But that assumes Diagonalizable

[delicate orchid]

no? it doesn't

[cobalt heath]

Hm, doesn't it

[delicate orchid]

the eigenvalues of any matrix raised to the nth power are the eigenvalues to the nth power no?

yeah I googled it, it's true

[cobalt heath]

I mean, same eigenvalue does not guarantee the same conjugacy class. Or does it

[delicate orchid]

yes, it does

well, it does here because everything is invertible

[cobalt heath]

Is it something that holds specifically in this case

[amber plank]

Also yeah this should be an isomorphism very easily

[delicate orchid]

similar matrices have the same characteristic polynomial

anyway this is way overcomplicating it

[cobalt heath]

But how about the vice versa

We need that when they have the same characteristic polynomial, then they are similar.

[delicate orchid]

true

but I stopped caring about 5 minutes ago

[delicate orchid]

this is still true

so the eigenvalues must still be the square roots of unity

(-1, -1) is central so it's off on it's own

[cobalt heath]

Yep, but there can be more conjugacy classes.

Like
(1 1
0 6)
Is this in the same conjugacy class as
(1 0
0 6)

[delicate orchid]

except everything in (1, -1) is diagonalisable because they have unique eigenvalues with multiplicity one each

[cobalt heath]

Argh. Right

Sorry

[delicate orchid]

it's ok, I took a while to realise that myself

very annoying question

[delicate orchid]

again the group theory is simpler

[amber plank]

the group theory was easy until the conjugacy class shenanigans

[delicate orchid]

thus everything non-identity, or -I_2 is conjugate

yurrr

[amber plank]

and once you get that you get everything else

[delicate orchid]

my preferred proof method ;3

this is kind of interesting

[cobalt heath]

.>

[delicate orchid]

can it be extended to groups of order pq^2 where p divides q-1?

I see no reason why not

[cobalt heath]

Yea it seems so

Oh, X^p - 1 is separable in F_q. Hm.

[teal vessel]

this seems too easy.... if g is in the normalizer of <x>, then gng^-1 is a map from <x> to <x> for all n in <x>, and for all members of the cyclic group of x, there exists some integer a such that n=x^a (technically infinitely many, but who's counting?), therefore, g being in the normalizer of <x> necessarily implies that there's some integer power a of x which gxg^-1 maps to, because of the definition of these two objects. Am I missing something?

(other than the second half of the question)

[teal vessel]

here's the second half, and I'm stuck on the final step, showing that for all 0<=i<n, the elements are distinct. What if a=0?

[delicate orchid]

conjugation is invertible

[teal vessel]

I know that means something but my brain just isn't taking it nicely

[delicate orchid]

it's a bijection

[teal vessel]

suppose gxg^-=1, then it must be that gx=g, and then therefore that x=1, so for any non-identity x, that would be a contradiction, ok.

idk why that was so hard

so then for any mappings that give you x^ak=x^am, k must be congruent to m mod n, which means that only one of k or m is in the desired range of 0<=k<n, therefore it must be unique. good

[sly crescent]

What are the biquaternion unitary groups? There are three choices for the involution: the biconjugate, the complex conjugate, and their composition. I know the first one gives Sp(2n,C). What about the other two?

[white oxide]

would we also generate M with linearity in the other components? i.e. we would also consider M to be generated by $(x_1, \dots, x_{i - 1} + x_{i + 1}', \dots, x_n) - (x_1, \dots, x_{i + 1}, \dots, x_n) - (x_1, \dots, x_{i + 1}', \dots, x_n)$ and then similarly for every other component

[cloud walrus]

okeyokay

[white oxide]

can somebody please help me understsand why the induced map takes the value 0 on N? or how f being multilinear implies it?

[delicate orchid]

N is pretty much defined to be the kernel of any multilinear map you put on M

[white oxide]

oh wait yeah i think i got it

i forgot that M --> G was a homomorphism

[night onyx]

I think it means for each i, lol a little poorly worded

assuming this is construction of the tensor product of the E_i's

[white oxide]

lol yea i figured

ye

also is this a definition for the action of a on x tensor y

or can it be proved

[night onyx]

you can prove it from the construction

[delicate orchid]

you can show that both of these elements are in the same equivalence class of the quotient you've just done

in fact the quotient was designed so that this property holds

[night onyx]

yep, specifically that you're quotienting the free module generated by the product, which are finite R-linear combinations of elements in the product of the E_i's

[white oxide]

so the elements of the tensor product $A \otimes B$ look like finite sums $a_1 \otimes b_1 + a_2 \otimes b_2 + \dots + a_n \otimes b_n$ right

[cloud walrus]

okeyokay

[white oxide]

because A x B generates this quotient module

[night onyx]

yep, not uniquely though

[white oxide]

hmm okay

this shit is hard to wrap my head around lol

do you know of any good videos which explain em?

actually i found a really good one it seems like

https://www.youtube.com/watch?v=KnSZBjnd_74

holy fuck this guy is phenomenal

[night onyx]

