groups-rings-fields

[delicate orchid]

Yea

[delicate orchid]

There’s a way easier formulation for you though

Just take x -> x^p

[static yew]

okay at least I grasp that part

I still don't understand how you can stuff that into an equation

[glossy crag]

He's right @static yew, at your present stage you're not gonna learn anything from this channel because 10 different people with 10 different opinions on presentation will bombard you with disconnected pieces of information that won't make any sense to you, so it's better to invest the time in a coherent and didactic exposition.

[delicate orchid]

You can have rings of functions

[echo gull]

yesterday I saw one of the most downvoted answers I've ever seen on stack exchange, saidly I flagged it and it was deleted 😦

[static yew]

unless - and + and 0 are Inside the field and not regular integers

[echo gull]

it was like a ChatGPT abstract algebra answer

[formal ermine]

it's more than 0

:(

[static yew]

do you guys have a recommendation on such a coherent and didactic exposition?

oh!

[delicate orchid]

Prove that the product is unique up to isomorphism

[static yew]

it's not a fortune

[formal ermine]

fuck you

it's trivial

trust

[delicate orchid]

Perhaps it will even be zero

[coral spindle]

https://math.stackexchange.com/questions/54839/good-abstract-algebra-books-for-self-study

[static yew]

prove that 1+1 = 2 up to isomorphism

[delicate orchid]

Yes! It is! So please provide me with a proof

[coral spindle]

get better material

[ivory trail]

many university students manage to pay zero dollars for their math textbooks

[static yew]

I'm not a university student

[coral spindle]

not that we'd ever recommend such a thing ;)

[static yew]

indeed

[glossy crag]

Since you're interested in fields, i'd recommend Stillwell's "Elements of Algebra", that's what got me first into the subject. It's not a standard fat algebra textbook that goes through the group->rings->fields procession and covers everything under the sun, but it's not like you're a college student taking an algebra course, and it's VERY elementary in its approach (starts all the way with natural numbers and division). It introduces all the relevant players, like group, rings, polynomials and fields, and builds towards the proof of insolvability of the quintic equation.

[delicate orchid]

I NEED my free copy of "Fusion Systems - An Algebraic Approach" Mr. Bobby fusions has left me and taken the kids but that HACK FRAUD DAVID won't GIVE ME HIS BOOK

what zero funding does to a mf

[coral spindle]

Damn you david.........

[static yew]

$51.90

the one advantage of being employed instead of being a university student: I can actually afford these books

you're probably genx like me

[delicate orchid]

ermmm aktually I'm a post graduate RESEARCHER not a studnet :theflamesofbabalyon:

I know less than most undergrads albeit

[coral spindle]

but when they have a student discount, what do you say?

[delicate orchid]

but that's besides the point

[coral spindle]

D:

that's me I'm the sandwich

[formal ermine]

maybe the library which also contains the first book of the bible has a copy of this book...

[delicate orchid]

the left emoji is :mock2: and the right emoji is :mock1:

[glossy crag]

This reminds me of that Futurama episode where Fry and Bender enlist to get a 10% military discount on pork chewing gum ("... that is unless war were declared" BWA "what was that?" "war were declared")

[formal ermine]

lmao

I love futurama

[static yew]

that sounds like exactly what I need

everything I know about the (polynomial ring?) of polynomials with coefficients in F_2 modulo an irreducible, I learned from Wikipedia and then worked out low-degree examples by hand

then I wrote scripts to do the math with higher-degree moduli and tested them against examples I'd worked by hand

it was easy cuz I could use test vectors for CRC calculations to check my work

[formal ermine]

too bad the new season is only on disney+

[delicate orchid]

it's funny how you use vectors so much but don't know what a vector space is

[glossy crag]

That one's one of the all-time classic episodes

The key to victory is the element of surprise.... SURPRISE

[delicate orchid]

what LINALG101 for CS does to a mf

[static yew]

they've got a lot of brains, and they've got a lot of... chutzpah

[delicate orchid]

I'm not reading all of that

ok maybe I am

ok I did

[glossy crag]

protestant work ethic

[delicate orchid]

I unironically agree with the first three

[formal ermine]

last week in physics class (10th grade)

[delicate orchid]

how tf did u know I was protestant you FED

[formal ermine]

our teacher asked us what a vector was

I answered with "an element of a vector space"

[static yew]

because I don't need to use them as vectors, not in the usual sense
once I have a primitive polynomial it's just (x^(e coprime to 2^k-1))^n to generate the nth element

[formal ermine]

he looked and me and said "you fish head"

and proceeded to ask another student

[glossy crag]

the catholic church has a Discord division now, we're keeping tabs

[agile burrow]

teacher owned with facts and logic 😂

[coral spindle]

illum you're also a king

but only for that

the rest is cringe and bad

[static yew]

https://tenor.com/view/oh-yeah-vector-minion-ooh-yeah-gif-5629902

[coral spindle]

algechill moment

[glossy crag]

should this all be moved to #math-discussion or is this a "mods are asleep, post X" situation?

