groups-rings-fields

[novel parrot]

a^2 + 11b^2 = 1, which only has solutions at a = +1 and -1

does that look like what you had in mind?

[summer path]

Yes

[delicate orchid]

More to the point, every element of that ring has a norm of at least 1, so you cannot make the norm smaller via multiplication - so it is impossible for anything other that -1, 1 to be a unit

Saves you having to think about a YUCKY ellipse

[summer path]

I mean you just observe that if b is not 0, then you have something bigger than 1 that you can't get rid off

There's not really much to thinking about the ellipse

[delicate orchid]

geometry=EW

I'm not actually saying there's anything wrong with their solution, it's perfect

I'm just providing an alternate way of thinking about it

[south patrol]

hm i mean wew what you're doing seems to just assume that there are no elements of norm 1 besides -1 and 1

Like okay you can't make it smaller via multiplication, but how do we know we can't just keep it as 1

[delicate orchid]

you can, that's why -1 is a unit

[south patrol]

No i mean

like if there were other elements of norm 1

[delicate orchid]

it's very easy to show that there aren't... if only there was some sort of... quadratic expression in two variables I could use...

what a world that would be

[south patrol]

Lol

[slim kayak]

If I have an abelian quotient group G/N and an epimorphism from it into the abelianization of G, then that map is an isomorphism. I can see why this holds for a finite group but what's the argument for infinite ones?

[chilly radish]

You're perfect

[lethal dune]

Not true for infinite ones

[slim kayak]

Oh? Counterexample tiem?

[lethal dune]

Countable direct sum of Z

[delicate orchid]

it's always the countable direct sum of Z...

[alpine remnant]

Ok so, if I pick two real numbers, a and b, with b being positive, and build the matrix

J=[a, -b;
b, a],

its minimal and characteristic polynomials are an irreducible polynomial of degree two (figured out by doing (a-bi)(a+bi)). When you do this for a bigger matrix:

A=

[J,0,
I_2, J]=

[a, -b, 0 ,0;
b, a, 0, 0;
1, 0, a, -b;
0, 1, b, a],

The minimal and characteristic polynomials are now f^2.

I could prove all of this by hand.

How would I generalize for a matrix of 2nx2n entries for f^n?

Since the matrix A is triangular by blocks, I know that its characteristic polynomial will be f^n

My issue lies in proving that f^n is also the minimal polynomial

As in, proving f^(n-1) is not null

which would mean all f, f^2, f^3, ..., f^n-1 are not null, so f^n is the only possibilityh

[dim widget]

I would just write out f(A) explicitly

Then if that matrix is nilpotent (it will be) you know exactly what power of f that you need.

[alpine remnant]

wait

f^n(A)=f(A)...f(A)

right=

ok i'm dumb

añsflkghañog

[alpine remnant]

how would I go figuring that out?

[dim widget]

you can figure out the minimal polynomial of a nilpotent matrix over a field by looking at "which diagonal" the nonzero entries lie on.

A matrix is nilpotent iff it's conjugate to a strictly upper triangular matrix over an algebraically closed field, in your case I think you get something actually upper triangular.

Or lower triangular rather

[alpine remnant]

I get one which is lower triangular by blocks

but not triangular

[dim widget]

specifically if N is an nxn nilpotent matrix with only the top right entry nonzero then it squares to zero, if the nonzero entries are on the next diagonal down from that then it cubes to zero, etc.

[dim widget]

f(A) will be strictly lower triangular.

[alpine remnant]

Ok so

let f=x^2+a_1x+a_0

f(A)=

f([J, 0, 0, ...;
I J, 0, 0, ...;
0, I J, 0, ...;
...])

= A^2+a_1A+a_0I

=[J^2+a_1J+a_0I, 0, 0 ...;
J+a_1I, J^2+a_1J+a_0I, 0, 0, ...
....]

Ok that notation looks like poop

[alpine remnant]

But A is not triangular

It's triangular by blocks

If you use the block I=[1,0;0,1] and J=[a,-b;b,a], then A is triangular

but if you don't you have chunks poking out

[dim widget]

Okay

I would write

A = D + N where D is block diagonal and N is upper triangular

[pliant forge]

hello can someone point where im making a mistake because this feels very weird

Let $0 \to A \to_f B \to_g C \to 0$ be an exact sequence. Then $C = Im(g) \cong B/Ker(g) \cong B/Im(f)$. But since $Im(f) \cong A/Ker(f) \cong A$ since $Ker(f) = (0)$ we have $C \cong B/A$.

[cloud walrus]

Eso

[dim widget]

Then f(A) = D^2 + 2DN + N^2 + a_1(D) + a_1(N) + a_0 = f(D) + f(N) + 2DN

f(D) = 0

f(N) is lower triangular

2DN is lower triangular

so f(A) is lower triangular

[pliant forge]

i believe my iso for modulo makes stuff odd

[dim widget]

that's it

[alpine remnant]

What are D and N again?

i'm confused by english notation sorry

well

terminology

[dim widget]

C is isomorphic to B/A that is correct, nothing weird about it, it's just the first iso theorem

[pliant forge]

the thing that gets me is the exact sequence where A = B = Z and C = Z/nZ

wouldnt that imply Z/nZ iso 0

[wraith cargo]

no

So the first map is multiplication by n

The image of A is nZ

so taking the quotient still gives you Z/nZ

[pliant forge]

i know but Im(f) \cong Z/Ker(f) = Z/(0)

[dim widget]

you are not thinking about this clearly.

The ideal of integers divisible by n (say n = 2 for simplicity) is isomorphic to Z abstractly

[wraith cargo]

you need to look at Im(A) as an ideal of B for that quotient to be valid

[dim widget]

but Z/(2) is not zero

[dim widget]

N is the lower triangular part of A and D is the block diagonal part of A

considered as 4x4 matrices

[dim widget]

Why not just try actually calculating what f(A) is yourself?

[alpine remnant]

I did I did

I have it on paper

The diagonal comes to blocks of 0

it is indeed lower triangular

[dim widget]

okay great!

[alpine remnant]

I was dumb and messing up a step

now I know this matrix must be nilpotent since f^n(A) is 0

I must demonstrate that indeed n is the index of nilpotence

[dim widget]

Also any lower triangular matrix with 0's on the diagonal is nilpotent

Is an entry of the form a_{i, i-1} nonzero?

[alpine remnant]

they all are

like, the diagonal just below the main one

[dim widget]

Then you should try to show that a matrix with those entries nonzero is nilpotent of maximal possible nilpotence degree

[alpine remnant]

can I do that by induction on the degree of the matrix?

[dim widget]

sounds good!

[alpine remnant]

okies

[alpine remnant]

you've helped me a lot every time i've come to ask questions, Thank you very much for all the help you've given me

you really have been a life saver

[dim widget]

no worries, glad to help 🙂

[chilly ocean]

Man rep theory is fun

[formal ermine]

indeed

[lethal dune]

[south patrol]

Reps r fun

Calculate all irreducible reps of Z/p over Fp

[lethal dune]

No

Give me example of an irreducible 2 dimensional representation of an abelian group

[dim widget]

Z/2 acts on F_2^2 by permuting the two basis vectors

[lethal dune]

Oh char 0, forgot to mention

[dim widget]

oh wait I thought you meant indecomposable

it is obviously reducible

lol

[lethal dune]

Char 0 you mean?

