groups-rings-fields

[next obsidian]

Do you know what the dimension of a ring is

[bleak crystal]

i don't understand this class

never have

i just sort of struggle by and manage decently high B's

every time i understand something in this class, it's like a goddamn nuclear lightbulb goes off in my head

[delicate orchid]

That’s why I like abstract algebra lol

[untold sapphire]

Commutative Algebra is hard to be fair

Especially module theory

[delicate orchid]

Non commutative rings

Modules are brutal but they’re very general

[untold sapphire]

I don’t even want to look at non commutative rings

Stuff is hard enough as it is

[next obsidian]

Okay well, my suggestion is to just show that every element of R is invertible.

You can do this inside of R[x], because is a•f(x) = 1, f(x) must be a constant, because if it contained a power of x, so too does a•f(x)

My suggestion for this, for any a in R, look at the ideal
(a,x)

[untold sapphire]

This proof is best as a contradiction IMO btw

[hidden haven]

Non commutative algebra is easier just because there's less you can say lmao

[untold sapphire]

Assume R is not a field, then what can we say about (x)?

[hidden haven]

At least that's the feeling I get

From the limited amount I've studied

[bleak crystal]

alright, a couple things

[untold sapphire]

(hint)

[next obsidian]

I like bdobba’s idea better

[bleak crystal]

1. I don't understand
2. i will never understand anything until 2 months after it's passed

[next obsidian]

That’s a good plan

[bleak crystal]

pretend i am a baby

[untold sapphire]

Ok

[bleak crystal]

goo goo ga ga I don't understand why a high school teacher has to take abstract algebra

[untold sapphire]

Well if R is not a field, then what can we say about the maximality of (x)?

[bleak crystal]

what is maximality

[untold sapphire]

I.e. whether or not it’s maximal

[delicate orchid]

a maximal ideal is one that is as large as possible without being the full ring

[bleak crystal]

so like

the opposite of principal?

[delicate orchid]

i.e. if I is a maximal ideal of R then there isn't a J such that $I \subsetneq J \subsetneq R$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

and no you can have maximal principle ideals

[bleak crystal]

i see

that makes sense

[untold sapphire]

You really need to learn what a maximal ideal is before doing this question

[delicate orchid]

for example (2) and (3) in Z are maximal

[bleak crystal]

i see

[untold sapphire]

It’s a fundamental concept in CA

[bleak crystal]

good thing I'm not taking ca

I'm taking "intermediate analysis" whatever that means

[maiden ocean]

this is... an analysis class?

[bleak crystal]

algebra

[untold sapphire]

Er what?

Lol ok

[strong valve]

limited amount

[maiden ocean]

but its named intermediate analysis

gigabrain

[chilly ocean]

(I thought CA means complex analysis and was so confused)

[bleak crystal]

no that's in the future

[untold sapphire]

no I meant commutative algebra lol

[bleak crystal]

wait does ca NOT mean complex analysis?

OH

lmao

[chilly ocean]

yeah but CA reserved for complex analysis

[untold sapphire]

Haha

[hidden haven]

Do you have any problems with that assertion?

[untold sapphire]

Shows how little of an analyst I am lmao

[strong valve]

cana reserved for complex analysis

[strong valve]

yes

[bleak crystal]

okay so

this

i don't think we've ever said "maximal" in this entire class for the record

[untold sapphire]

Oh right

[bleak crystal]

so probably don't assume that's allowed to be used

[maiden ocean]

ii here is maximality

haha

[strong valve]

(i) + (ii) is basically saying maximal ideal

[bleak crystal]

ah

[rustic crown]

basically

[bleak crystal]

alright

so let's say i got part 1 of this proof down, the "if R is a field then..." part

[strong valve]

an ideal is def to be maximal if R/I is a field

problem?

[untold sapphire]

No that’s a good definition IMO

[bleak crystal]

so now i need to start with suppose every ideal of R[x] is a principal ideal

[maiden ocean]

to show the reverse direction ya

[rustic crown]

yes v good def

[strong valve]

thanks

[maiden ocean]

i think you should assume R isnt a field, maybe let x be some non-zero non-unit in R and see where that takes you

[untold sapphire]

Yup

That gives you a 2 line proof

I think

[maiden ocean]

uh on second thought dont call it x

call it like

r

[strong valve]

1 line

[untold sapphire]

Lol yeah to be fair

[bleak crystal]

i feel like i need a line or two rn

fuck

[maiden ocean]

of cocaine?

[strong valve]

no reserve r for smol rings

[bleak crystal]

this is the second to last assignment of the class

okay so

assume R is not a field

[strong valve]

ok

[maiden ocean]

mhm

[bleak crystal]

there exists some r in R s.t. r≠0 and r has no inverse

[delicate orchid]

that's the definition of a field so far

[untold sapphire]

And think about whether the ideal (x) is maximal or not

[strong valve]

go on

[bleak crystal]

me?

I'm like 3/4 understanding this conversation

how does assuming R is not a field help me

[untold sapphire]

I think if I say anymore then I’ve just said the answer

[strong valve]

what do you need to show?

[bleak crystal]

oh contrapositive

duh

[delicate orchid]

if R was a field then every ideal would just be the full field

[maiden ocean]

yes

[bleak crystal]

so i need to conclude that if R is not a field, then it is NOT the case that every ideal of R[x] is principal

[maiden ocean]

mhm

[bleak crystal]

yes

how is this a two line proof?

[maiden ocean]

can you use r to make a non-principal ideal of R[x]?

[bleak crystal]

i must be missing something huge

[maiden ocean]

remember r is also an element of R[x]

[bleak crystal]

true

i feel like

that character in any B movie

the one who says something stupid and foreshadowy and then says

it's right behind me, isn't it?

[maiden ocean]

lol

[untold sapphire]

Lol

[bleak crystal]

it's right behind me, isn't it

[maiden ocean]

ok basically we want to use r and some other element of R[x] to make an ideal that cannot be principal

[untold sapphire]

Banishing your head against problems like these is a rite of passage

Banging*

[maiden ocean]

its true

AM

[bleak crystal]

i just want to teach calculus to 17 year olds

why must i endure

[maiden ocean]

ok think about (x)

what ideals in R[x] contain (x)?

[bleak crystal]

uh

[unique juniper]

are you becoming a teacher?