A good way to think about this is when you have two modules A and B, you want to form a bigger space where you can "multiply" elements of A and B together in a bilinear way, like how a(b + c) = ab + ac. The natural way to do this (lol it'll seem natural after a lot of experience at least) is that you take the free module generated by pairs (a, b), and then quotient in such a way that (a, b + c) and (a, b) + (a, c) are equal.

quotienting is literally like, imposing new relations on a structure

in terms of pairs (a, b + c) and (a, b) + (a, c) are different, so you just force them to be the same

[white oxide]

yeah that makes sense!!

he was just talking about that in the vid but i think ur explanation really reinforced it

[night onyx]

lol but you're right, it's definitely difficult to wrap your head around at first

eventually though this construction is like, THE way to form new structures with relations you want to hold

like you want to find some "universal" structure where certain formulas hold, so you form some free structure, and quotient by those relations

[white oxide]

i see, so for example we're just quotienting by distributivity

cuz a tensor (b + c) = a tensor b + a tensor c

similarly scalar multiplication

[night onyx]

exactly

[white oxide]

so that we can have equality in the bigger space

[night onyx]

yeah exactly

[white oxide]

got itt, def making more sense let's go

[night onyx]

lol and more generally basically every algebraic object is the quotient of some free object

[next obsidian]

Line bundles be like:

[night onyx]

cause like, you have some algebraic structure A, you form the free structure generated by A as a set, and then quotient out by exactly the original relations in A

which is pointless obviously lol, but it motivates the idea that if you want to find certain algebraic objects which have certain properties, you should form some free thing, and quotient out by the relations you want

the tensor product is arguably the most important, because a lot of other objects are formed by taking quotients of the tensor products, like alternating products, symmetric products, any thing where you multiply and want certain weird symmetries/antisymmetries to hold

[torn warren]

suppose r1, r2, ..., rn are roots for poly of degree n, then we can prove in general that Q(r1, r2, ..., rn) : Q(r1, r2, ..., r n-1)=1, right? because we have Vieta's formula.

[white oxide]

i see that makes sense, thanks!

so the tensor product kinda allows us to "multiply" vectors

[night onyx]

kindof, though technically it's the space of "multiplied vectors"

if you really want to form a space where you can do multiplication on vectors, like take a vector space V and be able to arbitrarily multiply them together, you need to start with the ground field K, add in all elements of V, add in all v \tensor w, add in all v \tensor u \tensor w, etc

then (ignoring tensor product symbols for a moment), you can form r, v, rv, vw, vwu, v^2u, all products like that

but it's a very large space lol

it contains V and all possible tensor products of V with itself

[abstract rock]

i always thought of it as the space of all possible multiplication-like operators between two vector spaces

[white oxide]

oh hm yea i'm about to go over ground rings

does this induced linear map T(f_1, \dots, f_n) arise from the universal property of the tensor product? or is it more or less a canonical map

[night onyx]

yeah but if you form V tensor W, that's like, JUST the space of two vectors multiplied. It's not a space in which you can multiply vectors, it's the space which contains the product of two vectors

[white oxide]

i see, i guess that makes sense

[night onyx]

iti doesn't contain any vectors in V, or 3 vectors multiplied together

[white oxide]

like would this map just be given by $(e_1' \otimes e_2' \otimes \dots \otimes e_n') \mapsto (f(e_1') \otimes f(e_2') \otimes \dots \otimes f(e_n'))$

[cloud walrus]

okeyokay

[night onyx]

It's the unique linear map T such that T(e'_1 \tensor. .... \tensor e'_n) = f(e'_1) \tensor ... \tensor f(e'_n) for all e'_i, yeah

[white oxide]

yeee

oh wait

it literally clarifies it in the rest of the page lmfao

[night onyx]

but actually, you need to be careful about just assuming you can define T on tensors and assume that properly defines a map, because tensors are equivalence classes

lol this is a subtle point, but you can't just say define T(a \tensor b) = .....

like you first have to define a bilinear map S: V x W -> X, and then use the universal property to conclude there exists a unique linear map T such that T(v tensor w) = S(v, w)

[white oxide]

yeah it seems like all of these proofs are relying on the universal property in order to show that two things are isomorphic lmfao

[night onyx]

yeah, that's really common

showing that X is isomorphic to V \tensor W often just means you're showing X satisfies the universal property of the tensor product in a clever way

[white oxide]

yeah that makes sense

[night onyx]

just to get your hands dirty, prove the tensor product of R^n and R^m is the set of linear maps R^n -> R^m

[white oxide]

you're so right

all i'm doing rn is verifications i don't have time for exercises

[night onyx]

lol fair enough!

[delicate orchid]

verifications are exercises!

[white oxide]

yeah but they're mostly tedious ones/not very enlightening...

like show such and such is a subgroup!

check that it's multilinear!

this shows that it's surjective since this bilinear map induces this linear map out of the tensor product right

the bilinear map that i'm referring to is the one in the last sentence, before "Gives rise"

[delicate orchid]

yeah so the universal property says that for each bilinear map out of E x F there is a UNIQUE map out of the tensor product, giving rise to a bijection