groups-rings-fields

[wraith cargo]

this

essentially it's an element larger then every other element

in some sense

[novel parrot]

gotcha

So in showing that the chain C has an upper bound in S

They are saying that the union of all sets belonging to C can be an UB. In the case where C was finite (C had n sets per say), C has a largest element trivially right

[dusty verge]

The entire module is a spanning set no?

Also, I've always heard in Zorn's lemma that the maximal element just wasn't smaller than anything else. The thing you're quoting seems quite a bit stronger than that

I'm pretty sure that quoting of Zorn's is wrong

Especially because they would just call it a maximum element if that were true

[dusty verge]

Yeah, your reading of this bit is correct. Also, I'll note that this definitely shows the phrasing of Zorn's lemma above was wrong - the maximal element is only comparable to linearly independent sets that happen to have the same elements as it

[novel parrot]

When we union two independent sets do we get an independent set?

[dusty verge]

Only if the subspaces they span are disjoint (except at zero)

Or to put into actual stuff and not just math words

(0, 1) and (1, 0) is a basis for R^2 but so is (0, 2) and (2, 0), but obviously their union is not independent

[novel parrot]

i see

[dusty verge]

It's best to think of this stuff in finite cases and see if it makes sense there usually

[novel parrot]

Yeah, i understand the finite case but with a different proof that doesnt use zorns lemma

[dusty verge]

A good skill to have is to reduce complicated abstract shit into simpler notions, but only if you can figure out when that does and doesn't work

Here's the way to look at the proof with Zorns

Its saying that since you can define a basis for each subspace of the thing, you can kind of stitch these bases together in such a way that you can define one for the whole thing

Let me go on an aside real quick

Okay, do you know how you can think of vectors as like (1, 0, 0) right?

This is because of bases. What the (1, 0, 0) notation is really saying is that the vector is that particular combination of basis elements

That means a good way to visualize bases (here at least) isn't as the individual elements, but as their entire spanning set, since that's kind of what they represent

[novel parrot]

yes

[dusty verge]

So these chains of bases are really just chains of subspaces of increasing dimension

Zorn's lemma is just brought in to do the heavy lifting in saying that the dimension of the entire space is well defined

But it's really just a formalization of the intuitive notion that if you have an increasing set of stuff, there's a biggest thing

(which is wrong, it's only maximal not biggest, but the intuition doesn't have to be precise, so long as you make sure to be precise when actually applying it)

[novel parrot]

its not the biggest because not everything is comparable in a partially ordered set right

[wraith cargo]

yes

[novel parrot]

got it

[untold cloud]

Hi, guys, in the category language, we define object of field extension of k is a non-zero field homomorphism f: k\to E, then for the degree of this extension, is it true that we should define the extension degree as dimension of E as f(k)-vector space?

[dusty verge]

Yeah pretty much

[tulip hawk]

But it is incompatible with bilinearilty of tensor

[frigid lark]

Wdym?

[dusty verge]

Hmm. That does make sense, intuitively, but I'm not sure as to whether it works

[frigid lark]

I would intuitively think it would be defined by the exact sequence $$0\to k \to E$$

[cloud walrus]

Parrot Tea

[frigid lark]

I.e. f is an embedding

But maybe I'm wrong

[dusty verge]

No that sounds right

Nope it's wrong. Field homomorphisms are always embedding

Every sequence is exact

[wraith cargo]

In corollary 1.4.24 do they mean that B is an integral closure of A in L?

[dusty verge]

Wikipedia says it's the dimension of E as a K vector space. not sure what you mean by f(k) vector space

[frigid lark]

Image of K

[dusty verge]

Oh right, then yes

[dusty verge]

I think they're the same right? Since the integral elements of K should be the same as those of A

I think the reason they use K is just because technically K is contained in L, but A isn't, even though it totally is

[wraith cargo]

Ah wait yeah

every UFD is integrally closed

[dusty verge]

Hmm sounds like I was right by accident

Oops

[wraith cargo]

see this confuses me because in the proposition before they do say integral closure of A in L

[dusty verge]

Have you read through the proof?

It might be that they show that the close of K is that of A in it or something

[untold cloud]

Thanks! what i am confused about is how we say the tower law, f: k\to E, g: E\to L be field extensions, then for the tower law, we should say [L: gf(k)] = [L: g(E)][g(E):gf(k)]?

[dusty verge]

That makes sense

Remember these are vector spaces and not modules, so you don't have to drive a spike into the back of your skull to understand them

[untold cloud]

Thanks!

[dim widget]

It is just a typo. The integral closure of K in L would be L by any reasonable definition.

[fiery berry]

can someone give me a direction on how to approach this? I haven't done algebra in a while and im really bothered that i can't solve this

i thought of something like Z^2 x Z/gcd(a,b)Z but i dont know, its probably wrong

[dusty verge]

Do you have ftofgag?

Or crt?

[formal ermine]

ftofgag ong

[fiery berry]

idk what that stands for but im assuming thats like structure theorem for finite abelian groups or smth?

[dusty verge]

It even sounds like you're gagging when you say it :)

Fundamental theorem of finitely generated abelian groups

Good old ftofgag

[agile burrow]

a special case of the ftofgmoapid

[dusty verge]

What's the a stand for?

[agile burrow]

a

[south patrol]

lol

[dusty verge]

Lmao

Fundamental theorem of finitely general modules over a principle idea domain?

[south patrol]

What's funny is like when I read jt out loud I just say fundamental theorem of finitely generated modules over a PID

Forgetting that PID is itself an initialism

[dusty verge]

Still, doesn't have the same ring to it as good ol 'gag

Yeah I had to force myself to write out pid

[dusty verge]

What's the subscript Z mean here? Just that it's a subgroup pf Z?

[fiery berry]

not sure, but it should be that the coefficients of the freely generated module are from Z

[dim widget]

ong ofg

[pastel cliff]

Topos_Theory_E-Girl

[dusty verge]

It's not always Z^2, since what if a = b?

Unless it'll be iso still at that point

Do you have any recently introduced theorems that seem suspiciously similar?

[pastel cliff]

today i am become algebra, destroyer of grades

[dusty verge]

Oh yeah? Name every finite group up to simple subgroups

Or quotients or whatever idfk

[pastel cliff]

name every group

[dusty verge]

I liked it more as grup

[fiery berry]

nah this is on the prerequisite exercises sheet for a homologial algebra class, but i think i just got another idea now, if i get it i'll post solutions here

[dusty verge]

Oh in that case, go read up on ftofgag. Good old meta analysis says that's the way to do it

[novel parrot]

What is the cokernel of a matrix?

[elder wave]

Codomain modulo image of it

[novel parrot]

we can represent homomorphism as matrix is some cases, what would the cokernel look like

[novel parrot]

how do we compute it?

is the coker(A) just another matrix?

R^s / Imf how can this be represented as matrix

[elder wave]

no coker(A) is R^s/im(f)

the module

[novel parrot]

yes

Coker(A) is not a matrix?

[elder wave]

"such that M is isomorphic to coker(f)"

M is a module

[dusty verge]

The kernel of the inverse, but generalized to what Timo said

Since not every matrix is invertible

[elder wave]

A is just the matrix representation of the map f

[novel parrot]

Okay, so any finitely genearted module over a noetherian ring is isomorphic to the cokernel of some matrix

[dusty verge]

It's almost like multiplication of subspaces

[elder wave]

finitely generated is equivalent to finitely presented over noetherian rings, yes

[novel parrot]

really

[dusty verge]

If M: A to B is a matrix, then M/ker(A) is a subspace of A right?

