groups-rings-fields

[steel light]

That’s very cool

[delicate orchid]

like polynomial multiplication but you just keep going

[void cosmos]

just like how you would think it is

[white oxide]

damn that's crazy

[void cosmos]

okay i tried to do a proof by like assuming some element is left-quasi regular

and then showing that it must be of the form a/b where a is even b is odd

[crystal turtle]

damn what a jump prime ideals straight to Nullstellensatz

[void cosmos]

do uwanna check it out? ( its prob wrong and trash )

[rocky cloak]

I can look at it, but I don't know/remember what left-quasi-regular means

[void cosmos]

okay

a left-quasi-regular element is an element a such that there exists r such that a+r+ra = 0

the jacobson radical is a left-quasi regular ideal which contains all left-quasi regular elements

[rocky cloak]

I mean the ring is commutative and unital, so shouldn't need this many adjectives

[void cosmos]

oh yea left is right but whatev

okay so suppose a = x/(2k+1) is quasiiregular , there exists r = x'/(2k'+1) such that x'/(2k'+1) + x/(2k+1) + xx'/(2k+1)(2k'+1) = 0

so we have x'(2k+1)+x(2k'+1)+xx'/(2k'+1)(2k+1) = 0

--> x'(2k+1)+x(2k'+1)+xx' = 0

solving for x we get x = -x'(2k+1)/(x'+1+2k)

we wish to show x is even so suppose -x' is odd

then x'+1+2k would be even which contradicts x' being in R?

so -x' must be even

and thats it?

[rocky cloak]

Makes sense

[void cosmos]

cool so now i just showed that if a is left quasi regular then its the claimed set

now i want to show the reverse inclusion correct?

[rocky cloak]

And I guess you're able to follow the same argument in reverse to just make it an iff

[void cosmos]

that is i want to show that an element of this form is quasi-left regular?

like im stuck here now

idk what argument should i do

oh but ig

the jacobson radical is the unique maximal quasi-regular ideal correct?

[rocky cloak]

That's right yeah

[void cosmos]

so i just need to show that

any element of this form

is quasi-regular

and im donzo?

[rocky cloak]

Yup

[void cosmos]

wait

[void cosmos]

how did u know

i just gave u the definition half a second ago

[rocky cloak]

Well, I extrapolate from the unital case

[void cosmos]

what

ur not human lmfao

take the test for me

[rocky cloak]

But yeah, should probably check it for nonunital rings as well

This ring is unital anyway, so...

[void cosmos]

yeah ig

yo do i just like

fuck it?

like

say a is in the claimed jacobson radical , a = 2k/(2k'+1)

then we wish to find r in R such that r +a +ra = 0 , r=L/(2L'+1)

then just like solve for r?

[rocky cloak]

That should do it

[void cosmos]

so r = -2k/(2k'+1)/[(1+(2k/(2k'+1))]

something lke that correct?

so now every element of this set is quasiregular

and every quasi regular set is in this set

can u just show why this leads to this set being the jac?

[void cosmos]

cuz if i were to use that the jac is the unique maximal quasiregular ideal then why wouldnt this be enough

[rocky cloak]

Yeah, that'll do it

I forgot what your stated definition of the radical was

[steel light]

So are the prime ideals of polynomial rings generated by linear functions in the ring?

[void cosmos]

theorem 2.3

[rocky cloak]

Depends. Over an algebraically closed field, yes.

[steel light]

Ah right

And if it's not closed you consider an extension

[rocky cloak]

Yeah, for example the ideal generated by x^2 + 1 in R[x] is prime.

[steel light]

I see

So the first half I understand fine up until V(I)

So I is an ideal in our polynomial ring (I understand this fine)

V(I) is the set of all x such that for any element of f our ideal (i.e. any function f in our algebraically closed polynomial ring), f(x) = 0

So all the roots of polynomials in our ideal

Is that right?

[rocky cloak]

Yes

x is in V(I) iff it is a root of every polynomial in I.

[steel light]

So Nullstellensatz says that if we have a polynomial which vanishes on V(I) (so one of its factors is an element of I), then some multiplicity of our polynomial is in I? O.o

[rocky cloak]

That's right

[delicate orchid]

wdym "some multiplicity"

oh some multiple

[steel light]

Yes

Sorry

[delicate orchid]

yeah

[rocky cloak]

Some power I guess would be the right word

[steel light]

Yeah that's so much better lmao my bad

That doesn't make sense to me intuitively

Nor do I see why it's such a strong result

[rocky cloak]

Maybe think in just one variable.

Take the polynomial generated by f(x) = x^2 . Where does it vanish? At x=0.

What other polynomial vanishes there? g(x) = x for example.

And g^2 = f

[steel light]

Yes

[void cosmos]

O.o

[rocky cloak]

The big take away is that if we restrict ourselves to these "square free ideals" (called radical ideals) we can recover the ideal from the set V(I)

[void cosmos]

points correspond to maximal ideals

O.o

[delicate orchid]

o_O

[formal ermine]

0-o

[rocky cloak]

So in this example V(I) = {0}, so we know that I must be generated by x^n for some n

Imposing that I is "square free" we get that it's generated by x

[steel light]

What do you mean by square free?

[delicate orchid]

in a way it "lifts you" to the largest ideal that generates that variety

[steel light]

Bruh is that what a variety is

Just the set of zeroes of an ideal

[delicate orchid]

what did u think V stood for

and these are a specific (very nice) type of variety

[steel light]

I don't know LMAO I just took it as notation 😭

[rocky cloak]

I'm sort of stealing terminology from integers. But formally I mean that whenever f^n is in I, then f is also in I

[steel light]

I see

[summer path]

V for very nice

[rocky cloak]

Intuitively this corresponds to squares not dividing the generators of our ideal.

[steel light]

Why are you saying squares specifically if it can be any natural n?

[rocky cloak]

But that intuition doesn't quite formalize in several variables

[rocky cloak]

Well x^2 divides x^n , so it's enough to check for squares

[steel light]

Nice

[void cosmos]

yo eren jageer

J(R) is the biggest quasi-regular ideal

[steel light]

AHHHH I see now

[void cosmos]

or wait i will wailt untill ur done

[formal ermine]

feather learning ag?

ayo

[void cosmos]

cuz i think this is a mathematical revelation

[steel light]

Not really

Just sorta curious

[void cosmos]

so i wont intterupt

[steel light]

Ok why is this useful in practice?

