groups-rings-fields

[restive quail]

The teeny tiny book collection

It is growing

[copper kestrel]

peak!!

[next obsidian]

Cursed af lmao

[copper kestrel]

i need to find a present day rep theory book

[next obsidian]

U should get like, basic number theory by Weil too

[restive quail]

My professor has some lecture notes we are going off

For basic rep theory

Theyre okay

[restive quail]

Ooh

Thats a good idea

[copper kestrel]

my directed reading program is using serre and my mentor said "his techniques and things are outdated"

[next obsidian]

Whatever

You’re probably just learning group rep theory no?

[restive quail]

Serre one is tricky for a first read as an undergrad

[copper kestrel]

yeah just linear representations of finite groups

nothing crazy

[next obsidian]

Yeah there’s only one way to make a wheel when you really get down to it

[restive quail]

Milne course notes on group theory have a chapter on rep theory of finite groups i thought it was good

[copper kestrel]

i agree, its my first grad level text im reading

[next obsidian]

The cavemen did rep theory of finite groups functionally the same that we do now

So it doesn’t matter

[copper kestrel]

fair fair!

[restive quail]

Brian hall a good text

[next obsidian]

Talk about semisimple modules

Prove that you split stuff off or whatever the fuck

Do that averaging trick which is why you say char F doesn’t divide |G|

Then talk some shit about matrices

[restive quail]

Hurr durr G-invariant

[next obsidian]

Then do character theory and prove a bunch of psycho shit for free

[copper kestrel]

chcracter theory

[next obsidian]

It’s easy

[restive quail]

We had to prove frobenius reciprocity just using linear operators and no characters or C[G] modules and it was tedious

[next obsidian]

You literally just fill in a sudoku grid and it tells you everything

[copper kestrel]

scary though

[next obsidian]

Nah it’s p fun

Ngl

[restive quail]

Character theory is lowkey the most user friendly part of rep theory for finite groups

[next obsidian]

And you can prove that groups of order p^nq^m are never simple

Which makes the hw you did 6 months earlier to show groups of a bunch of different sizes can never be simple using Sylow trivial

[south patrol]

Modular character theory

[next obsidian]

Shit the fuck up poraro

[south patrol]

Ok

[copper kestrel]

woag

hi potato!!

i need to figure out easy problems to do today

[south patrol]

Yo

[restive quail]

My professor is trying to get us to leap from finite groups to GLn(Qp) in the span of like a month

[swift root]

poraro

[next obsidian]

For what

Rep theory?

[restive quail]

Yea

[next obsidian]

Lol

[restive quail]

Its cuz of his research and stuff but ive been tryna decipher some notes on GLn(Qp) by another prof in the department and im just totally lost

Uhhh parabolic induction blah blah something something

[next obsidian]

Idk rep theory so idk

[restive quail]

I checked out linear algebraic groups by humphreys so hopefully that helps

🙏

[next obsidian]

That book is

Shit

Jk idk

But my rep theory class I never went to was from it and the prof was ass and it was Covid zoom

So I didn’t learn shit (my fault fam)

All I remember is like 5 different letters that you write in fraktur or something

Type A baby

[restive quail]

Sound like lie algebras

I always run into issues taking notes on lie algebras cuz i write the fraktur letters as just lowercase

But then its like

Let h be in h (fraktur)

😓

[next obsidian]

Wait yeah

What am I on

I think I rea linear algebraic groups for a reading course

That was decent

I don’t really know shit about algebraic groups still tbh

Zoom era reading courses be lik

But it’s cool they’re all smooth

[copper kestrel]

computation slop

[next obsidian]

Plug in 3 numbers

Lol

[copper kestrel]

wdym

is it not going to be a 9x9 table

[next obsidian]

It’s a quadratic it’s irreducible iff there’s no roots

You have 3 things to plug in

[copper kestrel]

oh no im talking about b

part a is easy

[next obsidian]

Oh

[copper kestrel]

sorry LMAO

[next obsidian]

Whatever

Sudoku

[copper kestrel]

yes computation slop is plugging in 3 numbers its true 💔💔💔💔

[quiet pelican]

Addition is trivial and if you know how to describe it abstractly, I wouldn’t even write it down unless it’s to hand in for homework
Multiplication only really has a 6x6 of non-trivial entries

[copper kestrel]

it is hand in unfortunately

but i agree it is trivial for addition

[quiet pelican]

And actually multiplication only has a 3x3 of genuinely non-trivial content
The rest is multiplying by scalars

[copper kestrel]

truth

[south patrol]

If it factored nontrivially in Z_3 then by reduction mod 3 it would factor over F_3

[quiet pelican]

😠

C_3 is a group
Z/3Z is a ring
F_3 is a field
Z_3 is an infinite profinite ring

[tall igloo]

I think you misspelled Z/3

[chilly radish]

Which is also a group when you want additive notation

C_3 is multiplicative notation only

[tall igloo]

ohhh that's a hot take but I like it

I have a confession to make, I sometimes write 0s at the ends of short exact sequences of noncommutative groups. just because my mind is evil like that

[swift root]

personally, i use Z3^-1 for the localization by 3

[tall igloo]

Z/3 = 1/3 Z = 3^{-1} Z

[karmic moat]

the only type

[dim widget]

just a small glimpse inside the mind of a twisted cycle path

[south patrol]

I have a confession that I often forget to add the 0s because of stable ∞-cats

[sudden condor]

my dumb ass thinking u meant felines rather than categories

[south patrol]

dw they are types of felines

[copper kestrel]

mrow

[knotty badger]

I am cat theory #1 fan

Lots of people have thought this meant only categories

But I like cats too

[quiet pelican]

🐱

[karmic moat]

which distinct primes ideals P and Q of Z[i] satisfy P \cap Z[15i] = Q \cap Z[15i]? (in other words, over which points of Spec Z[15i] is the normalization Spec Z[i] -> Spec Z[15i] not an isomorphism)

the 15 should play some role here but it's not clear to me how

[velvet hull]

I think it is useful to first understand Spec Z[i] better

[karmic moat]

yeah it's a PID and gaussian primes have one of three forms

[velvet hull]

what about the residue fields?

what can they look like?

[karmic moat]

probably some finite field

let me think about it

[velvet hull]

you can say something better than that

[karmic moat]

should split into three cases

[velvet hull]

because if you can figure out what the residue field looks like, you get an quotient of Z[15i] into said field

and that aleady tells you a lot about what the kernel should be, i.e. the contracted prime ideal

[karmic moat]

oh wait quotient field lol i was thinking about residue fields

the quotient field is just Q(i)

[velvet hull]

I mean residue field

[tough raven]

There's an ideal called the conductor (IIRC) such that it can only go wrong away from there. So you compute the conductor and check the finitely many primes dividing it.

[karmic moat]

oh

[tough raven]

Specifically you only have to check primes dividing the discriminant of (the minimal polynomial of) 15i.