[elder wave]

finite presentation (isomorphic to the cokernel of a matrix) always implies finitely generated

and here you are showing the reverse direction over noetherian rings

[dusty verge]

It's the part that doesn't get squished together by the matrix

The cokernel is the part of the codomain you have to squish together if you wanna invert your matrix

This helps compute it cause of degrees

[novel parrot]

Why does cokernel being 0 mean that the matrix is unit?

[dusty verge]

Because you can invert it

It's the first isomorphism theorem

It means that you have a surjective linear map.

[novel parrot]

Yes

Why injective?

[dusty verge]

Hmm, not sure

Maybe they're talking about it as if its from X/mod ker(A)

I.e. it's a unit via first iso thm

[novel parrot]

dont understand that

[dusty verge]

Oh I see, so think about it this way

[novel parrot]

We are talking about the map from R^s -> R^r right

[dusty verge]

You can make an inverse that sends it to a subspace of the domain, right?

Or rather, do you see how the first isomorphism theorem applies?

[novel parrot]

yes, R^r/ Kerf = R^s

Since the co kernel is {0}

[dusty verge]

So it's a unit cause you can invert it to send back to a subspace of dimension Dim(R^r/kerf)

[novel parrot]

So are you saying that the map f is an isomorphism

[dusty verge]

Wait no that doesn't do it

Nah, I'm saying it's an isomorphism between R^r/kef and R^s

[novel parrot]

but the text is saying that when coker = 0 then f is an isomorphism

[dusty verge]

Are you talking about the exact sequence?

In the first one?

[novel parrot]

nah

i understand the proof

im asking about the second screenshot

[elder wave]

where does it say that

[dusty verge]

You've made a classic blunder

[novel parrot]

😓

[dusty verge]

The implication arrow points the other way

It's just saying the cokernal is zero for infectives

[novel parrot]

A is unit => coker is 0

right?

[dusty verge]

Yeah

[novel parrot]

oof

[elder wave]

yes but the other way around doesn't really make sense

dimension wise even

[dusty verge]

It does. It's just saying injective functions are invertible

[elder wave]

?

[dusty verge]

Unless you mean the arrow pointing backwards

[elder wave]

yes

[dusty verge]

Oh, yeah, that's cause it's wrong

[elder wave]

it'd have to be a square matrix in the first place

[novel parrot]

A is unit => coker is 0, this is correct right? Since A is unit means f is isomorphism

[dusty verge]

Yeah, should be true. Although it doesn't have to be an iso I think

Ignore that

The second half, that's just me being dumb

It's completely correct

[novel parrot]

ok nice

[pastel cliff]

what is the overline denoting here

[delicate orchid]

looks like an equivalence class

[elder wave]

it is indeed one

[dusty verge]

It's the equivalence class of m in R/I

Oops. In M /IM

[elder wave]

there's a nicer way of showing this though

[pastel cliff]

with a sequence right

this is in KC notes

[elder wave]

Yes, using the exact sequence and right exactness

okay but that version is more elementary tbf

[pastel cliff]

this class technically hasnt mentioned sequences lool

only knw them from alg. topo

[elder wave]

crime

[pastel cliff]

[dim widget]

What does mean exactly? Is it like "I am dying/that's awful?"

^Yes mods this is technically an analysis question but I'm asking it here.

[pastel cliff]

i remember when i first joined i had to ask what sully meant

then i got sullied for it

[sly rain]

Let $ \pi : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} $ be epimorphism and $\sigma: Sub(\mathbb{Z}/n\mathbb{Z}) \rightarrow Sub_{m\mathbb{Z}}(\mathbb{Z}) $ with $ \bar{H} \mapsto \pi^{-1}(\bar{H})$.

Where the first set is the set of subsets of $Z/nZ$ and the second set is the set of subsets containing $nZ$.

I want prove that $\sigma$ is a bijection and I already established that both sets have the same cardinality , which is equal to the amount of the unique positive divisors of $n$.

I want to show its surjective so im picking arbitrary $kZ$ (where k is a divisor of n) so $\sigma(\bar(H) = \pi^{-1}(\bar{H}) = kZ$.

However I feel like im missing something or doing something which im not allowed to because its so simple. I wrote down an example for Z/4Z and thought about what element maps to what element, but im not using anythig of that sort here.

[cloud walrus]

aabb

[dusty verge]

Petition to get Mike wazowski as a reaction image please

[elder wave]

pretty much

[delicate orchid]

I absolutely do not believe those two sets are the same size, there are an infinite number of different subsets of Z that contain nZ, and the powerset of Z/nZ is of size 2^n, which is finite

[south patrol]

It feels like

[elder wave]

Doesn't look like it's passing

[south patrol]

The monke thing feels like someone going bruuuuh and squinting like wtf did i just hear

maybe i have misinterpreted it this whole time

[pastel cliff]

me when i thimcs

[sly rain]

oh well lol, let me thing about it again

[dim widget]

I kind of thought that was a use case for

[delicate orchid]

like in your Z/4Z example, you have 4Z u {5}, 4Z u {6}, 4Z u {7}, 4Z u {9}, ...

[dusty verge]

I feel like people are sullying it as a show of support mr mod guy

[rotund aurora]

that's what I understand too

[sly rain]

It is supposed to say subgroups in both cases, translation error. I was careless here

[delicate orchid]

ok now I recognise it as the correspondence theorem

[dusty verge]

I know it's a nonstandard use but please consider the context

[elder wave]

idk there isn't a single green mark supporting it

[pastel cliff]

dont "mr mod guy" timo

https://tenor.com/view/timi-gif-20112884

[elder wave]

okay lets keep this channel on topic now

my bad also

[dusty verge]

I'm new here

[elder wave]

Hello new here, i'm Timo

[pastel cliff]

okay lets keep this channel on topic now

[elder wave]

yes

[dusty verge]

I'm gonna pretend that the green checkmark is Mike winking

[pastel cliff]

the inverse map here is a result of first iso righ

hmmm noo

or maybe it iss

ii should think before speaking

[floral imp]

can anyone help with logic

[delicate orchid]

what's the sequence btw

I can't be bothered reading

[floral imp]

i love just seeing people cast spells on each other

[delicate orchid]

never do this

[floral imp]

with these fancy words

[summer path]

#proofs-and-logic

[lusty marlin]

Not the right channel

[rotund aurora]

ask Godel

[pastel cliff]

he's not wrong

[summer path]

What's wrong with aa today

[formal ermine]

@waxen lichen

fuck

<@&268886789983436800>

[delicate orchid]

wahh wahh

[pastel cliff]

IM -> M -> M/IM

[delicate orchid]

oh yeah tensoring an R-module with R is just the dude lol

[formal ermine]

my favorite ses is -> a -> A -> A/a ->

[delicate orchid]

that's all of them

[pastel cliff]

mine is 0 -> 0

[floral imp]

who is godel

[pastel cliff]

that's him

[floral imp]

is he good at logic?

?