[bleak crystal]

no i want to teach calculus to 17 year olds on the street corner of 12th and main

[unique juniper]

lol

[bleak crystal]

HAVE YOU HEARD THE WORD OF GOTTFRIED MY YOUNG FRIEND?!

[maiden ocean]

ok let me modify that: what principal ideals of R[x] contain (x)?

[bleak crystal]

the linears

I think?

[maiden ocean]

like (r + x) for r in R?

[bleak crystal]

yes? I think?

[delicate orchid]

oops sorry

[maiden ocean]

R is an integral domain

[bleak crystal]

right

[maiden ocean]

so can x = k(r + x) = kr + kx for some k in R?

[bleak crystal]

i don't know

[delicate orchid]

well can x = kr+kx?

just intuitively

[bleak crystal]

sure it could

wait but

[delicate orchid]

how? there would always be an r

[bleak crystal]

r isn't 0

ah

[maiden ocean]

right

[delicate orchid]

exactly

[bleak crystal]

and no zero divs

[maiden ocean]

ya

(for completions sake we can rule out k having any terms of degree 1 or greater because then we'd have an x^2)

(but then we've seen that k cant be an element of R either)

[delicate orchid]

you said k was in R not R[X] 😡

[maiden ocean]

yes i am justifying that

[delicate orchid]

grrrr

[bleak crystal]

I'm in the middle of the trees and i cannot see the forest

[delicate orchid]

I'm not gonna lie this isn't how I proved it when I saw this problem posted

[bleak crystal]

i like proof by contraposition

i don't get to do it often

but I don't understand at all lmao

so we have

there exists some r in R s.t. r≠0 and r has no inverse

[delicate orchid]

I just wrote out a general non-principle ideal and then showed it to be always equal to some principle ideal but we can continue with Moth's idea

[unique juniper]

this proof is actually quite involved

the one that i read atleast

[delicate orchid]

Should I write mine out?

[bleak crystal]

if i could get away with contraposition for part 2 i would be happy

I'm not seeing the jump though

from line 1 to 2

[maiden ocean]

ok basically what we are trying to do is assume that R is not a field and construct a proper ideal that contains (x) from it

and rn we are showing that no principal ideals contain (x)

which in turn implies that our proper ideal is non-principal

the second step is the meat of the process

but you have ruled out linear maps like (r + ax) already

[bleak crystal]

because r is not a multiple of x

yes

[maiden ocean]

right

do you see how the way we dealt with (r + ax) basically works for any polynomial with a_0 non-zero?

[bleak crystal]

yes

[maiden ocean]

and if a polynomial has a_0 = 0, it is divisible by x, right?

[bleak crystal]

yes

[maiden ocean]

so that would mean that if p is that polynomial, (p) is contained in (x)

[bleak crystal]

fuck i have to get to class

I'll be back in 75ish

fuck

[maiden ocean]

np

we are mostly done

to summarize we just showed that for p a polynomial:
-if p has a non-zero a term (p) doesnt contain (x)
-if p has a zero a term (p) is contained in (x)

meaning no principal ideal of R[x] contains (x)

which is really the majority of what we had to do

[unique juniper]

$[\mathbb{Q}(\sqrt{3},\sqrt{21}) : \mathbb{Q}(\sqrt{7})]$

[cloud walrus]

Yes

[maiden ocean]

[unique juniper]

the basis is just {1, sqrt 3 ,sqrt 21,sqrt 7} right?

[delicate orchid]

I'm pretty sure sqrt(21) isn't linearly independent of the others

I don't know what the colon means there so that could affect it

[unique juniper]

but

uh

sqrt 3 x sqrt 21 = 3sqrt 7

is that what you mean

[delicate orchid]

wait nvm I'm being stupid it's linearly independent over Q

[unique juniper]

oh

[latent ingot]

[unique juniper]

lol

[latent ingot]

yea that's the basis for ℚ(\sqrt{3}, \sqrt{21}) over ℚ

[unique juniper]

over Q ?

[latent ingot]

yes, but if you're doing it over ℚ(\sqrt{7}), then you should go ℚ(\sqrt{3}, \sqrt{21}) = ℚ(\sqrt{3}, \sqrt{7}) = ℚ(\sqrt{7})(\sqrt{3})

[unique juniper]

a +bsqr3 + csqrt21 + dsqrt21sqrt3

[latent ingot]

so its an extension of ℚ(\sqrt{7}) by \sqrt{3}

[unique juniper]

then let a b c d be from q(7)

[latent ingot]

but sqrt21sqrt3 and sqrt21 are already in ℚ(\sqrt{7})

uhh

[unique juniper]

um

[latent ingot]

no

they are if you factor out the sqrt(21) and do some more algebra

[unique juniper]

ok

[latent ingot]

but you should probably simplify things a bit before

[unique juniper]

but wouldnt it be a basis for that over Q(7) too?

[latent ingot]

it would be a generating set, because they are not linearly independent

(-sqrt(7))*1 + 1*sqrt(7) = 0

[unique juniper]

i see

lol

i thought we can just think of Q(root3,root21) as avectpr space over q(root 7)

[latent ingot]

you can

[unique juniper]

$= {a + b\sqrt{3} + c\sqrt{7} + d\sqrt{21} : a,b,c,d \in \mathbb{Q}(\sqrt{7})}$

[cloud walrus]

Yes

[unique juniper]

thats what i did

[latent ingot]

if you want to do it that way then yea, you take {1, sqrt 3 ,sqrt 21,sqrt 7}, but then you need to take out all the linearly dependent elements

[unique juniper]

right

[latent ingot]

to get a basis

[sly crescent]

Is there some sort of generalization of the central product that doesn’t require the subgroups to be central?

[unique juniper]

thank you

[latent ingot]

:)

[chrome hinge]

are you talking about direct product of normal subgroups?

[sly crescent]

I’m talking about merging subgroups.

[chrome hinge]

Oh, my bad. I dont know the answer then

[unique juniper]

anyone can help me show $\sqrt[3]{2} - i \in \mathbb{Q}(\sqrt[3]{2}+i)$

[cloud walrus]

Yes

[rustic crown]

yep

[unique juniper]

thanks

[rustic crown]

lol

[unique juniper]

:D

[rustic crown]

so say x = cbrt(2)+i

most likely

[unique juniper]

ok

[rustic crown]

so x-i = cbrt(2)

cubing both sides

[unique juniper]

so find the degree?

