groups-rings-fields

[south patrol]

Ah noice

That is big from the perspective of the course this was for lol

[dim widget]

It sounds like a fancy statement but it just comes down to Hensel's lemma 🙂

[south patrol]

I assumed so but this course doesn't use padics even xd

But ye it ok

But ye it ok

[dim widget]

can i get uhhhhhhh discord?

naw discord machine broke

[next obsidian]

There’s a lot of discord in Discord servers rn

[glossy crag]

I have a question about how the bijection between unramified and finite separable extensions is constructed.

The proof demonstrates how to lift simple extensions (this then is used inductively to lift arbitrary finite extensions): take an element a' of C' and its minimal polynomial f', lift them to a monic f and root a, and then show that K'(a') is the residue field of K(a) and that |K(a):K|=|K'(a'):K'| (i.e. the lift is unramified if a' is separable).

Now at first I wondered about the ambiguity of this, because there might be multiple roots of f in C that are lifts of the same a' and hence possibly multiple field lifts, but if we restrict to the separable case this falls away, since f' has no multiple roots (and hence each lifts uniquely), and using this we see that lifting and taking residues give a bijective correspondence between finite separable extensions of K' and unramified extensions of K.

However then it crossed my mind that there **still **is some ambiguity in how we lift fields: at one point we lift f' to a monic f in the valuation ring of K, what happens if we vary the coefficients of f with elements from the valuation ideal? We get polynomials whose reduction is still f', but they might have roots different to that of f and hence produce different lifts, no?

Doesn't that mean that the lifting map {F/K' finite separable} -> {L/K finite unramified} is not well-defined?

Discord is acting out on me, so apologies if some messages are doubled...

[grand cliff]

Could someone help me unpack this insane definition of a group:

I've mentally broken it up into 4 stages:
Stage 1: First we describe some set Omega (the set of cosets of the diagonal group).
Stage 2-4: Then we proceed to identify 3 different faithful actions on Omega

It's not clear to me why this is so special or how they did it. Surely if you just keep arbitrarily combining more and more permutation groups, you'll end up at the symmetric group or something. So there's something unique about this construction

[dim widget]

You have to show that the extension is uniquely determined by its residue field, if you change coefficients of the polynomial f by integers divisible by p then you don't change the extension generated by the roots.

[glossy crag]

You mean that an unramified extension is determined by its residue field? That's what step (e) of the proof does.

[dim widget]

So then it seems that the ambiguity you mentioned is already dealt with

[glossy crag]

I guess you're right (if L and L' are unramified and generated by roots of polynomials that reduce to the same polynomial, then their residue fields coincide and therefore they're equal).

[formal ermine]

how do I show that det - 1 is irreducible

[dim widget]

what is - 1? ♾️ - 1 = ♾️

[formal ermine]

determinant of a matrix

[dim widget]

What do you mean by irreducible? What does it mean to say that det - 1 is irreducible? As what kind of object?

[formal ermine]

f(n^2 variables that are the elements of a matrix) = det(those n^2 variables) - 1 where det(...) is viewed as a polynomial in n^2 variables

[dim widget]

okay

[formal ermine]

also I'm confused if the determinant of a polynomial in M_n is in n or n^2 variables. I believe it should obviously be n^2 but I've looked online and some people say n

[dim widget]

It is of degree n as you can see by row expansion

[rustic crown]

- 1

[formal ermine]

wait no not degree

I meant variables

oop typo sorry

[dim widget]

Okay well it obviously is a polynomial of n^2 variables since there are n^2 entries in the matrix

[formal ermine]

yeah

[rustic crown]

if you look it as a poly in n^2 variables, then there isn't much to say as it's linear in each variable, and linears have to irred

[spice whale]

lmao using for det

[formal ermine]

oh lmao I'm dumb

thanks

[rustic crown]

(i mean there is still a little check involved, but that's the basic idea >.< like 2x+2 is obv reducible but you know what i mean >.<)

[formal ermine]

can't we just view it wlog as a polynomial in one variable with coefficients in the ring over all of the over variables

[rustic crown]

yea but we'll have to show that the two coeffs don't share a commond divisor

wait now i think about it, it's not obvious anymore

[formal ermine]

wdym

[rustic crown]

like 2x+4 is reducible in Z[x] but irred in Q[x]

[dim widget]

I mean it's still fine

there are no constants which divide

Because otherwise the determinant of every matrix over Z would be divisible by that constant

And anyway the constant term is 1 so we don't have to worry about that even over Z

[rustic crown]

i think we still need to say a little more... and i don't see any way that doesn't involve a little computation. the problem is we want to show that they dont' have a common divisor in Z[all but that one variable]

also the computational argument isn't very sensitive to the constant term

the idea is this, if = a * b and say x and y are variables either in same row or same column, then both of them appear in the same factor. otherwise, if x was in a and y was in b, you'll have a term with xy in it, but does have that as they were assumed to be in same row/column.
but now by transitivity, all variables will lie in a single factor. now your point finishes it.

[dim widget]

Here is a non-computational argument: the vanishing set of det - 1 is the group Sl_n, and you can verify over R or C that it is smooth and connected

[formal ermine]

fwiw I'm trying to show that SL_n(C) is a variety lol

[dim widget]

It is maybe best to just show that it is an algebraic group and that it is smooth at the identity

[formal ermine]

what do those words mean

[dim widget]

okay nvm

Anyway the point is that Sl_n(C) being a connected manifold means that the variety Sl_n must be irreducible

[slim kestrel]

Let G be a group, and let H and K be subgroups of G. Prove that HK is a subgroup of G if and only if HK = KH.

To prove this statement, I need to show both directions right?

If HK is a subgroup of G, then HK = KH and If HK = KH, then HK is a subgroup of G

[prime sundial]

yes

hi det

[rustic crown]

haiii

[slim kestrel]

You saying yes to me or?

[rustic crown]

yes

[slim kestrel]

Ohk

How the fuck is this a level 3 question

[prime sundial]

what is a level 3 question

[slim kestrel]

Oh

umm

as the level increases the difficulty increases

its a thing in my country

[prime sundial]

[slim kestrel]

level 6 is hardest according to my knowledge

fuck i cant speak

Let G be a finite group of order n, and let H and K be subgroups of G such that |H| = m and |K| = k, where m and k are relatively prime integers. Prove that the intersection of H and K is the trivial subgroup {e}, where e is the identity element of G.

What does it mean by "identity element"

[formal ermine]

do you know the definition of a group?

[prime sundial]

the element e such that e*g = g*e = g for any g in the group

[slim kestrel]

I think..

[slim kestrel]

Oh

[slim kestrel]

so basically when the identity element is combined with any other element in the set its value doesnt change

[prime sundial]

think about how 0 is the additive identity in the real numbers

[slim kestrel]

Ah

[prime sundial]

for any number x, 0 + x = x + 0 = x

first part is straightforward
for the second part, is the idea to construct H_n and B_n using the kernel/images of the boundary maps

[slim kestrel]

I just need to show that the only element common to both H and K is the identity element e

Im sorry 😭

[prime sundial]

i think they are questioning why you're being asked to do a question like that before you know what the identity of a group is

[slim kestrel]

I skipped a lot of classes

trying to recover on my own so i might have skpped some stuff..

[dim widget]

Here is the most important thing to know about groups: If G is a group and H is a subgroup then the order of H divides the order of G.

[echo gull]

Hint: contradiction

[prime sundial]

counterpoint:

[white oxide]

why is the second part of the first paragraph necessary? didn't they show that using the assumption of =>, x - a is a common factor and it's obviously of positive degree?

[echo gull]

Moreover, giveaway hint: ||assume the intersection is nontrivial, consider the cyclic subgroup of the nontrivial element and apply Lagrange's||

[slim kestrel]

Lagrange's theorem right?