[wraith cargo]

factorize 15 into primes over Z[i] :)

[wraith cargo]

it can only go wrong with the primes dividing the conductor!

since the primes dividing the conductor are those for which the map I |-> I cap Z[15i] isn't an isomorphism

tho here finding the conductor should be pretty trivial

even I think you can do this from first principles

[karmic moat]

(1+i) should give Z/2Z
(a+bi) where a^2 + b^2 = p = 3 mod 4 should be F_p^2
(p) where p = 1 mod 4 should be Z/pZ
or something along those lines right

[velvet hull]

Z/p^2Z is not a field

[karmic moat]

sorry i meant F_p^2

[velvet hull]

you actually know a little bit more on the explicit structure of F_p^2 given by the isomorphism in this case

[karmic moat]

F_p[i]?

[karmic moat]

i've heard of this a long time ago i'll look at it thanks

[karmic moat]

3(1+2i)(1-2i)

[wraith cargo]

good now show the only ideals for which this is possible are those that are divisible by 3, 1+2i or 1-2i

[chilly radish]

I'm not sure what you mean by an embedding here

[velvet hull]

i meant to say quotient

[chilly radish]

Ah

[south patrol]

Aye

[karmic moat]

i guess the issue is i'm not really sure what p \cap Z[15i] looks like, like explicitly

for p = (3), (1+2i), or (1-2i), it's easy

but i can't seem to get a clean idea of what it should look like for other primes of Z[i]

[south patrol]

Le Dedekind(-Kummer) lol

[karmic moat]

i don't know ant 😔

is this my sign to read neukirch

[dim widget]

do you understand the story for Z[i]? As people have been saying, there will only be a meaningful difference when p divides 15 which you just covered

[karmic moat]

yes ^

[south patrol]

Lol why this Tteg

But yeah okay for Z[i] lmao not needed

[dim widget]

oh idk what Dedekind Kummer is

[south patrol]

For Z[i] I think u can just do this quite explicitly by hand viewing it as Z[x]/(x^2 + 1)

[dim widget]

I thought you were trying to make some kind of brain rot joke

[south patrol]

Ah no I mean like what I think is usually called Dedekind's theorem like

On how primes split for integral extensions of Dedekind domains

[dim widget]

So if p is a prime in Z[i] you’re having trouble taking p \cap Z[15i]?

that part is just linear algebra basically: if p = (a + bi) then p \cap Z[15i] can be calculated explicitly in terms of a, b

[karmic moat]

Yes, I was running into some issues with divisibility, but probably my approach was wrong

I gtg for now but I’ll be back later ty all

[dim widget]

I think the only complication is if 3 or 5 divide b or a

but even there you can show nothing actually changes

unless you want to find minimal generators for the ideals

[wraith cargo]

is this something like it's the element of (a+bi) with 15|b and having minimal norm?

[dim widget]

Yeah

or just (Nm(p), 15p) if p is split

[wraith cargo]

ah okay I wasn't sure if these norm arguments transferred over to Z[15i] cuz that shit ain't a Euclidean domain I think

[coral steeple]

This is $\langle\lambda,v\rangle=\lambda(v)$ for $\lambda\in V^*,v\in V$?

[cloud walrus]

person2709505

[coral steeple]

I've never seen angle brackets take objects of different types

[dim widget]

Everything works fine for ideals prime to 15 as you know

[wraith cargo]

Ik this result abstractly by appealing to the conductor but I was trying to see if I can derive everything from some notion of first principles

[dim widget]

well it’s easy to prove from first principles that R[1/x] is a UFD if R is, and this shows that one has unique factorization in R[1/15]

[wraith cargo]

Ahh okay

That's kinda cool

[lucid dagger]

has anyone heard of CFSG potentially having an unfixable gap https://mathoverflow.net/questions/114943/where-are-the-second-and-third-generation-proofs-of-the-classification-of-fin#comment1322743_217397 I have heard that experts in the field are now considering cfsg conjecture again

[wraith cargo]

Have you heard this from anywhere else other than an MO comment

[lucid dagger]

yes i emailed up a chain of people, though not all the way to the top. so far i have "testimony" from a somewhat famous professional mathematician about anonymous experts in the area who consider CSFG conjecture and do not use it anymore in their work

[wraith cargo]

this still seems incredibly suspect to me

[proud vigil]

https://math.stackexchange.com/questions/348539/how-confident-can-we-be-about-the-validity-of-the-classification-of-finite-simpl?rq=1

[chilly radish]

Kinda crazy to me that this has been kept under wraps if it's true

[karmic moat]

yes I suppose this is my problem, i'm trying to find minimal generators

i only have this so far but i can't think of minimal generators for the last two cases

[somber goblet]

wow that would be wild yeah

i wonder if such an error, if fixed, would somehow help us with inverse galois problem

[dim widget]

it’s hard to imagine how they’re related

unless the error was that everything was isomorphic to an iterated extension of PSL_2(F_p)’s and Sn and they missed it the whole time

that would be nice

[somber goblet]

idk, just thinking about how we more or less know how to put galois groups together and find an extension with a galois group of the resulting group extension

so i feel like a result about "all finite groups" would have to appeal to this classification somehow (?)

i don't know that much advanced group theory lol

[dim widget]

well I think the answer depends on what a, b are at a split prime

if a or b is already divisible by 15 then it’s clear what happens

[karmic moat]

hmm i guess it's probably not necessary to compute minimal generators then

[dim widget]

if not you get (p, x\pi), where x is either 5, 3, or 15

[karmic moat]

oh

[dim widget]

and \pi is a + bi

[wraith cargo]

in you case you only have the first and last case to consider :)

[karmic moat]

yeah because the residue field of (1+i) in Spec Z[i] and the residue field of the image of (1+i) in Z[15i] are both F_2?

so the normalization is an isomorphism on this point

[small yacht]

Oh I get this now
Although there’s a more intuitive explanation I thinks

[swift root]

maybe, but this is a nice picture to have in mind

though, what is your explanation?

[small yacht]

Okay so I think it’s best to start with the analogue of this in set

Well with sets

If you define an equivalence relation where
a ~ b iff f(a) = f(b)

Uh wait let me gather my thoughts

then for all functions f: A -> Z
such that equivalent elements have the same image, there exists a unique f': A/~ -> Z

Now in groups, this looks like a ~ b iff \phi(a) = \phi(b)
rewriting this, we see that it's equivalent to saying \phi(ab^-1) = e

[swift root]

I don't think this is more intuitive than my picture lol

at least, not when the picture is actually explained

[small yacht]

Because it's a subgroup of ker it's also normal
so yada yada yada qed

[small yacht]

well idk i thought making a connection to a probably more intuitive thing with sets is helpful

because it's easy to understand how the equivalence classes of similar elements make a unique homomorphism

[swift root]

the picture is about sets

[small yacht]

but not necessarily with groups

[swift root]

or algebraic structures in general

[small yacht]

Oh uhhhhh
Without quotients

[swift root]

sets have quotients

[small yacht]

idk i just don't think it's immediately obvious how that works with a quotient compared to that

[small yacht]

no i meant like connecting it to something more intuitive without quotients

idk am silly ikd

[swift root]

this is a quotient

A/~

[small yacht]

WELL

[cursive spindle]

[small yacht]

equivalence relation i mean

[swift root]

hai deltoi

[small yacht]

hii

Wait how do quotients with other sets work in set

[swift root]

you dont

you quotient by an equivalence relation

that is what my picture shows

the boxes are the equivalence relation ker a

[cursive spindle]