[rustic crown]

(x-i)^3 = x^3 -3ix(x-i) -i^3 = 2

i'm very slow with these calculations 😅

[unique juniper]

dont apologise lol :)

[rustic crown]

x^3-3x + i(1 -3x^2) = 2

hence i = (2+3x-x^3)/(1-3x^2)

hence cbrt(2) = x - i = x - (2+3x-x^3)/(1-3x^2)

so both cbrt(2) and i are in Q(x)

(i saw that 😶)

[unique juniper]

ooops

[rustic crown]

lol

?

[unique juniper]

lol

still thinking

sorry

[rustic crown]

lol

[unique juniper]

by this definition does it mean that Q(cbrt2 , i) contains cbrt2 and i ?

[rustic crown]

yep!

but you can't say that directly for Q(cbrt(2) + i)

[unique juniper]

oh

yes

[rustic crown]

its not very clear why that should contain both cbrt(2) and i

[unique juniper]

yes i see

[unique juniper]

how do we carry on from here then?

😅

[rustic crown]

we wnated to show cbrt(2) - i in Q(x)
we showed cbrt(2) in Q(x) and i in Q(x)

fields are closed under subtraction! 😮

[unique juniper]

but

i thought

we wanted to show its in Q(cbrt2 + i)

sorry im probbaly being dumb rn

[rustic crown]

We actually just proved that Q(x) = Q(cbrt(2), i)

[bleak crystal]

do you guys know the song u by kendrick lamar

[rustic crown]

i don't

[bleak crystal]

i feel like the first 5 seconds of that song

[rustic crown]

so we did that right

Q(x) contains (2+3x-x^3)/(1-3x^2) and x - (2+3x-x^3)/(1-3x^2)

which are i and cbrt(2)

[rustic crown]

AAAAAAAAAAAAAAAAA

[bleak crystal]

you see?

[rustic crown]

first 3 seconds

[chrome hinge]

Another aproach would be: calculate the minimal polynomial of x=cbrt(2)+i, it has degree 6. Since Q(cbrt(2),i) has degree 6 and trivially contains Q(x), one has Q(x)=Q(cbrt(2),i)

[rustic crown]

calculating min poly isn't an easy task

or do you have some nice idea to do it without finding one poly which it satisfies and then stuff

[chrome hinge]

yes

look

thats always a nice method

Nah but you can take powers of x and combine to get 0.

[rustic crown]

yea but that won't guarantee it being minimal right

solving matrices by hand is hard

[chrome hinge]

I know, you gotta check irreducibility

Im not saying it is a better aproach, its just an alternative :D

Probably your way is faster

[rustic crown]

the problem with my approach is it won't generalize nicely

but probably no one will ask you to do harder calculations by hand

is the only way to compute minimal polynomial by solving matrices?

(sorry for my bad english)

[chrome hinge]

Ive never calculated a minimal polynomial using matrices actually

[rustic crown]

yea mostly i just hope to find a polynomial and hope that eisenstein works somehow

[chrome hinge]

yea exactly that haha

Or check irreducibility somehow

[delicate orchid]

since there are infinite primes you are guaranteed to find one that works with eisenstein and if you can't find one then you just need to look at more primes

trust me

[rustic crown]

x^2+1 irred using eisentein?

[delicate orchid]

bro there has to be one out there that works cause there's infinite choices honest

[rustic crown]

oh i geddit

(i don't)

[chilly ocean]

[delicate bloom]

well (x+1)^2+1 = x^2+2x+2

so actually eisenstein works

[chrome hinge]

Nani

[rustic crown]

wait is it actually true that any irred can by translated to show irred by eisenstein over Z?

[chrome hinge]

Theres infinite primes but finite ones which divide a finite given set of numbers

[chilly ocean]

x^2 +1 =/= (x+1)^2 though?

[rustic crown]

but if x^2+1 = f(x)*g(x)
then (x+1)^2 +1= f(x+1)*g(x+1)

[chilly ocean]

I see

[delicate bloom]

what's this matter?

there's no product happening just pure translation

f(x) is irreducible iff f(x+1) is

otherwise you could just factor and then translate back

[delicate bloom]

that's a good question I'm sure I've seen something about this before but I can't remember idk maybe we can figure it out

[rustic crown]

x^3+x+1

i'm sure there would be ways to cheat

[chilly ocean]

i been writting it heisenstein for some reason lol

[delicate bloom]

are there other tricks on top of this to do too

[rustic crown]

yea

you can scale as well

[delicate bloom]

like you can also do x^n f(1/x) to reverse coefficients

[rustic crown]

by going into Q and then Gauss to switch back and forth

[rustic crown]

yea was thinking the same lol

[delicate bloom]

do these allow you to get everything? probably not but

[rustic crown]

there is an algorithm to factor polynomials over a Z

[unique juniper]

oh right

[rustic crown]

but not fun

[unique juniper]

thanks @rustic crown

[rustic crown]

it uses too much interpolation

[unique juniper]

[rustic crown]

another is just going mod p and trying to factor it there and then coming back up.

other techniques are rational root theorem which only work for deg 2 and 3 over Z

idk any others

rest are very weird like showing irreducibility of cyclotomic polynomials over Z

that is just a very weird idea imo

[old lava]

ya, that's basically it for the standard techniques

1. rational root thm
2. looking at the polynomial images in quotient rings
3. eisenstein

[delicate bloom]

I just go to the algebraic closure and forget about most of this 😎

[old lava]

but they aren't exhaustive

[rustic crown]

lol

[rustic crown]

is there a way to see this?

[old lava]

x^4 + 1 in Z[x] is irreducible, but none of those 3 can prove it

pretty sure

[chilly ocean]

I asked in discussion math at one point, a algorithm to reduce polynomial over Z

I'm not sure I got an answer

[delicate bloom]

eisenstein with translation by 1 does it

[old lava]

oh true, I forgot with translation

[rustic crown]

say the polynomial is degree n and you look at f(0), f(1), ...., f(n)

so if f = g * h

then g(i) is a factor of f(i)

[old lava]

I don't know if they aren't exhaustive if you include translation

[rustic crown]

so you interpolate g by bashing over all divisors of f(i)

and then checking if its an integer polynomial

[delicate bloom]

p-power roots of unity I just immediately go to translation so that it's p eisenstein

[old lava]

my intuition tells me that even eisenstein + translation together with the other 2

still isn't exhaustive

but I don't have an immediate counterexample

for it

[chilly ocean]

Ah nice, that makes sense

[rustic crown]

this can be very good sometimes actually

consider f(x) = (x-1)(x-2)...(x-n) - 1

but that's just cause we made it so

only divisors of -1 and 1 and -1

so then g(i) = -h(i)

[chilly ocean]

Probably just an edge case: What if you get f(i)=0?