[echo gull]

ye

[slim kestrel]

let x be an element in the intersection of H and K
so x is in both H and K

since x is in H, ord(x) must divide |H| = m by Lagrange's theorem
so ord(x) must divide |K| = k

since m and k are prime, the only positive integer that divides both m and k is 1

therefore, ord(x) = 1, So x is e

the intersection contains only e, proving that it is the trivial subgroup

[echo gull]

relatively prime but you have the right idea

[slim kestrel]

ohk

[white oxide]

is it ordinary for textbooks to introduce finite fields before splitting fields? im reading gallian rn and the techniques used in proving that there's only one finite field of a certain order up to isomorphism are so much cleaner using splitting fields than that of fraleigh

(who introduces splitting fields like 10 chapters later or smt)

[lethal dune]

and counting

[dim widget]

It varies

[echo gull]

this is one convention

[slim kestrel]

depends on the author

different authors may have different pedagogical approaches

[echo gull]

if you don't mind me asking, what coursework/study have you done that you are now catching up?

[slim kestrel]

as in

[prime sundial]

my class used fraleigh and my professor loved finite fields so we spent a lot of time on them before going to splitting fields

[echo gull]

what have you taken "above" introductory group theory?

[slim kestrel]

oh

Advanced Group Theory

[echo gull]

as in?

[white oxide]

wait i'm confused what did your advanced group theory classes and regular group theory classes call the identity for any group if you didn't know what it was

or did u just forget

[slim kestrel]

lemme check

[slim kestrel]

"1"

[slim kestrel]

idk how to explain the system

[formal ermine]

how do you define a group lmao

[slim kestrel]

structure consisting of a set of elements?

[echo gull]

that's not a group

[slim kestrel]

well fuck

[echo gull]

lots of structures with elements

the study of such things is called math

lol

[slim kestrel]

im contemplating to do my course again

but im in too deep

fuckkkkkk

[formal ermine]

also what does introductory group theory at your uni cover btw?

[white oxide]

i would probably consider doing the course again yea if you forgot the def of a group

[slim kestrel]

eh
definition and basic properties of groups, Group operations and properties, Subgroups, homomorphisms and isomorphism, Cosets and factor groups, Permutation groups, symmetric groups,

[echo gull]

that's fairly normal for introductory group theory

[formal ermine]

permutation groups and symmetric groups

wut

[slim kestrel]

oh also Group Actions

and cyclic groups

[echo gull]

my personal philosophy is that 1 semester should contain through group actions and at least 1st sylow

[slim kestrel]

perm is the study of symmetries and rearrangements of objects

[echo gull]

fuck nitro reactions they suck

[prime sundial]

we also went over this in my group theory class

[echo gull]

i think they know

[prime sundial]

i thought it was standard

[delicate olive]

S = {1,..,n}. S_n is the set of bijections f:S --> S or smth

[formal ermine]

we did group actions, sylow, normal groups, solvable groups and a bit of comm alg + field theory + galois theory + applications in one semester

[prime sundial]

that was my second algebra course

[white oxide]

wtf was this an introductory algebra course?

[echo gull]

my first class was 1/2 rings and fields and 1/2 groups

[slim kestrel]

OH YEAH SYLOW

[formal ermine]

it is but it's weird naming it as separate entries

[slim kestrel]

im doing my course again fuck this

[echo gull]

in that 1/2 of groups we managed sylows

[formal ermine]

when permutation groups are just subgroups of symmetric groups

[formal ermine]

first undergrad algebra course

[slim kestrel]

im telling by memory

[formal ermine]

ah

[slim kestrel]

maths is the only thing killing me inside out

[echo gull]

are you in undergrad?

[slim kestrel]

yes

[echo gull]

what was the syllabus for your advanced course?

[south patrol]

Mathematics is pain

[white oxide]

bruh that's hilarious my introductory algebra class didn't even get to extension fields/algebraic extensions

so basically we didn't even cover any field theory

groups & rings & fields

[slim kestrel]

uhh lemme remember

[prime sundial]

my first abstract course was baby group theory, no sylow or anything, and like 2 weeks of rings

that was it lol

[formal ermine]

algebra 2 is alg geo

[echo gull]

our alg 2 is galois

[prime sundial]

then algebra 2 was all the way up to galois theory, but we didn't do much of it

[echo gull]

i know some schools which lean into representation theory for alg 2 as well

it varies based on the abilities of faculty

[south patrol]

we don't number them

lol

Algebra is not linearly ordered

[slim kestrel]

advanced group theory, ring theory, field theory and what was that dudes name theory, module theory and homological algebra, category theory and a whole lot of other shit that i dont remember

what was that one theory ehhh

[rotund aurora]

wheel theory

[slim kestrel]

relationship between field extensions and group theory

whats that dudes name

[prime sundial]

galois

[rotund aurora]

lebesgue

[slim kestrel]

YES

that dude

[prime sundial]

newton?

[rotund aurora]

dirichlet I think uhh

[dim widget]

What did you guys do in algebra $\omega^\omega$

[cloud walrus]

Topos_Theory_E-Girl

[prime sundial]

something something riemann sum something

[echo gull]

cry

[south patrol]

Lol

[rotund aurora]

funny as hell

[slim kestrel]

i still dont understand the notion of a Galois group and its relationship to field extension

[south patrol]

Uh it is defined in terms of field extensions

[echo gull]

lmao

[south patrol]

Well or do you mean like

[prime sundial]

if you don't know what a group is i sure wouldn't expect you to

[south patrol]

You want to think more about the Galois correspondence

[slim kestrel]

i need to study the automorphisms of a field extension that preserve the base field and their behavior on the elements of the extension.

[prime sundial]

idk if its the usual way to think about it but we just went on about permuting roots of polynomials

[slim kestrel]

my brain is a complicated place

sometimes i understand and then sometimes i do fucking 1st grade mistakes

[echo gull]

definitions are inextricable from mathematics as a discipline, if you want to advanced work in any area you should be solid on definitions

[slim kestrel]

i know

too much shit to remember

why does Abstract algebra have to be part of the foundation of my course

[echo gull]

what course

[slim kestrel]

im currently doing bachelors in astrophysics

[echo gull]

oh alright

[slim kestrel]

imma do masters then PHD and then suicide

[echo gull]

conventional path for a lot of us in math

[rustic crown]

nuuu

[slim kestrel]

Bro

[rustic crown]

not allowed

[slim kestrel]

what is this question

fuck that

my professor is messing with us

i know it

[echo gull]

i know nothing about astrophysics but this seems like an astrophysics question a very confident and incompetent AI would produce

[slim kestrel]

its in the worksheet

¯_(ツ)_/¯

[rotund aurora]

sure its not enough with general relativity? like there is nuclear stuff going on

and black holes have charge

anyway no one cares, this is abstract algebra

[delicate orchid]

get this PDE nonsense OUTTA my channel

[echo gull]

i live in the terra nullius between analysis and pdes

[slim kestrel]

i wont be surprised if i search this up and find something out of syllabus

[slim kestrel]

true

why does category theory exist

okay imma do cat and idk where to start from

[hollow fjord]

can someone help me with linear representations of compact groups?

[echo gull]

in a naive sense category theory is a sort of unifying theory of mathematics***

[white oxide]

I'm confused as to how exactly the paragraph before Corollary 1 serves as a pseudo-proof, isn't $\text{GF}(p^n) \simeq \mathbb{Z}_p \oplus \mathbb{Z}_p \oplus \dots \oplus \mathbb{Z}_p$ if we regard $\text{GF}(p^n)$ as an additive group and not a field? How does this result follow if we're only considering the additive group structure?

[cloud walrus]

okeyokay

[slim kestrel]

can i learn the entirety in 7 weeks?