Idk if it's really helpful to jam in categorical nonsense

[small yacht]

uhh uhhh uhhh

[swift root]

and the FIT is the statement that the boxes are (naturally) in correspondence with the points in B (the image of a)

[quiet pelican]

The reason we’d usually write this sort of thing in other contexts is that if things are sufficiently nice, we can associate a “canonical” equivalence relation with any subobject, and get a “reasonable” quotient

[cursive spindle]

this is literally more than enough

[small yacht]

yuh i understnad that

ugh srory i dont knpw you're right

[swift root]

i guess i dont really understand how bringing the universal property makes it more intuitive

[cursive spindle]

[swift root]

like, youre right, if you understand what it does then the universal property is intuitive

in fact, that property is great and its super useful for constructing maps

[small yacht]

this is a universal propertty too isn't it

[vestal trench]

but it is not at all necessary for grasping the first iso thm

[swift root]

no, its a visual representation of the three maps of interest in the first isomorphism theorem

[vestal trench]

i would wager FIT is most peoples first example of the universal property

[small yacht]

oh i was just thinking in terms of this

[swift root]

thats different from my picture

[small yacht]

am so stupid

[swift root]

(in fact, the book its from has something against CT for some reason)

(lol, still a great resource from UA)

[small yacht]

but they look so similar

[swift root]

they do look similar

[cursive spindle]

For a first glimpse to algebra I agree with not jamming in CT

[small yacht]

what's CT

[cursive spindle]

cat theory

[swift root]

a course in UA is not a first glimpse to algebra

[small yacht]

ohh

[swift root]

it very much should warrant categorical notions being explored

[cursive spindle]

well yeah but what Joli is doing is not UA

[swift root]

the picture is from that book

thats what i meant

[small yacht]

yeah my textbook just really likes introducing stuff with cat theory hence why i guess im very cat theory pilled

[vestal trench]

i smell aluffi

[small yacht]

indeed

[swift root]

the difference is that in my picture, B is the codomain of a, while in your picture G' is some group over G

or under G, i guess if you want to use categorical terminology

[cursive spindle]

wait i recommended aluffi to someone

I will tell them not to

[vestal trench]

😭

[swift root]

aluffi is great

[vestal trench]

aluffi is a good book

[swift root]

but not for a first glimpse at algebra

[small yacht]

it was my first glimpse at pure math

[swift root]

i actually dont know what a good book for introductory algebra is

[vestal trench]

aluffi's undergrad book is also very nice

and not catpilled

[swift root]

undergrad boom

[azure cairn]

notes from the underground?

[vestal trench]

it is rings first

yeah

[small yacht]

dostoevsky reference

[azure cairn]

i was planning on reading the actual book by dotsoevsky at some point but then forgor

[small yacht]

i read it in 8th grade i think

it was pretty great i liekd it

[somber goblet]

notes from undergrad

[vestal trench]

he also does modules 😍

[azure cairn]

aluffi's book??????

[vestal trench]

society if undergrads learned about modules

[azure cairn]

oh wait the

nvm lol

[swift root]

modules my beloved

well actually

abelian categories my beloved

[vestal trench]

close enough

[somber goblet]

i got modules with my algebra

but like

at the end of vsp chapter

"btw if you relax field to ring nothing works anymore lolololol"

[small yacht]

no lol waht

i started reading that 2 years later

which is

right now

well i started it a whole ago

[swift root]

hey you still have the invariant basis number

(over comm rings)

[somber goblet]

to be fair they took rings to be non-unital

so

nothing works ever in general

[vestal trench]

has anyone actually learned their linear algebra through blythe

[swift root]

death

[vestal trench]

blythe module theory i mean

[small yacht]

qusi semi ring

[somber goblet]

quasimagma (just a set)

i know an ideal I of R is a R-module

right

or am i making that up

[swift root]

yes if unitalcomm ring

[somber goblet]

so you can recover kernels if you switch from talking about rings to R-modules

[swift root]

left (resp- right, bi) ideal if it is a left (resp- right, bi) submodule of R

[somber goblet]

if non-unital then ideals are rings anyway

[cursive spindle]

they do?

[swift root]

not here 💔

i bepieve

VELIEVE

okay

i will quit typing

[cursive spindle]

okay it's not mandatory here but in theory you can take a course

[somber goblet]

i learned about them as "dumb stupid idiot vector spaces also any abelian group is a Z-module also they have a structure thm"

[swift root]

torsion is what makes them interesting

also not being projective

so Ext is nontrivial for once

[cursive spindle]

tfw torsion part of cohomology of arithmetic groups

[somber goblet]

i cant think about torsion without thinking about that one gif of the cat going "my bals"

[cursive spindle]

I recommended Rotman to my friend

[somber goblet]

i know its when element in infinite thing has finite order

[swift root]

just finite order

[quiet pelican]

This is true
Just ignore all of algebra after it

[swift root]

or more generally being annihilated by some element

[vestal trench]

no one i know of does (in the US)

[somber goblet]

well yea but nobody cares about torsion if ur thing is already finite ☕

[vestal trench]

yeah you can just take a graduate algebra course but

its kinda silly that they dont teach them

[cursive spindle]

In my institution, abstract algebra is famously known to be the hardest class so it makes sense why most students don't consider taking algebra afterwards

[somber goblet]

inchesting

[cursive spindle]

If anything, most of math majors are going into education or f*nance

[vestal trench]

yeah

[somber goblet]

yikes

[cursive spindle]

There are a couple who try to get into quant but afaik nobody succeeded

[vestal trench]

my institutions default math degree has single variable real analysis as the capstone course 💀

[somber goblet]

damn

[vestal trench]

its built to churn out finance and engineering

[somber goblet]

one uni i was looking at had for its algebra track intro group theory as a senior class

they make you sit through matrices and diffeq and abunch of bs first

[cursive spindle]

It's so funny to see my peers struggling with abstract algebra and succeeding in numerical analysis

while I'm literally the opposite

[azure cairn]

the college im going to this fall has too many courses to choose from

[quiet pelican]

cries in geometric group theorist

[karmic moat]

just pretend all your rings are PIDs and all your modules are projective

add whatever wonderful conditions you'd like

noetherian artinian whatever

[cursive spindle]

Take algebra, analysis, topology, algebraic geometry, algebraic topology, differential geometry, representation and couple of number theory. Easy choice

[karmic moat]

oh well I guess PIDs are noetherian so that doesn't really mean much

[quiet pelican]

I recommend only taking geometric group theory, mapping class groups, low dimensional topology, trees & group actions, algebraic groups, group cohomology, topics in infinite groups, groups & computability and maybe a couple more group theory courses

[azure cairn]

i will Decide whenever i actually know what half of these courses are 😂

[cursive spindle]

lmao

Seriously though once you finish the core classes your heart will know what courses to take

[karmic moat]

take introduction to pottery, introduction to painting, introduction to sports management, yoga, and spin a wheel to decide what foreign language courses

[cursive spindle]

real

[quiet pelican]

Yeah partially I took an absurd courseload, but I didn’t really feel like I was making a choice beyond “do all the pure except maybe my 2 least favourite” in my Bachelors

[cursive spindle]