[dim widget]

No it's reducible (contains a nontrivial subrepresentation) in any characteristic

but it isn't a direct sum

[lethal dune]

Yeah just checked

Mind helping me in the diff geo problem

[dim widget]

SO(2) acting on R^2

[dim widget]

sure

[lethal dune]

I had Z on R2

Both similar

[dim widget]

That is another example which is reducible but indecomposable, unless you have some weird action of Z in mind?

You could have Z act by an irrational multiple of pi radian rotation I suppose

[lethal dune]

Rotation by irrational angle or any 2 jordan block

[dim widget]

(x 1 | 0 x) doesn't work since it has a subrepresentation

[lethal dune]

Yeah just realised

[south patrol]

Oh ye I forgot about indecomposable stuff

Hm I guess it's equivalent to irreducible provided stuff is over a semisimple ring right

[lethal dune]

Hm

[south patrol]

Like p sure any semisimple module admits direct complements for stuff

[dim widget]

yes by definition but it's generally harder to prove that a ring is semisimple than a module over it is semisimple without some kind of strategy

like in lie theory you have the casimir element and in finite groups you have the averaging operator.

These examples are basically showing that Q[Z] is not semisimple.

Which is not surprising since that's Q[X]_{x}

which is not zero dimensional

[south patrol]

Yee fair lol

[pastel cliff]

old q but does this look right

[dim widget]

It's a little unclear what you're doing, maybe it's good to say that the elements which are "algebraic over L" are closed under multiplication and addition

since it seems you're trying to use that

[pastel cliff]

[cloud walrus]

MyMathYourMath

[solar glacier]

I just took an algebra II final and wanna know if I did it correctly

[rustic crown]

yee that right

[solar glacier]

And is splitting field of $x^4-4x^2-5$ over the rationals just $\Bbb{Q}(\sqrt{5},i)$

[cloud walrus]

MyMathYourMath

[solar glacier]

Nice thanks

[solar glacier]

With degree 4

[solar glacier]

For this is subbed $y=x^2$

[cloud walrus]

MyMathYourMath

[rustic crown]

(x^2-5)(x^2+1) so yee

[solar glacier]

Cool

Another way of doing it ^^^^ lol

I made a substitution LOL

[rustic crown]

[solar glacier]

And to show the graph of a function in affine n space is an affine alg set can I argue it’s complement is infinite

Thus under Zariski top it’s closed

And thus and affine alg set

And the Zariski topology on any arbitrary fields affine 1 space is clearly compact right ? U take one of the open set… look at its complement which is finite pick an open set containing each of the finitely many elements and you have a finite union as your sub cover

He had us prove whether or not under Zariski top if affine 1 space is compact for arbitrary field k LOL on an algebra II final

[rustic crown]

wait n space?

or line?

[solar glacier]

Line

If the field k is finite then it’s clearly compact

[rustic crown]

so this not true right

[solar glacier]

Wait the graph of an affine n space is algebraic

So this set

[rustic crown]

i mean not every infinite set is open

[cloud walrus]

MyMathYourMath

[solar glacier]

Sets with finite complement are open in Zariski top

[rustic crown]

yea but you're saying the converse

[solar glacier]

Oh no sorry

Disregard that part

So to show the graph is an affine alg set

I only need to show it’s finite

[rustic crown]

the field is finite?

[solar glacier]

Someone told me the trick they used after the exam of showing it directly but I forgot what they said

[solar glacier]

No not necessarily

[rustic crown]

cause there are certainly infinite algebraic sets

like consider y = x^2 in A^2

or like literally anythign

[solar glacier]

Yes

[rustic crown]

A^1 itself

[solar glacier]

True

[rustic crown]

so yea, they need not be finite

infact the cardinality of the graph is just same as cardinality of the source

[solar glacier]

But under Zariski top we define the alg sets to be finite subsets

Or the whole space

Or empty set

[rustic crown]

that's true for the affine line, but the reason is because a polynomial only has finitely many roots

(assumign it's non-zero)

[solar glacier]

But not true for affine n space ??

[rustic crown]

(that's not called the zariski top, that's the "cofinite" top)

[solar glacier]

Zariski top and cofinite I thought were the same

[rustic crown]

for line, yee

[solar glacier]

Ohhh

Not for general n space though

??

[rustic crown]

yea 😦

[solar glacier]

Then I fucked that proof up lol

[rustic crown]

parabola for instance is closed but not finite/whole space

[solar glacier]

Ahhhhhh good example

Here’s a question none of use could get right : if u wanna@give it a shot

[rustic crown]

you define closed as stuff which are solutions to a bunch of polynomials

[solar glacier]

Let $V$ be an affine algebraic set in $\Bbb{A}_{\Bbb{R}}^n$ show there exists a polynomial $f \in \Bbb{R}[x_1,…,x_n]$ such that $V=Z(f)$ and show this is fals if we replace the reals w complex it seemed so counterintuitive since C is alg closed but R isn’t

[cloud walrus]

MyMathYourMath

Let $V$ be an affine algebraic set in $\Bbb{A}_{\Bbb{R}}^n$ show there exists a polynomial $f \in \Bbb{R}[x_1,…,x_n]$ such that $V=Z(f)$ and show this is fals if we replace the reals w complex it seemed so counterintuitive since C is alg closed but R isn’t

[solar glacier]

I chose my polynomial to be

[cloud walrus]

MyMathYourMath

[coral spindle]

Hint: if x, y are real numbers then x^2 + y^2 is usually positive

[rustic crown]

boytjie

[coral spindle]

heyo det

[rustic crown]

haiii

[coral spindle]

sorry I feel like I've made a terrible faux pas

let me try again

heyo det

[rustic crown]

hewwo

[solar glacier]

I think I get it so that doesn’t hold@over C

[rustic crown]

(your kids hungry yet?)

[solar glacier]

Because of i

[coral spindle]

We'll see...

[coral spindle]

That's right

This could be summarised as saying that R is an ordered field

[solar glacier]

Omfg I totally missed that

C isn’t totally ordered

[pastel cliff]

det moment

[solar glacier]

Wow

[pastel cliff]

[solar glacier]

I cannot believe I missed that on the exam

SMH 🤦‍♂️

[coral spindle]

you win some, you lose some

[solar glacier]

I got 6/9 problems completely correct

[solar glacier]

Even tery Tao says that in an interview its so true

He usually gives an A if u score over 50/55

[rustic crown]

(or being more fancy, as saying R is a formally real field)

[solar glacier]

Hey

It R is all rationals with odd denominator

To show the ideal (2) is maximal

It’s enough to show R/(2) is a field

Which it clearly is , right ?

[coral spindle]

OK fancy boy smh

[pastel cliff]

can i get a hint on this pls

[solar glacier]

If u mod out by (2) you still have all. Non zero elements have mult inverses

[rustic crown]

can you say something about Q(x)/K? is it alg?