[echo gull]

no

[delicate orchid]

no of course not

[slim kestrel]

fuck

can i get somewhat knowledge to pass and not fail

[echo gull]

there is no "entirety"

it is an active research area

[slim kestrel]

i mean

my syllabus

[delicate orchid]

if GF(p^n) is isomorphic to n copies of Z_p then the result is pretty much immediate from the definition of a degree of a field extenstion

[echo gull]

then still no probably not

[delicate orchid]

assuming I'm reading this right

[delicate orchid]

yeah it's doable, 7 weeks is like 2/3rds of a semester lol

[white oxide]

Yes but here the n copies of Zp is not a field

under addition at least

[slim kestrel]

Representation of physical systems, Mathematical foundations, Information flow and processes, Quantum mechanics and gravity, Cosmology, Data analysis and classification, Multimessenger astronomy, Quantum astrophysics, Topological analysis, Network analysis, Information theory and entropy

Are you sure

[south patrol]

Lol how are you defining GF(p^n)

[delicate orchid]

it's still just an order argumnet boss

as sets, one is n copies of the other

[echo gull]

category theory is deeply involved with some broad notion of mathematical structures; certainly you should build some more intuition for "simple" structures like groups first

[south patrol]

Oh nvm

[slim kestrel]

Imma get off discord and start studying

[south patrol]

Yeah lol, you can even just like

View GF(p^n) as a F_p vector space in the natural way

And then like

[white oxide]

oh right oops i suppose GF(p) iso to Zp lmfao

that was kinda obvious my fault

[south patrol]

$\mathrm{GF}(p^n) \simeq \mathbb F_p^{[ \mathrm{GF}(p^n) : \mathbb F_p]}$ as vector spaces over $\mathbb F_p$ by definition of dimension and index

[white oxide]

yea i guess it follows immediately for there being only one field of prime order up to iso

[south patrol]

That is unnecessary

[cloud walrus]

potato

[south patrol]

Now just take sizes

But yeah

[delicate orchid]

still do not know what GF stands for, galois field?

[south patrol]

The reason I say it like this is that it mirrors the proof that every finite field has order p^n for some n

Yes Wew

[rustic crown]

i was reading GF(p^n) as GL(n, F_p) and getting super confused

[white oxide]

is there any canonical way to think about GF(p^n) (in other words the elements of GF(p^n)? my first thought was to take any extension field of the form GF(p)/<p(x)> for any irreducible polynomial p(x) over Zp of order p^n, but that's a bit ambiguous

[rustic crown]

my mind only saw G, n, p and filled in teh rest of the details arbitrarily

[white oxide]

(because there may be multiple irreducible polynomials right)

[slim kestrel]

What channel should i go to for Mathematical modeling and representation of a physical system?

[echo gull]

#modeling

[slim kestrel]

ok

[rustic crown]

how do you think of complex numbers? R[x]/(x^2+1) or R[x]/(any irred quadratic)? does that make a difference? (you know everything would be iso in the end)

[south patrol]

The terminology GF(p^n) intrigues me tbh

[delicate orchid]

It infuriates me

[south patrol]

Do people just hate subscripts or smth lol

Or is it to be weird to stop people thinking $\mathbb F_n = \mathbb Z/n\mathbb Z$

lol

[cloud walrus]

potato

[white oxide]

ah yea i guess that's a good point

[south patrol]

But also like

Well okay nvm I guess actually

People often say like splitting field of x^{p^n} - x i guess but then that is also non-canonical lol

But it chill

I do think it's a good point though like how do you actually write out the elements of it lol

[rustic crown]

ig if you really want something canonical, then fix an algebraic closure of F_p and always work inside it

[south patrol]

I mean sure but then that's like making it canonical by making an arbitrary decision right xd

But yes

[rustic crown]

yea i mean only difference is that these are galois, so there is a unique copy inside the alg-closure as opposed to Q(cbrt2)

[south patrol]

True

Hot

Yeah I'd not thought about that actually lol nice

What is actually the easiest way to show that a finite extension of finite fields is always Galois

[rustic crown]

[south patrol]

I guess it'd just be like lol

The big one is splitting field of x^(p^n) - x for some n which is separable

But there are various ways I suppose

Well I guess you can show directly that |K:F| <= |Gal(K/F)| using Frobenius too and various other things

[white oxide]

ok this is probably a stupid question but do sylow theorems have any correspondence at all with finite fields? i'm assuming no since sylow theorems are only concerned with group structures, but each field of order p^n having a subfield of order p^m for a divisor m of n is quite similar to the first sylow theorem (but that's for groups)

or am i trippin per usual

[slim kestrel]

you can use the fact that finite fields are always perfect fields

[delicate orchid]

well you can use the fact that the order of a subgroup has to divide the order of the group to conclude that if a subfield existed if would have to be order p^m - doesnt' show existence but that's what pops into my head

it has nothing to do with sylow subgroups though, the additive subgroup of a field is a p-group

[hollow fjord]

how you define the regular linear representation of a compact group?

[white oxide]

yea makes sense

also how is F* of order 3, isn't is of order 15

do they mean "the three nonzero elements of this subfield must be in the cyclic subgroup"

[delicate orchid]

yeah that seems to be worded really poorly

what they're saying is it has to be a cyclic subgroup of F* that has order 3, and that F* = <x>

[slim kestrel]

"regular linear representation of a compact group is a representation that acts on the vector space of complex-valued functions defined on the group"

[south patrol]

Ye that's another

[white oxide]

oh yea just got it thx

[south patrol]

Tbh Sylow seems a bit irrelevant since for fields everything is commutative and stuffs

and for abelian groups sylow is chill

[delicate orchid]

this reads like a chat gpt answer

[slim kestrel]

yea

[slim kestrel]

wikipedia

i shud mention actually

[rotund aurora]

wikipedia enjoyers are at the pinnacle of wisdom

[dim widget]

they all have cyclic multiplicative group

[echo gull]

$F_q$ generally denotes the group of affine linear transformations on $\mathbb{F}_q$ for prime power q right?

[cloud walrus]

kodiak

[dim widget]

i have never seen it denote that

[south patrol]

If someone writes F_q i'd assume it means finite field order q lol

[echo gull]

i've seen it used in denotation of frobenius groups

hence my question

[dim widget]

i mean you can use whatever notation you like, it just has to be explained

If someone wrote $F_q$ I think that I agree with potato that my first thought would not be $\mathbb{F}_q \rtimes \mathbb{F}_q^*$

[cloud walrus]

Topos_Theory_E-Girl

[jagged galleon]

what is the square route to the obama pyramid times four to the pour of six divided by 7

[dim widget]

please post this in #1061337023413244018 this is the wrong channel

[echo gull]

it wasn't mine either, but groupprops was using it and my assumption was the only thing which made sense in context

[dim widget]

yeah I think it's not super standard but may be standard on groupprops

[echo gull]

👍

[hollow fjord]

There is a correspondence between the linear representations of a group G, and the linear representations of the algebra generated by de group (all over a field F). Can we do the same for compact groups?

[dim widget]

That's an interesting question.

It depends on which compact groups you're interested in, I'm not sure there's something which works uniformly for all of them. The natural candidate is the action of L^2(G) on a representation (V, \rho) of G when this makes sense

[hollow fjord]

Mmm, it would be the same to ask if i have a representation \rho of a compact group and find if i can extend it to a non continuous representantion of F[G]?

[glossy crag]

Maybe I'm being overly pedantic, but I'm still stuck on the minutiae of this.

The way I understand it is that the proof constructs for every \alpha' in C' a finite extension L(\alpha',\beta,f) (where f is a lift of f' and \beta is a root of f that is a lift of \alpha') s.t. the residue field is K'(\alpha') and the degrees coincide.

In the separable case the choice of \beta falls away since every root lifts uniquely, so it becomes L(\alpha',f), but if we want this to be an assignment {F/K' finite separable} -> {L/K finite unramified}, we still have to show these lifts are independent of choice of f and \alpha' (where \alpha' is a generator of F/K'), right?