[lusty marlin]

Yes

[formal ermine]

Pending G+

mfw

[dusty verge]

Don't be afraid of saying something wrong imo

Everyone's wrong until they're right

[formal ermine]

every time I say something wrong I get shouted at by timo

[pastel cliff]

that's what i use this discord for

[formal ermine]

"common illu L"

[floral imp]

whats pending g+

[pastel cliff]

to ask the things im too scared to ask my profs

[formal ermine]

I've never asked a prof anything

I just always use this discord

or ask the ta

or MSE

hey

mse is sometimes useful

[dusty verge]

That's lame. It creates an unrealistic perception where it looks like everyone else is right the first time, and then you have to look like you're right too or else you're behind

[floral imp]

what dies that mean?

[formal ermine]

some of my questions have yet to be answered though

[dusty verge]

Being wrong is just an opportunity to become right

[karmic moat]

i am terrified of asking my professors certain questions

[floral imp]

?

[dusty verge]

The fuck is an mse?

[lusty marlin]

Look at your roles.

[karmic moat]

like if im stuck on a problem that i feel like should be really straightforward they will look at me and be like "damn this guy stupid"

[lusty marlin]

Mathematics stack exchange

[formal ermine]

math stack exchange

[karmic moat]

even tho they are required to help me with the question

[summer path]

I mean I agree but I think I've used it like a total of 3 times and all of them were on a self study where my grad student mentor didn't know how to answer it

[floral imp]

what does that mean

what is pending g+

role

[summer path]

Bruh

[formal ermine]

https://math.stackexchange.com/q/4481781/943349 if someone wants to answer this

I'm still looking for the solution

[dusty verge]

Oh fuck that. Remember what I said about a culture of having to look like you're always right?

[karmic moat]

pending g+ = you applied for the graduate+ role and waiting for it to be accepted i think

[formal ermine]

nah

[floral imp]

oh

ok

[formal ermine]

you just clicked the "graduate+" thing

[karmic moat]

wait what is pending g+ then lol

wait yeah isn't that what i said

[formal ermine]

nah

[karmic moat]

applied but waiting to be accepted

[formal ermine]

you have to separately apply

[karmic moat]

oh

[dusty verge]

No one will think that. They'll think you missed something obvious the way they have a billion times before

[formal ermine]

I still don't know what classifies as g+

[karmic moat]

weird, last time i tried they had a built-in application thing

[dim widget]

I found this very unclear tbf

[karmic moat]

like u clicked the role and there was a form that automatically popped up

[formal ermine]

I applied but got denied

[karmic moat]

yea ik

as the story goes

[summer path]

You just gotta accept that you know nothing sometimes and just ask them for help, or just ask them the random question you've got in your head and allow intrusive thoughts to win

[delicate orchid]

welcome to the advanced channels

[dusty verge]

If you don't ask a stupid question .you get stuck with the question

It stays within you

[formal ermine]

I'm talking about this

[karmic moat]

a math professor, in a lecture, says "this proof is trivial," leaves the classroom for 15 minutes, and comes back and says "yeah it's trivial"

[formal ermine]

[karmic moat]

oh i meant from this

[dusty verge]

But just be wrong sometimes instead. It's freeing

[delicate orchid]

oh trust me, I'm nothing but wrong

[dusty verge]

Wait but then that sentence is wrong

[summer path]

Be wrong intentionally to lose ego

[dusty verge]

Don't paradox yourself in a math discord come on

[delicate orchid]

there is no paradox :thirdeyeopen:

[formal ermine]

question time

[delicate orchid]

if I see the word "galois" my fist is going through my monitor

[formal ermine]

what is an intuition behind the definition of a sheaf of rings on a prime spectrum

[dusty verge]

What's a sheaf?

[karmic moat]

a collection of stalks

[dusty verge]

What's a prime spectrum?

[formal ermine]

[delicate orchid]

spec(R) I presume

anyway #algebraic-geometry

[formal ermine]

I am scared of using #algebraic-geometry

[delicate orchid]

always makes me laugh that comm-alg is in there

[dusty verge]

What's spec(R)?

[karmic moat]

im terrified of using #diff-geo-diff-top

[delicate orchid]

yeah let me ask my tensor product question right after the 500 message long O_x-module discussion

[delicate orchid]

the set of prime ideals of R

[karmic moat]

sometimes i feel like i should ask my topology questions in #geometry-and-trigonometry because they are far more insignificant than the questions asked in #diff-geo-diff-top

[pastel cliff]

galios

[delicate orchid]

#point-set-topology

[karmic moat]

yeah that too

very terrified

anyways, can't believe galois died at like, what, 21?

[delicate orchid]

topology-alg-top isn't that bad

[karmic moat]

because of a silly duel

[summer path]

GaIois

[pastel cliff]

Galois

[dusty verge]

Can you give the definition exactly, please?

[formal ermine]

this one time I didn't know whether to ask my question in #geometry-and-trigonometry or #algebraic-geometry

[karmic moat]

imagine asking a question in one of the advanced questions and getting pointed to #geometry-and-trigonometry

that would be so insanely humbling

[delicate orchid]

this keith conrad guy is pretty swag

[karmic moat]

i love conrad's lecture notes

[dusty verge]

I mean geometry is scary

[summer path]

That's like once every 10 questions in that channel

[dusty verge]

But maybe not trigonometry

[formal ermine]

,, \mathcal{O}(U) = \set{s : U \to \coprod_{\mathfrak{p} \in U} A_{\mathfrak{p}} | s(\mathfrak{p}) \in A_{\mathfrak{p}} ~\text{and $s$ is locally a quotient of elements of $A$}}

[formal ermine]

ong

[dusty verge]

I wonder if you can actually do cool shit with trig at all

[wraith cargo]

fourier analysis is a big field

[summer path]

Keith conrad's notes are so nice for reviewing or catching up on stuff

[wraith cargo]

and that gives into harmonic analysis

[formal ermine]

IRONY

[elder wave]

Just be right

[formal ermine]

you ignored me in discussy 😭

[wraith cargo]

ello

[delicate orchid]

oh so it's just a coproduct :weed: coproducts are like adding things :blazing8s: so this is basically like a sum of functions :blazed2the9s:

[dusty verge]

Fuck oof I forgot forrier

[wraith cargo]

ahhh yeah I saw I was pinged but I couldn't find where the ping was

[formal ermine]

what's your battle tag

overwatch

[rotund aurora]

Holomorphic functions on cylinders

[formal ermine]

it's disjoint union or something idk

not coproduct

or I don't even know at this point

the text doesn't explain

[delicate orchid]

the disjoint union is the coproduct in set

[formal ermine]

YES

I KNOW

[delicate orchid]

these are rings

[dusty verge]

What's A_p in that definition?

[delicate orchid]

the corproduct in ring is not disjoint union

[dusty verge]

And what's a local quotient?

[formal ermine]

localization

[delicate orchid]

localisation of A away from p

[formal ermine]

fuck do I know at this point

what is it

[delicate orchid]

no fuckin clue

[formal ermine]

tensor?

[dusty verge]

What does that mean?

[formal ermine]

read a comm alg book

[delicate orchid]

rude

a localisation is like

[dusty verge]

Read my question

[formal ermine]

localization*

[delicate orchid]

you add some fractions in

[dusty verge]

Did I ask for a bedtime story?

[karmic moat]

me when

localize R_S is fractions of form r/s

[pastel cliff]

abstract chillgebra moment

[delicate orchid]

I'd take a bed time story rn

"The coproduct of nonzero rings can be the zero ring; in particular, this happens whenever the factors have relatively prime characteristic"

............... lol?