I took a couple of foreign language courses

[karmic moat]

i took two courses in classical chinese and chinese calligraphy

those classes were so fun

[cursive spindle]

I made the dumb choice of doing cs and math

[karmic moat]

it was nice to have a break from math in my schedule

[cursive spindle]

I should've done physics/math or just math

[azure cairn]

i have to take 8 humanities courses so ill have to take 1 per semester till 4th yr lop

[cursive spindle]

Actually since there are not many pure classes in my uni, math only would be my downfall cuz that will mean I have to take A LOT OF STATS

[somber goblet]

epic

what are good freshman courses to take undergrad

[cursive spindle]

Finish your calculus slop then right away start with analysis and algebra

[quiet pelican]

All the group theory courses they offer :3

[cursive spindle]

My honest reaction

[somber goblet]

blegh i wish i could skip calc slop

[karmic moat]

whatever your uni tells you too

[vestal trench]

you can skip analysis too

[karmic moat]

get all the stupid annoying stuff out of the way

[cursive spindle]

My math degree consists of 2 years of just pure calculus slop

[somber goblet]

thats so nasty

[cursive spindle]

In your third year you start analysis and algebra

[karmic moat]

i didnt have to take any B)

[somber goblet]

maybe if im lucky theyll take calc bc for a calc 2

[karmic moat]

my school did

i only had to take calc 3 which i got waived after a while

so i didnt take calc 3

i took it in high school tho

[somber goblet]

bro if i have to take an intro proofwriting

its over

[quiet pelican]

I love being in a European country
Doing only maths and less mathslop

[cursive spindle]

genuinely one of the worst classes you can take as a math major

its so ugly

[karmic moat]

i thought "intro to proofs" was just "intro to analysis"

[quiet pelican]

It’s more often like “intro to elementary number theory, set theory and combinatorics”

[karmic moat]

hmmm

[somber goblet]

if i have to prove fermat's little theorem without lagrange im going to turn into the joker

[karmic moat]

in the us we call that "discrete math"

[quiet pelican]

But like, I mean the sort of all of those that every mathematician has to know

[cursive spindle]

elementary number theory is so ass

[somber goblet]

if i have to learn all of elementary number theory after slaving away on rings its actually over

[karmic moat]

you could just talk to the department head

or whoever is in charge of undergrad studies

and ask them to waive some classes and/or just let you take the final

[somber goblet]

hmm perchance

[karmic moat]

that's what i did

¯_(ツ)_/¯

[cursive spindle]

I did ask the department head and they said no

like 2 years ago

[somber goblet]

gulp

[karmic moat]

did you try begging

[cursive spindle]

yeah lol many times

These were core classes so it's not possible

[karmic moat]

like groveling and orz and the whole nine yards?

unlucky

[cursive spindle]

on the bright side I had to learn a lot of math on my own since many classes I wanted to take were not offered

[somber goblet]

idk

i hope what i do in uni is actually a continuation of what im doing now

[azure cairn]

some colleges (mine is one) start designate intro analysis as the proofwriting course

[small yacht]

oh the next chapter literally talks about FIT

[cursive spindle]

that could go a lot of ways

[azure cairn]

yeah not sure how it is

i think peopple willingly volunteer to help people with proofwriting in that class

[cursive spindle]

my proofwriting class was so miserable it was genuinely hard to keep track of all the things the prof wanted us to write

like they wanted a very specific format of proofwriting which was not my thing at all

on the same semester I was taking representation theory and it was really hard but so much bearable than the proofwriting class

[chilly radish]

Cringe. Giving these things names is useful, they come up all the time

Even if you don't number them you at least need some way to refer to them, and it's useful to differentiate exactly which type you're using

[small yacht]

this is the theorem he's talking about for context

[frail shoal]

homomorphism theorem my beloved

[somber goblet]

is this a different formulation of first iso thm

[small yacht]

a universal property of... something?

[next obsidian]

The quotient

[somber goblet]

oh that

oh yeah ok cool i guess

[small yacht]

oh right because H is a subgroup of kernel it's normal

so it's any quotient

[next obsidian]

The quotient represents the functor “maps from G to -, such that H is sent to the identity”

[somber goblet]

subgroup of normal subgroup is normal

totally forgot

[next obsidian]

Which gives it a universal property

[small yacht]

no i got it i just forgor that kernel <=> normal

[next obsidian]

Its initial in some stupid category which is things under G for which the structure map annihilates H

[small yacht]

also i don't know what a functor is so your words mean nothing

[next obsidian]

That’s fine

It’s just a way to formalize the concept of universal property in this case

[small yacht]

it's initial in the category of morphisms from A such that for f:A -> Z, object is (Z, f) and a ~ b <=> f(a) = f(b)

or something wait it's formulated better

[next obsidian]

There’s in general for any category the colon and co-colon category I think is the name

But I call them over and under

In this case it’s the under category and it’s just objects A along with a specific map G -> A

Called the structure map

And morphisms of these objects are maps A -> B such that the relevant triangle commutes

The full subcategory of objects for which the structure map sends H to the identity has an initial object, G/H

[small yacht]

you're very smart chines monkey

[next obsidian]

I’m just a chair monkey

Also I’m Japanese

[small yacht]

iis this how you got postgrad
by knowing a lot of stuff about abstract algebra

[small yacht]

oh i recall tyou saying that after last time i called you chcinese monkey

[next obsidian]

Uh

I mean at the time j got t for knowing a lot of AG

[small yacht]

what's AG

[next obsidian]

Algebraic geometry

[small yacht]

ohhh

i see i see

[next obsidian]

Maya the jolii

[small yacht]

Chill monkey

[next obsidian]

Chill monkey

😎

[small yacht]

Maya the jolli

[chilly radish]

This is absolutely false.

[chilly radish]

No, you have to assume that H is normal

[next obsidian]

No it’s sometimes false

[chilly radish]

As a universally quantified statement it is false

[next obsidian]

[small yacht]

In groups I mean

[chilly radish]

I'm not sure what you mean

[somber goblet]

wait what

[chilly radish]

Not every subgroup of the kernel is going to be normal

G is normal in G but for any non-normal subgroup H, H isn't normal in G

[somber goblet]

if H is a subgroup of N and N is normal in G then H is normal in G

[chilly radish]

No

[next obsidian]

Not even a normal subgroup of a normal subgroup is normal in general

[karmic moat]

one of the most devastating facts ever

[somber goblet]

wait

no

omg

i thought about it

[next obsidian]

A characteristic subgroup of a normal subgroup is normal

[chilly radish]

Counterexample is S_4 has Z/2 x Z/2 as a normal subgroup which has Z/2 as a normal subgroup which isn't normal in S_4

[small yacht]

Oh, I see what you mean

The kernel itself is normal

But subgroup isn’t

Necessarily

[chilly radish]

I got the generators wrong

[chilly radish]

Right

But in the theorem he takes H normal

[small yacht]

Oh well quotients of groups are only groups if H is normal

So it’s still the universal property of quotients

[somber goblet]

ring theory quotients are constructed in RMod right?

for commutative rings

[next obsidian]

Wut

I mean you can do that and then see it’s still a ring or whatever

[somber goblet]

like ideals aren't rings

[next obsidian]