[glossy crag]

And here we use the statement from part (e) (E/K finite extension, E'_s\supset F\supset K' finite separable subextension, L lift thereof => L\subset E) to show that L(\alpha',f) is the same for any generator of F/K' and choice of f, thus producing a true assignment.

Am I being overly pedantic or do these questions actuall have to be considered?

[south patrol]

let ß be a root of X^4 + 1, then calculate the degree [F(ß): Q] in two ways @willow geyser

That's how I'd do it at least

i.e. show that [F(β): Q] = 12. If you want another hint ||it suffices to show divisibility by 3 and 4; the fact the order is <= 12 is obvious||

Yes it is

Do you know another expression for [Q(β): Q]

Yes

Fun fact: X^4 + 1 is reducible mod p for all p lol

idk if you've seen that

Nah dw ^

Also just realised I accidentally used ß instead of β lol

Rip

So true except i'm a σ

[rustic crown]

🥔

[summer path]

[white oxide]

im confused, aren't all sylow p-subgroups conjugate by the second sylow theorem so there's not that much to prove here?

[hollow fjord]

But H is a p-subgroup, not a Sylow p-subgroup

i.e it is not maximal

[white oxide]

oh yea misread

thanks for catching that

oh wait this just follows from the first sylow theorem i believe

[rustic crown]

unless your sylow-1 statement is super strong, it probably requires a lil more work.

(for example, the question above can be thought of as a generalization of sylow-2)

[white oxide]

this is what my book reads, my idea was that the order of P is p^n for some integer n and where |G| = p^nm, and since H is a p-subgroup it is of order p^i where i < n, then I can just form a chain of subgroups H < H1 < H2 < ... Hn = P because H is a subgroup of H1 of order p^{i + 1}, H1 is a subgroup of order p^{i + 2}, so on and so forth until Hn-1 is a subgroup of order p^{n - 1} of P, don't know if this works or not

[white oxide]

yea that was my first thought

[rustic crown]

i think you wanted to start the index at H = H_i

but notice that H_n need not be P, though it is a sylow p-subgroup again

[white oxide]

wait wdym i was just defining it as such

sorry im confused my fault

[rustic crown]

it's not a big thingy... just that in your notation H_n would have order p^{i+n}

[white oxide]

you're right about the indexing though it's so shit

[white oxide]

yea lmao i was just writing it down and it's not readable at all or economical

[wraith cargo]

TIL there's a proof of this using group actions huh

[white oxide]

but thx for the help!

[glossy crag]

all proofs of Sylow use group actions afaik

[rustic crown]

(i wouldn't call that a construction )

[dim widget]

@glossy crag it depends on how detail oriented you want to be. I would adapt the proof of part e) to show that any two unramified extensions with the same residue field are equal

[glossy crag]

i haven't read it through entirely yet, but the setup looks like a group action by right multiplication on the set of subsets of a fixed size

[dim widget]

Then you know that the extension you construct depends only on the residue field

And doesn’t depend on any other choices

[glossy crag]

is that doable though? i haven't considered it yet, but the proof uses the lifts i talked about, and that's very much choice-dependent

[ripe basalt]

cant you use the fact that sylow p-subgroups are conjugate to each other and also that every p subgroup is contained in a sylow p subgroup?

[dim widget]

@glossy crag use Hensel’s lemma

[ripe basalt]

(my response was to okeyokay sorry if i interrupted something)

[white oxide]

ah yes that was the answer they provided LOL

first thought was second sylow theorem but idk first worked for me so oh well

[rustic crown]

you used both first and second right

[white oxide]

how is this an adequate proof of every basis of G having the same number of elements? couldn't we have taken G/3G, G/4G, etc..

[white oxide]

nah only first

that's fine right

[rustic crown]

so like H_n need not be P. by second sylow it's a conjugate of P

[white oxide]

ohhhh right oops

ok well tha tmakes sense

because any two sylow p-subgroups are conjugate

[glossy crag]

we could have

2 is the simplest option

[white oxide]

ah yea that's true

[white oxide]

so just checking that my reasoning here is correct, H = H1 \leq Hn, and Hn = xPx^-1 for some x in G, and so we get H \leq xPx^-1 and the rest is immediate

lol why do we refer to the length of a basis of a free abelian group G as the rank and not dimension

[glossy crag]

How about this:
Let L,L' be finite unramified extensions with the same residue field. Since the residue field is separable, it's Kbar(x). Taking any monic lift f of the minimal polynomial of x, by Hensel there is a unique y in L lifting x. Since the valuation ring of K and y are contained in the valuation ring of L we get |L:K| = |Kbar(x):Kbar| \leq |K(y)bar:Kbar| \leq deg f = |K(y):K| \leq |L:K|, so L=K(y) and f is the minimal polynomial of y (in particular this implies L/K is separable), same with L'=K(y'). Now if E is the composite of L and L', then it's complete and again by Hensel there is a unique z\in E lifting x, therefore by uniqueness z=y=y', and L=L'. This implies that the procedure of starting with a separable extension F/Kbar, taking a generator x (by primitive element theorem), lifting that to a monic f and root y, and getting an unramified K(y)/K is actually independent of the choices made along the way (since all the lifted extensions have the same residue field F).

[dim widget]

Yep that seems good, but you have to show that compositum of unramified is unramified

I was thinking you can just show that a lift of y is in L’ and a lift of y’ is in L so they contain each other

[glossy crag]

Why do I need that, I did not use E being unramified in any way.

E is finite, therefore complete, so by Hensel it has a unique lift z of x in it. But being the composite it already has 2 lifts, y and y', so all 3 are equal, implying L=L' (since we've already shown L=K(y) and L'-K(y')).

[dim widget]

Yeah I guess that’s fine as is then

[glossy crag]

Is this last part correct?

This implies that the procedure of starting with a separable extension F/Kbar, taking a generator x (by primitive element theorem), lifting that to a monic f and root y, and getting an unramified K(y)/K is actually independent of the choices made along the way (since all the lifted unramified extensions have the same residue field F).

[dim widget]

Yep

Changing f doesn’t change the residue field, and since f is separable mod p there is no choice of root involved

There is a unique root congruent to x mod p

[glossy crag]

I thought I don't need to take this into consideration and could just use "unramified, same residue field => equality" directly.

[dim widget]

Sure if you like, just trying to unravel some of your confusion

[glossy crag]

understood

[boreal inlet]

Let S and T be two submodules of an A-module M.

Prove that the quotient modules (S + T)/S and T/(S intersection T) are similar upto isomorphism.

What I did was

showed that both (S + T) and (S intersection T) are submodules of M

Now I am having trouble in constructing the isomorphism between these two

<@&286206848099549185>

[glossy crag]

You have a natural map S+T -> S+T/S and since T is contained in S+T, you get a natural map T -> S+T/S (t goes to t+S). Show that this map is surjective and find its kernel, then use the 1st isomorphism theorem.

[glossy crag]

Again, I think I'm being overly pedantic, but would you mind taking a look at my slightly modified write-up? I want every detail to be perfect and I'm also wondering if a lot of what I'm doing is redundant somehow (and there isn't an easier way of doing things).

[boreal inlet]

The previous natural map is surjective yes, but I don't get how the T -> (S+T)/S map comes from

what I was thinking is like ->

If we can define a surjective map from T to S + T

then we can use the first isomorphism theorem, as the composition of two surjective maps is surjective

but I have zero idea on how to do that

I only know that S + T is the smallest submodule containing both S and T

[glossy crag]

The map T->(S+T)/S comes from restricting the map S+T->(S+T)/S to T, which is contained in S+T.

[boreal inlet]

bruh...

yeah got it

kernel is S intersection T because first of all domain is T so the point needs to be in T, and also it needs to be in S because for any element s in S, s + S = S

[glossy crag]

i'd phrase it as "the map is t->t+S, so if t+S=0, then t\in S\cap T, the reverse inclusion is clear"

[boreal inlet]

true

far better

now to show it's actually surjective

hmm

if we take a coset of form x + S

x can be in anywhere in S + T

if x is in T, the coset has a clear preimage

but if

x is in S, then the coset is just the identity element in the quotient module

so, x must be in Kernel, which is S cap T

so x is in T anyway

.....is this sufficient?