I mean

ig it means like

[formal ermine]

spotted the analysist

[steel light]

If we see p^n

We can recover the smallest factor p

Which is a (the?) generator of our ideal?

And we like generators for obvious reasons

Did I just speak English or nonsense?

[void cosmos]

u did not speak u typed

[summer path]

How do you know they're not speech to text

[steel light]

If what I said is correct

Is there some linear algebraic formulation / relation to linear algebra? The concept of "smallest" ideals (prime ideals, generating polynomials, whatever the technical terminology is) reminds me of the concept of vector space bases

I've not done any algebra beyond goop theory so forgive me if I sound like a nimrod

[rocky cloak]

Guess it's time to Google Gröbner basis

[delicate orchid]

any goopers?

[steel light]

Or topological bases if you take union as a sort of product

[steel light]

Whaattttttt

This is exciting

I've never heard of that

Dude literally fuck Loch so much

I've turned into what I despised the most

[delicate orchid]

are you using his notes lol

[steel light]

Well not really but I feel like I'm walking down that road

[steel light]

No he just got me interested in AG

which I had next to no interest in before and then one stupid conversation with him and now I'm into it

and it wasn't even directly related to AG iirc it was about elliptic forms and he transitioned into AG

[formal ermine]

feather you should read geometry of schemes

become a real hsag

[steel light]

What is HS here

[delicate orchid]

high school

[void cosmos]

yooooooooooooooooooooooooooooooooooooooooooooo

the sum of two nil ideals is nil

[steel light]

WTF IM LITERALLY THE GOAT?!?!?

[void cosmos]

is actually an open problem

[steel light]

The way I just predicted mathematics 🤯 watch out Riemann

[formal ermine]

next grothendieck

[steel light]

Nah I want my $1m first then I donate it to all the poor grad students here

[delicate orchid]

thanks

[rocky cloak]

If there is a God out there

[void cosmos]

bro its actually cool

check it out

give me like 3 more weeks

[rocky cloak]

Or you know if all rings are Artinian

[void cosmos]

and i wills olve it

its called the koethe conjecture

[steel light]

So what is the correct entry to AG

[delicate orchid]

how the hell is that a conjecture

[void cosmos]

lmfao its open from 1930

[void cosmos]

i asked the same question some time ago

what is your background

[formal ermine]

geometry of schemes!!!111!

well

first do a+m

[steel light]

I'm skimming it

[delicate orchid]

insane

[formal ermine]

or eisenbud comm alg

then geometry of schemes

[void cosmos]

yea dude

i actually like it

[steel light]

Introductory real analysis and surface knowledge of everything except combinatorics, category theory, and number theory

Lmao

[void cosmos]

maybe i solve it who knows

haha

[summer path]

ive always wondered, what does wew actually work in

[steel light]

I forgot

He's no longer active

[rocky cloak]

Because some rings are noncommutative

[long geyser]

what is a+m

[formal ermine]

fusion systems

[steel light]

at least in discussy channels

[void cosmos]

try to get better at algebra as in like commutative algebra using either AM or an easier book

[formal ermine]

atiyah macdonald

[void cosmos]

thats just imo

[long geyser]

oh

[void cosmos]

and also learn point-set topology

[steel light]

Yeah I think that's what Tterra recommended

I know enough PS top to get by

lems may disagree

[formal ermine]

ps?

[steel light]

[void cosmos]

yea and i think some knowledge of like diff geo should be cool too

[long geyser]

yeah I do

[steel light]

point set

[formal ermine]

you need like 0 topology for algebra lmao

only the basics

[delicate orchid]

sure.... whatever u say....

[void cosmos]

for basic AG tho

[formal ermine]

that's teh basics

you don't need to know whatever the fuck ibsen does

[void cosmos]

i remember when i tried to do AM and one of the exercises were like "draw Spec(Z)

[formal ermine]

Real

[void cosmos]

i instantly threw it out and went back to munkres

[steel light]

bruh lumi

[formal ermine]

you should've thrown it out when it told you to draw spec R[x]

[steel light]

why does this book talk about sheaves idk wtf sheaves are

[void cosmos]

yea it had all of those haha

[delicate orchid]

that's a shit exercise ngl

[formal ermine]

it literally defines them in section 1

[rocky cloak]

Better than drawing spec(Z[x]) at least

[void cosmos]

yea i would have been a tenured AGeometer now but i threw it out .. 😦

haha

[formal ermine]

a sheaf is a way to put data on an open set in a topology

like for every open set you assign it some data

[void cosmos]

a sheaf is someone who cooks food but written wrong

[formal ermine]

then there's conditions so you can combine them

go from one to another

glue them

stuff like that

[delicate orchid]

that's why.... you're reading.... the book....

[steel light]

I'm on pg 8 and I don't see it there's nothing before this but one page and blank pages + title

[formal ermine]

what

[steel light]

The Geometry of Schemes

[void cosmos]

feather maybe try to check out the textbook called ideals , varieties and algorithms

it has a low starting point

[formal ermine]

Yes lemme open my copy

[rocky cloak]

https://pbelmans.ncag.info/blog/2011/05/25/a-latex-version-of-mumfords-impression-of-spec-zx-or-some-tikz-tricks/

[formal ermine]

@steel light Bro

page 8 is the index

[steel light]

what

[void cosmos]

yea

anyways

[steel light]

https://www.maths.ed.ac.uk/~v1ranick/papers/eisenbudharris.pdf

[formal ermine]

Oh

you're reading that short intro

[steel light]

Yes

[formal ermine]

where it just flies over whatever the fuck it does

yeaH

[void cosmos]

the jacobson radical is unique in the sense that it is the biggest left-quasi regular ideal correct?

[formal ermine]

every book has that lol

they just talk about shit to feel superior

[steel light]

LOL

[formal ermine]

I.1.3

[void cosmos]

so if i show that every quasi-regular ideal is contained in this set, and show that this set is quasi-regular then im done right?