[delicate orchid]

abstract algebra is my channel I rule this domain

[coral spindle]

nah it's just algechill

[formal ermine]

anyway guys and gals whom are we sully bombing next?

[glossy crag]

Like I said, the book is not one of those 700-page bibles that are used for college courses, but given that you lack some of the requisite knowledge (e.g. not knowing \varphi^2=\varphi\circ\varphi and minor stuff like that), I think it might actually be helpful to you. It's the most elementary and "chatty" introduction to the subject that is known to me.

Also it's very short, like 200 pages tops.

[glossy crag]

guys and Gal(L/K)s

[formal ermine]

short introduction
200 pages

[glossy crag]

Artin, a standard reference, is 500+ pages. Others are just as much if not more

[formal ermine]

[static yew]

I have severe ADHD (one of the reasons I dropped out of college) I ain't gonna be able to stick with a 500 page book no matter how much amphetamine I stuff into my system

[delicate orchid]

ADHD gang

[formal ermine]

Have you tried cocaine

[glossy crag]

also the book's margins are pretty narrow as i recall, so 2-3 pages of it correspond to 1 page of a larger book

[delicate orchid]

what do you think adhd medication IS

[static yew]

isn't that bad?
it's what led to the fermat problem being unsolved for centuries

[formal ermine]

fermat never had a prove and I can prove it

[glossy crag]

but the character limit of a Discord message restricts me

[formal ermine]

so true king

[delicate orchid]

fermat didn't have a proof because he didn't know what swagggg was

[static yew]

can only sets with prime-power cardinality be a field?

[glossy crag]

a finite field, yes

[proud zenith]

What proof did Fermat THINK he had...

[static yew]

discord max message size = 2000 characters
not a prime power
therefore: discord messages cannot be used to construct a field

do they constitute a vector space

[delicate orchid]

nope

[static yew]

(asking the real questions here)

oh? why not

[delicate orchid]

vector spaces over finite fields have to have order p^k

[static yew]

the vector spaces themselves have to have order p^k?

[delicate orchid]

yeah, like the set underneath them

[static yew]

hm

[rustic crown]

i mean that's the reason why finite fields are also p^k

[glossy crag]

I'm not an expert, but I think the historical consensus is he had something similar to Lame's faulty proof (who thought Z[\zeta_n] is a UFD for all n until Kummer came along).

[rustic crown]

cause they're vector spaces over F_p

[proud zenith]

would make sense

[static yew]

what if Fermat's proof actually disproved the axiom of choice
and Zermelo and Fraenkel went back in time to put him on ice

[proud zenith]

Fermat was missing ... a lot of thought about number fields

[delicate orchid]

you can't disprove an axiom

[static yew]

sure you can
you need another axiom for it, though

[next obsidian]

disproved

[delicate orchid]

*disprooven

[rustic crown]

you can, by working in a system where you assume the negation of it

[static yew]

exactly

oh!

[delicate orchid]

differetn system = I donm#t caRE

[next obsidian]

You quite literally cannot disprove AOC in ZF

[glossy crag]

The achievements of Kummer in that regard absolutely astound me to no end, more than a lot of other great proofs.

[delicate orchid]

I could do it if I wasn't warm

[static yew]

Fermat's proof:

AXIOM OF INTEGRAL EXPONENTIATION: there are no positive integer solutions to a^n + b^n = c^n for n>2

QED

[proud zenith]

I am continually astounded by the height of knowledge in the 1800s

[glossy crag]

How he was able to prove his regularity theorem without the modern algebraic framework (they didn't even have ideals then, Dedekind introduced that later to clean up Kummer's work a little) is truly a marvel.

[delicate orchid]

it was fun

[glossy crag]

Same, I feel like history of mathematics is a little underappreciated.

[rustic crown]

Fun Things are Fun!

[delicate orchid]

as someone who's doing research in an incredibly niche, pure field

[rustic crown]

https://tenor.com/view/kon-fun-things-are-fun-anime-cute-yui-gif-16787421

[static yew]

people probably got high

"dude, what if--- what if we redefined plus and times?"
"duuuuuuuuuuuuuuuude!"

[delicate orchid]

you don't need a reason other than "WHAT THE FUCK IS GOING ON HERE??? WHAT IS HAPPENING?!??!"

[proud zenith]

they knew they were on to something

[glossy crag]

In Kronecker's case, faith.

[static yew]

it's weird that a bunch of the theorems about elliptic curves predate the transistor

[rustic crown]

wait for homotopy type theorists to enter

[static yew]

hold on, lemme just craft a quick isomorphism from N to R
Goedel be damned

I call it the "on-demand isomorphism"
you give me an element of R, one at a time
and I will give you an N that it maps to

no, you can't see the whole thing
I have a piece of paper attesting that it's there though
just like Tether's balance sheet

[delicate orchid]

you give me an element of R one at a time
nice axiom of choice stinky

[static yew]

and people wonder why I distrust AC

[glossy crag]

@static yew I checked and Stillwell's book unfortunately doesn't have a section on finite fields, which seems to be the motivating factor for you, but everything else I've said about the book holds.