[pastel cliff]

x is transcendental no?

[rustic crown]

over Q, yep.

what about over K

[pastel cliff]

no

[solar glacier]

Nope

[rustic crown]

so can K/Q be alg?

[solar glacier]

It’s alg over K

[pastel cliff]

it can be

[south patrol]

Tower law moment

[solar glacier]

That rule is so useful

[pastel cliff]

what does tower do here

[rustic crown]

Q(x)/K alg and K/Q alg => Q(x)/Q alg

[solar glacier]

Tower your given sets

You have 3

[pastel cliff]

oh so x being transcendental over Q has no impact on extending by it?

[rustic crown]

yee

(in fact every such thing is purely transcendental)

(there is this cool theorem by Lüroth saying that any intermediate field of k(x)/k is simple)

[pastel cliff]

wdym

[rustic crown]

so if you pick any intermediate thing in k(x)/k, say E

then as long as E is NOT k you have that k(x)/E is algebraic

so E/k would be transcendental

the theorem says it's also simple

[pastel cliff]

simple = adjoin one elt right

[rustic crown]

so E = k(alpha) for some rational function alpha

[pastel cliff]

wait dumb question

what's the index of Q(x)/Q

[rustic crown]

infty

[pastel cliff]

ohhhh

[coral spindle]

guess what the basis is

[rustic crown]

(also i had a typo, i meant \neq)

[pastel cliff]

it's Q(x)/Q(x/x^3+1)

which is finite right

[rustic crown]

yee

simple, alg = finite

[pastel cliff]

and Q(x/x^3+1)/Q is also finite

[rustic crown]

nup

[pastel cliff]

oop

[rustic crown]

that would make Q(x)/Q finite

[pastel cliff]

hmm then im not sure i see the tower

oh wait you can be infinite but still alg

right?

[rustic crown]

yea

but does that matter in this situation?

only alg extension in sight is finite

which is Q(x)/Q(alpha)

[pastel cliff]

ok so to see that $x$ is in fact algebraic over $\Q\big(\frac{x}{x^3 + 1}\big)$, we first observe that $\Q\big(\frac{x}{x^3 + 1}\big)/\Q(x)$ is a finite extension

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

sorry other way around

$\Q(x)/\Q\big(\frac{x}{x^3 + 1}\big)$

[cloud walrus]

not sebbb not stμ₂dying

[rustic crown]

umm

just show it directly, no?

show x satisfies a polynomial over Q(alpha)

that's what you'll do anyway to show it's finite

[pastel cliff]

polynomials over an extension adjoined x are just the fraction field right

[rustic crown]

went above me

[pastel cliff]

wdym

[rustic crown]

didn't understand whut you wrote >.<

[pastel cliff]

ok then what do polynomial over Q(x/x^3 + 1) look like

since we're adjoining a polynomial in the first place

[rustic crown]

T^2 +alpha T + (alpha^3 -2)

[delicate orchid]

(2x/x^3+1)t^2+(x/2x^3+2)t+3984575423

very standard polynomial

[pastel cliff]

[delicate orchid]

don't let letters in maths scare u now

too late for that

[rustic crown]

(hint: write alpha = (...)/(...) and recall that Boytjie's kids hate denominators)

[coral spindle]

They truly do

||in the end, this isn't that different from finding a min. poly. for a rational number over the integers||

[pastel cliff]

how the fuck do i find this min poly

[delicate orchid]

god knows

[pastel cliff]

wew please i am at my limit

https://tenor.com/view/an-emoji-disintegrating-in-pain-gif-22507163

[rustic crown]

okie let's start slow

what' the min poly of alpha = 1/x

[pastel cliff]

x...?

[delicate orchid]

no I legitimately don't know I've never learnt galois theory

[rustic crown]

that's the min poly of 0, and certainly alpha doesn't look like 0 :p

for instance if you remove the intersection point in alpha it disconnects into 3 components, while 0 stays connected

(sowwy :p)

[pastel cliff]

hmm

sorry im splitting my attention rn

[rustic crown]

(also write the min poly in terms of T, to not confuse it with x)

so you're allowed to use T and alpha and numbers such that when you plug T = x, the equation is true

[pastel cliff]

did you mean what's the min poly over Q(1/x)?

[rustic crown]

yea

[pastel cliff]

oh that makes more sense

[rustic crown]

(and no denominators)

[pastel cliff]

so what polynomial with coefficients in Q(1/x) has x as a root

[rustic crown]

great question

:p

[pastel cliff]

yeah just restating it clearly for myself

[rustic crown]

hehe :p

[pastel cliff]

,ti det

[cloud walrus]

72 members found matching det!

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[rustic crown]

,ti

[cloud walrus]

The current time for det is 01:22 AM (CEST) on Fri, 12/05/2023.

[pastel cliff]

oh lord det

i wanted to bother you more too :(

[rustic crown]

you can

[pastel cliff]

okok focusing on this first

full attention

[rustic crown]

just assume that i'm asleep if i stop replying

[cloud walrus]

Member selection timed out.

May 11, 2023, 11:23 PM

[glossy crag]

Dumb question maybe, but is there any notion of uniqueness of vector space structures?

What I mean is, if V is a K-vector space, do we know that it's uniquely so?

[chilly ocean]

Whats this

Because of me?

[pastel cliff]

no bot just timed out

[chilly ocean]

O i thought maybe because i pressed the reaction buttons here

[glossy crag]

I was thinking about how uniqueness of prime power decomposition for finitely generated modules over PIDs is proved.

The proofs I know use R/(p)-dimension of p^kM/p^{k+1}M to characterise the prime powers appearing in the decomposition of M intrinsically

[pastel cliff]

t^2-1?

[rustic crown]

1/x^2 -1 = 0 then?

[glossy crag]

And I was kind of wondering whether any ambiguity whatsoever in this situation is possible, i.e. if you might get different R/(p) dimensions based on how you define the R/(p)-VS structure on p^kM/p^{k+1}M

[pastel cliff]

uhhhh

[rustic crown]

(reminder: you're allowed to use alpha)

[pastel cliff]

oh i thought we were doing min poly of x over Q(1/x)

like besides alpha

[rustic crown]

we're doing min poly of x over Q(alpha) where alpha = 1/x

[pastel cliff]

ohhh

i thought you mean old alpha

[sage lodge]

trying to understand the proof of (c). why is the last statement true?

[pastel cliff]

agh im so confuse

this shouldnt be this hard

[rustic crown]

there is an ambiguity in general, but doesn't matter here.

[rustic crown]

yee overthinking makes life hard

[pastel cliff]

is f(t) = t - x valid

[rustic crown]

you're not allowed to use x

[pastel cliff]

only 1/x

[rustic crown]

only alpha, t, and numbers

just write it lol, don't think more :p

[pastel cliff]

f(t) = t(1/x) - 1

[rustic crown]

for an example, consider the k = Fp(t), then you can define k as a vector space over itself using the map k --> k being identity or the frobenius. one had dimension 1 other has dimension p

[rustic crown]

right, alpha * t -1

so you start with alpha = 1/x

you cross hate denominators so you do alpha * x = 1

and make this a polynomial

alpha * t - 1

[pastel cliff]

cross hate?