@rocky cloak

[formal ermine]

page 11

defines what a sheaf is

[void cosmos]

im just trying to wrap my head around the argument

i had it as a plan but then i forgot what i am trying to do midway lol

@formal ermine i think you need not like trivial category theory to understand these definitoins right?

like a sheaf is a functor ig

[steel light]

Can someone give me an example of a zero ring

[void cosmos]

0

[delicate orchid]

0

[formal ermine]

you can define a sheaf as a functor

but you need not

[rocky cloak]

I may have lost the thread on what you're proving

[delicate orchid]

there's only one zero ring it's a terminal object

[void cosmos]

yea mb

1 sec

[steel light]

Oh I see

I was about to ask for nontrivial examples lulw

[void cosmos]

problem 7

[delicate orchid]

zero pretty much always means trivial

[void cosmos]

they should make

some kind of structure

that has only 1 but not 0

[delicate orchid]

a monoid?

[void cosmos]

yea but it has additive structure too

likke both additive and multiplicative

but no identity

xd

[summer path]

just remove it then

[delicate orchid]

so a non-unital commutative ring with distribution reversed

[summer path]

wait

does 2Z work?

[rocky cloak]

Yes, the Jacobson radical is the ideal of all quasiregular elements as in the image you posted, and you already showed that the quasiregular elements where exactly those described in the exercise

[delicate orchid]

work for what

[summer path]

oh wait no it has 0 but not 1

[delicate orchid]

no ideal of any ring will work for this as they all contain 0

[summer path]

2Z+1?

[void cosmos]

yea

thank you so much

[rocky cloak]

Like the natural numbers?

[delicate orchid]

yeah natural numbers work

[summer path]

i forgot negative numbers exist

[formal ermine]

Real

[rocky cloak]

So sort of like a semiring / rig

[topaz solar]

@delicate orchid sat through a mf talking about quiver reps and grassmannians

[rocky cloak]

Sounds like a Tuesday

[topaz solar]

Something something explicit descriptions of rigid representations of Kronecker quiver

[delicate orchid]

oh boy I can't wait to find the reps of the category with one object and two morphisms! (clueless)

send a link to the talk

[rocky cloak]

Let me know when you have a classification, it will be a wild ride

[karmic moat]

i'm having trouble showing the (ii)

[delicate orchid]

moderators ban this user for a terrible pun

[karmic moat]

i know that, by exactness, f(A) = ker(g)

[steel light]

Given a commutative ring A and ideal a, A/a is a subset of a right?

[delicate orchid]

no? it's a quotient ring

[karmic moat]

but i'm not sure how to show that ker(g) \cap j(C) = {0}

[topaz solar]

In person so doesn’t exist

[delicate orchid]

https://media.discordapp.net/attachments/1025970402662563866/1123062919618367667/goofy3.gif

[topaz solar]

But uhh basically he classified every sub rep of rigid reps or smth

[karmic moat]

since g o j = 1, j has a left inverse, thus j is injective

so C \cong j(C)

[delicate orchid]

makes sense to use a grassmannian for that ig

[steel light]

My thinking was that A/a = {a'a : a' in A} but by definition a'a is a subset of a since a is an ideal

[topaz solar]

And -> cell decomposition of the associated grassmanian

[steel light]

I just Wikipedia'd what a quotient group and coset are sorry if misunderstand

[topaz solar]

Which was known to exist by Zielinsky or however it’s spelled

[delicate orchid]

I think there's a typo here - j should be C -> B

[karmic moat]

yea that's a typo

sorry should've specified

[delicate orchid]

a'a is a set

[steel light]

Yes

[karmic moat]

but yeah idk how to show (ii) still

[delicate orchid]

I'd use the universal property of a free modul- oh you've done (i)

[steel light]

But for fixed a' and all elements x of a, a'x is in a, so why is a'a not a subset of a

[karmic moat]

isn't A/a = {a' + a}, where a' \in A and a is the ideal

[delicate orchid]

oh right I see

[delicate orchid]

yeah this

[topaz solar]

@delicate orchid https://meetings.ams.org/math/fall2023e/meetingapp.cgi/Paper/25937

[steel light]

Oh

[delicate orchid]

you mentioned quotient groups for some reason and it crossed my wires

[topaz solar]

Here ya go

[steel light]

OHHHH because the ring is an additive group

[rocky cloak]

So rigid means without self extensions, or does it mean something else here?

[steel light]

So we need to use the addition operation

[delicate orchid]

yeah ideals are just a special type of additive subgroup

[topaz solar]

Bro said “im not defining it”

Unironically

[delicate orchid]

then who CARES

[steel light]

Okayyy I see

[topaz solar]

Anyway link above

[delicate orchid]

I nearly studied cluster algebras instead of fusion systems

[topaz solar]

Since it was a more informal variation of the talk

[steel light]

But the ideal is only closed under multiplication so of course this makes no sense to automatically assume subset under addition

[topaz solar]

This should give a better idea

[karmic moat]

uhh yeah all i can think about is like

[delicate orchid]

in fact there's a fairly obvious counter example - consider the ideal generated by 0

[summer path]

What's a fusion system to a child?

[steel light]

(In the ring of integers yes?)

[delicate orchid]

in any non-trivial ring

[steel light]

Oh

Let me think

[topaz solar]

Anyhow I’ll look and see if I can find an arXiv link

[delicate orchid]

||if we're not in the zero ring there's another element x out there somewhere, and x+(0) = {x} definitely isn't (0) = {0}||

[karmic moat]

take $v \in f(A) \cap j(C)$. by exactness, $f(A) = \ker(g)$ and so this is equivalent to considering $v \in \ker(g) \cap j(C)$. now $v \in \ker(g)$ implies $g(v) = 0$, but im not sure why this implies that $v$ itself is 0. maybe something like $g \circ j = 1_C$ so $j(v) \in \ker(g) \cap j(C)$ implies $g(j(v)) = 0$, implies $v = 0$?

[delicate orchid]

ok wait, you're doing AG without knowing what a ring is? that's cracked

[steel light]

Yeah

[karmic moat]

idk how right this is tho

[steel light]

I know what a ring is I just don't know anything about them :^)

[cloud walrus]

ana(functor)mono(morphism)

[delicate orchid]

g has an inverse so it's bijective

[karmic moat]

:(

aha yeah you're right lol

wait

[topaz solar]

Doesn’t exist it seems

[delicate orchid]

sorry, no

[formal ermine]

we have the embedding 1 + pZ_p ---> Z_p^times, if we now reduce mod p^n we get G --> (Z/p^nZ)^times where G is the image of 1 + pZ_p under Z_p^times -> (Z/p^nZ)^times, how do I show that the restriction of the reduction 1 + pZ_p -> (Z/p^nZ)^times is continuous?