[formal ermine]

omw asking in #category-theory 🏃

[delicate orchid]

it's analogous to free groups according to wikipedia.co.uk

[wraith cargo]

I believe this has to do with the commutative diagram associated with the coproduct
So it can only factor through the 0 ring

[delicate orchid]

this makes sense

yeah wouldn't the coproduct need to have characteristic equal to the gcd of the characteristics of the rings in it

hence coprime char => trivial

[summer path]

Category theory scary

[dusty verge]

So what does "locally a quotient away from elements" mean?

[formal ermine]

lemme send the part where this stuff was defined

[dusty verge]

Afraid of a little arrows?

[delicate orchid]

take ur ring R, and a multiplicatively closed set S, then the localisation at S is {r/s : r \in R, s \in S}

[formal ermine]

akchkckhkchually

[delicate orchid]

shut the fuck up R is an integral domain

[dusty verge]

Okay, it's fractions but only nearby?

[delicate orchid]

quotenited by the equivalence relation r/s ~ r'/s' <=> rs' = r's

[formal ermine]

akkchkchually

[delicate orchid]

this equivalence relation is messier if R is not an integral domain but I don't care

[pastel cliff]

all rings are integral domains

[karmic moat]

lame localization vs based quotient field

[dusty verge]

r's what?
What does r own?

[formal ermine]

this is what rust does to a mf

[delicate orchid]

it's called the localisation because the result is a local ring when u do it "away" from a prime ideal

at least that's my justification

by "away from a prime ideal" I mean you take S to be everything outside that ideal in R

[dusty verge]

I think I've seen something similar before

[formal ermine]

algebraic number theory is sooo fun but my homework is just

[delicate orchid]

for example, Z localised away from (2) is the subset of fractions in Q who's denominator is coprime to 2

[pastel cliff]

last q abt this but

that last part is a bit confusing

[dusty verge]

Anyone familiar with leavitt path algebras?

[pastel cliff]

what's he doing with the elementary tensr

[delicate orchid]

https://tenor.com/view/its-kinda-conchfusing-conchfusing-its-as-shrimple-as-that-shrimple-its-actually-very-clampicated-gif-26109078

[pastel cliff]

yes lads

[delicate orchid]

which bit specifically

[formal ermine]

a field extension is integral iff algebraic right

[pastel cliff]

sum of tensors has that form implies all tensors are 1 x m

[summer path]

Relatable

But at least it's fun

[formal ermine]

https://cdn.discordapp.com/attachments/929807268994744340/1093978547565105182/unknown.gif

[delicate orchid]

you can write r \otimes m as 1 \otimes rm for any r and any m - apparently - so we write our tensors in our sum as 1 \otimes m_i for some m_i and then just use linearity to combine them

[chilly radish]

[chilly radish]

Yes you can divide through by the leading coefficient always

[formal ermine]

ye

ok epic

[pastel cliff]

forgor abt linearity

https://tenor.com/view/kaboom-boom-nuke-nuclear-bomb-gif-20734003

me when i forgor

[delicate orchid]

that's the WHOLE POINT

[wraith cargo]

this was my reaction when I learned how sheaves can be used in complex analysis

[pastel cliff]

shhh

[formal ermine]

ong

[dusty verge]

The intuition is that the sheaf of rings is the set of functions that associate prime ideals with places that they're local

So it's probably used for taking about "where" a part of a ring is "localish"

Hmm, maybe not quite

[formal ermine]

@delicate orchid coproduct in ring is tensoring over Z

[dusty verge]

Let's see, so looking at Wikipedia, the closed sets are sets of prime ideals containing a given ideal

Which is analogous to prime factorizations, so the open sets are analogous to stuff not divided by a given prime yeah?

[dim widget]

Geometrically it's more like localizing "at" that prime ideal.

[formal ermine]

@dim widget are you a phd student

[dim widget]

Naur

I'm just a discord degen

I should be getting my PhD in that sometime soon though

[delicate orchid]

I got absolutely no knowledge of geometry so I'll take ur word for it

[delicate orchid]

nifty

[formal ermine]

what is a polynomial called whose resultant with its derivative is non zero? my nob prof calls them generic but apparently no one else calls them that. like I know you can just call them separable, but that feels wrong

[delicate orchid]

wonder if I can prove that in a single diagram

[dim widget]

You can't just call your prof a nob

It's rude

[formal ermine]

newton okounkov bodies

[elder wave]

bodies whom?

[dim widget]

I would call it "nonsingular"

[formal ermine]

your mum

[formal ermine]

hm

[dusty verge]

Okay, so what's a local quotient?

[formal ermine]

@dusty verge

[pastel cliff]

"graduate +"

[elder wave]

That role isn’t necessarily tied to the status as a graduate student

[dusty verge]

That doesn't define a local quotient there, does it?

[formal ermine]

what even are the requirements for the role

[formal ermine]

it does

"to be precise, ..."

[dusty verge]

NVM I saw my mistake

[dim widget]

Nice try FBI

[dusty verge]

They're quotient rings lmao

Or whatever

[formal ermine]

why does my prof have to use this really DISGUSTING background that is just hard to read text on

[dusty verge]

Wait, why does it say a, f in A?

Oh wait nvm

[summer path]

That doesn't even look that bad

Maybe except for the orange in the middle of the screen

[formal ermine]

oh

that wasn't my image

these are my profs notes

[summer path]

Oh

Never mind

Also that is one fat marker

[dim widget]

Cringe

[karmic moat]

blue light filter paper

[dim widget]

poor prof, imagine being german unironically

[formal ermine]

couldn't be me

[elder wave]

Real

[formal ermine]

it's really weird

I had the same prof in algebra 1

we did like

a fuck ton of stuff in one lecture

this is algebra 2

and we've barely done anything in a week

[elder wave]

Hat gemerkt dass ihr alle skill issues habt

[formal ermine]

schwöre aber

sind nur noch 4 leute in der vorlesung

ist eine connection zwischen klassischer algebraischer geometrie und konvexer geometrie

[elder wave]

That’s a good size

[formal ermine]

bin der einzige der weiß was eine varietät ist

[formal ermine]

we learnt like only two things in a whole week lol

[summer path]

[formal ermine]

resultant and symmetric polynomials

[dim widget]

I mean those probably take longer than that to understand well.

[formal ermine]

my favorite result is that the invariant ring is equal to the elementary polynomials

[dim widget]

A good thing to think about is what the invariant ring is over Z.

[formal ermine]

what

[dim widget]

Like invariants in Z[x_1, \dots, x_n] unless you already did that in class.

[formal ermine]

oh

I meant

,, \bC[x_1, \cdots, x_n]^{S_n} = \bC[e_1, \cdots, e_n]

who needs fields other than C anyway, amiright

[dim widget]

It is a beautiful result

[dusty verge]

Okay, I'm starting to get it

For a fixed p, the sheaf will give a set of fractions that are equal to any element of p for some neighborhood around it

Generalizing this to all of U, it gives the set of fractions that are equal to elements of U, for some neighborhood around that element

Basically you can think of different localizations as different ways to write the elements of a ring as fractions. Seems like sometimes it'll be the same, and sometimes different

[pastel cliff]

how trivial is that first result

[dusty verge]

This sheaf sends prime ideals to the places where it's the same, right?