[static yew]

does it do vector spaces?

cuz a lot of people here tell me I gotta understand those

[glossy crag]

What you could do, if you're REALLY interested in the subject, is reading Lidl/Niederreiter's "Introduction to Finite Fields" after collecting the requisite knowledge.

It's a pretty elementary intro and assumes only the basics of groups, polynomials and fields (which they lay out in the introduction).

[static yew]

"Introduction to Finite Fields and their Applications"?

good grief it's $181

[delicate orchid]

jesus

if only we could find a method to.... reduce the price...

[glossy crag]

I don't think so and I don't think any algebra book does, all of them assume prior acquaintance with linear algebra.

[proud zenith]

I remember about a couple of years ago springer were giving 1000s of full book pdfs away for free. I got some of them but I didn't know then what I would have looked up in the future, then they stopped it

[static yew]

btw was there an answer on whether or not there was a factorization of 40 that involved sqrt(-13)?

when I tried I thought the answer was no

[proud zenith]

I'm going to try that in sage

[static yew]

sage is cool

it's like free matlab

[proud zenith]

sage: K.<a>=NumberField(x^2+13) sage: K.ideal(40).factor() (Fractional ideal (2, a + 1))^6 * (Fractional ideal (5))

[static yew]

I used it to help me debug my F_p(p>2) polynomial division routine last week

[delicate orchid]

do itttttt

[proud zenith]

haha

I never never use a computer

[static yew]

please explain what that means, I don't know what K.<a> even does

K dot less than a greater than a

what kind of binary relation R is that

[proud zenith]

ok. K is a number field. a is the "primitive element". here it stands for sqrt(-13)

Q is the field of rational numbers

Q(sqrt(-13)) = K

[static yew]

is K automatically a number field?

[proud zenith]

the polynomial that defines it has to be irreducible over the base field

[static yew]

wait
can't you have Z and add more than one irrational though?

like Z[sqrt(2), sqrt(3)] or something

[proud zenith]

not the same, because Z[sqrt(-13)] is not a field, it's not closed under multiplication

[static yew]

wait why isn't it closed

[proud zenith]

in fact you have had your first glance at an order in a number field

and the 'maximal order' is the ring of integers

[static yew]

I thought Z[sqrt(-13)] was Z plus (a + b*sqrt(-13))

for all a,b in Z

or maybe N

[proud zenith]

the way you write it is

Z[sqrt(-13] = Z + sqrt(-13)Z

[coral spindle]

You may have meant to say something else where

[static yew]

the way I write it is probably wrong

[coral spindle]

It is indeed closed under multiplication

But it is not a field due to lack of inverses.

[proud zenith]

sorry yes

I should think 2 seconds more before I type

[static yew]

is Z[whatever] an integral domain

[slim kayak]

Like, whatever in C?

[wraith cargo]

Yes

[proud zenith]

That is not where my brain would be at at the beginning of number fields, Z is an integral domain, would have to ... think

about Z[whatever]

[coral spindle]

typos happen, dw

[static yew]

I can only assume the name "integral domain" came from Z

[proud zenith]

and then you have to be able to recall the definition to check it

[coral spindle]

It's more that it came from C

[static yew]

this sounds like more people-were-high stuff

[coral spindle]

ok

[static yew]

"dude what if pi was an integer"

[proud zenith]

the most important thing a the start of number fields is what is it, and what is the ring of integers

[static yew]

ring of integers == Z with the "usual" addition and multiplication, right?

[next obsidian]

No

[static yew]

gdi

[next obsidian]

It’s everything “algebraic” over that

[static yew]

every time I think I've got a grasp on this stuff

[proud zenith]

Z is the ring of integers in Q

[slim kayak]

Z[epsilon] as the subring of the dual numbers deffo isn’t integral but any subring of an integral domain will be integral,

[next obsidian]

really you need to say “integral” meaning it satisfies a monic polynomial with integer coefficients

[proud zenith]

Z[sqrt(-13)] is the ring of integers in Q(sqrt(-13)) .... or is it let me check

[ivory trail]

it is

[proud zenith]

sage: K.<a>=NumberField(x^2+13)
sage: K.ideal(40).factor()
(Fractional ideal (2, a + 1))^6 * (Fractional ideal (5))
sage: OK=K.ring_of_integers()
sage: OK
Maximal Order in Number Field in a with defining polynomial x^2 + 13
sage: OK.basis()
[1, a]

[static yew]

is Q(sqrt(-13)) different from Q[sqrt(-13)]?