[glossy crag]

no, because as you yourself said "x is anywhere in S+T", so "x is in S"/"x is in T" doesn't cover all the possibilities (S+T isnt the union of S and T, it is the set of all sums x+y with x in S and y in T)

an arbitrary element of S+T/S is of the form z+S, where z=x+y with x in S and y in T

therefore z+S=y+S and y is mapped to z+S

[boreal inlet]

how is again addition of cosets defined

(x + S) + (y + S) = (x + y) + S

this right?

oh okay so

z + S = (x + y ) + S = (x + S) + (y + S) = S + (y + S) as x is in S

S is identity of addition

so z + S = y + S

y is in T, so we have a clear preimage

thank you

this makes a lot of sense

[glossy crag]

or z-y=x\in S, so z+S=y+S by definition of cosets, same thing

[flint cave]

How can I show that $trace > 2$ elements in $PSL(2,\mathbb{R})$ are conjugate to a scalar matrix representative with $\sqrt{\lambda},\frac{1}{\sqrt{\lambda}}$

Oh wait should I post this in LA

[cloud walrus]

A free isometry

[agile burrow]

It's prob fine here

I'm thinking uhh trace > 2 implies real eigenvalues right?

det = 1 implies reciprocal eigenvalues

[flint cave]

Oh crap

Didnt try the most straighforward thing

[agile burrow]

Maybe there's something to be said about existence of eigenvectors, but yeah I think that should give it to you after change of basis

[flint cave]

Yes it should be fine

[agile burrow]

Why is the square root there

Oh, right you can choose your representative to have the positive eigenvalues

[flint cave]

Yeah I was learning some stuff in hyperbolic geometry

[agile burrow]

Fun fun fun stuff

I too will be thinking about hyperbolic geometry over the next few months

[flint cave]

Oh cool what exactly will you be learning?

[agile burrow]

GOOD question

I have no idea

[flint cave]

LOL

Im trying to learn Fuchsian groups because my prof said horocycle and geodesic stuff has a lot of cool dynamics in them

[agile burrow]

Ooh yeah discrete subgroups of PSL(2, R) seems really neat

I'd be interested to hear about the dynamical side of it sometime

[flint cave]

Same lol, havent gotten to it

The reference I was suggested is a book titled "Geodesic and Horocycle trajectories", Dal'bo if you are interested

[glossy crag]

Proof that local fields are locally compact. Why can we pick a_2 so that a_2\equiv a_1\mod\pi?

[glossy crag]

NVM: if a_i is a repr. system of R/pi R and b_j a repr. system of pi R/pi^2 R, then a_i+b_j is a repr. system of R/pi^2R and a_i+pi R is the disjoint union of the cosets a_i+b_jpi^2R, so if all a_1+b_jpi^2 R were finitely coverable, so would a_1+pi R be => there is a b_j with a_1+b_jpi^2R not finitely coverable and obv. a_1+b_jpi\equiv a_1\mod\pi.

[abstract spear]

What book is this

[glossy crag]

Herstein's Topics in Algebra.

[primal tusk]

If you have a set of n integers, is there some sort of algorithm which could give you a unique number given some set of integers?

For example if you had {2,1} is there some sort of way to uniquely represent this set when compared to other sets with just two integers?

[formal ermine]

what?

[primal tusk]

If you were given two numbers, is there a way you could tell the other people exactly what numbers you have using only a single number.

Without showing them

Like for example maybe you could add the two numbers together and then multiply that by 4 and somehow that would be unique

(It wouldn’t be)

But I’m just wondering if there is a way to do it

[formal ermine]

n_1 + reverse(n_2)/10^(length of n_2 + 1)

[primal tusk]

Thanks did you just come up with that?

[formal ermine]

yes

uh okay the reverse part might be ambiguous

what I'm saying is that you should reverse the second number (including 0s)

and then put it as the decimal part

because otherwise you might get stuff like it cutting away the zeros at the end

[primal tusk]

Oh true

That’s really clever

[spiral anchor]

Need help in 27.3 quick

[primal tusk]

A L G E B R A M O M E N T

[formal ermine]

#prealg-and-algebra

[spiral anchor]

Isnt it algebra?

I dont see function

Channel

[formal ermine]

this is advanced algebra

[spiral anchor]

Oh ok

[formal ermine]

this falls under #prealg-and-algebra

[spiral anchor]

Sorry

[formal ermine]

the entry barrier to this channel is knowing what a group is

[primal tusk]

He’s got a looong way to go lol

[echo gull]

explain group like I’m 4

[formal ermine]

symmetry go brrrr

[echo gull]

Thank you, I am now prepared for homological algebra

[formal ermine]

you're welcome. just never forget that the reduction of a projective scheme over an algebraically closed complete field is noetherian in the étale cohomological topology

[echo gull]

it trivially follows

[formal ermine]

so true

everything follows from the definitions

[long nebula]

2^n1 * 3^n2 * 5^n3 * ...

If the numbers are all positive

This type of procedure where you assign a natural number to everything is called "Gödel numbering"

It enjoys wide use in computer science now because to computers, everything actually just is a long number, but it was of great use to logicians studying proofs far before computers were even invented and in mainstream use!

[primal tusk]

Thanks for telling me that because this was actually for a computer science problem

I was basically seeing if there was an easier way to see if two arrays had the same numbers without looking at each slot and comparing or something

[long nebula]

hahahaha

Well if you have them as arrays, the best way is just to check each slot

Converting them both to numbers and compating the two numbers would take longer

@primal tusk

[primal tusk]

Yea that’s what I was just thinking lol

[white oxide]

can somebody help me here? how did they get from the first equality to the second equality?

oh

they just rearrange dit

[alpine island]

Is it possible for F[x] to be a field for any F? My assumption is no, and you would run into issues with whatever you define 1/x to be, but idk how to formalize this

[rustic crown]

use that degree is a thing

and if one of the factors is monic, then deg(fg) = deg f + deg g

moldi

[white oxide]

what exactly do they mean by minimal such nonzero value |k_i|? does it mean select the basis that has the lowest value of |k_i| for some coefficient of some k in K?

[tribal moss]

Take all the possible bases and all the possible integer combinations whose values lie in K, and the absolute value of all the coefficients of all those combinations. You get some subset of N+. It is not empty, so it has a minimal value. For that minimum value, pick one basis and one linear combination where this minimum occurs as a coefficient.

[white oxide]

ah that makes a lot of sense, thanks!

how would you use the minimality of d tgo see that h1 is a multiple of d1? what does subtracting by a suitable multiple of d1x1 even do?

[tribal moss]

What subtracting a multiple of d1x1 doesn't do is take you outside K.

[white oxide]

ye i know that

[tribal moss]

So if h1 were not a multiple of d1, you'd be able to find an element of K whose x1-coefficient were strictly between 0 and d1.

But that contradicts the choice of d1 as the absolutely smallest coefficient of anything in K under any basis.

So h1x1 is actually the multiple of x1d1 you want to subtract in order to get rid of the x1 term.

[white oxide]

got it thank you

[white oxide]

just a quick sanity check, [GF(729): GF(9)] = 4 right

[rustic crown]

3^6, 3^2 so 6/2?

[white oxide]

oh right i subtracted

thx

[summer path]

detuwu

[white oxide]

is there any reason why they're writing $\mathbb{Z}_2[a]$? that doesn't necessarily refer to the splitting field for f(x) right

[cloud walrus]

okeyokay

[south patrol]

Not sure what you mean

They mean the ring/field Z_2[a]

Or are you just unsure what that notation means?