[rustic crown]

oops

wanted to use the old story again

was writing cross multiply

but then thought i'll write hate denominators

anyway, let's do alpha = x/x^3+1 now

[rustic crown]

okie it does matter, but whenever you say consider the R/I module structure on M/IM, you mean the obvious one.

[rustic crown]

another example is k = Q(t1, t2, .....) and k --> k being the map identity or shift these ti. one will have dim 1, other has ifninite dim

[pastel cliff]

hmmm but 1/x^3+1 isn't in the field right

[rustic crown]

try copying what i did here

[pastel cliff]

oh

[rustic crown]

(you're really overthinking lmao)

[pastel cliff]

so t^3 + 1 / t (alpha) - 1

wow

[rustic crown]

whut whut

that doesn't look like a polynomial in t to me

[pastel cliff]

sorry

$f(t) = (t^3 + 1)\alpha - t$

[rustic crown]

yep!!

[cloud walrus]

not sebbb not stμ₂dying

[rustic crown]

hopefully you see now that it didn't depend at all on what alpha was

[pastel cliff]

is it easy enough to see that it is in fact the min poly?

like smallest irr.

[rustic crown]

sort of?

but we don't care about that

all you wanted to show was Q(x)/Q(alpha) is algebraic

[pastel cliff]

(and find min poly of x over K)

[rustic crown]

oh oops

i forgot the question :p

[pastel cliff]

but it is minimal right

[rustic crown]

yep

directly showing that might be hard, so just eisenstein it up

with the UFD D = Q[alpha]

as alpha is transcendental, this is isomorphic to Q[X] to a pid

and alpha is prime in D

alpha divides the constant term, but its square doesnt :3

[dull cobalt]

Basically I'm trying to understand a series of papers titled "Algebraic Approach to Signal Processing" where the author goes about generalizing the concepts of signals, filters and Fourier transform etc. using objects from Abstract Algebra... So I'm looking for beginner recommendations to the topic from an applied math perspective, basically something which is not too heavy on the formal rigor. What do you folks recommend? (If it helps I'm very fluent with Linear Algebra)

[glossy crag]

Why?

[pastel cliff]

can i get a hint on this one too

[glossy crag]

I mean, partly because we use the usual R/(p) structure on the module (if I\subset Ann(M), then M is automatically an R/I module), but how come something weird can't happen.

[rustic crown]

sowwy >.<

[glossy crag]

Just remove denominators in the minimal polynomial of v over F(u), that should work (I don't even see how v being transcendental is necessary; NVM, it's so that the equation doesn't turn to 0 and u really ends up being algebraic over F(v)).

[rustic crown]

that's a nice example

[glossy crag]

Interesting.

[rustic crown]

okie i go sleep

i have 5ish hours if i sleep now

[summer path]

,ti det

[rustic crown]

,ti

[cloud walrus]

The current time for det is 02:04 AM (CEST) on Fri, 12/05/2023.

User cancelled member selection.

May 12, 2023, 12:04 AM

[summer path]

[rustic crown]

oyasumi tubu

[summer path]

Bai bai det

[glossy crag]

what are all these bots for

[pastel cliff]

it's only one bot

[pastel cliff]

does this look right

[stone fulcrum]

I like it!

[pastel cliff]

long time no see kaynex

[pastel cliff]

small bump

[stone fulcrum]

You're noting that:
1/2(√3 + √2) + 1/2(√3 - √2) = √2
Which I think is a bit more consise but up to you, this is solid

[pastel cliff]

yee haw!

[stone fulcrum]

Anyway hi! Long time no see

[pastel cliff]

good to see you around here again for sure

[stone fulcrum]

This was a reply to that bump

[pastel cliff]

oh i missed that

this is what i had started

[lapis trail]

I'm going to try to work through as much of dummit and Foote as I can this summer. Any advice on how to balance breadth with depth in self study

I need a hyperbolic time chamber

[pastel cliff]

mood

[stone fulcrum]

I don't love dummit and foote for self study, imo grab a different book

It's a fantastic reference though

[pastel cliff]

KC notes

[pastel cliff]

is this the right start btw running on fumes a bit so maybe im not thinking clearly

[stone fulcrum]

KC notes?

[pastel cliff]

keith conrad

[stone fulcrum]

I don't actually know how to do this one, haha. I'm really spotty when it comes to field extensions

[pastel cliff]

ah shooot

[stone fulcrum]

I'm sure someone here does

[lapis trail]

@stone fulcrum can you recommend me something? How is Artin?

[frigid lark]

Lang Chapter 5 and some of Chapter 6

If you can learn from Lang

[lament dawn]

Im a huge believer in Lang Bad but i actually quite liked Lang for the field theory

[rustic crown]

awuffi is uwu too

[hidden haven]

Still waiting for the day det becomes a normal human didn't expect that trauma to last this long

[rustic crown]

(it has gotten worse btw, ever since #1055183210444763205 has existed)

[prime sundial]

so i was playing around with my rubik's cube and found out that the rotation RU has order 105
is there any intuition as to why multiplying 2 elements of order 4 would give something so "random" as 105, or does it just work out that way for this very specific group

[hidden haven]

It just be like that for that group

[rustic crown]

moldi is expert at every fun thing

[prime sundial]

figures

i was always too quick to assume the orders of products of elements, but now it just seems so random

[hidden haven]

Ye it can be completely random

[prime sundial]

that's super frustrating haha

[hidden haven]

You can have elements x, y with orders m, n > 1 respectively such that xy has any order you want

[frigid lark]

You'll be waiting for at least the size of the monster group seconds

[prime sundial]

this is very unsettling lol

[hidden haven]

Construct an example as
< x, y | x^m, y^n, (xy)^p > for any p

[prime sundial]

i was under the impression that groups behaved really nicely, but the more i learn about them the more i lose grasp on that

[rustic crown]

yea they be whack

[prime sundial]

i believe you that this is a group

[hidden haven]

Groups were hit in the head as a child

[rustic crown]

(idk the meaning of that word tho >.<)

[frigid lark]

This is why we study fields

[prime sundial]

i don't like any algebraic structure

[hidden haven]

Based

[frigid lark]

And by fields I mean only Z/2Z

[prime sundial]

best field

[frigid lark]

Anything else is to hard for my bird brain

[rustic crown]

same uwu

but i have brain

[hidden haven]

Adding fractions is the worst math ever invented

Det stop it

[rustic crown]

okie >.<

[grand cliff]

The following exercise is from a past paper

suborbits are the induced orbits of a stabiliser

An orbital is the orbit of H's action on Omega x Omega

An orbital is considered self-paired if (x ,y) in orbital => (y, x) is in the orbital

If I can somehow show that the order of H must be 2p, I'm guessing the solution follows relatively quickly from there

Since H must then either be a product Z/2Z x Z/pZ or D_{2p}

• I'm using the convention |D_{2p}| = 2p in case that's not obvious

[tribal moss]

It's probably easier to see intuitively if you consider permutation groups. If you have the two permutations
a = (1 2)(3 4)(5 6)...(n n+1)
b = (2 3)(4 5)(6 7)...(n-1 n)
then each of them taken individually has order 2, but if you do one after the other, the net effect is to move every odd thing two places to the right and every even thing two places to the left (except at the end of the line where there's a thing one of the two permutations doesn't touch) -- so overall the product of the two permutations is a cycle of all n+1 things, which order n+1.