[proud zenith]

here's a nice trick

Q(anything) means the smallest field containing Q and anything

[slim kayak]

One is meant to be the fraction field of the other

[proud zenith]

Z[anything] means the smallest ring containing Z and anything

[static yew]

so theY are different

[ivory trail]

people don't really write Z(x)

it's slightly nonsensical

[static yew]

my travels have taken me to math.stackexchange where I see Z(t) for transcendental t
whereupon I close my eyes, stick my hands in my ears, and say LALLAALA and move on

[ivory trail]

transcendental here is the same as some unknown x

because they both satisfy no algebraic relations

basically Z[pi] and Z[x] are isomorphic

[static yew]

but Z[sqrt(-13)] and Z[x] are not?

because sqrt(-13) can be turned into -13

[ivory trail]

yeah basically

[slim kayak]

Sqrt(-13) is algebraic, so not transcendental

[ivory trail]

not actually a proof but that's the idea

[static yew]

as pi is not the root of any polynomial with (real?rational?integer?) coefficients
there's no way to turn pi into an integer
so once you add a pi into an expression, there's no way to eliminate it except for (a) dividing it by pi, canceling it to 1, or (b) subtracting pi, canceling int to 0

[glossy crag]

On a completely unrelated note, it crossed my mind today that R_1\times\cdots\times R_m\cong S_1\times\cdots\times S_n with the R_i and S_j simple rings implies m=n and R_i\cong S_i. There's simply no getting away from unique factorisation in math, is there...

[static yew]

it crossed my mind that you said "simple rings" and the word "simple" does not appear to apply here at all

[ivory trail]

in other words, pi is not the root of any polynomial with integer/rational coefficients

[static yew]

yeah that

[ivory trail]

so it is transcendental over the rationals Q

[static yew]

In contrast, anything that is a rational exponent of a rational is algebraic, right?

[glossy crag]

wait till you find out how simple "simple groups" are

I wonder, do we have anyone in the channel of the regulars who is capable of understanding the proof of the classification?

[static yew]

if I can't generate the group with 1, a, a^2, a^3, ... a^(n-1) for group order n, it ain't simple enough for my needs

C is a field right?

[formal ermine]

those groups are simple iff n is prime

the irony

[glossy crag]

it's better to leave out the "irreducible" part on first acquaintance, no?

[static yew]

wait what if a is a GF(2) polynomial modulo an irreducible poly? the multiplicative group, so not zero

but it's not simple?

cuz the group order is 2^n - 1, which is almost always not prime

trust me, I studied the hell out of this for some project at work

I needed pseudorandom number generators with very specific properties
and I used galois fields to build them

[formal ermine]

what is your job

[static yew]

I'm a software engineer

[formal ermine]

where

[static yew]

I work in credit card processing

[formal ermine]

ah

does that mean you can add free money to my card?

[static yew]

It does not, sorry. Only the bank can do that

[formal ermine]

big scam

[static yew]

There's actually a way I could do that, buuuut it's obviously illegal and I'm not going to jail over it

[glossy crag]

Have you seen Mr. Robot? I always wondered if the bit about how by getting some kind of "scanner" device close to a key card they can copy its information.

Not exactly credit cards, but somewhat related.

[static yew]

hotel key cards?

[glossy crag]

not exactly, but close

[static yew]

like access badges?

[glossy crag]

yeah, exactly

forgot the word

[static yew]

There is. Did that once, on a lark

Depending on the system design it may be possible

Note that you can't build a universal one

[formal ermine]

do you know if I can carry an nfc tag around with me that will automatically charge $50 from nearby credit cards?

[static yew]

you cannot

because the money isn't on the card

the money is in the bank

the card is just an identifier.

And the chip in the card uses RSA-based signatures to prove its legitimacy, so you can't just copy the output unless the credit card terminal screws up and the challenge isn't random enough (fun fact: when the chip cards first came out, a lot of credit card terminals did screw up and always used the same challenge, so you could replay a captured signed challenge to effectively clone the card)

we don't stream anything to the various world governments

oh, internet hub

they probably stream everything

that doesn't sound like a great trade

[elder wave]

What the fuck is this conversation

[static yew]

we're trying to construct a vector space over discord messages

i.e. rings of integers to groups of idiots

the group operator + is:

nonidiot + nonidiot = nonidiot
idiot + nonidiot = idiot
nonidiot + idiot = idiot
idiot + idiot = idiot

it's basically F_2

idiot * idiot = nonidiot

F_2 but with + and * swapped
why? because idiots

[ivory trail]

who is going to say it

[slim kayak]

2 is my favourite odd number

[slim kayak]

(Hint: a number n is odd iff n mod 2 is 1)

[coral spindle]

This has been the worst algechill moment of all time, I think

[summer path]

#1100503863586455632

Oh wait

[static yew]

does that mean no numbers are odd in F_2 cuz there is no 2 and mod 0 is undefined?

[slim kayak]

I was trying (without success apparently) to joke about you writing idiot + idiot = idiot, which translates to 1+1=1, which is the same as saying 2 is odd

I guess you meant nonidiot =1 and idiot=0 with addition and mult swapped looking at it again, so that works unless I overlooked sth

[summer path]

Well if idiot=0 and non idiot=1, you still don't get 0 + 0 = 1

[solemn garden]

what do steps 5 and 6 do?