@white oxide

[white oxide]

well ig im a bit confused because prior they would write Z_2(a) to emphasize that it's a field but now they're putting brackets, but i know that Z_2[a] is not not a field

idk if that makes any sense

[coral spindle]

?

Z_2[a] is a field

[delicate orchid]

spooky!

[rustic crown]

Z_2 >.<

[coral spindle]

Just like how Q[sqrt(2)] is a field 👻

Oh SORRY det I meant F_2

[south patrol]

If $K/F$ an extension and $a \in K$ algebraic over $F$ then $F[a] = F(a)$

[cloud walrus]

potato

[coral spindle]

Clearly we are talking about the 2-adics :P

[white oxide]

no yea i know it's a field

but why tf would he randomly put brackets

that's what fucked me up

[south patrol]

Idk lol

[white oxide]

shit got me fucked up 😭

[delicate orchid]

wait are we not talking about the 2-adics

[south patrol]

Lmao

Tbh Z_2 is cursed notation

Idk why people would use it here

[delicate orchid]

the second centre

[south patrol]

And Z_2 in the context of fields bruh momentum

[rustic crown]

how do this? me forgot linear alg

A is arbitrary commutative ring

idk how to deal with these things when not in some integral domain

[south patrol]

Oh lol

Is this from Matsumura

[rustic crown]

yesh

chmuwu told me to read it and become superdet

[south patrol]

Yeah I dind't know anything about this and ignored it lol

Cause idk about minors really lol

[coral spindle]

minors on the math discord server?

instaban

[rustic crown]

i have seen an alternate proof of IBN, but i want to know is it that easy that matsumura doesn't even give a hint about it >.<

det no know det

[delicate orchid]

simply consider the determinant

I'm assuming "are zero" means "has zero determinant" rather than "are literally zero in M_n(A)"

[rustic crown]

minors are already det-ed

[delicate orchid]

that's not standard

I'd call the determinant of a minor the cofactor (up to a sign 🙄 )

[rustic crown]

valid >.<

okie wikipedia says minor is det-ed

and cofactor is (-1)^... det after deleting

[delicate orchid]

fairs

I've never seen it used like that though but then again I've only seen minor matrices pop up in computational algebra stuff rather than pure algebra

different convention perhaps

anyway this question sounds fun

ok I've done it

[rustic crown]

how do >.<

[delicate orchid]

use the fact that EVERY r x r minor is 0 to ||show that the r+1 x r+1 minors must have linear dependency somewhere||

and then induct on dimension

[zenith hollow]

guys

is latex part of discord?

[delicate orchid]

no

[zenith hollow]

but you guys usually just type something here and it generates the math letters

[delicate orchid]

we have a bot that compiles it

[zenith hollow]

ah ok so it depends on the server?

[delicate orchid]

no dude

it's a markup language

it is not part of discord it has been around for 30+ years

[zenith hollow]

Yeah that's what i meant

[coral spindle]

#latex-help

[zenith hollow]

thx

[zenith hollow]

[white oxide]

could i get a hint for this question pls

i feel like im hella overthinking it

[rustic crown]

me no understand >.<
you probably meant the other way (r+1)x(r+1) are 0 so that induces a dependence on rxr minors? but that's like inside A and any two things here are dependent, so don't see how that would help at all

[delicate orchid]

no, you can do it the other way det

consider two overlapping r x r minors

[rustic crown]

(lol i was considering r = 1 )

[delicate orchid]

it's a cubic extenstion over a field of size 2

[delicate orchid]

well in that case everything has to be LITERALLY zero

[south patrol]

You just need to show it splits after u add a single root

[delicate orchid]

true though my answer did not include that case

[rustic crown]

wdym

[south patrol]

Which I assume you have done above already

[dim widget]

No the condition is that the r+1xr+1 minors vanish but at least one rxr minor doesn't

[rustic crown]

we have a n x 2 matrix where every 2x2 minor is 0

need to show the columns are dependent

[delicate orchid]

n x 2? why not 3 x 3?

[rustic crown]

yea that's the simplest non-trivial case right

if it works for r = 1 and n x 2, then it works for all (r = 1) and n x m

i tried experimenting a little with cauchy-binet

but if it's any more complicated then idk why matsumura says it's easy

[delicate orchid]

you only need to consider square ones I think

[dim widget]

Here's a slightly higher level approach: by padding we can assume that $m = r+1$ and that $n \geq m$. Then the matrix M induces a map $\Lambda^mM: \Lambda^mA^m \to \Lambda^m A^n$ (this is just the matrix given by the minors of $M$ in whatever bases we've chosen)

[delicate orchid]

my overlapping idea should still work for non-square ones though but I'm just confused why you're considering them

topos please fix the latex or I'm going to shake my head back and forth really fast for 15 seconds and get dizzy

[cloud walrus]

Topos_Theory_E-Girl

[south patrol]

Yes lol

[rustic crown]

yee makes sense

[dim widget]

Now you can show that if $T: M \to M'$ is a map of free modules with $\text{rk}(M) \geq \text{rk}(M') = m$ then $T$ being injective implies that $\Lambda^r T$ is injective for $r \leq m$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

This is because the exterior power of a free module injects into the tensor power, and by flatness we win.

[rustic crown]

does it inject even if non-zero char?

[south patrol]

This reads like stacks project

[dim widget]

Yes! but you are remembering correctly that there is no retraction!

[dim widget]

.

[south patrol]

By Topology 146.124175097.1025871598615.120597195871 we win.

[dim widget]

Lol I like Johan's "we win" thing

[south patrol]

Me too

[dim widget]

Johan is best girl

[rustic crown]

[south patrol]

I really like Johan's style in general tbh

Fun things are fun

<3

[dim widget]

Anyway yeah the map that takes a wedge to the alternating average of the corresponding tensor is always injective for free modules

[rustic crown]

the other way right? there is a surjection from tensor to wedge, but not necessarily a section?

[dim widget]

There are both natural maps from wedge to tensor and tensor to wedge

[rustic crown]

ah oops

[dim widget]

But the composition is multiplication by some factorial

[white oxide]

ye but i'm still a little bit stuck on why it's a splitting field, i know it def has one zero but i'm trying to see why the other zeros are contained in it

[south patrol]

Hm what is the wedge -> tensor one in general characteristic

[rustic crown]

so just like sending a wedge b to a⊗b-b⊗a without the division?

[south patrol]

Oh lol

Makes sense

[south patrol]

Isn't that the content of question 6

[dim widget]

This is why it's better to always have divided powers 🙂

[white oxide]

i skipped it lol

[south patrol]

llmao

*lmao

Anyway yes so like the remaining step follows from q6

So you should maybe do that 😎

[delicate orchid]

a nation divided against something something cannot something

[south patrol]

I wonder if there is any particulrly nice way to see it splits after appending a single root - I'll think

I mean I have an idea lol

Oh wait no yeah it works fine

||Frobenius|| lol

I think it was tteg who helped me understand finite fields a lot better lmao

[rustic crown]

this is so clever

[south patrol]

Like I asked you about smth once and it changed how i think about their Galois groups etc

[white oxide]

ok lol

i feel like this is fucking trivial but im just slow

[delicate orchid]

u could always just add the roots in and then count the elements

[white oxide]

frustration has ensued

[south patrol]

I'm honestly not sure how else to do it besides the thing I gave so maybe you can look at it as a hint if you'd like

[dim widget]

This was this thing we talked about about how the Galois group being procyclic is related to there being a unique field of each degree is related to every extension being Galois 🙂

^hint for okey

[delicate orchid]

god I wish I cared about galois theory even slightly

[south patrol]

What does procyclic mean lol

[white oxide]

what language is that

[delicate orchid]

profinite but just cyclic groups right

[dim widget]

An inverse limit of cyclic groups or any finite quotient is cyclic

[delicate orchid]

yussss

[south patrol]