[old lava]

that sounds extremely cringe to do lmaoo

[old lava]

pick another i

[rustic crown]

(ig you don't pick i that case 😶 )

[delicate bloom]

divide out the factor

[chilly ocean]

But I'm happy know there is an algorithm

[delicate bloom]

of (x-i)

[chilly ocean]

Ah ok right

[delicate bloom]

and start over

[rustic crown]

oh lol

xD

[old lava]

also if f(i) is 0, then it's obviously reducible isn't it

[delicate bloom]

😛

[old lava]

so you're just done?

[delicate bloom]

yeah

if you're trying to show it's irreducible you know it isn't lol

[chilly ocean]

Although, if we are over a finite field, I think this algorithm fails for =0 issues. But in that case the problem is finite anyway

[urban acorn]

I like the fact that an F[x]-module is a vector space over F together with a linear operator on it.

[old lava]

it's very nice

F[x] modules are extremely beautiful

[urban acorn]

Generalizing this to choices of multiple operators is a possible way to motivate noncommutative polynomial rings.

[rustic crown]

PIDs are always nice

[old lava]

structure theorem of modules over PIDs applied to F[x]-modules

gives very nice results

[rustic crown]

and very nice if they happen to be Euclidean

[old lava]

(rational and jordan canonical forms)

[chrome hinge]

Im starting to study modules on thursday yaay

[urban acorn]

Euclidean domains basically just feel like constructive PIDs

[rustic crown]

yea lol

imma sleep nyow

[urban acorn]

good night

[old lava]

the only PIDs I have seen so far that aren't Eucliden domains are like Z[sqrt(d)] for some weird d

night det

[rustic crown]

good night

yea that Z[1+sqrt(-19)/2] is weird

[urban acorn]

I'm pretty sure that there's a generalization of the idea of a euclidean norm that the existence of is equivalent to an integral domain being a PID

[old lava]

ya that one is one example

[urban acorn]

https://en.wikipedia.org/wiki/Dedekind–Hasse_norm

I didn't really get into how it works, just heard of it

[old lava]

ya, d&f mentioned that

[urban acorn]

but it should be useful if one wants to understand what's up with PIDs that aren't euclidean

[old lava]

I dont wanna work with cringe PIDs that aren't euclidean

[urban acorn]

lmao

[chrome hinge]

lol

[old lava]

only reason to use PIDs in proofs is because you don't actually need to be constructive in 99% of proofs involving PIDs

so might as well prove things for PIDs to be slightly more general

but for any actual things, having the euclidean algorithm is just too nice to not have

[urban acorn]

by the way, this sort of reminds me, so if n,m are coprime, by chinese remainder theorem a residue class mod nm is just a choice of reside classes mod n and mod m, and I think the proof of that typically given isn't constructive, but there's a cool way to find the class mod nm that you want

[chrome hinge]

Z is nice

[old lava]

every ring is Z

actually

[urban acorn]

you just write an + bm = 1, and then x_1an + x_1bm is equivalent to x_2 mod n, and x_1 mod m

[delicate bloom]

good thing every field is a ring, now we know all fields are Z

[old lava]

true

[urban acorn]

yes and hence R is countable

take that cantor

[chilly ocean]

R doesnt exist, the largest ring is Z20

[urban acorn]

yes, every ring is an integral domain because zero divisors are cringe, and it's finite since it's no bigger than Z20, and therefore it is a field

• wilderberger has entered the chat *

[jagged dune]

Hi guys. I was wondering how to describe ring homomorphisms from Z6 to Z10.

[chrome hinge]

From the previous messages you can conclude Z6 and Z10 are basically Z

Thats a huge step

[urban acorn]

the image of 1 determines the entire homomorphism, so you just need to check which ones work

for it to be an additive group homomorphism, you want the additive order of the image of 1 to divide 6

since it needs to also divide 10, it has to divide their gcd which is 2, and thus be either 1 or 2

so that gives you either 0 or 5

and they both define additive group homomorphisms

if you want a ring homomorphism to be unital, then 0 doesn't define one, but 5 defines one either way i think

{0, 5} in Z/10Z is a subring isomorphic to Z/2Z, and clearly you can collapse Z/6Z into Z/2Z

(5^2 = 25 = 5)

[jagged dune]

Okay then. Thank you!

[urban acorn]

Here's something I just thought of. Let R be an integral domain. Inductively define $N_{n}$ by $N_{0} = {0}$, and $N_{n+1}$ = {$x \in R$ | for every $y$ in $R$, $y = qx + r$ with $r \in N_i$ for some $i \leq n$}. Then define $N = \cup_{i=0}^{\infty} N_{i}
$. \newline
I'm pretty sure that this is well defined, and $R$ is a euclidean domain iff $N = R$, in which case the norm taking $x$ to the minimal (but no longer unique) integer $n$ such that $x$ is in $N_n$ is a euclidean norm.

[cloud walrus]

Intel

[delicate orchid]

hold on lemme just

[urban acorn]

it's a nightmare to get everything to format properly

[delicate orchid]

$N_{n+1} = {x \in R-N_n : such : that \forall y \in R, ; y = qx+r such that r \in N_i : for : some i \leq N}$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

ah ok so you're seeing what N_n approaches?

[urban acorn]

I just realized I defined it incorrectly.

I don't just want to exclude N_n, I want to exclude N_i for all i at most n

[delicate orchid]

by R-Nn do you mean R\Nn i.e {x in R not in Nn}

[urban acorn]

yes

[delicate orchid]

got you

[urban acorn]

okay, I made a change to the definition that should also work

[next obsidian]

tfw ignoring classification of f.g. modules over a PID

[delicate orchid]

safe bet that q is just some element of R cause you're messing around with euclidan domains right?