[pastel cliff]

that that tensor is 0

[agile burrow]

tensoring with Q kills torsion elements, so each factor in the product is 0

[pastel cliff]

oh literally just bc Z has inverses in Q

[agile burrow]

that's right

[delicate orchid]

omg!! profinite completion!!

[pastel cliff]

what do those words mean

[delicate orchid]

wholesome 9001

[pastel cliff]

i love you 3000

[dusty verge]

Let p be a prime ideal. Then for each x in p, there's some fraction that it's equal to in some range around p. The sheaf sends all the ideals in U, to all those fractions, in all combinations

It'd be better to define it as a function from the disjoint union of the ideals tbh

Well, maybe not. You might not want that association later on

Yeah the sheaf is all the fractions that are equal to elements of p in U at p

[formal ermine]

heh?

[dusty verge]

The neighborhood bs is just to be more general

[pastel cliff]

another silly question - what does it mean for that tensor to be exact

[slim kayak]

What is a condition for the direct sum of images being equal to the image of an internal direct sum?

[agile burrow]

if you have an exact sequence of Z-modules (i.e. abelian groups), then tensoring the sequence with Q still yields an exact sequence

[delicate orchid]

preserves exact sequences, i.e. Q is flat as a Z-module 🙄

weird saying the tensoring is exact rather than just the module is flat but w/e

[dim widget]

It's a little hard to tell what you mean, but it seems like the condition is just that they have parwise 0 intersection.

[pastel cliff]

ohhh i didnt think abt it like a sequence

[dim widget]

Can mods edit our comments?

[formal ermine]

nah

[elder wave]

no

that'd be funny though

[delicate orchid]

I would be swiftly assassinated via implantation of FALSEHOODS

[formal ermine]

"it is clear from the local nature of the definition that O is a sheaf"

[slim kayak]

$\bigoplus f(M_{i}) = f(\bigoplus M_{i})$

[cloud walrus]

Kerr

[slim kayak]

I meant this

[dusty verge]

It is. A sheaf describes local properties, i.e. properties on open sets.

[pastel cliff]

a nonzero amount of trolling

[dusty verge]

Break it down by first looking at O(p) is. It's the set of fractions elements can be equal to at p

[agile burrow]

Lol I think it's interesting that some other books define the structure sheaf on the basic open sets, and then it's a nontrivial amount of work to show that this does actually form a sheaf

[dusty verge]

What are sheaves used for?

[delicate orchid]

making me cry

[dusty verge]

It can't be that hard, come on

I'm sure way easier stuff could make you cry

[pastel cliff]

the contradiction here then is that G x Q has a nonzeroo subring and is therefore non zeroo right

[slim kayak]

it holds trivially when f is injective i.e. the kernel is trivial but like projection from R^3 to R^2 with regards to the standard basis for example still preserves that sort of "splitting". In the general case I am not sure what condition would imply uniqueness of expressing any element as a sum of elements from f(M_i)

[dusty verge]

Oh after some googling about what sheaves are, think of it as the shape of the space of the fractions the things can be

[agile burrow]

I usually just think of functions on spaces, but go off ig

[delicate orchid]

I know exactly what a presheaf is that's enough for me thanks

[lethal dune]

which one's the first?

[dusty verge]

Well I'm still figuring this out, so let me know if I'm wrong, but it seems like it's breaking down global things like the set of continuous functions into local things, right?

[pastel cliff]

i got that ryu it's okay

[delicate orchid]

tensoring with Q

[lethal dune]

https://tenor.com/view/caveman-fixed-angry-beard-man-smash-gif-17065028

[dusty verge]

Like a sheaf takes the set of continuous functions from A to B and turns it into the set of sets of continuous functions from open sets of A to B?

[agile burrow]

Right, yeah. Sheaves are useful for formalizing how to study functions on a space where you have some notion of functions being defined by their local behavior

[dusty verge]

Or do these smaller functions need to be compatible?

[lethal dune]

In general torison free modules over PIDs are flat

(In fact over dedekind domains it's true)

[dusty verge]

Oh oh oh yeah that makes sense right

[formal ermine]

nonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexistnonflatmodulesdontexisteverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflateverymoduleisflat

[dusty verge]

Oh, so it is compatible continuous functions , you won't have an element of the sheaf that contains who functions equal to different things at the same place

Right?

[formal ermine]

non exact functors are cringe amiright

[delicate orchid]

u gotta stop doing this

[lethal dune]

flat is justice

[delicate orchid]

name one exact functor

[formal ermine]

LOCALIZATION

[delicate orchid]

he's good

[elder wave]

mate

[delicate orchid]

by jove he's good

[elder wave]

stop posting shit like this

[pastel cliff]

okay let's get this channel back on topic

make me mod already

[elder wave]

okay let's get this channel back no topic

small typo, i win

[delicate orchid]

I'm goofy!! I'm insane!!

[pastel cliff]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[agile burrow]

Right, but that's like the definition of a function lol. If you read the sheaf axioms, the notion of identity and gluing reflect the idea that if you can define functions locally in a compatible way, then you ought to be able to glue them into a function defined globally

[delicate orchid]

so chat can you help me compute a little group theory question

[formal ermine]

https://tenor.com/view/preposterousgrammaticalerror-gif-25846191

[agile burrow]

That's why the classical examples of sheaves are like smooth functions on a manifold and stuff

[delicate orchid]

actually stealing my gifs and then getting banned by posting them

[formal ermine]

what is an manifold?

[elder wave]

#point-set-topology

[delicate orchid]

locally homeomorphic to R^n :pack:

[agile burrow]

Honestly illu, you might benefit from learning some stuff about smooth manifolds before trying to read more Hartshorne

[formal ermine]

I don't want something that is locally homeo to R^n, I want a manifold!

I know the first couple of pages of a diff geo book

[agile burrow]

I mean, Zariski tangent space stuff would be very abstract and unmotivated

But who am I to stop you

walter

[delicate orchid]

is there a p-subgroup S of GL_p(F_p) such that for all x \in GL_p(F_p) - S, |S \cap xSx^{-1}| is coprime to p

[lethal dune]

no

[delicate orchid]

remains to be seen

[formal ermine]

obviously

[delicate orchid]

ok what is it then

[formal ermine]

it's really trivial

I'm too lazy to write it out

[delicate orchid]

nerds

[lethal dune]

feels true for p≥3 actually

[dusty verge]

Makes sense

[lethal dune]

ok wait

[lethal dune]

is it not true that N_G(P) > P for a P subgroup?

unless it's maximal?

[delicate orchid]

that sounds familiar

[dusty verge]

The intuition for your sheaf thing is that it maps out open sets as fractions that "continuously" deform into each other

Don't know why I responded to that

[novel parrot]

If they are R-linearly independant => Z-linear independant way. How do we do this?

[delicate orchid]

I think it's only true if G is also a p-group

it's used in the proof that every finite p-group is nilpotent iirc

[fleet pelican]

some argument about order and divisibility iirc

[dusty verge]

Okay, I understand ring sheafs

[lethal dune]

yes so doesn't it sound kinda impossible unless you take maximal p subgroup

but even then, the intersection is p-subgroup which might be nontrivial

[delicate orchid]

need to read more about these strongly embedded p-subgroups tbh

[agile burrow]

Here's a fun exercise: if you have a ringed space (i.e. a space X equipped with a sheaf of rings) and a global section f, show that the set of points where the germ of f is invertible is an open subset of X.