[rocky cloak]

I see, rigid objects should correspond to cluster variables so that makes sense.

[karmic moat]

it only has a right inverse

[delicate orchid]

exactly

[karmic moat]

yeah

[steel light]

Holy fuck this channel is too busy

[karmic moat]

so we have that g is surjective (by exactness) and j is injective (since g is its left inverse)

[steel light]

😭

[topaz solar]

Yeah he just said “im not defining it” since time constraints so idk

[delicate orchid]

oh wait

[karmic moat]

so then C \cong j(C) by injectivity

[delicate orchid]

if v is in the kernel of g then g(v) = 0

so j(g(v)) = 1_B(v) = v = 0

[karmic moat]

yea

oh yeah haha i forgot that j(g(v)) = 1_B

true true

thank you wew lads tbh

[delicate orchid]

no problem "ana(functor)mono(morphism)"

[karmic moat]

yeah and then this just follows since A \cong f(A) and C \cong j(C) since they're both injective

yeah yeah okay yeah

[delicate orchid]

yuh yuh yuh yuh

[FL studio built in "yeah" sfx]

[karmic moat]

need to use the restroom can y'all watch my laptop for a second

[formal ermine]

@crystal turtle

[crystal turtle]

Like how do you know G --> (Z/p^nZ)^\times is continuous?

[steel light]

ker(f) = {x in A : f(x) = 0}
Let a in A and x in ker(f). Then f(ax) = f(a)f(x) = 0f(a) = 0, hence ax in ker(f)
This is enough to show ker(f) is an ideal of A right?

I'm a bit unsure about the last step 0f(a) = 0, from the ring axioms I don't see anything about rings necessitating/implying a multiplicative 0 element?

yes yes I know I'm lacking prerequisites just bear with me

[delicate orchid]

u need it's closed under addition as well

[steel light]

Oh right

Needs to be a subgroup

[delicate orchid]

no no no an ideal

[crystal turtle]

Should just be by restriction to a subspace, if that's what you mean

[delicate orchid]

not a subring

[karmic moat]

erm actually 🤓 f(a)f(x) =/= 0 f(a)

[long geyser]

needs to be a subgroup under addition yes

[steel light]

Yes the ideal needs to be a subgroup under addition

[karmic moat]

but it's actually f(a)f(x) = f(a)0 🤓

and then u get 0f(a) 🤓

[delicate orchid]

0 is always in the centre of a ring

[steel light]

How about the statement I made at the end here

[karmic moat]

shut up youre ruining my moment

[steel light]

How do we know that there's a multiplicative 0 element

[delicate orchid]

rings are always additive groups so they must have a 0?

oh you're asking why 0a = 0

prove it

[steel light]

Commutative ring loser

[steel light]

LOL

Okay

[delicate orchid]

it's a fun exercise

[steel light]

I'll blackbox it for now and tryto show additive subgroupness of kernel

[karmic moat]

ah lame

[topaz solar]

I wouldn’t call it fun but it is an exercise

[karmic moat]

boooo boooo throw tomatoes at this loser boooo

🍅

[long geyser]

ran from the grind

[delicate orchid]

it's one of those nifty ones like showing the identity is unique

[topaz solar]

I’ll go further and say uhhh

Do this

[delicate orchid]

2 and 3 are so annoying bro

[topaz solar]

Literally the same argument

[delicate orchid]

yeah which is why they're both annoying

[steel light]

Let f: A -> B be a ring homomorphism

ker(f) = {x in A : f(x) = 0}

Let a in A and x, x' in ker(f).

Then f(ax) = f(a)f(x) = 0f(a) = 0, hence ax in ker(f)
Also, f(x + x') = f(x) + f(x') = 0 + 0 = 0, which is in ker(f), so it's closed under addition
I'll pretend the rest of the group axioms are true (or maybe they're inherited? Too lazy to think and figure it out rn)

Is that right?

[crystal turtle]

[topaz solar]

Literally my class’s homework

[delicate orchid]

you don't need to show the rest of the group axioms - cause you're showing this is an ideal

[crystal turtle]

hopefully not this years HW

[delicate orchid]

those are the two properties of an ideal

[steel light]

But I need to show it;s a subgroup under addition

[topaz solar]

I got it yesterday

[coral shale]

pretend

[steel light]

Which requires closure under addition and the rest of the axioms no?

[crystal turtle]

I would cry tbh

[long geyser]

hopefully you've taken a group theory course before, so you know that a ring hom is a group hom under addition so the kernel is a normal subgroup already

hopefully

[delicate orchid]

the rest of the axioms suck balls

[steel light]

I never got that far lems <3 now I know!

I'll probably forget though

[delicate orchid]

ok you need to stop doing AG right now

[steel light]

But it'll set in with time lols

[coral shale]

pretend the 0 field is a field

[steel light]

NOOOOO not you toooooo

[topaz solar]

Learn the basics mf

Pls

[steel light]

Boring :(

[topaz solar]

Don’t be ilum

[delicate orchid]

mf you are attempting to read ancient french literature without knowing what a "je suis fatigue" is

[steel light]

I'll learn as I go, I'm happy with surface-level understanding :c

[delicate orchid]

this leads to crankery

[steel light]

LOL

TRUE

[topaz solar]

You’re fine but don’t just intuit group theory

[steel light]

Okay

[long geyser]

I know like 10 times more algebra than u and I'm not even close to begin AG lol

[delicate orchid]

and group theory is cool too I promise...

(HELP)

[steel light]

yeah but you care about understanding it properly, I don't :^)

[coral shale]

ive never heard anyone say this to anyone

[delicate orchid]

numberphilepilled

[steel light]

LOL

Man I do sound like a crank

[topaz solar]

“I just intuit real analysis” has changed me

[steel light]

zadge

[delicate orchid]

that's based though

[long geyser]

I wonder if I can begin AG after finishing Rotman's first volume

[delicate orchid]

real analysis is literally just the most boring topology known to man

[steel light]

truers

[topaz solar]

Real

[delicate orchid]

do not ask me to show that e^-x converges to 0

[topaz solar]

But it leads to struggles with why ln converges as a power series

[steel light]

It's enough to show that 0a = 0 in a commutative ring since the right multiplication follows from commutativity right

[long geyser]

I mean...