[pastel cliff]

omg moldi

[pastel cliff]

this look ok?

oh wait i didnt finish ir

[solid olive]

how do you go about proving if G is abelian and H is a subgroup of G then the left and right cosets are equal?

[frigid lark]

What is a left coset explicitly?

[pastel cliff]

i dont see what this min poly is

maybe im overthinking again

[chilly radish]

This isn't necessarily true unless the subgroup is normal. Unless you meant the set of left cosets is the same as the set of right cosets, which is true

[frigid lark]

G is abelian

[chilly radish]

Oh my bad

I glossed over that

[frigid lark]

It is a non trivial relationship b/w the powers of v over f(u) of smallest degree

Well not really but close enough

[pastel cliff]

wdym

[solid olive]

yes sorry thats what I meant

[frigid lark]

Cause X^2 - 3 is the minimal polynomial of sqrt(3) over Q

Not sqrt(3)^2 - 3

[pastel cliff]

ooh

i meant for the old question

[pastel cliff]

this onoe

[frigid lark]

Well there's gonna be some relationship $$a_0 + a_1v + ... + a_nv^n=0$$ for $a_i\in F(u)$

[cloud walrus]

Parrot Tea

[frigid lark]

But those a_i's are merely quotients of polynomials in F[u]

So just multiply by the lcm of the denominator of all the a _i's

And you get a non trivial relationship b/w the powers of u over F(v)

[pastel cliff]

[solar glacier]

To show if P is a prime ideal containing ideal I then it contains radI you use a minimality argument right

[wraith cargo]

Uhh no you use primality iirc

[solar glacier]

Let’s say m is the least integer for which a^m \in I which is contained In P then aa^m-1 \in P but P prime so either a or a^m-1 in P

Couldn’t you used the argument I just wrote

[wraith cargo]

yeah if a^m is in P then a is in P for P a prime ideal

[solar glacier]

Cool cool

[wraith cargo]

that's what I meant

[solar glacier]

Thanks

[south patrol]

Well taking radicals preserves inclusions

As is fairly clear hopefully

And the radical of a prime is itself

That's another way to thjnk about it in terms of like just the radical operation

[prime sundial]

this helped. thank you

[pastel cliff]

whats an example of a finitely generated module over an integral domain $R$ which is not a direct sum of cyclic $R$-modules.

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

does $R \times R$ work?

[cloud walrus]

not sebbb not stμ₂dying

[next obsidian]

Those are cyclic

[pastel cliff]

[next obsidian]

You have to be over a not-PID

[dim widget]

You can be over a PID as well

oh wait

[next obsidian]

The classification tells you it’s impossible over a PID

[dim widget]

I see what you mean

[pastel cliff]

woops

[dim widget]

sorry I thought he just wanted a non-cyclic module

[pastel cliff]

im not sure then tbh

[dim widget]

As a hint: examples of this are generic once you are not a PID

So maybe think of some rings which are not PID's and modules over them

[next obsidian]

Unfortunately it’s difficult to come up with generic examples as a human

[pastel cliff]

Z[x]?

[next obsidian]

My first inclination are always cyclic ones or products thereof

[dim widget]

Z[x] will work!

[pastel cliff]

yeah that's my only default example

[dim widget]

But you need a module

another good one to keep in mind is K[x, y]

[pastel cliff]

K is a field?

[next obsidian]

These are the same thing

Sort of

[pastel cliff]

so a module over Z[x]

[dim widget]

They have some formal similarities, for the purposes of the problem they are equivalent 🙂

[pastel cliff]

im still not sure i see the example though

[dim widget]

What is an example of a module that would exist for a non-pid but not for a PID?

Hint: it is a submodule of R

where R is your ring

[pastel cliff]

oh an ideal of Z[x]

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

kinda guessing

[dim widget]

okay but probably you want a non-principle ideal right?

[pastel cliff]

ok then $\ang{2,x}$

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

ok i get it but not really

[dim widget]

You should try to show that if you take any generators (f, g) for (2, x) then there is a pair r, s \in R such that rf = gs so this cannot be a direct sum of cyclic modules

[pastel cliff]

but it is that ideal right

[dim widget]

yes that ideal is an example of the kind of module you want

[chilly ocean]

it is an ideal example

[pastel cliff]

,av TTerra

[cloud walrus]

TTerra#5291's Avatar

Click here to view the image.

[pastel cliff]

not quite what im used to tbh

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

most accurate expression of how i've been feeling these past two weeks

[rustic crown]

TTewwa

[chilly ocean]

i plan to switch it back at some point

[pastel cliff]

no dont

[chilly ocean]

is this the dude who voiced gabe from ultrakill

lmao

[pastel cliff]

yes lol

[river nebula]

Hi, I was watching a proof of Laskar-Noether theorem and this guy was going through a lot to prove that if 0 ideal is irreducible then it is primary, isn't it straightforward that if 0 ideal is irreducible then it is prime?

[south patrol]

No

e.g. (0) is irreducible in R:= k[x]/x^2 as the ideals are just (0), (x) and R

But it isn't prime as R has zero divisors

And note in this case R is obviously very well-behaved in some ways (only 3 ideals lol)

[formal ermine]

okay so

I have an exact sequence

( \bC \xrightarrow{\phi} \bC^2 \xrightarrow{f} \bC )

[formal ermine]

I know phi, how can I find a polynomial expression for f?

[coral spindle]

f is not uniquely determined by phi and exactness

[formal ermine]

I'm not looking for a unique solution

just for any cuz then we can just break it up into irreducible elements and pick out the necessary ones

[dim widget]

what is \phi?

is it a linear map?

[formal ermine]

in my case it's t -> (t^2, t^3 + 1)

[dim widget]

?

Oh you mean on coordinates

[formal ermine]

yes

[dim widget]

(Y-1)^2 - X^3 is the minimal f, you should try to prove this

[formal ermine]

hmmm

[white oxide]

Can somebody give me a hint for the following question? I'm assuming G is a semigroup and trying to prove that if the equations ax = b and ya = b have solutions in G, then G is a group; I'm a little concerned about a priori reasoning since I'm not sure if I can introduce a^-1b or ba^-1 (I'm trying to prove the existence of the left inverse and left identity or right inverse and right identity, etc. which will lead to G being a group)

like am i justified in saying that x = a^-1b is a solution (it exists) iff a^-1 in G

[coral spindle]

No, you have not justified that

In the first place, you need to say what you mean by a^-1, which is not a priori defined since G does not have an identity.