Oh okay sure

Are you talking about the absolute Galois group then sure

which is like

Wait lol

Was I correct lol in saying profinite completion of Z

Well you get dat limit of cyclics

[dim widget]

Yeah it's just what people call $\widehat{\mathbb{Z}}$ which is the profinite completion of $\mathbb{Z}$

[cloud walrus]

Topos_Theory_E-Girl

[south patrol]

Yeah I just thought I was misremembering smth lol

Hm so potentially silly question

Why do we have like $\mathrm{Gal}( \mathrm{colim}{n} F{p^n}/F_p) \simeq \mathrm{coliim}n \mathrm{Gal}(F{p^n}/F_p)$ here then

[cloud walrus]

potato

[south patrol]

Actually I guess this is kinda just clear

Like

$F_p$-linear map $\mathrm{colim}n F{p^n} \to \mathrm{colim}n F{p^n}$ is equivalently a compatible collection of $F_p$-linear maps $F_{p^n} \to \mathrm{colim}n F{p^n}$ and the image of each of those is just $F_{p^n}$ anyway

[cloud walrus]

potato

[rustic crown]

arigatou tteg

[dim widget]

Yeah I think in general you could make this some kind of category theory theorem: like if you have $X_n$ such that for all transition maps $X_n \to X_m$ restriction gives a map $Aut(Y) \to Aut(X_n)$, then $Aut(\text{Colim}_n X_n) = \text{lim}_n Aut(X_n)$, maybe this needs more rigidity but something like this.

[cloud walrus]

Topos_Theory_E-Girl

[south patrol]

Noice

Fhank

[dim widget]

I guess the real thing is that you want a.) X_n is preserved under automorphisms of objects containing it in the inverse system b.) the X_n are compact objects in the category

[remote nymph]

idk if this is the right channel but i am so lost with this. I found this paper only available in italian and im trying to translate it (with google) but I am getting very confused by this guy's notation. I've attached my translated version and the original. S_(4, 3) is just F_3^4. Do any of you guys have any clue what this means?

like why are there 4 points on a line? shouldnt it be only 3?

[formal ermine]

did someone say absolute galois group

[rotund aurora]

What is $\Z[[x]]/(3x-2)$ ?

[cloud walrus]

Croqueta

[rotund aurora]

like 3x-2 is prime and so admits a fraction field. The fraction field is just $\Z_2$ right, becuase we can invert 3 first and it will commute with the quotient

[formal ermine]

my intuition would tell me R but that's probably wrong lol

[cloud walrus]

Croqueta

[rotund aurora]

Z[[x]] is power series ring

ok maybe you dont know it

Z[[x]]/(x-p)=Z_p

[formal ermine]

yeah that makes sense

but if you plug in 2/3 then all of them should converge, no?

[rotund aurora]

to what?

there is no topology

[formal ermine]

the power serieses

[rotund aurora]

there is no topology

you cannot talk about convergence here

like you are plugging 2/3 yes, but these things dont converge in R or anything even if you imposed some topology

this is like the 2/3-adics wtf

[formal ermine]

(p, q)-adics

[rotund aurora]

ok I clowned here because when you invert 3 you get the ideal x-2/3

[rotund aurora]

Is this a thing

[formal ermine]

unaware

[rotund aurora]

I dont understand this meme but OK

[rustic crown]

det no understand most memes >.<

[hot lake]

isn't it the 2-adics ?

if you send x to 2/3 in Z2

[rotund aurora]

right 3 is a unit in Z_2

yeah, like 3(1-x)=1, so 3 is also a unit in the quotient above

ring theory is so cool

[summer path]

[rustic crown]

[rotund aurora]

can you obtain R (real numbers) from Z[[x]] somehow

[delicate orchid]

False

[delicate orchid]

Depends how generous your definition of “somehow” is ig

[rotund aurora]

I was trying to see if the real numbers arise as a special case of the p-adics when using the definition Z_p=Z[[x]]/(x-p) with p prime

[delicate orchid]

Do the dimensions as Z-modules line up 🆙

Z[[x]] has a countable basis

[rotund aurora]

yeah right obviously not a quotient

but was asking if there was something else

like all quotients in Z[[x]] are iso to products of Z_p lol

or iso to Z

Z corresponds to the trivial valuation I think

and R should be like quotiening by P(x)=1/x, because P(0)=infinity. But this doesnt make sense right now

[slim kayak]

If G is the free product of groups A and B, do 3 pairwise distinct conjugates of A or B in G generate a free group of rank 3?

[feral umbra]

Let G be a finitely generated group and H a subgroup of finite index. Suppose H has a subgroup H' of finite index such that its commutator subgroup [H',H'] has infinite index in H' (and is also finitely generated). If [H',H'] is virtually nilpotent does it follow that G is virtually solvable?

For context see this post: https://math.stackexchange.com/questions/4713266/understanding-proof-of-lemma-in-gromovs-paper-on-polynomial-growth

[stray saffron]

Idk if I can ask it here

But when calculating D=b^2-4ac

Can the x^2 be negative for example -3x^2+3x-4 OR do I need to do this -(3x^2-3x+4)

[void cosmos]

u should ask in #prealg-and-algebra

[stray saffron]

Thank u

[slim kayak]

You just set the a, b and c to the values they should be. So a=-3 and so on

[stray saffron]

Oh so I dont do -()

It doesnt matter if the x^2 is negative or positive as a= when calculating D?

[slim kayak]

But since you are on #groups-rings-fields you have to prove that the Galois group of the splitting field of that particular polynomial is solveable as a rite of passage

[stray saffron]

No too much math

Do it without theory by remembering solving :c

[formal ermine]

that's not math

[south patrol]

Lol I remember comments talking about stackexchange overcomplicating stuff like that

Does that even happen often that they overcomplicate stuff a lot

[slim kayak]

Before or after claiming your question was already stated 12 years ago? (it wasn't)

[south patrol]

Lmfao

[echo gull]

as a stack exchange contributor math SE is notoriously pedantic

[south patrol]

Okay maybe this just means I am a pedant lol

Cause I don't find it pedantic

Lmao

[echo gull]

lots of very qualified, very helpful people look at simpler questions and will point out things which while true and not particularly helpful

[rotund aurora]

I dont find SE pedantic either

[echo gull]

maybe pedantic is not the optimal word choice but I find lots of people prefer to be correct over being genuinely helpful

this isn't everyone of course

[slim kayak]

Probably really depends at what level your question is, if your question demands the constant level of pedantry present you won't notice it

[echo gull]

that is probably true but the wide range of mathematical ability on the forum is conducive to some of this behavior certainly

[white oxide]

here, would it be sufficient to just observe that both are fields of order 9 and thus isomorphic?

i mean they're both fields, the rings is kind of throwing me off

[void cosmos]

yes

wait no

show that when u quotient by a different irreducible polynomial u still get the same finite field

maybe try to adjoin the roots

and then think of a map

@white oxide

[south patrol]

Yes

[white oxide]

wdym

[void cosmos]

wouldnt he have to prove that

[white oxide]

if they're of the same degree then they're isomorphic regardless right

(the irreducible polynomial)

since they're both finite fields of the same order

[void cosmos]

yea

but i thought u have to prove thata

[south patrol]

It's probably best to construct an explicit isomorphism yes

[void cosmos]

i thought thats the actual point

of the exercise

like its not a true or false question

hahaha

[void cosmos]

if u have some finite field F_p and two irreudcible polynomails

F_p mod first irr poly is the same as F_p mod the second

if they are

of the same

deg

[white oxide]

well i mean i could just send alpha in the first field to beta in the second field where alpha is a zero of the first polynomial and beta is the zero of the second polynomial

just writing out the field elements of both fields i think that's right

[void cosmos]

try to do it in general

i dont think this is a situation where an explicit example would help

[white oxide]

Wdym

[void cosmos]

like just try this setup

[white oxide]

aren't the field elements of this form

[void cosmos]

yes u are right

but i just meant as a better exercise

or as a better way

[white oxide]

ah yea i got you

[void cosmos]

for more understanding

but ur right 100%

[white oxide]

okay i'll try that thx

[void cosmos]

like try to see it in general

[south patrol]

Hilbert theorem 90

[void cosmos]

its similar to that if u like

quotient over an irred prolynomial

u say that this is adjoined one of the roots right?

and then if u adjoin any different root

they are still the same field

or same ring 😄

same idea i thik

[void cosmos]

😠

[south patrol]

What lol

[woven obsidian]

You can't map alpha to beta

They satisfy different relations

[void cosmos]

😠

[white oxide]

Wdym

[woven obsidian]

Well if you write down a map you need to show that it is well-defined and a homomorphism

Your map can't be because alpha^2+alpha+2=0, while beta^2+2beta+2=0

As a homomorphism commutes with polynomials, the image of alpha will satisfy the same relations as alpha

So if beta was the image then simplifying yields beta =0

[white oxide]

hm okay i'll redo it then

thanks!