So what

[somber goblet]

oh i guess it doesnt matter

you phrase in terms of homomorphisms anyway

[next obsidian]

I mean it depends on what you mean by quotient

It won’t be a cokernel

Maybe you can phrase it as one from a product of R if the ideal is finitely generated

Not too sure and don’t wanna work it out

[chilly radish]

They are, just not with unity

Don't let "them" lie to you

Rings don't need a 1

[cursive spindle]

[chilly radish]

We can just add it back anyways if we want to

[somber goblet]

ok wait an ideal is a submodule of R as a R-module

if R is commutative

[karmic moat]

sure

all the submodules of R as an R-module are the ideals

[chilly radish]

It's a bimodule in the noncommutative case

[karmic moat]

do you need commutativity tho?

sniped

[chilly radish]

Well, depends how you define ideal

A left ideal is a left module, a right ideal is a right module, a two-sided ideal is a bimodule

[karmic moat]

what do people usually mean by "ideal" in the non-comm case

i always thought "ideal" meant "two-sided ideal" but i guess now that i think about it, it could just mean any of the three

[chilly radish]

Usually ideal not preceded by an adjective means two sided

[karmic moat]

yea thats what i meant

nice

[cursive spindle]

I hate non commutative land. Some papers say rings but they don't include one and some use rngs which is fine and straightforward

[chilly radish]

But if you are only working with, e.g. left ideals, you usually say that at the start and then just use "ideal" unless you need to clarify

[karmic moat]

fair enough

[somber goblet]

i like commutative algebra

rings are commutative and have 1

life is good

[chilly radish]

Noncommutative rings with 1 are still very nice

Without 1 is kind of the horrors

But uhhh

The theory of the jacobson radical is cool!

You need to define like a pseudo identity to even make sense of the definition

[karmic moat]

aren't there non-commutative rings which are noetherian on one side but not the other

that's pretty horrifying to me

[chilly radish]

Yea

That's cool!

[somber goblet]

awful

[cursive spindle]

Okay noncommutative with 1 is good stuff

[chilly radish]

It's interesting

[somber goblet]

matrix algebras come to mind

[cursive spindle]

Rings without 1 is just annoying

[chilly radish]

Matrix algebras are extremely well behaved

[somber goblet]

yeah matrices are nice

matrix algebras are like symmetric groups

[cursive spindle]

I remember my prof yapping about them being important for compact support

Rings without 1

[chilly radish]

Noetherian, artinian, semisimple, central simple

How lovely

[azure cairn]

i only needed to read 1 chapter to be able to meaningfully contribute in #real-complex-analysis , its been 6 with artin and i Still have no idea what goes on in this channel 😂

[cursive spindle]

CSA my beloved

[somber goblet]

what

[chilly radish]

Central simple algebras

[somber goblet]

ohhhhh

[chilly radish]

The abbreviation is unfortunate, but it is standard

[somber goblet]

👍

[cursive spindle]

Huh what is the other thing you thought of

[chilly radish]

First word is "Child"

[cursive spindle]

[azure cairn]

child sexual abuse

[cursive spindle]

bruh

[chilly radish]

/assault

[cursive spindle]

I need to be careful when I'm around non mathematicians then

[azure cairn]

when i read it i thought it stood for "computer science a"

[somber goblet]

anyway

[chilly radish]

Removing commutativity makes things interesting

[somber goblet]

yes

removing 1 makes things dumb

[chilly radish]

For example, now you have two Jacobson radicals! Are they identical? Yes, but that's fairly nontrivial to show

[azure cairn]

is that what is called a nonunital ring?

[somber goblet]

yea

or a rng

[azure cairn]

🎉

[cursive spindle]

I don't have a very good reason to care about dropping 1 in my rings but that might change

[chilly radish]

I think C* people care about nonunital stuff

[somber goblet]

rn i'm trying to pick up commutative algebra

[chilly radish]

Von Neumann algebras

[azure cairn]

comm alg looks interesting

[somber goblet]

so life is all well-behaved

[chilly radish]

Approximate identity

Idk

[azure cairn]

Exactness, direct limits, tensor products, Cayley-Hamilton theorem, integral dependence, localization, Cohen-Seidenberg theory, Noether normalization, Nullstellensatz, chain conditions, primary decomposition, length, Hilbert functions, dimension theory, completion, Dedekind domains.
this is what the course decription is for comm alg at the college im going to

[somber goblet]

well now it's all about integral domain, UFD, PID, dedekind domain...

so maybe i have more to contend with

[chilly radish]

I hope by Nullstellensatz they mean the full Jacobson ring one

Cuz otherwise that's way too late

[somber goblet]

is that an undergrad course or

[azure cairn]

the prereq to comm alg is tw ocourses that cover the entirety of artin

[azure cairn]

no it's a grad course

[somber goblet]

oh thank god

[azure cairn]

there's nothing after the abstract algebra sequence besides grad courses at my college though

[somber goblet]

i was worried bc i havent touched most of that lol

[chilly radish]

Lowkey should at the very least be a 4th year ug course

[azure cairn]

a grad course is basically a requirement for the math major

[chilly radish]

Suspicious lack of primary decomposition or flatness though

[cursive spindle]

Wtf is Cohen Seidenberg theory

[chilly radish]

Amd by extension discussion of Tor

[azure cairn]

like a pure math major needs to choose 1 of 4 core courses (the math major, is very lax in requirements tho though) that are grad courses (functional analysis, analysis on manifolds, or fourier analysis)

[chilly radish]

Going up and going down, apparently

[cursive spindle]

Oh

[chilly radish]

I thought it was gonna be about the cohen structure theorem

[azure cairn]

Analyzes theories of gender and politics, especially ideologies of gender and their construction; definitions of public and private spheres; gender issues in citizenship, the development of the welfare state, experiences of war and revolution, class formation, and the politics of sexuality. Graduate students are expected to pursue the subject in greater depth through reading and individual research.
this course is a prereq to the abstract alg sequence

[cursive spindle]

?????

[chilly radish]

Also no discussion of completions it seems?

[chilly radish]

I love how there's one algebra course and 3 analysis ones and 0 geometry ones

[somber goblet]

do you pigeonhole principle the proof for finite-dim integral domains over a field being fields

[azure cairn]

here are the requirements wait

[cursive spindle]

gender and politics is a prereq for abstract algebra

[somber goblet]

that's a funny discrete math course description

[chilly radish]

No need, you just use the fact multiplication is linear

[azure cairn]

diffeqs, real analysis, algebra 1-2 (the intro algebra sequence), one of fourier analysis/functional analysis/analysis on manifolds, a seminar in some overarching topic (seminar in analysis, topology, nt, geometry, idk), and two >100 level courses

very lax obviousl y

[somber goblet]

its an injective endo therefore auto?

[vapid vale]

they should actually make that a prereq to abstract algebra

[cursive spindle]

Jesus christ I've been laughing about this stupid prereqs

[vapid vale]

make things better

[chilly radish]

Yup

[azure cairn]

btw i was kidding if it didnt come across correctly 😭

[somber goblet]

or trivial kernel

[cursive spindle]

It's a good joke

[somber goblet]

more directly

for vector spaces, $W \oplus V/W \cong V$, right?