(Or more specifically, you have not proven that an identity exists)

[white oxide]

hm ok

maybe it has to do with an equality of some sort?

ie ax = ya

bruh i feel like such an idiot

idk where to go from from just saying that ac = b is a solution

oh

i can just consider the equation bx = b since it's for all a b in G i think

oops

[tribal moss]

Well, yes, that's what you need to do before thinking about inverses. But you need to prove the same x works for all b.

[white oxide]

oh sorry the hypothesis said that the equations ax = b and ya = b have solutions for all a and b in G forgot toinclude that

oh wait no

that's not what you're saying

my fault

so basically i have to show that they have unique solutions

[tribal moss]

The hypothesis is that for each pair of a and b, there are x and y such that ax=b and ya, but these x and y will depend on a and b.

[white oxide]

ah ok i see

thanks

[tribal moss]

Hint: ||if you have an e such that ea=a, find an x such that ax=b and use associativity to conclude eb=b||

[white oxide]

i got it with that hint, thanks!!

[chilly ocean]

<@&268886789983436800>

[pastel cliff]

https://math.stackexchange.com/questions/575171/galois-group-of-sqrt2-sqrt2-over-mathbbq

not sure i get the answerr here

why do we get to use sqrt{2- sqrt{2}} to show that that thing itself is in K?

or is it just bc it's technically 1 so we can ignore that for a bit

det button

[lapis trail]

What's the difference between What's being done in the video and a compass and straightedge construction? They show how using a ruler and compass how to draw a polygon of any number of sides. But they only measure once, then use the compass to find the rest of the lengths, so what additional tool are they using here?

https://youtu.be/5tl1_7lhzdw

[pastel cliff]

dont we need an assumption about finiteness of the extension here?

for each of these i mean

tterra button...? 🐺

[chilly ocean]

i do not know field or galois theory

[pastel cliff]

pain

but the button worked

det button

if you spam enough maybe a wild det will spawn

[dim widget]

no you don’t need any extra assumptions

[pastel cliff]

what's the topos theory egirl button

[dim widget]

[pastel cliff]

i was thinking we might because conrad's notes use it

for example

[dim widget]

Part of being galois is that you are algebraic in your definition?

[pastel cliff]

ohh i think so, maybe conrad doesn't explicitly say that but my notes might

[dim widget]

if so the infinite case reduces to the finite case I think

[pastel cliff]

my prof's notes

[dim widget]

yes okay

[pastel cliff]

normal and separable give finiteness right

[dim widget]

No just algebraicness but you only need that

What do L_1, L_2 correspond to

[pastel cliff]

wdym

[dim widget]

Like using galois thory

[pastel cliff]

oh subgroups

i haven't read conrad's proofs i was just wondering about the difference in assumptions

actually this doesnt really use finiteness anyways

https://kconrad.math.uconn.edu/blurbs/galoistheory/galoiscorr.pdf

full link

galois implies it's a splitting field for some poly right

oh wait claim is slightly dif

[pastel cliff]

button

this too shitty of a proof

[pastel cliff]

omg ryu

what's the ryu button

[agile burrow]

I like it

[pastel cliff]

,av walter

[cloud walrus]

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[pastel cliff]

,av walter#1555

[cloud walrus]

walter#1555's Avatar

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[pastel cliff]

when will this be an emoji

[next obsidian]

NaughtyWalterWhite

,av Walter

[cloud walrus]

NaughtyWalterWhite#2828's Avatar

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[pastel cliff]

perfect

[cloud walrus]

Member selection timed out.

May 13, 2023, 5:05 AM

[pastel cliff]

Determine, up to conjugation, all matrices in Mat$_{4 \times 4}(\Q)$ whose order is exactly 6.

[cloud walrus]

not sebbb not stμ₂dying

[next obsidian]

Owned

[pastel cliff]

are there 256 of these? or which ones dont work

invariant factors have to divide t^6 - 1

$$(t^2 + t + 1), : : (t-1), : : (t^2 - t + 1), : : (t+1)$$

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

breaks up into these if ive done it right

but not every combination of these things works right?

[next obsidian]

Normal forms

[pastel cliff]

chmonkey does you know

[next obsidian]

I have no idea

[pastel cliff]

pain

[next obsidian]

I am better at module theory than vector spaces

[pastel cliff]

ok then something simpler

i can do this with lin algebra but like

[next obsidian]

You can do this with duality and the definition of a determinant using exterior powers for “free” I think

[pastel cliff]

[next obsidian]

Actually that’s so based

I’m a genius

[hollow shore]

hey guys how do we tell if a system of congruence is odd or even?

[pastel cliff]

https://tenor.com/view/mig-dead-chat-gif-26330667

can i get a hint on this pls

[rustic crown]

what's your guess?

[pastel cliff]

ngl i dont really know

Q(2^1/3 , 3^1/3, zeta_3)/Q?

or

bleak eevee when modss

[pastel cliff]

also bump to this if you know anything det

@rustic crown

ping me if you see this det, tabbing out

(or anyone )

[rustic crown]

@pastel cliff oopsie

me bacc

[pastel cliff]

im not sebbb not studying

but hi det

[rustic crown]

.<

[pastel cliff]

[rustic crown]

.<

[pastel cliff]

any ideas on this one?

[pastel cliff]

also how much proof does this need

i wanna just it

[rustic crown]

not much

cause you know all roots of that poly

[pastel cliff]

it's just those three things right

[rustic crown]

nah, first two and multiply them with the last (twice)

[pastel cliff]

okie one sec

that other one scary so i'll type this up

wait but that's the field, what's the group

im used it being something easy like D8 lol

Z/3 x Z/3 x Z/3?

[rustic crown]

describe what can happen to the generators

z3 has degree 2 over Q, while c2 and c3 have degree 3

you can send z3 to either z3 or z3^2

and ci to ci*z3^j for i = 2,3 and j = 0,1,2

and this gives you a weird group of order 18

you can do some work and see what it's iso to lol

[pastel cliff]

da frick

[rustic crown]

ofc for this one needs to show that [Q(c2, c3):Q(c2)] = 3

[pastel cliff]

by Tower Property

[rustic crown]

no

like you know x^3-3 is the minimal polynomial of c3 over Q

you wanna show it remains a minimal polynomial over Q(c2)

[pastel cliff]

c3 = cube root of 3 right

[rustic crown]

yee

[pastel cliff]

im so tired det

[rustic crown]

z3 isn't a root

[pastel cliff]

oh but the products are

[rustic crown]

,ti not sebbb

[cloud walrus]

The current time for stμ₂dying is 05:22 AM (EDT) on Sat, 13/05/2023.
det is 6 hours ahead, at 11:22 AM (CEST) on Sat, 13/05/2023.

[pastel cliff]

https://tenor.com/view/hng-gif-4501621

[rustic crown]

[pastel cliff]

finals

im finishing up corrections for my algebra hw's

but corrections kinda also mean just doing them in the first place lol

this week has been a long month

and number theory final soon too

in any case

[pastel cliff]

those prod's are still roots right

[rustic crown]

yep, cause when you cube them up you get either 2 or 3

[pastel cliff]

i still dont see what they generate though

[pastel cliff]

Z/3 x Z/3 x Z/2?