[woven obsidian]

Np, if you want to do it explicitly then you need to map alpha to another element that is a root of x^2+x+2

It's still tedious to do it explicitly, show well-definedness etc

Unless you use that mapping a root of an irreducible polynomial to another root gives a homomorphism but at that point you might as well just use that fields of the same finite size are iso

[formal ermine]

what does $\left( \bF_p^\times \right)^2$ mean?

[formal ermine]

I was thinking just like the regular cartesian product

but then the context doesn't make sense

namely that it's contained in Fp^times

is it the ideal of Fp^times squared?

if that makes sense?

[woven obsidian]

The cross just means it's the group of units, i.e. all nonzero elements

[formal ermine]

yes, I know

[woven obsidian]

What's the context?

[formal ermine]

definition of a quadratic residue

[woven obsidian]

Ah

Yeah then it's probably just all squares of elements in F_p^*

It's not an ideal though

[next obsidian]

Yeah it is (when p = 2)

Oh wait I guess it isn’t an ideal even then

Kekw

[formal ermine]

yeah but what does the notation imply like

it's not $\bF_p^\times \times \bF_p^\times$

[next obsidian]

It’s literally

Apply ^2

To every element of F_p^x

It’s like when you write down HK < G for subgroups

It’s like F_p^x•F_p^x

[formal ermine]

yeah so like with the ideal thing, maybe I worded that badly lol

thanks

btw does toric geometry go in #algebraic-geometry?

[rustic crown]

compare it to the notation 2Z inside Z

it not ideal tho, since stuff is abelian group

[formal ermine]

I was trying to make an analogy or something like that

[rustic crown]

that included all pairs >.< (not just the diagonal ones)

[next obsidian]

Fuck

I knew something was off…

[white oxide]

Why would $x$ not be a generator of $(\mathbb{Z}_3[x]/<x^3 + 2x + 2>)^$? $|\mathbb{Z}_3[x]/<x^3 + 2x + 2>|$ = 8, so $|(\mathbb{Z}_3[x]/<x^3 + 2x + 2>)| = 7$, and therefore every nonzero element in $\mathbb{Z}_3[x]/<x^3 + 2x + 2>$ is a generator, and clearly $x$ is nonzero

[cloud walrus]

okeyokay

[rustic crown]

you mean Z_2 or 27?

[white oxide]

wait wut

[glossy crag]

your quotient ring has size 3^3=27, not 8

[white oxide]

ohh right

3^3

oops

thx

[formal ermine]

"a lattice N is a Z-mod isomorphic to Z^n for some n. define N_Q through Q tensor N iso Q^n"

what exactly are we defining N_Q as?

[elder wave]

Q tensor N?

[rustic crown]

N ⊗_Z Q

[elder wave]

[rustic crown]

[formal ermine]

it doesn't say "as" but "through" though

that's why I'm confused

[elder wave]

how did you do that symbol det

[rustic crown]

unicode u+2297

i memorized 2295 and 2297

⊕, ⊗

[elder wave]

is 2297 direct sum

ah

[rustic crown]

5 is sum

7 is prod

[chilly ocean]

what's 2296 then

[rustic crown]

⊖

[chilly ocean]

[rustic crown]

i should memorize other useful ones, like wedge, a few greek letters

[formal ermine]

wait what key combination do you have to press, I'm trying alt + 2296 (numpad) and it just gives me ¨

[rustic crown]

i'm using gnome, for that it's <ctrl+shift+u><number><space>

Γ Δ

393, 394

[glossy crag]

yet another reason to switch to Linux

man i oughtta do it

[dim widget]

prepositions are hard for 🇩🇪 's 😢

[rustic crown]

pretty sure you can as easily do it elsewhere >.<

there is also a cute graphical application

𝔭

[elder wave]

¨

hmm

[elder wave]

ë è

for me

[formal ermine]

binbows gang

[rustic crown]

these are in hex ig

[elder wave]

⊗

[rustic crown]

[formal ermine]

how did you do it

[south patrol]

Fr

[delicate orchid]

○_○

[formal ermine]

⊗⊕

got it

had to edit something in my registry

[delicate orchid]

just (x) (+) ;lollollo

[elder wave]

Huh

What did you do

[teal vessel]

is it sufficient to say that the dihedral group of order 24 and the S_4 are not isomorphic because the highest order of an element of the dihedral group is 12, which is far higher than any member of S_4? knowing that isomorphisms imply that |x|=|f(x)| for all x.

[rustic crown]

yee

[south patrol]

Noice

Feel like this is probably the quickest way tbh lol

[uncut cloud]

Another quick way to see it is to count the number of order 2 elements.
In S_4 they are 8 while in D_12 they are 13

[south patrol]

Ok I hope I'm not gonna be told to go to #elementary-number-theory lmao but proving that if $\tau := \sum_{n \in \mathbb F_p} \zeta_p^{n^2}$ then $\tau^2 = (-1/p)p$ is hard right lol

[chilly ocean]

go to #elementary-number-theory

[cloud walrus]

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

[south patrol]

LOL

Idk cause like I was doing an exercise and it just told us to do this calculation without hints lol

I saw proof a while back but have forgotten

But like lol checking I'm not being dumb for being stuck

/ any hints

[uncut cloud]

Indeed it is not that easy. If you write $\tau^2$ brutally, what do you have ? And wich terms can you erase ?

[cloud walrus]

ChristopherNcL

[south patrol]

Well I just get like $\sum_{i,j} \zeta_{p}^{i^2 + j^2}$ - we could alternatively write this as $\sum_{r=0}^{p-1} n_{r} \zeta_p^r$ where $n_r = # { (i,j) : i^2 + j^2 = r}$ and if $(-1/p)=1$ it's not too hard to see $n_r = \begin{cases} 2p-1 & r = 0 \ p-1 & \text{otherwise} \end{cases}$ which gives the result, but it's the $(-1/p)=-1$ case that intrigues me

[cloud walrus]

potato

[south patrol]

Hm well I suppose n_r is constant for all r not 0 though still (hm not sure)

[uncut cloud]

you are absolutly right at the beginning. But indeed you should express n_r in another way

[south patrol]

Hm okay sure thanks

[glossy crag]

I don't get why L=K(zeta). Since {0} \cup {zeta^j : 0\leq j<q^n-1} is a representative system for o_L/m_L (valuation ring/ideal), i see how every element of L is a rational expression in zeta with coefficients in o_L, but that's not the same as being a K-rational expression. Is there some trivial way of seeing zeta has degree n over K?

[west sinew]

What exactly is a sparse vector space (as opposed to just a vector space)? Learning to use SAGE, and it's telling me that something is a 'Sparse vector space of degree 3 and dimension 1 over Rational Field'.

[coral spindle]

I've never heard of sparse vector spaces in the context of algebra. This sounds like a computational tool. People might know in #computing-software

[west sinew]

ah alright I'll repost there