[urban acorn]

Here's something I just thought of. Let R be an integral domain. Inductively define $N_{n}$ by $N_{0} = {0}$, and $N_{n+1}$ = {$x \in R$ | for every $y$ in $R$, $y = qx + r$ with $r \in N_n$}. Then define $N = \cup_{i=0}^{\infty} N_{i}
$. \newline
I'm pretty sure that this is well defined, and $R$ is a euclidean domain iff $N = R$, in which case the norm taking $x$ to the minimal (but no longer unique) integer $n$ such that $x$ is in $N_n$ is a euclidean norm.

[next obsidian]

Anyway, having a ring be a PID is actually miraculous

don't knock PIDs

[urban acorn]

yes

[old lava]

I mean PIDs are very nice and cool

[cloud walrus]

Intel

[old lava]

I agree that it is miraculous

but like if you're already a PID, just the little nudge so that it's euclidean

is very nice

[hot lake]

do you think sometimes you might want to extend that construction to some infinite ordinals until it stabilizes ?

[old lava]

and not that rare in the situations I've encountered

[next obsidian]

Euclidean domains are overrated, they exist to show something is actually a PID

[hot lake]

ah wait euclidean norm has to have image in N

?

[next obsidian]

the definition I know does

[delicate orchid]

I'm pretty sure N1 is just {1} although a second opinion would be much appreciated

[hot lake]

N1 would be the set of units

[urban acorn]

yes

N2 is universal side divisors

[delicate orchid]

wait but you'd need y = qx for all y in R?

[next obsidian]

This is weird, you're like doing valuation ring type stuff but with a different sorta norm

[delicate orchid]

which only works if x = 1 surely

[urban acorn]

and that's exactly true for units

[delicate orchid]

ohhh

so if x is a unit you'd set q to yx^-1

that's what I wasn't seeing

[urban acorn]

yeah

[delicate orchid]

got it now

[urban acorn]

anyway, @hot lake, I think that your idea about infinite ordinals might just work

but i'm not sure how helpful it will be

[next obsidian]

This is a chmonkey statement, but anything that's true for 1 is true for any unit

usually

[hot lake]

well I'm not sure if you ever need it

[next obsidian]

A unit is just 1 with an extra step

[delicate orchid]

very big "usually"

[urban acorn]

yeah, me neither, but I think you could prove everything that you can do with an actual euclidean domain about a "euclidean domain" with a norm into an ordinal as you could with a norm into the integers

makes me wonder if there's a ring where that matters

[old lava]

I usually just think of units as like a generalization of 1

that's how I kinda thought about units when I first saw them

[urban acorn]

I usually just think of math as like a generalization of linear algebra

[wraith obsidian]

I think of units as the irrelevant factor that gives you the same generated ideal and the same notion of divisibility anyway so you might as well ignore it

[urban acorn]

that's a good perspective

another one is that they're the "field-like" elements of your ring

[delicate orchid]

I just think of them as elements with inverses lmao

I haven't internalised rings to that point yet

[urban acorn]

an non-zero element a of R is not a zero divisor precisely when x -> ax is injective, and it's then a unit when it fully succeeds to define a bijection

[urban acorn]

literally lmao it takes an extra "step" by u^-1 to get to 1

[old lava]

I mean that's literally the definition

but it's nice to think about units in terms of their consequences

and what it means to be a unit

rather than just their explicit definition

[delicate orchid]

oh yeah of course

that's true with everything in maths

[wraith obsidian]

it means that body and soul are very well aligned
and something something crown shakra

[delicate orchid]

I'm just not well versed enough with them yet

[urban acorn]

I just think of units as * insert long and incomprehensible categorical jargon *, it really simplifies it

[next obsidian]

I think it's literally just so many things are true if (some statement about 1) if and only if (same statement about units)

Like an ideal is equal to R iff it contains 1 iff it contains a unit

In a valuation ring, the units are all things with valuation 0

[urban acorn]

btw, speaking of ideals

they sort of measure the way a ring fails to be a field

[next obsidian]

I think honor is a good one

that's a new one

I don't know if I necessarily agree

[delicate orchid]

yeah I can see that

[next obsidian]

I mean... in some senses

[urban acorn]

i mean, one way to fail to be a field is to have zero-divisors, and that's a different matter, but suppose we're looking at just integral domains

the thing that sets apart non-field domains from fields

[next obsidian]

I guess I agree in a way

[old lava]

I can see it, that ideals measure how a ring fails to be a field

in some way

[urban acorn]

is that in non-fields, you can multiply by something and then you can never go back

[next obsidian]

Like a DVR is pretty much as close to a field you can get

and they have only two primes, and all ideals are a power of the maximal one

besides 0

[urban acorn]

and an ideal, is precisely a set that you "can't multiply out of"

[delicate orchid]

isn't it as close as you can get by definition?

[urban acorn]

and then it's really awesome when you motivate ideals that way, and then independently show that it happens to be the right object to quotient a ring by

[next obsidian]

eh

[urban acorn]

cause it behaves "like 0"

[next obsidian]

I mean you can look at like, reduced local rings blah blah

I had some example I was looking at some time which seemed even closer than a DVR imo

Maybe it's just a special kind of DVR lol

[old lava]

[delicate orchid]

I mean you could literally just define a ring where each element has an inverse except for a single one I suppose

[next obsidian]

No you can't

[delicate orchid]

watch me

[next obsidian]

If x isn't a unit then ux is not a unit for all units u

[delicate orchid]

oh yeah

[urban acorn]

non-zero element

[old lava]

DVRs seem to be the closest to being a field, also like there's only "one" field substructure in a DVR, while any ring with multiple maximal ideals would have like multiple field substructures in it as quotients

I suppose

[next obsidian]

I mean any local ring has that latter property

also I think that's not true

[urban acorn]

yeah, btw, there are pretty cool actions of the group of units on a ring

[next obsidian]

I think you can have R/m1 and R/m2 isomorphic

just not canonically

[old lava]

I mean ya, but I still think of it as different structures

somewhat

[next obsidian]

¯_(ツ)_/¯

[old lava]

like in a supposed lattice of ideals or something, they would be different sublattices?