[rustic crown]

nah, going to be non-comm

prolly semidirect product of Z/2 and Z/3 x Z/3

[pastel cliff]

pain

but thank you det

[cloud walrus]

det

[rustic crown]

just stare at this diagram and figure stuff out

[pastel cliff]

rebump to this now then

[pastel cliff]

[rustic crown]

all extensions in that are galois, so not so bad

[pastel cliff]

oh technically cant use galois here

this is an old thing from before we learned that

[rustic crown]

and to do the final part, it's equivalent to determining all intermediate groups contained in Gal(whole thing/Q(z3)) which is the order 9 group Z/3 x Z/3

[pastel cliff]

ok i see that

im so close det

6 more questions

[rustic crown]

[pastel cliff]

2 on JCF adjacent stuff 4 on Galois

[rustic crown]

need to say a little more, making sure that not all of them divide t^2-1 or t^3-1 and reducing the order

[pastel cliff]

hmmm

[rustic crown]

(is that an "okie i see that "?)

[pastel cliff]

i gots to think abt it

[rustic crown]

you want the order to be 6

but if the elementary divisors were like (t-1), (t-1) and (t^2+t+1) then you'd be sad because the order is 3

[pastel cliff]

ohhhh

ok i see that

just need to generalize

[pastel cliff]

thiis should be (t-1)^2 no

wait

actually idk maybe i dont see it

[rustic crown]

does that divide t^6-1

[pastel cliff]

ah

[rustic crown]

(since char Q is not 2)

maybe i'll spell out the details a lil more

say you got a nice operator T which had order 6

so you see that T acts on a vector space of dimension 4, let's call it V

you can turn V into a Q[t]-module by defining t * v = T(v)

now theory tells you that isomorphic Q[t]-modules structure on V correspond to conjugate matrices

so up to conjugation, you wanna identify the iso class of the module V

by structure theorem of f.g mods/pid

they decompose into cyclic modules

if Q[t]/(f) was a summand, then you notice that as you've been given T^6 = 1, the element (t^6-1) in Q[t] must be in the annihilator

therefore each elementary divisor f (which is power of some irreducible) will divide t^6-1

that's why it reduces to understanding irred powers dividing t^6-1

but that got no repeated roots

so just irreds dividing t^6-1

another way to think is via the invariant factor decomposition. the last invariant factor is the actual annihilator of the module, so you don't want that to be dividing t^2-1 or t^3-1, which means it has to be divisible by the irred (t^2-t+1)

uwu?

[pastel cliff]

reading through rn

det i might give up

im at my limit

[rustic crown]

,ti not sebb

[cloud walrus]

The current time for stμ₂dying is 05:58 AM (EDT) on Sat, 13/05/2023.
det is 6 hours ahead, at 11:58 AM (CEST) on Sat, 13/05/2023.

[pastel cliff]

and those galois questions are scary

[rustic crown]

get some sleep

[pastel cliff]

i'll finish these two module questions and beg for extra time on the galois ones

they were due midnight

[rustic crown]

https://tenor.com/view/mala-mishra-jha-pat-head-cute-sparkle-penguin-gif-16770818

[pastel cliff]

soon det

not yet

https://tenor.com/view/goku-gif-21807429

[rustic crown]

you prolly don't have to give all the reasoning here, just write down the possible invariant factors/elementary divisors

[summer path]

why is det g+ and also pending g+

.<

[rustic crown]

https://tenor.com/view/inugami-korone-hololive-thinking-anime-confused-gif-17902545

[pastel cliff]

done

[old zenith]

just to be clear, the galois group of an irreducible polynomial in Q(x) of degree n is S_n right?

and then the galois group of any arbitrary polynomial is the product of the galois groups of its irreducible factors?

[wraith cargo]

this is not true generally

This is true when the polynomial is of prime degree and has exactly two non real roots

[old zenith]

damn i thought i was starting to understand galois groups

what is the general procedure for determining the galois group of an arbitrary polynomial then

[south patrol]

So basically you can view the Galois group G of a degree n poly as acting on the n roots, which gives an embedding of G into S_n. But often there will be other restrictions that mean you don't get the whole thing.

[old zenith]

ye and i am struggling to see those restrictions

[dim widget]

You could look at Lagrange’s theory of resolvents if you want a more computable criterion

[south patrol]

Often the point is you take a splitting field of the polynomial and break it up into subextensions which let you construct elements of the group. Then getting some bound on its size will help you know when you're done

[old zenith]

but this holds right? as long as the roots of each of the factors of the irreducibles are linearly independent

[dim widget]

It will only really be helpful in low degrees though

[south patrol]

It's good to work w some examples

[old zenith]

so in general finding the galois group is "pretty hard"?

[south patrol]

So also like examples of other restrictions that stop you from getting S_n like

It could be that where all the other roots go is determined by one of them

[old zenith]

obvious example would be cyclotomics right

[south patrol]

Exactly what I was gonna say lol

[old zenith]

cos u just change zeta_n and then that fixes where zeta_n^k go

[south patrol]

Yup yeah

Another way of thinking about this uh

The size of the Galois group of a poly f over Q is the degree of the extension K/Q with K the splitting field

[old zenith]

and then naively you assume that the galois group is Z/nZ but if you change zeta_n to zeta_n^k where gcd(n, k)>1, then it ceases to be an automorphism or smth like that?

like in Q(i) you cant just map i to -1 cos then you dont have anything mapping to i

is that what is happening in general?

[dim widget]

No I don’t think that’s the point

[wraith cargo]

automorphisms preserve order of elements

[south patrol]

I don't really think that's the key point like

To me the sort of entry point is like

Note that if K,L are field extensions of F and φ: K -> L is a map over F, then for any a in K with min poly f, φ(a) must be a root of f as well

So for example, for the cyclotomics case, we can only send roots of the cyclotomic polynomial to other roots

In general you can generalise the theorem I gave and it allows you to "build up" elements of the Galois group by sending generators of K (over F) to appropriate elements of L

[old zenith]

ok so in this case K=L since we're discussing automorphisms?

dont rlly see how this holds for K not a superset of L since like what if phi goes from Q(sqrt(2)) to Q(pi)

[south patrol]

The point is that there are no such maps over Q

Since it'd have to send sqrt(2) to a root of x^2 - 2

and there are none of those in Q(pi)

But, for example, there is an isomorphism Q( 2^{1/3}) -> Q(2^1/3 w) over Q where w is a primitive third root of unity

[old zenith]

ah
ye i see what you mean, you can't map sqrt(2)->pi because then phi(2)=phi(1+1)=phi(1)+phi(1)=2
but then also phi(2)=phi(sqrt2)phi(sqrt2)=pi^2 which breaks the field structure
i was thinking of the spaces as vector spaces mb

[south patrol]

But yes it's hard to talk about this more generally without just getting into actual theorems lol

galois theory fun

[old zenith]

ye

also if you dont mind

could you give a basic rundown of how these ideas can then be applied in differential galois theory? just the basics lol

[south patrol]

Idk anything about differential galois theory lol

[old zenith]

rip

anyways thanks for teaching me basic galois theory

[glossy crag]

Is this a valid proof of "M free of rank n over PID, N<=M => N free of rank <=n" (been a while, wanted to reproduce it):
n=1 clear, let M be free of rank n, then the span M' of the first n-1 basis elements is free of rank n-1 and N'=N\cap M' is free of rank <=n-1. N/N' embeds into M/M' which is free of rank 1, so N=N'\oplus Ax and N is free of rank <=n.