[topaz solar]

Yes

That’s commutativity

[steel light]

Hm

Yes

[topaz solar]

But you can show it without that too ofc

[delicate orchid]

holding up my hands in two L shapes in front of my screen rn

[coral shale]

ok by like why care cmon

[steel light]

Yes

[coral shale]

im sure its like a few more lines

[steel light]

Why care about what

Oh

Yeah

LOL

[coral shale]

0a = a0 = whatever

[steel light]

I guess I can give it a shot (if I can figure out one direction)

[topaz solar]

Literally one extra addition

[delicate orchid]

moment

[white oxide]

yo boys, could i have a hint on this one pls

[delicate orchid]

oh now THIS is swag

I hate how this is true for k[[x]] but not k[x]

[long geyser]

have you investigated what the ideals in this ring look like

[topaz solar]

What do they look like

[coral shale]

I

[white oxide]

i'm fucked then

[long geyser]

lmfao

[white oxide]

i mean kinda? my dumbass just took some elements of k[[x]] and looked into the principal ideal generated by them

[delicate orchid]

well what do you think the maximal ideal should be

[coral shale]

Have you seen in this ring elements like 1-x have an inverse

[white oxide]

idk i tried the ideal generated by 1 + x + x^2 + x^3 + ... for some reason lol

then attempted the argument sent by chmonkey

[delicate orchid]

hint: the quotient has to be a field

[white oxide]

this one

[white oxide]

yea

o

[long geyser]

okay so I'll give a couple hints

1. ||if your ideal contains x^n, by the ideal condition it has to contain...||
   2.||conversely, given any ideal, consider the element of minimal degree in the ideal and try to form some kind of converse to hint 1||

[steel light]

Let A be a ring and a in A

Then
0a = (1 - 1)a = 1a - 1a = 0?

[delicate orchid]

yeah that works

[coral shale]

ok.

[steel light]

Nice

I remembered a similar trick from linear algebra for the zero vector

[coral shale]

then prepend a0 =

[long geyser]

the cleaner way imo is this tho

and what most ppl have in mind

[steel light]

I don't know what you mean Shuri

[delicate orchid]

prepend
holy nerd emoji

[delicate orchid]

add a0 = to the start

[long geyser]

0a = (0 + 0)a = 0a + 0a

[coral shale]

a0 = 0a = (1 - 1)a = 1a - 1a = 0

[steel light]

Oh

Yes LOL

[long geyser]

and follows by cancellation

for groups

[coral shale]

ooh.

[steel light]

Yes I like this better

[coral shale]

so this is neater because it doesnt rely on the ring having a 1

right?

works for more general rings

[steel light]

But a ring has to have a 1 no?

[coral shale]

oh sorry, u heard nothing

[long geyser]

no

[coral shale]

[steel light]

what

[long geyser]

so feather

you will realize

throughout time

[coral shale]

rngs which some people call rings dont have 1.

[long geyser]

that there are like 5 different definitions of what a ring is

[steel light]

i see

And some are more general than others? Lol

[long geyser]

it is indeed true that not every definition will assume the existence of 1

[solar vessel]

y'all ain't ready for felds

[steel light]

Ah ah ok this is like manifolds

[long geyser]

I think most general is that you are an abelian group under addition, semigroup under multiplication and distributivity

[delicate orchid]

the most general that's generally used is that a ring doesn't have 1 and isn't commutative

[steel light]

What is a semigroup

[coral shale]

skew feld time ?

[long geyser]

associative binary operation

don't assume anything else

[steel light]

Wtf

LOL

[long geyser]

actually a common term in PDE

[steel light]

I see

[steel light]

I don't know what this means

Man fuck engineering

[delicate orchid]

non-commutative "fields" are called skew-fields

[steel light]

Oh

[delicate orchid]

like the quaternions

[crystal turtle]

those don't exist

[steel light]

I see

[solar vessel]

in PDE it's more homomorphisms from [0,infty)

[steel light]

I smell so fucking good

[long geyser]

I've never actually explored semigroup methods in PDE but I assume it takes its name from some kind of operator like the flow map in diffgeo

[topaz solar]

Double monoid

[coral shale]

in any case feather - the earlier point was, the fewer axioms u have to use for your proof the cleaner/better

cus it then holds in more general structures

[steel light]

Right

[delicate orchid]

the best proof is the first one I come up with because then I can stop thinking

[coral shale]

just come up with the right one

[solar vessel]

a semigroup usually refers to a (continuous) semigroup homomorphism of [0,infty) into bounded operators on a Banach space

[delicate orchid]

continuous semigroup
I need to take a breather

lie semigroups

[topaz solar]

Which semigroup

[solar vessel]

I said semigroup maybe I meant monoid

[topaz solar]

lie algebroids

[topaz solar]

I was joking under * or +

[topaz solar]

That’s a monoid yes

[solar vessel]

it's kinda like 1-parameter subgroup

[topaz solar]

Sion would call it a semigroup

[solar vessel]

it's a homomorphism not a subgroup

[karmic moat]

ok here's my attempt

$\ker\varphi \subseteq \ker\varphi^2 \subseteq \cdots \subseteq \ker\varphi^i \subseteq \ker\varphi^{i+1} \subseteq \cdots$ as submodules
But $M$ Noetherian implies there is some $n$ for which this ascending chain stabilizes. So then $\ker\varphi^n = \ker\varphi^{n+1}$ implies $v \in \ker\varphi^n \iff v \in \ker\varphi^{n+1}$ implies $\varphi^n(v) = 0 \iff \varphi^{n+1}(v) = 0$. But $0 = \varphi^{n+1}(v) = \varphi(\varphi^n(v)) = \varphi^n(v)$. Since $\varphi$ is an $R$-module homomorphism, we have that $v = \varphi^n v$ and thus $v = 0$?