[urban acorn]

that's literally a field

[delicate orchid]

but one element doesn't have an inverse so it can't be a field

check mate

[next obsidian]

0

[delicate orchid]

yup

[urban acorn]

that element is 0

[old lava]

(I don't even know if a lattice of ideals is a real thing like it is for lattice of subgroups, I'm just kinda using it to have some basis for my imagination)

[delicate orchid]

no inverse so it's not a field

[next obsidian]

It is

you get the same isomorphism of lattices for ideals above I and ideals of R/I

It's just given by \cap and +

for meets and joins

[old lava]

and in this supposed lattice of ideals, a field is just a single line segment, while a DVR would be a few line segments connected end to end I suppose

which is the closest to the lattice for a field

as it gets

this is just how I try to think about it, because having some visual basis helps me I feel

[next obsidian]

I'm not certain that just having a lattice structure like that means a DVR

[urban acorn]

does the element of the field with one element have an inverse?

[delicate orchid]

ok I hear you 👂

[next obsidian]

not a field

not a field

not a field

[old lava]

I wasn't trying to say that, I didn't mean to say it's an iff relation between the DVR and the lattice

[urban acorn]

i'm not talking about the zero ring

[next obsidian]

That "field" has two elements tho

[old lava]

I just meant that I use the lattice to kinda ground the concept more

[next obsidian]

sadge

[urban acorn]

the "field with one element" is a theoretical entity

[next obsidian]

I know

[urban acorn]

then that "field" doesn't have "elements" so to speak

[delicate orchid]

pretty sure the field axioms require 1 to be distinct from 0 unfortunately

because they hate fun

[old lava]

[next obsidian]

cuz it would be annoying to write "except for the 0 field"

over and over

[urban acorn]

they do, and that's a really good thing

[next obsidian]

and "assume the map is non-trivial"

[urban acorn]

and that's not what we're talking about

[next obsidian]

Did you know Bourbaki doesn't require fields to be commutative lmao

[old lava]

wot

[urban acorn]

LMAO

[next obsidian]

They call those commutative fields

[urban acorn]

[old lava]

is that why there was a question posted here in this channel a while back where a textbook asks them to prove a ring is a field

but the ring was obviously noncommutative

[urban acorn]

oh my god i fucking love that

[delicate orchid]

what would be the benefit of a non-commutative field lmao

[next obsidian]

I doubt someone was using Bourbaki Algebra I as their text

but idk, maybe

[urban acorn]

i mean i hate it, but i love it as a source of comedy

[old lava]

nah, but a tb that was inspired by bourbaki

[urban acorn]

normal human beings call them division rings

[old lava]

not bourbaki as the primary source

yes

[urban acorn]

like, im pretty sure the hamiltonian quaternions are an example

[delicate orchid]

yeah I know but I meant like what would be the benifit of CALLING them fields

[old lava]

I think it was actually exactly that

[delicate orchid]

I know division rings are swag

[next obsidian]

It's easy to show finite rings without zero divisors are division rings

but they're also commutative

[old lava]

the question was to prove that the hamilton quaternions were a field

lmao

[next obsidian]

O_O

that proof is hard, idk it

[sly nexus]

F_1 is a meme

[urban acorn]

there's a really cool generalization

[next obsidian]

it's literally just

x -> yx

ooo injective

oh wait bijective too cuz finite

[sly nexus]

doesnt that just require a simple action from one element

[urban acorn]

if F is contained in R where F is a field and R is an integral domain, and that "field extension" is of finite degree, then R is a field and it's actually a field extension

[old lava]

lmao

[next obsidian]

showing this implies commutativity is hard... I think?

[sly nexus]

ah

ye

that thing

[next obsidian]

you don't need finite degree, just that it's integral

it also goes the other way

[urban acorn]

i'm pretty sure that's not true

[next obsidian]

no it is

[urban acorn]

like consider the minimal ring in R containing Q and pi

[next obsidian]

if R < S are integral domains

and it's an integral extension S is a field iff R is

I can give you a source

[final pasture]

In french it's actually common to still call division rings "fields"

[urban acorn]

(or any other transcendental number)

[final pasture]

(well, french's translation of fields, which is "corps")

[next obsidian]

cuz France is based on Bourbaki

Is my take

[urban acorn]

consider this extension of Q

[final pasture]

what I read @next obsidian

[sly nexus]

thats because rings are commutative

[old lava]

a commutative division ring is a field, so maybe there's some way to show that there's no proper ideal other than the 0 ideal by looking at kernels of homomorphisms or something?

[final pasture]

Oh and also, it's starting to change

[next obsidian]

@urban acorn Matsumura Commutative Ring Theory, section 9 Lemma 1

[final pasture]

Like newer generations are more likely to call division rings "corps gauche" (~= left (hard to translate properly) fields ?)instead of just "corps" (= fields)

[next obsidian]

"Let B be an integral domain, and A < B a subring such that B is integral over A. Then
A is a field <==> B is a field"

[urban acorn]

anyway, the proof of this is analogous to the proof that a finite integral domain is a field, x -> ax is a linear map over F, and injective, and over a finite dimensional vector space, and therefore surjective

linear algebra "makes it finite"

[old lava]

doesn't that mean "left field" literally

(it's been a while since I took french)

oh wait I can't read

you did say that

that's some weird naming

[urban acorn]

"such that B is integral over A" is the crucial hypothesis

[next obsidian]

Yeah I said that you don't need finiteness, just for it to be integral?

[urban acorn]

ohhhh

[next obsidian]

I meant integral wrt the extension

not just an ID containing a field

[urban acorn]

i thought you meant it just needs to be an integral domain

[next obsidian]

no haha

Yeah then that's clearly dumb like

K[x]

is clearly not a field haha

[urban acorn]

exactly

I don't know why I didn't name that counterexample that way

I said "minimal subring of R containing Q and pi"

which is isomorphic to Q[x]

[next obsidian]

lol

I was confused as to what you were going on about because that isn't integral

[urban acorn]

lmao

[next obsidian]

Like pi being transcedental means it's definitely not integral over Q

[urban acorn]

yeah

i don't like that I used pi specifically, I should've said "some transcedental number"

circles have nothing to do with this

[old lava]

why is K chosen to be the letter chosen for arbitrary fields typically rather than F

is there like some historical reason behind it

[next obsidian]

From German

[old lava]

ah

[scarlet estuary]

early indoctrination into k theory

[next obsidian]

Korper

which means body lol

(umlaut over the o tho)

[urban acorn]

[old lava]

I see

d&f has always used F for fields so far

[next obsidian]

American

[scarlet estuary]

you run out of letters eventually

[old lava]

and I noticed that online it was always k

[next obsidian]

Conversely, German people wonder why sometimes people use F

[old lava]

so I was wondering why

[next obsidian]

lmao

This isn't a joke btw

[scarlet estuary]

F in some sources implies finite field

[next obsidian]

that's pretty cringe

[old lava]

also I see lowercase k used a lot

rather than uppercase K

[next obsidian]

You should just write K with finite amount of ink

[old lava]

which I also found odd

[urban acorn]

can someone ELI5 k-theory?