[cloud walrus]

lexi

[azure cairn]

sounds right

well yeah of course, W and V/W have no common elements except 0 and the dimensions add up properly

[somber goblet]

maybe need finite-dim

[azure cairn]

ah

uhh

no i dont think so

[somber goblet]

"dimensions add up properly" assumes finite-dim

[azure cairn]

yeah i know but i think it holds ture in infdim

nah wait

i am usign scrap rn

ok let $B_W$ be a basis for $W$, then note you can extend it to a basis $B_V = B_W \cup B'$ for $V$. then the basis for the quotient is ${b + W : b \in B'}$, so the basis for $W \oplus V/W$ is ${(w, 0): w \in B_W} \cup {(0, b + W): b \in B'}$. define $\phi: W \oplus V/W \to V$ by $\phi(w, 0) = w$ and $\phi(0, b + W) = b$, then $\phi$ is an isomorphism

[proud vigil]

||start with a basis of W, extend to a basis of V, let U be the span of the added basis vectors

then V is the direct sum of W and U

now you can show that U is isomorphic to V/W. define the map phi: U -> V/W as phi(u) = u + W

this is injective: phi(u) = 0 -> u is in W -> u = 0

this is surjective: take any coset v + W, decompose v = w + u, then w gets absorbed into W, leaving us with a valid choice of u

so we have an isomorphism between U and V/W, and since V is W oplus U, it follows that V is W oplus V/W||

[chilly radish]

Not really

[chilly radish]

Yes, since they're free modules

[somber goblet]

i guess we never assume basis is finite here

[chilly radish]

If you want to do dimension arithmetic computations you just need choice

Otherwise it gets unwieldy

[somber goblet]

ZL says every vsp has a basis 👍

[proud vigil]

yeah the goal was to not assume dimension at all, just allowing for generic basis extension

[chilly radish]

But the general argument for why free modules are projective works here

[azure cairn]

how does dimension counting work if the dimension isnt necessarily finite

[chilly radish]

Well like I said you need choice to make cardinal arithmetic behave in any way, but then you just do cardinal arithmetic, which is given by

dim_F(V)+dim_F(W)= max{dim_F(V),dim_F(W)} if they're both infinite

I guess it also works if at least one is infinite

[restive sparrow]

i'm struggling to understand the point behind this induction

so presumably the base case is when |G|=1, since in that case the statement that there exists a Hall pi-subgroup conjugate to all other Hall pi-subgroups is vacuously true

but i'm struggling to understand the way you're supposed to handle when p is in vs not in pi

like if p is in pi, you take the quotient and presumably induct on the fact that G/M has a Hall-pi subgroup, which you lift via the 4th iso thm

but why can't you do that when p is not in pi

[azure cairn]

when p isn't in pi the preimage includes the unwanted p-group M and so it's too large

[restive sparrow]

wait how does that make a difference 😅

like, if you're trying to show that a hall pi subgroup exists

you care about the primes in pi

OH i get it

coz if p is not in pi the hall pi subgroup in G/M will lift to something that isn't a hall pi subgroup

@azure cairn is this correct

[azure cairn]

ye

since the preimage necessarily contains M, its order will be divisible by p, which ruins it being a pi-group if p isn't in pi

[restive sparrow]

yep

[somber goblet]

euclidean domains are so nice

[restive sparrow]

wait hang on

how do i reduce to the case where |G|=p^a n if we don't even know that a Hall pi-subgroup exists for G at this point

[restive sparrow]

ok no got it

[balmy python]

what are some nice ways to show something is an irreducible element of an integral domain

[maiden crater]

really silly

So, I got and verified with wolfram the gcd is 1

BUT

the 2nd part is confusing me

[wraith cargo]

you have to use the euclidean algorithm

[maiden crater]

like a liner combination menns 1= ka(x)+ qb(x)

right

[wraith cargo]

essentially the idea is to trace is back from the end of the algorithm to the front

[quiet pelican]

Where k and q are polynomials

[wraith cargo]

and this is how you'd get the two polynomials

[maiden crater]

okay, that makes much more sense

that's v easy here

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

direct conseqeunce of division algo

[quiet pelican]

I think the signs should be the other way around

[wraith cargo]

ts ain't true

[quiet pelican]

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

[maiden crater]

oops, wrong way round

huh

[rocky cloak]

Doesn't this seems to feel 3 at x=0

[quiet pelican]

Bleh I can’t maths

Ok yeah this is non-trivial

[maiden crater]

I just manipulate this right

I got 1 instead of -3 , which explains a bit

[rocky cloak]

Yeah, just divide by -3 and you're done

[maiden crater]

cool, thanks

[tardy hedge]

For S < T mult closed subsets of R, T^-1(S^-1R) (image of T in S^-1R) should be isomorphic to T^-1R right? Just universal property right

[maiden crater]

Let F be a finite field. Prove that F[x] contains infinitely many primes.

I have an answer

(x^n)

[south patrol]

None of those are prime except x

[maiden crater]

mhm

I suppose I wnna use the fact that F[x] is a UFD?

[fickle dirge]

It is similar to the usual proof of infinite primes iirc

try to copy that and see where you go

[maiden crater]

Let F[x] have a finite number of primes. Then consider p_1p_2...p_n+1. If p_1p_2...p_n+1 is prime we have a contradiction. If not consider p_i| p_1..p_n+1, which would then mean p_i \nmid p_1p_2...p_n which is a contraidction

[south patrol]

Proof becomes a little easier if the field is infinite lol

[maiden crater]

nx would all be prime would it not

so I saw a solution to this I don't fully het

[south patrol]

What is n

An element of the field F?

[maiden crater]

N_F

natural field elements

[south patrol]

What's that lol

I wouldn't say it's standard

[maiden crater]

well, every infinite field has a subfield in bijection with Q

right

so it follows there's a subset in bijection with N

[south patrol]

Well every infinite set has a subset in bijection w N

I think saying "in bijection w Q" is somewhat misleading here. It's definitely not the case that you have a subfield isomorphic to Q, which is what I suspect you might have in mind

Indeed there are infinite fields of positive characteristic

Anyway this would work to give different primes if you take cx for c different nonzero things in the field. But this is not too interesting because they are the same up to units

[balmy python]

is 3 not implied from 2?

ahhh it's not because not every element of a ring is invertible right

[chilly radish]

It is definitely not. The most you can say is that

f(1)=f(1*1)=f(1)^2, so f(1) is an idempotent. It is very possible to have idempotents that are not 1 or 0

[balmy python]

yep

[chilly radish]

As for an example, consider the map R->R^2 given by 1 -> (1,0). This satisfies 1 and 2 but clearly does not satisfy 3 since the identity of R^2 is (1,1)

[copper kestrel]

looking at problem 16, i found that pi^2 is algebraic but having a lot of trouble coming up with irr(pi^2, Q(pi^2)). i think x^3 - pi^6 is irreducible since it has no zeros in Q(pi^3)? so then deg is 3?

[wraith cargo]

yep

[copper kestrel]

peak

[round anvil]

i'm with a small issue about S_n: what's the main ideia to prove that S_n isn't abelian for n >= 3?

[knotty badger]

you can just find two transpositions that don't commute

[rocky cloak]

S3 doesn't have that many elements, so just try some

[south patrol]

I would say it is also intuitive like if you swap some stuff round, it matters what order you do it in.