[delicate orchid]

ur a bit of an algebroid

[cloud walrus]

ana(functor)mono(morphism)

[delicate orchid]

wait what we doin again

[chilly ocean]

lie algebroids are pretty nice

the name might sound funny, but there's a lot of differential geometry you can do with them

[karmic moat]

so then the submodule \ker\varphi^n = 0, and thus every submodule contained in it is {0}, and thus \varphi has trivial kernel, thus injective

injective + surjective + R-mod hom implies bijective

something seems off about my proof

idk if i can conclude v = \varphi^n(v) like that

[delicate orchid]

could you elaborate a bit more on why $\phi^n(v) = v \Rightarrow v = 0$

[cloud walrus]

Wew Lads Tbh

[karmic moat]

because we have that $\varphi^n(v) = 0$

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

since $v \in \ker\varphi^n$

[cloud walrus]

ana(functor)mono(morphism)

[delicate orchid]

right of course

[topaz solar]

Why is v = phi^n(v)

[karmic moat]

yeah idk that's what im tryna figure out

[topaz solar]

[delicate orchid]

this is definitely the right idea though I remember that chain of kernels

[topaz solar]

What was the original problem

[karmic moat]

^

yeah i mean im pretty sure the main idea for this proof is to show that the "stabilized" kernels are just {0}

[topaz solar]

Ah

[karmic moat]

my original idea for v = \phi^n(v) was to use that \phi(v) = \phi(n) implies v = n

but i def didnt use that right

if at all

[delicate orchid]

ur morphism is surjective not injective

[karmic moat]

oh shid

aw balls

[steel light]

Uh is this a typo

[proud spindle]

ngl i've become so overwhelmed by the sheer amount of concepts in aa that i might just do elementary algebra and trig for a while

[steel light]

Should it say x neq 0

[topaz solar]

Stabilize phi - id?

Idk

[steel light]

Because 0 is a zero divisor of Z otherwise then no?

[delicate orchid]

I think they literally used \neq to mean the phrase "not equal to"

lazy bastards

[steel light]

LMFAO

LMFAOOO

That's awful writing smh

I'm starting to see why math nerds prefer books like this though

With less exposition

The longer books are starting to wear me out with all the talking

[topaz solar]

Lurie

[delicate orchid]

can u find an example of a ring with zero divisors?

[topaz solar]

Uhh it’s surjective so phi^n surjective right

[steel light]

Hm

Let me think

[rocky cloak]

This is similar to fittings lemma:
For f: M -> M, if M is Noetherian there is an N such that f is injective when restricted to im(f^n) for n > N. And if M finite length then M is the direct sum of ker f^n and im f^n

[topaz solar]

But kernels stabilize

[delicate orchid]

yur

[topaz solar]

So the kernels stabilize, but what about phi(v) -> kernel of phi

[delicate orchid]

for all m there exists an m' such that phi(m') = m then just brbrbrbrbrbr all the way through the composition

[topaz solar]

Which has to exist such a v since surjective

And we know 0 -> 0, so if x neq zero is zeroed, what about y in phi^-1(x)

[steel light]

I'm not reading that

I don't want to psoil myself

[coral shale]

wikipedia at least thinks 0 is a zero divisor

[topaz solar]

So we should always add another kernel element

[steel light]

I came up with what I think is example and I'm trying to prove to myself that it's a ring with a zero divisor

[coral shale]

on the other hand wolfram mathworld thinks it isnt

[steel light]

Can you delete Shuri or spoiler it lols

Ty

[coral shale]

oh sorry didnt see the 1st line was a spoiler

[ashen heron]

delete Shuri?

[long geyser]

yea I've seen resources that considered 0 a zero divisor

and some defs exclude 0

[coral shale]

nlab thinks 0 is a 0 divisor

therefore 0 is a 0 divisor

[long geyser]

ring theory gotta be the messiest terminology battleground in math

or at least gives foundations a run for its money

[ashen heron]

limit points, cluster points etc in topology too

[topaz solar]

Im fine with 0 as zd but

[long geyser]

really I think those are pretty agreed upon no?

[topaz solar]

Makes some lemmas need uh “zd but not 0”

[coral shale]

non-trivial zero-divisor

[topaz solar]

Ye exactly

[solar vessel]

[delicate orchid]

the differential geometry in question:

although I suppose that's more notational than terminology

[steel light]

Aww the thing I wanted to use as an example doesn't work

[coral shale]

what was the thing

tbh we should ask you what examples of rings u already know

[long geyser]

differential geometers don't understand their notations enough to be able to fight over it

[steel light]

I wanted to consider neighborhoods of a topology tau under ring addition = set union and ring multiplication = set intersection

[coral shale]

this sounds like a mad ring wtf is this

[steel light]

BUt there's no additive inverse

[topaz solar]

That’s close to a Boolean ring

[steel light]

Integers and polynomial rings that's it

[delicate orchid]

ok those won't get you any zero divisors

[coral shale]

all fields are rings ||but no 0divs||

[delicate orchid]

since you don't know quotients

[steel light]

Yeah oops lol duh

[steel light]

I know quotient spaces that's it

[coral shale]

which examples of.

[topaz solar]

What about other things you know about with multiplication

[steel light]

But now I know what quotient rings are I think

[topaz solar]

And addition

[steel light]

You explained earlier

[delicate orchid]

wtf is ur pedagogy bro

[topaz solar]

Like, idk, what about functions?

[solar vessel]

what is this about

[coral shale]

[topaz solar]

Linear ones?

[steel light]

None Wew, I'm an engineer lol

[delicate orchid]

anyway you know another type of ring that has zero divisors

[topaz solar]

Bro you took LA

[delicate orchid]

no quotients needed

[delicate orchid]

this

[topaz solar]

I will clown you for this one tbh

Anyhow, it’s a good example

[steel light]

Zero formal math experience beyond an introductory real analysis course, the rest of it is just "self-taught" me skimming through books and articles and chatting with people here and learning stuff as I go

Uh ok let me think of what LA example you're thinking of

[solar vessel]

what's the question

[topaz solar]

Since the LA one that I’m thinking of has a few fun things

[coral shale]

hunting for 0divisor example

non-trivial that is.

[steel light]

Don't spoil pls

[topaz solar]

Bro use your thinker

[steel light]

I want to be a better math student

[solar vessel]

ok

[steel light]

Except for the part where I do my prerequisites properly but everything else

ANyways

[solar vessel]

||what u clicking for goofy ah||

[topaz solar]

What in LA

Has addition

And multiplication

[steel light]

Vector addition and scalar multiplication

[coral shale]

sharp trapping the fish

[steel light]

But there's no nonzero zero divisors no?

[coral shale]

unfortunately the multiplication has to act on the same set

[steel light]

Okay

[coral shale]

scalar multiplication takes a field element

[steel light]

Dot product is what my mind immediately goes to

[coral shale]

and a vector space element

[steel light]

But let me check

[long geyser]

dot product is not a binary operation either

[topaz solar]

You are an engineer

[steel light]

I am an engineer Dr Sharp :(

[topaz solar]

What do engineers use

[long geyser]

need to think of "multiplication" as a binary operation

[steel light]

eigenstuff

[topaz solar]

Not just vectors

[long geyser]

takes two things and returns thing of the same kind as the two things you put in

easy

[topaz solar]

And you can multiply them

[ashen heron]

dot product is bilinear?