[final pasture]

German

[urban acorn]

i have no idea what a vector bundle is

[scarlet estuary]

read quillen

[urban acorn]

and i remember looking it up in wikipedia

[scarlet estuary]

oh dont read quillen

[final pasture]

oh chmonkey already said it

oops

[scarlet estuary]

read grothendieck

[next obsidian]

So Grothendieck wanted to prove Grothendieck Riemann-ROch

then he invented K0

then Quillen went

asjfoadsjpoifjadsopivjasdiop fjlwfjodpaskfj dasokfj dsa

and now you have K-theory

[scarlet estuary]

chmonkey i think you might be slightly biased by me being a quillen stan

i mean okay quillen probably contributed the most of anyone

but leaving out atiyah is

pretty yuck

he basically invented k theory

[next obsidian]

I wrote Atiyah in invisible text

[scarlet estuary]

rather than just leaving it as a weird thing groth did once

[urban acorn]

but what is a vector bundles, and how do these generate rings?

[next obsidian]

Vector bundle is a locally free sheaf

[scarlet estuary]

a vector bundle is where you attach a bunch of vector spaces onto another space

its like vector fields but instead of vectors you parametrize spaces

kinda

thats a bad definition but thats the motivation

[next obsidian]

Take a space B, and then for a vector space V you can consider the projectino B x V -> B

[scarlet estuary]

now as it turns out we want to figure out the structure of these since like

[next obsidian]

To me it looks like, a cylinder

mapping down onto the disc

[scarlet estuary]

lots of shit like "space of continuous functions passing through a point"

[next obsidian]

the "height" comes from V

[scarlet estuary]

can be interpreted as vector bundles

if you assign it to each point

[next obsidian]

A vector bundle is a map say... X -> B which locally over B looks like B x V -> B

[scarlet estuary]

in practice thats not the kinda thing k theory looks at

(in fact its slightly wrong)

but that might help you visualize why we care

[old lava]

I need to learn some topology sadge

[next obsidian]

No u don't

you can just learn shneaf theory

[scarlet estuary]

now it turns out that you can relate SESes of vector bundles through black magic

[next obsidian]

[urban acorn]

so, like, should I think of it as a generalization of maps X -> V where X is some topological space and and V is some topological vector space?

[scarlet estuary]

and eventually you get a group structure

[next obsidian]

no, it's the other way around sort of

[scarlet estuary]

this is K_0

K_n for larger n is a mess

and requires you to read multiple long ass papers, one of which was coauthored by a spirit medium, to figure out fully

(K_1 and K_2 arent that bad but the general case blows)

[next obsidian]

What are K_n

[scarlet estuary]

higher alg k groups

[next obsidian]

are they related to SESs of vecotr bundles in some way

just in a fuck way

or does it kind of go off into space

[scarlet estuary]

im sure thats what quillen tells himself at night

i mean

you still deal with SESes

[urban acorn]

i still don't even have a vague intuition of what vector bundles are, and I don't know what SES stands for or how the fuck you end up with a group structure

and why does it create rings?

[scarlet estuary]

but of morphisms rather than the bundles themselves

and the morphisms are like

[next obsidian]

short exact sequence = SES

[scarlet estuary]

of morphisms of morphisms of morphisms

idk what you mean by "create rings"

[urban acorn]

oh okay

[scarlet estuary]

[in fact the quillen Q construction is a category of SESes kinda]

[urban acorn]

K-theory is, roughly speaking, the study of a ring generated by vector bundles
wikipedia

[scarlet estuary]

[and your k groups are loop spaces]

oh i see what you mean

[next obsidian]

wait

do you get a graded ring

by summing all K_n

[scarlet estuary]

but you usually already have a ring and study it by looking at vbundle invariants

no

[next obsidian]

or something

sd

[scarlet estuary]

you dont get anything nice

[next obsidian]

that would be cool if u did

[scarlet estuary]

its worth noting that the quillen approach is

very grothendiecky

in that it never really looks at concrete objects

just morphisms

[urban acorn]

what I take away from this conversation is that this is all black magic

[scarlet estuary]

so its hard to explain

[urban acorn]

and that all researchers in this field cry themselves to sleep every night

[scarlet estuary]

and theres like 4 big papers that you have to read before you can consider yourself familiar with it

2 of them by atiyah, quillen 1, and thomason trobaugh

they give the historical development which is nice

but on the downside

quillen 1 STILL isnt texed

so its in that terrible early-70s typewriter math format

typewriter comm diagrams are not fun to try and read

[next obsidian]

this is true

[scarlet estuary]

yeah basically

k groups naively are motivated in a similar way to pi_n almost

theyre weird but theyre invariant under things we care about

[next obsidian]

how did you even get into K-theory?

[scarlet estuary]

undergrad advisor needed a slave to do k theoretic computations for his math phys paper

[next obsidian]

lmao

[scarlet estuary]

i had a bit of experience in combinatorial AG at the time

so he figured id be able to pick it up vaguely enough to work

[next obsidian]

like flag varieties?

[scarlet estuary]

thats how im technically a published physicist 😎

[urban acorn]

lmao

[next obsidian]

wowza

then u decided to run with it?