[dusty verge]

Say that two ideals are similar if their localization contains the same fractions. The sheaf of an open set consists of collections of ideals such that you can draw a path through the ideals such that each is similar to it's neighbor

[dusty verge]

What's a germ?

[formal ermine]

equivalence class of functions that agree around an open neighborhood

[agile burrow]

So sheaves assign, say functions, to open sets. It would be nice to talk about how functions behave at a point, but in general points won't be open sets in a topological space. The stalk over a point can be defined by taking all sections over all neighborhoods of that point and quotienting out by an equivalence relation that essentially sets two sections to be equal at that point if they agree on some neighborhood of the point. The equivalence class of a section is called the germ of the section at that point

Really some of the most important properties of sheaves arise from the interaction between sections and stalks so you should read more about it if you're interested

[dusty verge]

What's a global section?

[agile burrow]

If F is a sheaf on X, then for any open set U, F(U) is called the sections over U. A global section is a section over X, so an element of F(X)

[novel parrot]

What does it mean that N is generated by the rows of the matrix?

The image of the map that A represents?

[delicate orchid]

I think so yeah

<x^2, x^3, x^5+x>\oplus R \oplus <x> ig

[novel parrot]

N is guaranteed to be finitely generated. Why will it have a basis?

$<x^2, x^3, x^5+x>\oplus R \oplus <x> ig$

[cloud walrus]

ActiveChapter

[delicate orchid]

Q[x] is a pid and submodules of free modules over pids are free

[novel parrot]

gotcha

[pastel cliff]

i am no longer algebra

bros...

[dim widget]

How did it go?

[celest furnace]

Wdym sebbb?

[pastel cliff]

it's okay

we're so back

https://tenor.com/view/were-back-its-over-gif-24400541

[dusty verge]

So a global section is just an element of the sheaf?

[agile burrow]

I don't understand what you mean. A sheaf assigns sections to each open set of the underlying space

[dusty verge]

Oh okay, makes sense

[dim widget]

It's best not to get into the habit of thinking of sheaves as a set of sections, the whole point in inventing them was to move beyond that perspective.

[dusty verge]

Hmm, how do you mean?

[dim widget]

I am too tired to explain but basically while you can for the first sheaves you encounter think of them as being collections of functions valued in some big space called the "espace etale" eventually you might encounter sheaves for which this is not true, so it's better to stick to the definition as a functor, which is maximally flexible.

[pastel cliff]

halliday is that yuor name or is it bc of the RP1 character

[dusty verge]

No

Its cause I needed a username so I stole it from the author of a book nearby

[pastel cliff]

ah

every element has that form bc of the same property as earlier right

in R/I x R/J we have that r' x r'' = r'r'' x 1 and then torsion...?

[dusty verge]

The germ here being the functions that are equivalent at said points?

Wait no that can't be it, oh invertible as in like multiplicatively?

[agile burrow]

The germ refers to the equivalence class, yes.

[pastel cliff]

kid named germ:

[agile burrow]

So if you have a ringed space, the stalk over a point also has the structure of a ring. I mean that the germ is an invertible element of this ring

[pastel cliff]

wait not torsiono

just the fact that it's a ring so r'r'' = soome other element

[dusty verge]

What's multiplication for this ring?

Or addition, even

Oh the ideals are prime, so is it just the rings operations?

Yeah of course it is, okay cool

[agile burrow]

I think maybe it's worth reading about these things in a book, I'm not sure where these ideals are coming from and I'm afraid you're still talking about the specific case of the structure sheaf on the spectrum of a ring

[dusty verge]

I am

[agile burrow]

If that's the case, that you're asking specifically about Spec R, then it is true that the stalk over a point p is isomorphic to R_p, the localization of R at p

[dusty verge]

Yeah, which would make your question not make sense there

Right?

[agile burrow]

No, it also holds for Spec R

It holds for any ringed space

[dusty verge]

It's going to be invertible nowhere, right?

[agile burrow]

Why would that be true?

[dusty verge]

I think I'm wrong yeah

I was going off this, and thought that f not in q implies that a/f can't be invertible, but that can't be true cause otherwise you couldn't have identity right?

[pastel cliff]

waltuh is this right

im so tired of looking at tensors

i wanna do galois stuff already

[agile burrow]

Yeah, that looks right to me

Sorry, I didn't see it on the first pass

[dusty verge]

For the structure on the spectrum of a ring, I'm pretty sure I have it. Let me look up the more general definitions

[pastel cliff]

just duoble checking, this ok?

[dusty verge]

Is this a locally ringed space, or a ringed space in general. If it's the second, what would the codomain of a structure be?

Would it just be elements of the ring?

No it wouldn't be, would it?

[pastel cliff]

Suppose $M_1$ and $M_2$ are finitely generated $R$-modules. Show that $M_1 \otimes_R M_2$ is also finitely generated.

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

is it enough to just say like

M1 and M2 are finitely generated

so there will only be a finite amount of elementary tensors

[celest furnace]

Any hints on this exercise? I've tried quite a few things. I was able to get it if deg f >= deg g, since you can prove in that case that deg fg <= deg h - 1, but that conjecture is NOT true in the case where deg g > deg f (take, say, f = x^2, and g = (x-1)^3). It seems like this should be really obvious, but I'm just not seeing it. I also wonder if there is a smarter choice of A,B, and C.

[pastel cliff]

https://tenor.com/view/ryan-gosling-its-over-over-blade-runner2049-blade-runner-gif-24854081

[dusty verge]

Ignore all previous questions, I understand enough to tackle the question now

Sorry for the barrage of questions, btw, I'm new to this subject

[agile burrow]

This essentially works, you can formalize it though. M being finitely generated is equivalent to there being a surjection f: R^n -> M for some n. In particular, you have surjections R^m -> M_1 and R^n -> M_2. Can you construct a surjection from a free module onto M_1 \otimes M_2?

[dim widget]

Tfw you were the shadow of a waxwing slain by feigned remoteness in the windowpane.

[pastel cliff]

i cannot comprehend that sequence of words but

it's okay we're fucking back

[pastel cliff]

this is what i had

[dim widget]

Yeah that's fine, you are saying that a tensor product of surjections is a surjection which is true.

[pastel cliff]

last check but does this look right

[delicate orchid]

actually making me think about the cross product

[pastel cliff]

https://tenor.com/view/pokemon-sadism-suffering-happy-pin-kiv-gif-19470753

[dim widget]

How are you ordering the basis lol

Are you doing e_i \otimes e_i first

you are perverse

[formal ermine]

hey seb

[delicate orchid]

count in base 3 u noob

e_1 o e_1, e_1 o e_2, e_1 o e_3, e_2 o e_1, etc

[formal ermine]

eating ben and jerry's brownie dough ice cream right now but there's no dough ):

[delicate orchid]

this is the canonical (in the biblical sense) ordering

[pastel cliff]

biblically accurate tensor product

[dim widget]

It's e_i and e_j not e_i and e_i !

[formal ermine]

ok now there is

[pastel cliff]

ngl i sharted this out pretty fast

[elder wave]

#groups-rings-fields ?