[steel light]

Yeah how is dp not binary

[rocky cloak]

Maybe we should ask Morpheus

[coral shale]

u dont land where u came from

[long geyser]

binary operation is not the same thing as a bilinear map

[steel light]

Oh

Right

[long geyser]

they are

[coral shale]

well unless its the dot product in R

[long geyser]

very different things

[coral shale]

your vs being R

i mean

Your vector space is over itself ******* (this is a pretty useless tangent)

[steel light]

Agh I have class now and this professor is pissy about being off topic lol I'll come back to tihs later, will think about it in class though

[delicate orchid]

screaming and crying rn

[topaz solar]

Real

[rocky cloak]

If you can't figure it out you may have to ask an Oracle

Or pray to the holy Trinity

[coral shale]

breh, sharp led them down the trap

but yes worth clarifying that cant be ring multi

[topaz solar]

I was trying to point towards ||matrices||

[solar vessel]

that mf ||nonabelian||

[delicate orchid]

yeah that's where my mind went too

[topaz solar]

Yeah and

[delicate orchid]

:uponthewitnessing:

[solar vessel]

gross

[delicate orchid]

gimme an abelian one that isn't a quotient

[topaz solar]

cereal2 even

[solar vessel]

real ones would ||function||

[crystal turtle]

:cereal4:

[solar vessel]

everything is a quotient

[coral shale]

we can pick special ||matrices that commute cant we, or dis fail||

[delicate orchid]

oh yeah true

[delicate orchid]

||matrices commute if and only if they're simultaneously diagonalisable so this looks like a product of ur base field||

actually there you go there's another one

[white oxide]

dumb proposition; M = the set of all power series with zero constant term works

[topaz solar]

R^2

[solar vessel]

||tfw nondiagonalisable||

[coral shale]

||isnt this an answer||

[solar vessel]

||yes||

||but it's basically my answer||

[delicate orchid]

I came up with this naught but a few moments ago

[crystal turtle]

so much [redacted] in the chat today huh

[coral shale]

||shhhhhhhhh then||

[topaz solar]

Well is it the only maximal?

[delicate orchid]

CIA hours

[topaz solar]

Well, the problem statement says yes but

[delicate orchid]

that is maximal yes

[white oxide]

i will attempt to prove this and either arrive at yes or no.

[formal ermine]

huh wdym

[topaz solar]

Well, what are your units?

[coral shale]

this is ||principal|| right

probably irrelevant to the original Q

[delicate orchid]

||k[[x]] is a pid||

||so yeah, but find out what the generator is||

[white oxide]

in k[[x]]? just the series with nonzero constant term right

btw i got that from wikipedia lol

[coral shale]

k is a field...?

[white oxide]

yea

[delicate orchid]

nah it's a letter

[topaz solar]

What are your units

[white oxide]

uh so i'm guessing that's a no...

[coral shale]

i thought it was a yes...

[delicate orchid]

yeah those are the units

[coral shale]

a0 + a1x + a2x^2 + ...
think how to cancel this

[karmic moat]

i am racking my brain over that noetherian kernel blah blah blah question

math sucks i gotta change majors to like art history or something

[topaz solar]

Oh I read it wrong

Ye that’s right

I misread it as the problem again

I do not know how I did this

[topaz solar]

It’s indeed the units

What do you know about local rings

[delicate orchid]

oh shit this is clever

[topaz solar]

See I can be big brain from time to time

[crystal turtle]

🤯

Sharp to save the day

[topaz solar]

Idk commalg well but

[hollow mica]

classic gradeschool teacher moment

[topaz solar]

I at least know baby stuff :^)

[delicate orchid]

it's just trying to remember all the baby stuff

[topaz solar]

4 what? 4 apples? 4 lads? 4 curbstomps?

[crystal turtle]

4 applies

[coral shale]

4 imaginary units.

[hollow mica]

A spacecraft FAILED because someone didn't specify units

[delicate orchid]

mars climate orbiter

[white oxide]

no clue what that is

[delicate orchid]

rings with a unique maximal ideal

[delicate orchid]

1 stomp this time!

slideeee to the left

[hollow mica]

if only they worked in a field

[delicate orchid]

ok so u probs don't know the thing sharp is using

[topaz solar]

If you don’t know properties about these, consider [set of nonunits] and [set of units]

Hopefully you see the relation of the former with your ideal

Consider a ring hom into a field k[[x]] -> K

(Which you know occurs for a maximal ideal quotient)

What then

[delicate orchid]

||or we just show that the ideals form a chain (x) \subseteq (x^2) \subseteq (x^3) \subseteq...||

either or at this point

[topaz solar]

Eh that seems more meh imo

[formal ermine]

frick u

[white oxide]

sorry i don't lol, is it just that k[[x]]/M is a field so k[[x]]/M iso K --> there exists a homo from k[x] --> K induced by this iso? i have no clue

[coral shale]

well arent we done with the previous observation of units. The ideal contains all non-unit elements, hence it contains all other non-proper ideals

[topaz solar]

Well there’s obviously a map k[[x]] -> K

Ya know

First iso

[topaz solar]

Yeah basically

[delicate orchid]

yeah sharp I have absolutely no idea why you introduced this map

I mean I get where you're going but it's a bit out there

[coral shale]

thinking practice

[karmic moat]

too radical for this moderate groups-rings-fields channel

[topaz solar]

Idk it’s the first that comes to mind imo

[white oxide]

is it just that if N is another maximal ideal not equal to M, then there exists some power series with constant term nonzero in N which is a unit and hence N is then equal to k[[x]]? is it just that?