[scarlet estuary]

nah it was like, a summer transfer project thing at the fields institute

[urban acorn]

i'm gonna do that someday

[scarlet estuary]

nothing got published

but we got a small article in a summary book

[urban acorn]

just make a weird non-physics contribution to a physics paper

and write on my resume i'm a published physicist in an elite university

[next obsidian]

Shamrock is a published PL person

[urban acorn]

by the way some 4chan anon once solved an open problem

[next obsidian]

about fucking

haruhi suzumiya right

on /a/

[urban acorn]

about sperpermutations

super*

[old lava]

damn, how is like 1 of the 4 major papers of the field not texed yet

[urban acorn]

i think there's a framing of the math problem that has to do with orders of watching that anime

anyway

so they wrote a paper and cited "anonymous 4chan user" as a coauthor

[next obsidian]

bruh

A lot of like important AG stuff is stuck in SGA and crap

which are like... hard to find, and all in French

[urban acorn]

and therefore anyone who uses 4chan is an anonymous 4chan user, and thus a published mathematician

[old lava]

I mean being hard to find or being in french isn't particularly as egregious

[next obsidian]

I don't know if they're texed either lol

[urban acorn]

french is the language of demons

[old lava]

since translating stuff and tagging stuff is a pretty damn hard job

[urban acorn]

any language which isn't english is the language of demons

[old lava]

but texing is literally 0 iq

I think I could tex a ton of things I don't even understand

it's literally just copying things

[next obsidian]

http://library.msri.org/books/sga/sga/pdf/sga4-1.pdf

here's SGA 4 for example

[old lava]

translating and tagging take a ton more effort

[old lava]

I think I could tex that

without too much issue

would I understand jack shit?

no

[next obsidian]

Yeah, SGA is not texed

and there's like

[urban acorn]

could you check that this diagram commutes without much issue?

[next obsidian]

literally thousands of pages

[old lava]

I don't even know the first thing about the diagram

[urban acorn]

this diagram checks that chuck norris commutes

[old lava]

but I feel texing papers isn't even hard

[scarlet estuary]

then tex quillen

i mean its technically illegal because copyright

but no one will care

[old lava]

lol

[next obsidian]

Tex SGA for me

[old lava]

I shall make myself known in the math community not for contributing any math knowledge

but by texing papers from the 70s

[next obsidian]

Fuck quals bro

every PhD student has to tex 50 pages of an important work which isn't tex'd

[old lava]

how long is quillen

if it's like <200 pages

[urban acorn]

math hasn't been at all meaningful for a long time

[old lava]

I would unironically not mind doing it

if my boredom levels go high enough

if it's like a thousand page monstrosity

[next obsidian]

lmao

[old lava]

then I might be a little not up for it

[next obsidian]

Why not like

learn AG or something instead

:^)

if ur that bored

[old lava]

because that would mean I have to use my brain

[next obsidian]

gggggggggggg

AG doesn't require using ur brain

it requires 1000 hours

[urban acorn]

math is sane, why do you ask?

[next obsidian]

I mean this is all done by a computer I'm sure

[urban acorn]

no, i'm not doodling, mom

[next obsidian]

lol

[old lava]

at first glance it doesn't look like anything

[urban acorn]

lmao

at first glance, that simplicial complex looks like a fish

[old lava]

your brain has some serious worms if you see that and immediately go "hmm, that's a trott curve"

[latent ingot]

what curve is it 👀

[next obsidian]

Idk, but it kinda sexy...

[urban acorn]

sexy curves

[old lava]

[urban acorn]

[old lava]

also I feel like these diagrams could be improved a bit (a lot)

I wish there was like a summer intern position where your sole job is to full time tex papers

I would love it

[urban acorn]

Ramanujan: hears hardy telling him about 1729
The worms in Ramanujan's brain: "hmm, that's the smallest number that can be expressed as a sum of two different cubes in two different ways"

[old lava]

I would not mind texing papers for 8 hours a day for the greater good of eyes all around the world

[next obsidian]

You are

the most interesting human being on earth

You somehow seem to not value your time on earth

[next obsidian]

I knew Muf did that react without checking

[latent ingot]

that would be time valuably spent

[old lava]

yes of course

[latent ingot]

cursed

[urban acorn]

i think that's like the cayley graph of S6 or something

it looks like something like that

[sturdy marsh]

perhaps one day I will actually learn what a cayley graph is

[next obsidian]

[sturdy marsh]

geometric group theory

[next obsidian]

💤

What should I do

[urban acorn]

ah, yes, this is what mathematicians do in their free time, of course

[sturdy marsh]

motivic cohomology

[latent ingot]

was this typeset in word?

[old lava]

probably not

[chilly ocean]

What are homology/arrow people gonna do when their diagrams start being 4 dimensional?

[maiden ocean]

when

[old lava]

just project them

down

and draw them

[sturdy marsh]

draw multiple diagrams

like draw each "level" separately

[maiden ocean]

hmm

does localization commute with all ideal sums or just finite ones

[sturdy marsh]

probably always

elements in an infinite sum of ideals are still finite sums

[maiden ocean]

yea id think so

the book only showed that formation of fractions commutes with finite sums of ideals but i dont see why it would not commute more generally

[sturdy marsh]

it should

[chilly ocean]

Pretty

[next obsidian]

prove it via universal property is my guess (I mean this is maybe one way to see it. Probably not the best way tbh)

Using Brofibrations observation that everything in an infintie sum is still made up of finite sums, you should be able to do it

[urban acorn]

btw til some people denote group multiplication "from the right"

i.e. ab means a then b

so then instead of left actions being the natural ones, right actions are

I guess the advantage is you can read it left to right and that's the order in which it happens, which is more natural

but it basically means you'll want to write f of x like (x)f

now that I think about it, why don't we do that?

[latent ingot]

(x)f looks wrong, also it's usually "f of x"

[old lava]

I swear this comes up every few weeks

someone advocating to write f(x) as (x)f

[latent ingot]

i'll sooner start writing f : B <-- A

[old lava]

https://tenor.com/view/not-happening-sorry-regina-george-rachel-mc-adams-mean-girls-gif-4670417

this is my reaction

every time

[latent ingot]

f : C^B, g : B^A, (g∘f) : C^A?

[snow flint]

it only looks wrong because uve been g r o o m e d

[latent ingot]

[chilly ocean]

I've been groomed?

[latent ingot]

but i'm usually the one grooming children

[next obsidian]

Commutative Algebra time

[latent ingot]

(x)f still reads wrong anyways

[chilly ocean]

Have you been groomed @next obsidian ?

[next obsidian]

Let A be a finite type subalgebra of B, a finite type k-algebra which is an integral domain

Show that for a prime p of B, that the coht_B(p) <= coht_A(p\cap A) (dis not true)