[pastel cliff]

so it's probably wrong

[formal ermine]

fuck wrong channel

[delicate orchid]

I mean u clearly know wtf is going on

u look at it as a linear function on the elementary tensors and then do the thing

[pastel cliff]

yuh

[delicate orchid]

computation errors are irrelevant

[pastel cliff]

fuck it we ball

[delicate orchid]

• I'm not looking at a 3x9 matrix it's not even sqaure

[dim widget]

Please do not swear in #groups-rings-fields there are children here

[pastel cliff]

no more tensors for me until my final

[formal ermine]

I feel offended

[delicate orchid]

that's probably because ur being insulted

[dim widget]

This has to be wrong, there should only be one nonzero entry in every column and each row should have exactly one 1 and one -1

No matter how you are ordering the basis.

[zealous forum]

What are the prime ideals of F_p[x] for prime p? I think we have (0) as F_p is an integral domain and also (f) for any irreducible polynomial, but do we have something else like (q, f)?

[dim widget]

In general for any field F the prime ideals of F[x] are just irreducible polynomials and 0.

Because F[x] is a euclidean domain.

[formal ermine]

I feel legally obligated to mention hilbert's nullstellensatz

[delicate orchid]

useful thing to remember is that F]x

fuck

F[x] is a PID for any field F

[formal ermine]

akchchually it's an iff

[delicate orchid]

who cares

[formal ermine]

k[x] pid iff k[x] euclidean iff k field

[barren sierra]

ok so one of my HW problems is this

solved it (the explicit isomorphism is ugly to work through but I did it)

however I'm curious and not too sure where to look since my representation theory is shoddy to say the least

but what exactly characterizes nonisomorphic groups with isomorphic complex group algebras

[delicate orchid]

two abelian groups of the same order have isomorphic group algebras for one

they're both isomorphic to C^|G|

which I assumed is what you did a specific example of using artin wedderburn

[prisma ibex]

what about as Hopf algebras

[south patrol]

lol

[dusty verge]

Feels like it would be dependent on the stalk of X being noetherian

[delicate orchid]

who is this ^

does anyone know who this is?

[prisma ibex]

[dusty verge]

The picture at the end? ur mum

[south patrol]

big up Hopf

[barren sierra]

yea that's exactly what I did

but I was wondering more in general

[delicate orchid]

I'm sticking to complex group algebras because I do not want to think

[barren sierra]

I figure it depends on field so yea complex group algebras

[agile burrow]

It doesn't. It's moreso an exercise in understanding what it means for the germ to be invertible

[delicate orchid]

if the multiplicity of the degrees of their irreducible characters are equal the group algebras should be isomorphic, again by artin wedderburn

I think

[barren sierra]

What about the other way?

[delicate orchid]

I think it works the other way as well

[barren sierra]

Like say CG \simeq CH for groups G and H. What more info do I need to conclude G \simeq H?

oh I see

[delicate orchid]

they won't be isomorphic

character tables do not determine groups up to isomorphism

and character degrees definitely don't

[barren sierra]

yea but what more info would I need I guess is the question.

Or is this a massive bucket of worms

[delicate orchid]

I'm not sure

I know that there are non-isomorphic groups who's group algebras are isomorphic over EVERY field

[barren sierra]

oh what

strange

[coral spindle]

This is a massive bucket of worms yeah lol

Shit's hard

[barren sierra]

I'll look into this once I figure out this last HW problem 🙃

[dusty verge]

Okay, I thought it had to be noetherian because I didn't know direct limits existed, and so I assumed it would only make sense to talk about a homomorphism into the stalk of an element if you could have the ring be noetherian.

[lethal dune]

Say R is a Noetherian ring and I is proper ideal, then is it possible that R ≅ R/I?

as rings

[dusty verge]

Is 0 an ideal?

[barren sierra]

what do you think?

[lethal dune]

"proper"

[dusty verge]

Proper means smaller

Is zero smaller than the ring?

[dusty verge]

I think nothing. My mind is a vast empty expanse

[lethal dune]

by "proper" I mean "such trivialities aside"

[dusty verge]

Then no. Otherwise you could chain the homomorphism with itself and get an infinite ascending chain I bet

That's a complete guess, though, and so might be completely wrong

[lethal dune]

could work

[dusty verge]

Wait no that should work. If you didn't have infinite ascent, then the ideal would be trivial

[cloud walrus]

Spamakin🎷

[delicate orchid]

if we're ok with just citing literature I will google

[barren sierra]

I mean sure I can try to figure it out from there but I'm just stuck lol

[delicate orchid]

that diagonalisation thing seems right but I can't remember if it's that simple or if it's some jordan block nonsense

[barren sierra]

it's jordan block nonsense according to someone else in this channel who helped me

[delicate orchid]

it does just have to be diagonalisable

over the algebraic closure ;3

[barren sierra]

Hmmmmmm

[delicate orchid]

https://math.stackexchange.com/questions/767975/textv-is-semisimple-as-a-kx-module-if-and-only-if-a-in-operatornameen

[dusty verge]

Let $x$ be such a point. Then the stalk of $x$ is the direct limit $\displaystyle \lim_{x \in U} \mathcal{F}(U)$. Let $\phi$ be the canonical homomorphism from $\mathcal{F}(X)$ to this limit. Then there exists $g$ such that $\phi(fg) = 1$. In particular, this means that there's an open set $U$ such that $fg = 1$ in $\mathcal{F}(U)$ (should work just by basic properties of the direct limit). But this means that the restriction of $F$ to $U$ is a unit, and since the image of a unit is a unit, every element of $U$ is. Thus the set of invertible elements is the union of a set of open sets, and is thus itself open

That do it?

[cloud walrus]

Halliday

[celest furnace]

Can anyone help on this

[dusty verge]

Also, I'm sorry if my notation is trashy here, I don't really know the standard for htis topic

[agile burrow]

You've got the right idea, some of the details are a little rough around the edges. Yeah, so let x be a point where the germ of f is invertible, say there exists a germ g_x such that f_x * g_x = 1 (where f_x denotes the germ of f at x). Then there exists a representative g in F(U) for some neighborhood U of x such that f * g = 1 in F(U). Then f_y is a unit for all y in U, which shows that the set of points where the germ is invertible is open

[dusty verge]

Awesome, that makes sense. What took me a bit to figure out was how to get that neighborhood where it was invertible. Again, that's why I wanted it to be noetherian - because then the direct limit would (maybe, I didn't actually check) correspond with the homomorphism whose kernal was the top of the chain

It's also super easy if you use this definition

Because you can show the converse by applying the fact that f (the one in the picture above, not your f) is not in q.

[next obsidian]

This is only for schemes

What Walter said is true more generally for locally ringed spaces (I do think you need local @agile burrow) which aren’t covered by spectra of rings

I can’t figure out why you’d need local

SMH my head

[agile burrow]

So the neat thing is that you don't need locality to show that the set of points where the germ doesn't vanish is open

But when you specify to locally ringed spaces, the point is that if the germ isn't in the maximal ideal, then it's invertible

So the set of points where the global section doesn't vanish in the residue field is an open set

[dusty verge]

What's a scheme?