[crystal turtle]

anyway your continuity thing you sent (idr where/what it was exactly) should just be resriction of the domain of some continuous map. idk if you saw my earlier response

[delicate orchid]

yeah actually it is just that

[topaz solar]

Basically yeah

[white oxide]

oh okay cool

[topaz solar]

Or it’s a sub ideal of M

And that’s anti - maximality

You can’t send a unit to 0

Since 1 —> 1

So it falls out immediately

[crystal turtle]

i can do whatever I want, thank you. Especially if I don't require 1 --> 1

[topaz solar]

True and good

But yeah if you don’t then it’s 0 gg

[steel light]

Oh

[topaz solar]

It collapses very quickly

[steel light]

Ring of matrices under matrix addition, multiplication

[topaz solar]

Yes

[steel light]

LOL

[solar vessel]

now find a commutative one

[steel light]

[coral shale]

ayyy hello

[steel light]

It took me that long to come up with this

Fine

[coral shale]

so whats our 0 divisor tho lol

[steel light]

I'll come up with a commutative one

Oh

Let me do the math lol

One second

I just went off intuition

[void cosmos]

can someone give an exmaple of a left primitive ideal that is not maximal?

[formal ermine]

we have the embedding 1 + pZ_p ---> Z_p^times, if we now reduce mod p^n we get G --> (Z/p^nZ)^times where G is the image of 1 + pZ_p under Z_p^times -> (Z/p^nZ)^times, how do I show that the restriction of the reduction, i.e. 1 + pZ_p -> (Z/p^nZ)^times, is continuous?

how do I know that Z_p^times -> (Z/p^nZ)^times is continuous lmao

[crystal turtle]

oh lmao

How are you defining the topology on (Z/p^nZ)^\times? Is it not the subspace of quotient topology?

[formal ermine]

doesn't matter what topology it has

as long as we can make the map continuous

[crystal turtle]

motherfucker what

[coral shale]

cmon discrete

or indiscrete

idk which

[topaz solar]

Just give it the topology to make it continuous then??

[coral shale]

surely just...

[topaz solar]

[crystal turtle]

how the fuck do you expect to have a continuous map without topology

[coral shale]

[formal ermine]

Yeah discrete topology

whatever

[topaz solar]

Bro intuited it

[coral shale]

[hollow mica]

The intuit strikes again

[coral shale]

bruh im so confused

how can u have a meaningful statement

using the discrete top

why care about continuity then

[topaz solar]

If it’s discrete, you want the fiber of every point to be open

That’s it

[formal ermine]

I need to show that the kernel of that map is an open subgroup in its domain

so I thought about just finding a topo where it's continuous

[crystal turtle]

Give Z/p^nZ the quotient topology under the projection Z_p ---> Z/p^nZ

[coral shale]

all sets will be open yeah?

[crystal turtle]

This will make it continuous

Now restrict the domain/codomain as needed. It will stil be continuous under the subspace topologies

[crystal turtle]

generally if you want to make a map continuous, you don't give the codomain the discrete topology

[formal ermine]

Thanks

[crystal turtle]

That greatly restricts your options for continuous maps.
discrete = easy to map out of, indiscrete = easy to map in to

[crystal turtle]

I had though you already knew the map was continuous in my first response lol. Sorry 🙃

but yes this should work?

[formal ermine]

so just to clarify we first restrict to Z_p^times ---> (Z/p^nZ)^times and then to 1 + pZ_p -> (Z/p^nZ)^times yes?

and the initial projection is continuous because it's a quotient map right (just repeating how we defined it)?

[crystal turtle]

Give (Z/p^nZ) the quotient topology, restrict the (co)domain to the map Z_p^times ---> (Z/p^nZ)^times, then precompose with the inclusion 1 + pZ_p ---> Z_p^times

[formal ermine]

Yeah

[crystal turtle]

Yeah, giving it the quotient topology just means giving it the finest topology for which it's continuous

[formal ermine]

Right

[topaz solar]

It’s continuous kinda by definition, trick is does it get your kernel open?

[formal ermine]

yeah

preimage of 0

wait

is { 0 } open in this topo lmao

[topaz solar]

Is {0} open

[crystal turtle]

idk if that does the trick, but hopefully?

Gotta check {0} is open lol

[topaz solar]

GL on that one

[delicate orchid]

SL on that one

[crystal turtle]

(I think this should give discrete topology on Z/p^nZ^times anyways? Not confident in that)

[formal ermine]

if we assume discrete on Zp^nZ^times then {0} is easy but the continuous thing might be harder ig?

[crystal turtle]

Yeah might not be continuous then

[formal ermine]

argggh

[topaz solar]

{0} being open is kinda exactly your og problem

It’s the same statement ya know

[crystal turtle]

0 is open I'm pretty sure

[topaz solar]

Well, it is if the kernel is open (which it is)

But proving that is ya know

The issue

[crystal turtle]

Under quotient map Z_p --> Z_p/p^nZ_p (which is iso to Z/p^nZ), the preimage of {0} is p^nZ_p, which is just an open ball lmao, hence {0} is open in the quotient topology

[formal ermine]

this whole thing is to show that the kernel is p^n Z_p

[topaz solar]

Huh

what

[crystal turtle]

then you do yout weird restriction shit and it should give the soution

[indigo ridge]

who here is a wizard at latex?

[crystal turtle]

also what?

[formal ermine]

I was gonna go kernel is open -> is nbhd of 0 -> closed set -> must be p^nZ_p

[topaz solar]

Why would you do it like that

[topaz solar]

Bro tf is the definition of the map homie

[crystal turtle]

It's a projection onto a quotient bro

[topaz solar]

Tf the definition of your p adics are you on crack

[crystal turtle]

of course the kernal is what you're quotienting by?

[topaz solar]

This is like

The definition of the map?

That this is the kernel?

[crystal turtle]

Unless I am misunderstanding your intentions here

[formal ermine]

I see it when we look at the original quotient map

but not at the restricted one

[void cosmos]

yo

this open?

[steel light]

Is there even a general formula

[crystal turtle]

but the kernal of Z_p --> Z_p/p^nZ_p is exactly p^nZ_p. Conjugate by whatever isomorphism you need to get Z/p^nZ instead

[void cosmos]

...

[coral shale]

huh no. well maybe. but i meant find just one.

[steel light]

Oh

[formal ermine]

"conjugate"?

[steel light]

Well give me a nonzero matrix

[crystal turtle]

the kernal of 0 after you restrict will just be p^nZ_p intersected with your subspace

[coral shale]

ofc finding them all for say 2x2 matrices

should be pretty doable. i hope.

[coral shale]

1 0
0 1

[steel light]

Sure

[coral shale]

why did u want that...

[formal ermine]

yeah

[crystal turtle]

maybe not the right word idk. Just identify Z_p/p^nZ_p and Z/p^n with each other is what I had in mind

[formal ermine]

so I need to show that p^nZ_p is contained in my space

[solar vessel]

let him land