[slim kayak]

Idiot * idiot = idiot makes more sense anyways

[solemn garden]

who knew, idiots are not isomorphic to a field

[dreamy fiber]

what has this conversation devolved into lol

[slim kayak]

Cutting edge mathematics

[dreamy fiber]

yeah i am sure it's at the frontier of mathematics

[rustic crown]

boowean awgebwa
idiot = 1 and + = or

[summer path]

Not in z/2z though

[solar glacier]

😂

[rustic crown]

a or b = a+b+ab

[slim kayak]

How to correctly pronounce awgebwa tho

[rustic crown]

aww-geb-bwa

[slim kayak]

So it does sound like talking about course work during lunch with a full mouth

[solar glacier]

When I tell my non math friends I’m taking algebra they think it’s the same as high school alg and they’re all like “oh I took algebra” so I gave up n say group and ring theory

[slim kayak]

"Oh linear algebra? Yeah I know y = mx + b"

[solar glacier]

LOL

[summer path]

Basic algebra :)

[sonic coral]

i got into that argument until they showed me their transcript where their highschool course was actually called linear algebra

[south patrol]

i just used linear algebra to do a group theory problem

feel cool

[wraith cargo]

If this happens just say:
"man those Cohen-Macaulay rings are a real fucker to get a grip on right"

[solar glacier]

😂

[south patrol]

How many subgroups of each order does C2 x C2 x C2 have

[wraith cargo]

And then just go with the awkward flow

[slim kayak]

Imagine you are saying you are working in algebraic geometry and some high schooler says that is easy and they are doing even analytic geometry

[solar glacier]

Yeah they think it’s plane geo and elementary algebra

[wraith cargo]

If I tell someone I'm doing commutative algebra and they tell me all HS algebra commutes so what is there to be done I will strangle them on the spot

[dreamy fiber]

They can only occur in order 1, 2, 4, and 8, try to see what groups occur there

yeah i mean analytic geometry do be hard 😂

GAGA

[wraith cargo]

how exactly does that relate to y=mx+b

In other news
man Eisenbud is so good

[slim kayak]

I think it has to do with parabolas

The comm alg book?

[wraith cargo]

ye

a certain chmonkey in the server dislikes it

[slim kayak]

I wonder which one

[wraith cargo]

800 pages of pure joy

[wraith cargo]

the same person reads matsumura 🤢

[slim kayak]

800 pages?

[wraith cargo]

yea

Eisenbud is longggg

like very long

it's memed that it contains all of comm alg in a single book lmao

[slim kayak]

I thought atiyah and McDonald's cute size was average for what you usually need

[wraith cargo]

I have heard that you need more than that for stuff like Hartshorne

[rustic crown]

it's enough to get you started with AG

but it doesn't have more homological stuff

[slim kayak]

Homological algebra gets grouped with comm alg often?

[rustic crown]

are there more than one chmuwu's?

[wraith cargo]

no it's more that homological algebra techniques are very useful in comm alg

[rustic crown]

no like the stuff in comm alg that requires hom alg... stuff about depth, proj/inj dim, cohen-macaulay etc

det hasn't read AM so me no know what all it has >.<

det will read matsumura tho

cause chmuwu recc it

[summer path]

Homowogy

[wraith cargo]

it's very shallow and terse
me no likey

[rustic crown]

chmuwu's wisdom

[wraith cargo]

agree

(I still haven't learned group coho for some reason )

[rustic crown]

(same >.<)

[wraith cargo]

I will let Number theory motivate me

[rustic crown]

me too

[wraith cargo]

my new math objective is to understand Wiles' proof of FLT at some point in my life

[rustic crown]

[wraith cargo]

I am growing closer

[delicate orchid]

cohomology is everything, except homology

[rustic crown]

[slim kayak]

Just turn your page upside down smh

[wraith cargo]

I need to learn more about modular forms mumble mumble

-<

[rustic crown]

det know supew basics >.<

[delicate orchid]

no you gotta flip it turnways

[wraith cargo]

I know they're connected to like alg geo in some interesting ways

[delicate orchid]

ok but super basics to you (det) is probably like

cohomology of chains of functors

[wraith cargo]

I mean I know what a modular form is lol
I don't know what the motivation for them is tho

[delicate orchid]

analytic functions
do not care

[slim kayak]

Does that even make sense?

[rustic crown]

nu me know a lil stuff about the ramanujan tau function, that's all

[delicate orchid]

I've seen notions of chains of functors I don't recall exactly how it works

[wraith cargo]

they're weakly modular of weight k >-<

[slim kayak]

Cohomology... things out of nat transforms?

[delicate orchid]

gimme the uhhh irreducible characters of height 0

[delicate orchid]

lemme get my notes

[slim kayak]

Yay

[wraith cargo]

yeah that's like the basic example but I don't think it's very interesting
Eisenstein series are where it's at
They actually somehow solve the four squares problem???