[next obsidian]

Didn’t you say here invertible tho

For any ringed space

[agile burrow]

Yeah, but in general you can just ask the germ to be in the group of units of the stalk over a point

[next obsidian]

A locally ringed space covered by Spec A for various A

[agile burrow]

oops yeah

I misspoke

I meant invertible

[next obsidian]

Oh I see yes okay okay

Yes I see

A unit in the stalk doesn’t need local

But local tells you “zero at x” means those in m_x

Which is the same as invertible

[agile burrow]

That's right, yeah

[next obsidian]

And that’s now open

Okay right, swag

I convinced myself the local part of LRS was to say “nonzero at a point means invertible in a nbhd”

And that part does need locally, but that’s cuz you said nonzero, not invertible

[agile burrow]

Yeah

[next obsidian]

Swag

[agile burrow]

I think this is where my intuition for residue field is coming from

[next obsidian]

Chmag

[agile burrow]

Sort of

[next obsidian]

Yeh

It’s the proper space for which the function takes values

[agile burrow]

Oh I guess it's like the same thing for manifolds, residue field is R

[next obsidian]

Yah

[agile burrow]

and set of points where smooth function doesn't vanish is just preimage of R - {0}

epic

[next obsidian]

Z -> Q

[dusty verge]

Hmm, I've followed none of this

[dusty verge]

Is this the part that needs locality? I looked up the answer after posting mine, and mse has something that sounds vaguely similar to this

[agile burrow]

Yeah, you need locality for that part, I misspoke

[dusty verge]

Yeah no worries, I'm just attempting to decipher what the hell you guys were talking about

What exactly is locality?

[agile burrow]

A locally ringed space is a ringed space in which each of the stalks is a local ring, meaning it has a unique maximal ideal

For instance, Spec R is a locally ringed space since the stalk over a point p is the localization R_p, which is a local ring

[dusty verge]

That's all it means? What's that have to do with pretending you have fractions?

[agile burrow]

this lol

Do you see what the unique maximal ideal is?

[dusty verge]

Hmm, not yet lol

Just to make sure I'm not being an idiot, this ideal is proper yeah?

Oh maximal ideals are defined to be, sounds familiar

I'm guessing its just the prime ideal divided by other stuff yeah?

[agile burrow]

Yeah, you take the extension of the prime ideal in the localization

[dusty verge]

Let me prove that to myself real quick

Because I went from an example of the prime ideal being (2) in Z

[solar shore]

i can clearly see at least one subgroup, the cyclic subgroup of rotations in d_6

but

[dusty verge]

Use the presentation for D_6 I bet

Or D_k

[solar shore]

(which is what, we've never been showed it)

is it the uhh

no i dont remember

[fleet pelican]

how about D_m for m|n?

[solar shore]

i was thinking d_3 was a subgroup of d_6

[fleet pelican]

Is it?

[dusty verge]

ab = ba^-1, b^2 = a^6 = 1 iirc

[solar shore]

geometrically, i can see it i guess?

[dusty verge]

I.e. the group whose generators are a flip and a spin

[solar shore]

rigorously showing it, i wouldnt know though

[fleet pelican]

How about write explicitly and prove it?

[dusty verge]

What happens if you double your rotations?

[solar shore]

then it would equal the single rotation in d_3

[dusty verge]

So use the homomorphism

[solar shore]

ah

im trying to not use homomorphisms because this part of the book hasnt done it yet

[dusty verge]

Or is that a homomorphism, idfk

But try and translate your intuition

It's a very important skill to practice

[solar shore]

because what im thinking

if i inscribe a triangle inside a hexagon, then two 60 degree rotations on the hexagon is one 120 degree rotation on the triangle

similarly, a flip on the triangle corresponds to a flip on the hexagon

but

[dusty verge]

Write down as letters

[solar shore]

for $r \in D_6$, then $r^2 = r'$, where $r' \in D_3$?

[cloud walrus]

blanket

[solar shore]

oh wait

im starting to get it now

ok im chillin

thank you

[dusty verge]

Hmm, yeah, why is it maximal?

[dusty verge]

Any time

[agile burrow]

So here's a hint, if you have an ideal I in R which isn't contained in the prime ideal P, then it intersects the complement of P

Now consider what happens to an element in the intersection once you pass to the localization

[wraith cargo]

Where did they pull that isomorphism at the bottom from lol

that would require that s-1 is in a for all s in S right

[rotund aurora]

was there an algorithm for finding primitive roots?

[dim widget]

No it would just require that s is invertible mod a

[rotund aurora]

modulo some prime

in Z/pZ I mean

[dim widget]

Nothing efficient

[wraith cargo]

interesting
Still I don't think that's something we assumed here tho

[dim widget]

a is maximal

[wraith cargo]

does that make it invertible?

Ah wait

fuck

[dim widget]

yep

[wraith cargo]

hmmmm

[wraith cargo]

OK so generally this holds
firstly wouldn't a requirement just be that all the s' are conguent to 1 mod a?
I mean if they were included in the ideal then it would be trivial but nothing guarantees that

[dim widget]

Not to 1 but to a unit.

[coral spindle]

I mean, one certainly exists (namely, just check every element) but as TTEG says, they're not easy to find by any means.

[wraith cargo]

OK I mean still

[dusty verge]

Sorry, was eating. Oh yeah that makes sense thanks

When you say unique maximal ideal, does that mean it's the only maximal ideal?

If so, how would you show that no others exist?

[dusty verge]

Units are just 1 with extra steps

[dim widget]

But not necessarily a unit in the ring by the way, just that s is invertible mod a.

[wraith cargo]

I figured it out

I calculated that the localization of a field is the same field

[elder wave]

[wraith cargo]

(except at 0)

[dusty verge]

(unless the field is 0)

[elder wave]

[glossy crag]

If m is maximal and disjoint from S you get that S'A/S'm\cong A/m (you can do it directly or using that isomorphism we were discussing the other day). Here b is S'a for a prime disjoint from S, so the iso holds.

[wraith cargo]

I have the feeling you're saying the iso holds because it holds but I'm not sure

anyways I sorted it but thanks

[glossy crag]

wdym

You can use the maximality of m and its disjointness from S to directly show the natural map A/m -> S'A/S'm is an iso

Alternatively you can use the iso we discussed yesterday, since S'(A/m)=A/m.

[pastel cliff]

is this a typo? should that be algebraic over C?

or can alpha be algebraic over Q

https://math.stackexchange.com/questions/2419674/abi-is-algebraic-over-mathbbq-iff-a-and-b-are-algebraic-over-mathb nvm ig

[south patrol]

No, it should be over Q

(Every element of C is algebraic over C, trivially)

[south patrol]

Yes, polynomials over Q have roots over C xd

[gilded wigeon]

How tf do u work with the commutator operation

I have no idea how to work with the premise $[G, G] \in Z(G)$

[cloud walrus]

francis n

[gilded wigeon]

Trying to prove that that’s equivalent to a group G being nilpotent of level 1

[void cosmos]

A-->B-->C-->0 is exact

Hom(A,M) --> Hom(B,M) --> Hom(C,M) -->0 is not always exact

so now if A is injective

then would ker(Hom(C,M))/im(Hom(B,M)) = 0?

[trail stump]

If N & H are normal in G. Is (NH)/H also normal in G?

I know NH is also normal in G, but don’t know how to proceed to NH/H is normal in G

[pastel cliff]

is using F(x) to denote the fraction field standard notation

F being a field ofc

[pastel cliff]

https://math.stackexchange.com/questions/2419674/abi-is-algebraic-over-mathbbq-iff-a-and-b-are-algebraic-over-mathb

im a little confused about getting a,b algebraic from a+bi being algebraic here