[round anvil]

I really thought it would be easy, but my mind went blank.

This happens quite often. Thanks everyone.

I had seen this idea months ago and forgot about it

[south patrol]

Np

[copper kestrel]

how does one show pi^2 is not in the simple extension Q(pi^3)

i know its algebraic

[lusty marlin]

Suppose π² is in the extension. Then π³/π² = π is in the extension. Which means that π can be written as a ℚ-linear combination of integer powers of π³, contradicting the fact that π is transcendental.

[copper kestrel]

ohhhhhhh

[coral steeple]

$\sigma\cdot e_{i_1}\otimes\cdots\otimes e_{i_d}=e_{\sigma(i_1)}\otimes\cdots\otimes e_{\sigma(i_d)}$, where ${e_i}_{1\leq i\leq n}$ is a basis for $V$?

[cloud walrus]

person2709505

[chilly radish]

Yes

[coral steeple]

Thanks

[chilly radish]

You don't need to define it on a basis tho

Wait

[chilly radish]

Sorry that's not correct. It just moves the factors around

[coral steeple]

Hmm

[chilly radish]

Unless sigma(i_k) means i_sigma(k)

In which case it's correct

Remember sigma is a permutation on 1,...,d

[coral steeple]

Oh whoops

So it should be $e_{i_{\sigma(1)}}\otimes\cdots\otimes e_{i_{\sigma(k)}}$?

[cloud walrus]

person2709505

[coral steeple]

But it doesn't need to be on a basis

[chilly radish]

But you don't need to define this on basis elements.

$\sigma \cdot v_1\otimes \cdots \otimes v_d= v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(d)}$

Right

[cloud walrus]

ShiN

[coral steeple]

So the action is defined on elementary tensors, and then extended by linearity?

[chilly radish]

Yea. And of course this is well-defined since it's multilinear

[coral steeple]

Ok thanks

The appendix of Fulton and Harris' rep theory book is not the ideal place to learn multilinear algebra the first time lol

[chilly radish]

Definitely not lmao. Dummit and foote is not bad though they do gloss over some details for my liking

[balmy python]

if R is a commutative ring and x is nilpotent, u is invertible, how can you prove x+u is invertible?

[quiet pelican]

It suffices to prove this for u = 1 (why?)
In that case, try to find the inverse explicitly

[balmy python]

hmmm

Thanks for the hint!

will try rn

let v be the multiplicative inverse of u

Suppose y is nilpotent. Then, yv is also nilpotent.

(y+u)r = 1 iff (yv+uv)r = v iff (yv + 1)ru = 1

pretty sure thats the reason for the why

[vocal pebble]

Maybe u=-1 will be easier to think of explicitly an inverse for

[balmy python]

my idea is
(x+1)r = 1 implies (1-r)^n =0 (xr is nilpotent for some n)

[swift root]

the binomial theorem might be useful

[balmy python]

i was thinking that

but it felt useless 😭

[vocal pebble]

Most naively, an inverse would look like 1/(x+1), but maybe we could make that work (there is a trick)

[balmy python]

hmmm

im guessing its geometric series or smth but that would stop at finitely many terms

wait

[vocal pebble]

And that's exactly what we want

[balmy python]

an infinite geometric series stops at a finite number of terms

wtf

[vocal pebble]

Now check that it is indeed an inverse

[balmy python]

i guess you cant represent it in the closed form though right?

well (x+1)^-1 is a closed form

[swift root]

it stops at a finite number of terms because x has a very nice property

[balmy python]

yep

[vocal pebble]

I mean it doesn't matter whether 1-x+x^2-...+(-1)^nx^n is a closed form, the important thing is that this is an element of your ring

[balmy python]

ah yeah

that is really nice

damn

is this a common trick?

or like something that you gotta see once to know its use?

[quiet pelican]

Using formal expansions like that to create elements you need is fairly common

[balmy python]

ah okay

any other similar tricks which may be good for me to try to work on?

maybe a similar problem

my university is kinda weak with algebra 😭

[vocal pebble]

One that comes to mind is:
Find the nilpotent elements of R[[x]] (the power series ring). R is noetherian, commutative, has unity. More specifically, give a characterisation of all nilpotent elements of R[[x]] in terms of the coefficients of the elements.

[balmy python]

this is the first time im hearing of a power series ring

is it just the set of up to infinite degree polynomials?

[vocal pebble]

Yeah, up to and including the "infinite degree polynomials" also called as power series (so elements like 1+x+2x^2+3x^3+... Are included)

Finding the units in the power series ring is also a nice one

[somber goblet]

i should prove that notherian rings have all finitely-generated ideals

i have no idea how to go about that

the two definitions seem totally different

[quiet pelican]

Try to prove every non-finitely-generated ring has an infinitely ascending chain of ideals

[kind temple]

is that true? if R is notherian, the formal powerseries ring R[[x]] is also notherian, but R[[x]] is not finitely generated

[quiet pelican]

Oh yeah

[somber goblet]

ideals are modules...

so a chain of ideals is a chain of submodules

and a finitely-generated ideal is a finitely-generated R-module?

maybe?

[rocky cloak]

Did you mean to say that finitely generated rings are Noetherian?

Or perhaps that a ring is Noetherian iff all ideals are finitely generated?

[somber goblet]

oh wait

[somber goblet]

"R is notherian iff all its ideals are finitely-generated"

[rocky cloak]

Then yes, you should prove that

[swift root]

this is actually not a fact fundamentally about rings at all!
This is a fact about general (algebraic) closure operators

a closure operator is algebraic if the closure of some A is the union of the closures of all finite subsets

so you can do this for e.g. groups too

[kind temple]

so the closure operator here is to send a subset of a ring to the least ideal containing it?

what is the statement in general?

[swift root]

yis

[swift root]

Call a closure operator Noetherian if every ascending chain of closed sets stabilizes. Call a closed set finitely generated if it is the closure of a finite set.

The statement then becomes: An algebraic closure operator is Noetherian iff every closed set is finitely generated.

[next obsidian]

Whatever.

[frail shoal]

bro is whatever

[swift root]

A proof is as follows:
Let C be an algebraic closure operator.

Lemma: If X0 < X1 < ... is an ascending chain of closed sets, then its union is again closed.
Proof: Let X be the union of all Xi. Then if X' ⊆ X is finite, we have X' ⊆ Xi for some Xi, and so
C(X') ⊆ C(Xi) = Xi ⊆ X
thus X is closed, by C being algebraic.

Now for the proof of the theorem:
(=>) Suppose C is Noetherian, and BWOC suppose that X is closed and not finitely generated. We define the following sequence of sets:

Pick some x0 ∈ X and let X0 = C({ x0 })
Now for some i let xi ∈ X \ Xi, and define Xi+1 = C(Xi ∪ { xi }).

Notice that, as X is not finitely generated, there will always exist such an xi. However, this contradicts the Noetherianness assumption, as we have constructed a non-stabilizing ascending chain.
(<=) Suppose now every closed set is finitely generated. Consider a chain X0 < X1 < ... By the lemma above, its union X is closed. This is finitely generated, so there is some finite subset A ⊆ X with C(A) = X. But it is finite, so A ⊆ Xi for some i. For all j ≥ i we then have
X = C(A) ⊆ C(Xj) = Xj ⊆ X
indeed, the chain stabilizes.