It's like in how many ways can you write a number as the sum of four sqaures

and the solution can be found through Eisenstein series

[rustic crown]

oh yee i rememebr a lil, they in fact give you a formula for number of solutions as well

[proud zenith]

I have a 600 page book on FLT from a conference after the proof was announced. I glanced at it when I knew nothing, now I know little bits more and I have practice reading lots of papers so I could probably get through more of it now. But the problem is if it starts assuming things from e.g. the Hartshorne book, then I am back to clueless (I have not read Hartshorne)

[delicate orchid]

lol

[summer path]

I only know one 50 minute lecture worth on modular forms

[rustic crown]

tubu kawaii

[summer path]

detuwu

[ivory trail]

there is no way it doesn't

[delicate orchid]

nvm I was chatting shit

an exact sequence of functors is some A -> B -> C such that for all objects X in the appropriate abelian category, A(X) -> B(X) -> C(X) is exact

[wraith cargo]

(the details are like that certain Eisenstein series generate $\mathcal{M}2(\Gamma_0(4))$, namely $G{2,2}(\tau)$ and $G_{2,4}(\tau)$ so you can get the solution from that because the modular form associated with the four square function is a modular form of weight $2$ with respect to $\Gamma_0(4)$)

[cloud walrus]

ironyincarnate

[slim kayak]

Imagine you could introduce all necessary concepts and build up the theory within 600 pages and fit the actual proof in the remaining margins.

One final flex on Fermat

[wraith cargo]

Man I think just knowing Ribet's theorem is already very deep
But Wiles used some wild comm alg stuff in there too

like Gorstein rings

[ivory trail]

i don't know how people get into these things

the commutative algebra that is beyond what is needed for basic algebraic geometry

[proud zenith]

I know a good exercise: understands Ribet's proof of the epsilon conjecture (you beat me to it there Irony), apparently it uses 'the full strength of Grothendieck's algebraic geometry'

[ivory trail]

i guess to prove FLT

[wraith cargo]

I'm slowly learning stuff abt Langlands since my new institution is very langlands focused so I'm getting into FLT related waters

[delicate orchid]

the funny schemes

[slim kayak]

Working in AG or related fields and you overtime learn the necessary material?

[wraith cargo]

Read Eisenbud bro

[ivory trail]

i read it

i took 200c

[wraith cargo]

lmao you read Eisenbud for 200c

[ivory trail]

nah professor had his own notes

but i consulted eisenbud and atiyah-macdonald

[wraith cargo]

the class the covers like 5% of the book

[slim kayak]

200c?

[wraith cargo]

class at our uni

[summer path]

Is it even possible for a class to cover all of either of those books?

[wraith cargo]

grad algebra sequence

[ivory trail]

sorry i just doxed myself

[slim kayak]

What do the numbers mean Jason

[wraith cargo]

Well our class covered a vast majority of atiyah-macdonald

[summer path]

[wraith cargo]

but I think you need a pure comm alg class that's very long and deep to cover all of Eisenbud

[ivory trail]

professor skipped primary decomposition iirc so i did it on my own

i don't remember any other big topics i covered outside the class

the resolutions and stuff is whatever

[wraith cargo]

did you do Grobner basis stuff

[slim kayak]

Ew, computations

[ivory trail]

lol no

i skimmed the algorithm

[wraith cargo]

wdym that's like ch. 15 of Eisenbud lol

(to kerr)

it's comm alg grrrr

-<

I can't think anymore need sleep

[ivory trail]

actually sorry i was lying

wikipedia:

Universally catenary rings ⊃ Cohen–Macaulay rings ⊃ Gorenstein rings ⊃ complete intersection rings ⊃ regular local rings

gorenstein is the only one i don't know here

LMAO

also about cohomology: i try to read the nlab page every few months to see how much more i understand

[rustic crown]

i'm a det

[delicate orchid]

no human being will ever understand an entire nlab page this endevour is fruitless

[ivory trail]

https://ncatlab.org/nlab/show/morphism what about this one

[delicate orchid]

unpossible

[slim kayak]

Wdym, you don't instantly understand how cohomology is related to the homotopy category of hypercomplete (inf,1)-toposes???

[rustic crown]

det slow >.<

[slim kayak]

Maybe one day I'll be able to read 10 words into a nlab page without having to open 6 tabs

[ivory trail]

they're invariants of elliptic curves

sections of some bundle on the stack of elliptic curves or something like that

technically modular functions are the invariant ones

but you can make modular functions out of modular forms

i watched (half of) borcherds's lecture series on modular forms lol

https://www.youtube.com/playlist?list=PL8yHsr3EFj51HisRtNyzHX-Xyg6I3Wl2F

that is motivation enough for me

but maybe you're talking about generalizations

[white oxide]

hey I’m taking a class with that dude next sem

[echo gull]

that's dope what's the course?

[glossy crag]

THE Eisenbud? lucky you

[white oxide]

It’s called groups rings and fields, it’s a graduate course iirc

[glossy crag]

If D is a division ring, how do I see that the homomorphism D^op->End_{M_n(D)}(D^n) via a\mapsto (x\mapsto xa) is surjective?

Said otherwise, why every group homomorphism D^n->D^n compatible with left matrix multiplication is right multiplication by a scalar.

[rustic crown]

wasn't this some jacobson density thingy?

i don't remember the details

tubuwu

[glossy crag]

Hm, actually you're right, I could use it probably.