[rustic crown]

yee that works

and you can do the same thing transfinitely

[formal ermine]

huh wdym

[south patrol]

Hm

Is there any nice way to compute idempotents of a group algebra in general? And I suppose more generally a semisimple ring lol but I imagine that's too hard

For example I did a question where we had to compute idempotents of \R[\Q_8]

Oof

[formal ermine]

how did you find this (I am aware that you can just find it by 'looking at it', but is there a better way)? do you have a hint for me?

[upper pivot]

right so i thought about this a bit

so my idea is for C[G] mainly

but right like we know its a sum of matrix algebras if G is finite, and so if we have an explicit description of this breakdown then we know the answer as we know the idempotents of a matrix algebra

this is over any field rn ofc

but heres where C comes in lol

you can use some von nuemann algebra theory to decompose C[G] = \oplus p_i C[G] where p_i are projections in the center that are mutually orthogonal and sum p_i = 1 ofc. The center of C[G] correspond to conjugacy classes of C

so this kinda reduces to finding explicit projections in the center like this oof

I have been unable to find this

(p_i C[G] can be seen to be isomorphic to matrix algebras)

(also from von nuemann theory lol)

maybe the spectral projections can help out here hmm

[south patrol]

Tbh this makes me realise idk how to find like all projections in any particularly nice way lol

for a matrix alg

But also interesing hmm

[upper pivot]

yeah id just describe them with subspaces kek

btw the kaplansky zero divisor conjecture (that for any group G, F[G] has zero divisors if and only if G has torsion when F is char 0) is proved using von nuemann algebras too!

gtg for a bit but ill come back and think about this

[formal ermine]

ok this seems way too easy so I'm susing it

what is Spec C[x]/(x^2)? what I did was

by this one theorem it's all prime ideals in C[x] that contain (x^2)

by hilbert's nullstellensatz all prime ideals in C[x] are of the form (x - a) for some a in C

so the only ideal that contains (x^2) will be (x)

thus Spec C[x]/(x^2) is just the origin

is this correct

[tribal moss]

A prime ideal must always contain all nilpotent elements, so (x) has to be there.

If you add anything in addition to that, you immediately get (1), which is by definition not prime.

So yes.

[south patrol]

Nullstellensatz for this is way overkill

This is a PID

The only divisors of x^2 are (up to units) x, 1 and x^2

0 isn't prime (the ring isn't a domain), and the whole ring is ruled out by definition

[dim widget]

In terms of "how do you find this" it's just the first thing you write down. In terms of "how do you prove this is the minimal equation" there are many ways

[lapis trench]

This is somewhat a general question, but suppose that you have a finitely presented group. Are there useful principles to be able to classify the quotient G/G' where G' is the commutator subgroup of G?

I'm trying to work with the generators but am struggling to make any progress.

[lapis trench]

The question has been resolved

[lapis trench]

I forgot that you have the row-Hermite normal form representation for these presentations

That's what I used

[noble hedge]

Is there a nice way to find all Sylow p-subgroups of a group? Take D_6 (order 6, not order 12) as an example and look at all Sylow 2-subgroups, of which there are 3. There's only 6 elements, so brute-forcing all conjugations of a subgroup of order 2 would be fine, but what about somthing larger?

GL_2(F_5) would have an unreasonable amount of elements to brute force (at least by hand), so there's probably some method to find conjugations, right?

If $S_1, S_2, S_3$ are the 3 Sylow 2-subgroup of $D_6$, then we should be able to partition $D_6$ into the sets: $N_G(S_1) = {g\in G \mid gS_1g^{-1} = S_1}$, ${g\in G \mid gS_1g^{-1} = S_2}$, and ${g\in G \mid gS_1g^{-1} = S_3}$, right?

[cloud walrus]

(𒀭)

[coral spindle]

It is tricky in general

You can use specific methods from different fields. For example, there are algorithms for permutation groups.

For the specific case of GL_2(F_5), there are techiniques from the theory of algebraic groups. We look at GL_2(F), where F is the algebraic closure of F_5, after which GL_2(F_5) is a particular group of fixed points under a Frobenius automorphism.

There are various powerful results regarding this kind of construction in certain nice cases.

Tl;dr not really, no.

[noble hedge]

Ah, alright

That's pretty cool, seems like I have some stuff to look into

thanks!

[dim widget]

A good exercise is to find the p-sylow subgroups of Gl_2(F_p)!

Though it has little to do with finding p-sylow subgroups in general

since there are special tricks

[coral spindle]

If you do try this exercise, it's good to mention that you should aim for a description of just one of them rather than all of them. It can be hard to actually write down a nice description of some of the conjugates of the nicest one

Or well, not hard per se, just not very illuminating or nice.

[noble hedge]

I got to the point where I can show there should be p+1 of them, just based off the index of the normalizer and size of Gl_2(F_p), but I wasn't sure exactly how to find which elements I could conjugate one of the subgroups by to find the remaining subgroups

That exercise is actually what sparked this question lol

I had thought that partitioning the group into sets of elements that conjugate one of the Sylow p-subgroups to the rest of them would probably be something I could find with orbits, but got stuck

[dim widget]

Do you see an obvious subgroup of order p?

[noble hedge]

(1,1,0,1) should generate that, right?

[dim widget]

What is the normalizer of that subgroup?

[noble hedge]

All matrices where the third (or bottom-left) element is 0, right?

[dim widget]

Yes. So how many conjugates are there?

[noble hedge]

p+1

but which elements conjugate the Sylow p-subgroup generated by (1,1,0,1) to the remaining subgroups?

[dim widget]

It doesn't really matter for your problem but you can get p subgroups using (1 0 | 1 1) and the last one from (0 1 | 1 0)

Also all of this generalizes to Gl_n(F_p) and probably to certain other cases as well.

[noble hedge]

neat

So aside from knowing how many Sylow p-subgroups there are, brute-force (for sufficiently small p I guess) is just the most reasonable way to find all Sylow p-subgroups of GL_2(F_p)?

Aside from the stuff Boytjie had mentioned earlier, which I definitely don't know enough of to comment on

[dim widget]

I think you can, to some extent, just use the Sylow theorems which is in the vein of problems which ask you to classify all groups of order pq or pq^2 where p, q are small primes. After that you have to actually do some work

I consider this Gl_2(F_p) example the opposite of brute force, but we did start off by observing that there's an obvious natural candidate

I also think that what I led you through was similar to what Boytjie was referring to, but disguised.

[noble hedge]

Once you have the first subgroup generated by (1,1,0,1), there's p elements that you can conjugate the subgroup by to find the remaining subgroups, right?