[swift root]

bro's just hating on anything

[karmic moat]

where did the monkey go

[frail shoal]

im cyan mary rn

monk key will return soon

[karmic moat]

i cannot wait...

i am so excited for the monkey to return

[frail shoal]

noted

[next obsidian]

Idk I just think saying “whatever.” Is funny hahaha

[frail shoal]

you're the only person i know who actually rocks the foot pfp modifier

never got a chance to point that out so i might as well say it randomly

[karmic moat]

it's lowkey awesome right

[frail shoal]

true

[kind temple]

this is neat. i am assuming that there is no constructive proof for the forwards direction

[chilly radish]

It needs dependent choice in general

You can do it constructively if e.g. R is countable

[next obsidian]

One girl randomly appeared in advanced lounge recently who rocks it

[frail shoal]

damn

[chilly radish]

(since instead of an arbitrary element you can just choose an element minimal w.r.t. the enumeration)

[next obsidian]

I commented on it and she didn’t respond

Well I innately have class-sized choice capabilities so I can actually do that proof without any choice assumptions built in

I’m good af at picking stuff

[karmic moat]

is it genetic?

or is it just something divine

[next obsidian]

Not genetic

I just have it

I’m built diff

[karmic moat]

woah

[chilly radish]

I don't know how to tell you this... You are the choice assumption

It's true I once constructed a model of ZF and Chmonkey wasn't there

[somber goblet]

ok every ideal being finitely generated

every submodule is finitely generated

[karmic moat]

is Zorn's lemma like chmonkey's cooler older sibling or stupider younger sibling

[somber goblet]

wait does "every ideal" include the entire ring

for notherian condition

[chilly radish]

R=(1)

[somber goblet]

so notherian rings are finitely generated??

[next obsidian]

Oh… oh fuck…. Oh no oh no oh no

[next obsidian]

Over itself

But all rings are generated by 1 overitself

[somber goblet]

oh duh

nvm

[next obsidian]

A finitely generated ring means one that’s a finitely generated Z-algebra

So there’s a finite set of elements where everything is a polynomial with integeral coefficients in those elements

[somber goblet]

yeah

[kind temple]

i think you should have xi in X \ X_{i - 1}, btw.

could you also explain how X is closed in your lemma? can't quite see why nvm i got it

[somber goblet]

that's different from R being a finitely generated R-algebra

[next obsidian]

Yes

[chilly radish]

Zorn's lemma is the younger sibling, choice is the older sibling and well-ordering is the estranged older sibling from a previous marriage

[karmic moat]

hmm i see

so zorn is dewey, choice is reese, and well-ordering is francis

like malcom in the middle

[next obsidian]

I’m Malcolm?

[karmic moat]

i thought you were choice

so you're reese

idk probably "every vector space has a basis" is the malcolm

[somber goblet]

does notherian mean every strictly chain of ideals is of finite length

or just that it terminates

[karmic moat]

finite

usually "terminates" means "terminates after a finite number of steps"

[next obsidian]

https://tenor.com/view/terminator-sunglasses-shades-t800-arnold-schwarzenegger-gif-27448475

[karmic moat]

arnold schwarzenoether

artin schwarzenoether

artin scholzenoether

[next obsidian]

Hey, schwartz is also a guy

Idk if I’m spelling it right

Cauchy Schwartz

[karmic moat]

yea but didnt he do analysis right

[next obsidian]

Idk probably

[somber goblet]

ok thanks

hmm ideals form a poset

in which all chains are finite (in notherian land)

[karmic moat]

there are infinitely descending chains of ideals in noetherian rings

e.g. in k[x], the chain (x) \supset (x^2) \supset (x^3) \supset ...

a ring in which every descending chain of ideals terminates is called "artinian"

artinian is equivalent to noetherian + dimension 0

[kind temple]

is dimension the krull dimension?

[karmic moat]

yeah

[chilly radish]

Yes

Dim 0 just means every prime is maximal

[somber goblet]

oh

ascending chains are finite

[karmic moat]

yeah

the condition "all ascending chains of ideals terminate" is also called the "ascending chain condition on ideals"

similarly the condition "all descending chains of ideals terminate" is called the "descending chain condition on ideals"

is it "on ideals" or "of ideals"? i dont remember

one of those two

[somber goblet]

another example of this in Z is (n), (n^2), (n^3), ...

right?

[karmic moat]

yeah

[somber goblet]

ok every ascending chain in this poset terminates

this is tricky

wait is this along the lines of the "every vsp has a basis" proof using zorn?

where you make some ascending chains and find them an upper bound

[karmic moat]

are you proving ascending chains terminate implies ideals are finitely generated?

or the other way around

in either direction i don't think you need zorn/choice/any of these equivalents

[somber goblet]

forward

i know i don't need zorn

[karmic moat]

suppose ascending chains terminate but there exists an ideal which is not finitely generated

[somber goblet]

but i'm wondering if it's a similar proof technique

[karmic moat]

i claim you can choose elements of this ideal and create an infinite ascending chain of ideals using these elements

[somber goblet]

oh wait

[chilly radish]

You need dependent choice for one direction

[somber goblet]

can you just take (a), (a, b), (a, b, c), (a, b, c, d), ...

and then if this terminates after a finite number of inclusions you've found a finite generating set?

[chilly radish]

Yea, notice though that you're making an arbitrary choice at every step which strictly depends on your previous step. This is called dependent choice and it's weaker than regular choice but stronger than countable choice

[somber goblet]

err wait not sure if this must be generating

i guess it is if every new element is unique

[karmic moat]

oh yeah true

[somber goblet]

hence dependent choice?

[chilly radish]

You want the ideal to get bigger at every step, so you need to choose something outside of it

[somber goblet]

yea

so the existence of such a sequence is dependent choice?

[chilly radish]

Yea, unless you have some other way to choose the elements. Like I said above, if R is countable you can choose the next element to be minimal with respect to an enumeration s.t. it's not in the ideal

Then you don't need dependent choice since you can constructively choose an element each time

[karmic moat]

dependent choice tells you that every uncountable set contains a countable subset?

[somber goblet]

if it's countable you pick a bijection with N then use well ordering?

like "set of elements not in ideal has a least element, so pick that one"

[chilly radish]

Yea

[somber goblet]

i like this stuff way more than group theory

[chilly radish]

Yea, and I think a weaker form of choice (at least countable choice) doesn't necessarily

[karmic moat]

hmm i see

i've never really thought about these size issues before lol

[somber goblet]

ok, now for reverse direction

an inclusion of ideals provides inclusion of generating sets

and if this inclusion is strict then there is a strict inclusion of generating sets (i think)

so an ascending chain of ideals implies an ascending chain of finite generating sets

so if you want this set to stay finite the chain needs to be finite as well

is my idea

maybe you take contrapositive

[karmic moat]

i don't think this is true. for example, the generating set of a principal ideal (a) is just {a}. but like (4) is a subset of (2) but {4} is not a subset of {2}

[somber goblet]

yea i think you need contrapositive somehow

or make an order argument

hmm

lowkey idk