JD is a little different tho, it says R->End_{End_R(M)}(M) is dense.

and M_n(D) is not End_D(D^n), M_n(D^op) is

nvm, it's End_D^op(D^n), therefore M_n(D)

but...

shit, all these ops confuse me

gimme a sec

[rustic crown]

D = k = k bar

[glossy crag]

I feel like the last part isn't convincing (the bit about how End_Dop(D^n)\cong M_n(D) => End_End_Dop(D^n)\cong \End_M_n(D)(D^n)). Is there a better way of doing this?

[barren sierra]

how is it not convincing?

[distant chasm]

Why is there no Z[i]-module with 12 elements?

[formal ermine]

wait are you the same person as yesterday

[tribal moss]

As an abelian group it would need to have a subgroup of order 3 -- that is, {0,a,-a} with 3a=0.
Now what is i·a? It can't be 0 because i^4·a = a.

It can't be a or -a, because then -1·a = i²·a would be a itself, but a doesn't have additive order 2.

But if i·a is outside the subgroup, then it and a generates a subgroup of order 9, which is impossible.

[distant chasm]

Yes, exam today

[distant chasm]

I don't think i fully understand but i will think about this, thank you

[tribal moss]

More abstractly, one can appeal to the structure theorem and the fact that the size of a quotient of Z[i] is always the sum of two squares.

[distant chasm]

I was thinking about this but i wasn't sure if is it always or just it can be sum of two squares, i think it is easier to check like this, thank you!

[tribal moss]

Z[i] is a PID, and for reasons of geometry, Z[i]/<a+bi> must have |a+bi|² = a²+b² elements.

[distant chasm]

Thank you again

[tribal moss]

(Except when a+bi=0, of course).

[distant chasm]

One last question(probably is a stupid one), if someone can answer. If i write directly a module as a sum of indecomposable modules, does it count as indecomposable?

[rustic crown]

each individual summand is by definition indecomposable

but if there are more than one summands, then the whole thing is again by definition decomposable

[distant chasm]

So do you know another Z[i] indecomposable module with no torsion other than itself?

[rustic crown]

How about Q(i)?

you'll need to go infinite-generated else structure theorem of mods over pid takes over

[rustic crown]

(notice that any two non-trivial submods will intersect here, so can't decompose)

[distant chasm]

I think that's it, thank you!

[devout dirge]

What are the finite commutative local rings with residue field F_p, such that the maximal ideal is principal? I suspect they're only Z/p^k and Fp[t]/t^k but I am unable to prove it.

[dim widget]

@devout dirge what about the totally ramified extensions Z_p[x]/(x^n -p)?

Modulo various powers of p

Mod p^k these are (Z/p^k)[x]/(x^n - p) and have maximal ideal x with the correct residue field.

Any proof of the correct version of this statement will use some universal mapping property for regular local rings to Artinian rings with fixed residue field, cohen structure theorem, and nakayama’s lemma and be very easy to prove though

[rocky cloak]

Seems there could be a few. If I'm not mistaken Z/9[sqrt(3)] and Z/9[sqrt(6)] are not isomorphic for example.

[dim widget]

@rocky cloak there is a square root of 2 modulo 3^2 because sqrt(2)=sqrt(3 - 1) and that power series converges

By which I mean the series sqrt(x - 1) is well behaved in characteristic not 2

[rocky cloak]

Is there? 0^2 = 3^2 = 6^2 = 0,
1^2 = 8^2 = 1,
2^2 = 7^2 = 4,
4^2 = 5^2 = 7

[dim widget]

Whoops I suppose only sqrt(-2) exists

Forgot about the constant term

Which is i

[rocky cloak]

My guess would be that all such rings are of the form (Z/p^k)[x]/(xf(x) - p, x^n) where f is a polynomial. But this should be "over counting".

[dim widget]

@rocky cloak I don’t think you need the x^n and your sqrt(6) example shows that they need not all be of that form. When p is not zero they’re certainly all O/p^k where O is obtained by adding the root of an Eisenstein polynomial to Z_p, that’s easy to prove.

[rocky cloak]

My example is of that form, it's isomorphic to Z/9[x]/(2x^2 - 3).

You need the x^n, but it could probably be avoided by putting some conditions on f. For example Z/9[x]/(3x - 3) is not finite.

[dim widget]

I think you mean 1/2 x^2

…..

[rocky cloak]

Yeah, right 5x^2 - 3

[dim widget]

I was assuming we wanted to write them with f monic to avoid huge redundancy

Eg writing them with an Eisenstein polynomial is (up to the problem with units) a complete classification

[rocky cloak]

Maybe having leading coefficient between 0 and p will be a good compromise

[dim widget]

In that such f modulo p^k parameterize exactly a choice of such an R and a generator for the maximal ideal

Without adding any bizarre extra data

For fixed p^k if one desired one could then quotient the space of such f(x) by the action of the units in Z/p^k by rescaling and obtain exactly the moduli space of such R

When p is zero things are much easier.

[devout dirge]

Thanks alot! Do you have maybe a criterion in mind such that the answer would be indeed only Z/p^k and Fp[x]/x^k?

[rocky cloak]

Not sure how natural it is, but your condition + p having degree being in the degree 1 part of the associated graded ring.

It's a bit silly, since p=0 in the latter example.

Actually, scratch that. Might still not be strong enough.

[static yew]

Bad news everyone, I'm back

Is a prime order subgroup of an elliptic curve over F_p a vector space?

It looks like it satisfies all of the axioms

[delicate orchid]

what would the scalar multiplication be

[static yew]

"Point multiplication"

[delicate orchid]

does that satisfy a(x+y) = ax+ay?

[static yew]

Hrm but wait

When I tested earlier, g^n was 0

[delicate orchid]

I'm having trouble visualising it

[static yew]

Are vector spaces allowed to have ab=0 for a nonzero a,b

[next obsidian]

If you have a group of order p, it is a vector space over F_p

Abelian group

[static yew]

Or wait no
I'm mixing up 0:1:0 in the group with 0 in the field

[formal ermine]

how do you multiply vectors

[static yew]

And if group order is composite it's a module?

[next obsidian]

No

I mean it’s a Z-module always

But that’s literally another word for abelian group

It can still be a vector space over F_p if its order is infinite or p^n for some n

But you can’t guarantee it

[delicate orchid]

oh prime order I read that as prime index

[static yew]

Sorry a+b

I learned on RSA and the related diffie hellman, which is based on multiplicative group mod N

So my mental model is "group operation is multiplication"

[next obsidian]

your equation is saying b = -a, so yes

[static yew]

Ok lemme try to rephrase

If a group is of prime power order q, I can easily turn it into a vector space with field F_q

But if the group order is not a composite, it cannot be a vector space, merely a module?

[next obsidian]

A module over what

any abelian group is a module over Z

It’s another name for abelian group

And it’s not the case that a composite order group can’t be a vector space. You can be a vector space over F_p with order p^n, or infinity

[static yew]

A module over a ring of order n?

[next obsidian]

It won’t work if any other prime divides it and it’s finite though

This isn’t a meaningful thing to consider

But

Yes

It’s a module over Z/nZ

This is just saying that in a group of order n that n•x = 0

[static yew]

Its meaningful to me cuz it lets me verify my understanding of things

[next obsidian]

The important part with being a vector space is that vector spaces are way better than modules

And the utility of saying abelian group = Z-module is that you can apply module theory which is kinda like weaker linear algebra

But insofar as moving from Z-module to Z/nZ module for n composite, I think practically speaking it affords basically nothing

[static yew]

Like: if all fields of same cardinality are isomorphic, then any can't any finite group of prime order be turned into a field by mapping elements to 0..p-1 and defining the "multiplication" operator in terms of Z/pZ?

[formal ermine]

prime order or prime power

[next obsidian]

Yes, but this is artificial

[static yew]

Yeah that

[next obsidian]

This thing you propose basically reduces to this question: “what is 1?”

[delicate orchid]

well, there is only one finite group of prime order lol

[next obsidian]

we know what 0 is, its 0 in the group

[static yew]

I get it

I'm just trying to verify that I understand things properly

[next obsidian]

But you can actually pick anything which isn’t 0 to be your “1”

[static yew]

g, or maybe a

[next obsidian]

That’s just a letter

My point is that everything in G\{0} could be 1

Everything that’s nonzero in a prime order group generates the group

[static yew]

Which means I can choose any 1 I want 😄

[next obsidian]

Sure

But then multiplication is pointless

Like, it’s just a thing you artificially put onto the thing

Like…

If anything can be 1, then the concept of 1 is pointless

The value of having extra structure is when it came to you with that structure already

[static yew]

Well

They cant all be 1 at the same time

[next obsidian]

Yes but it doesn’t do anything

Whatever you do depended on your choice

Like I can turn {,,} into F_3

By saying = 0

= 1

And =2

But this doesn’t say anything about my original set

I’m just studying F_3 now

[formal ermine]

F_1 = { }

[static yew]

Ok I think this leads into the thing I'm currently trying to understand: Schoof's algorithm for counting the points in an elliptic curve subgroup in polynomial time

You can easily construct F_p by building off of Z/pZ

Slightly more formally, the set of integers 0..p-1 with addition and multiplication defined from the Peano axioms, reduced modulo p

Schoof's algorithm involves applying the Frobenius endomorphism to the curve points over Fp-bar

[coral spindle]

Bro just say F_p. You don't need to invoke peano whenever you want to talk about the integers.

F_p is actually equal to Z/pZ, fwiw.

[static yew]

Isomorphic

[coral spindle]

That is the typical definition, so no, I mean equal.

[formal ermine]

mfw peano

[static yew]

What about F_p^k though?

[coral spindle]

That is harder to construct.

[static yew]

That's what I'm having issues with

Since x^p = x in Fp, the algebraic closure of Fp obviously cannot use the "usual" addition and multiplication

[coral spindle]

OK, F_p^k is F_p[x]/(Phi_k(x)), where Phi_k is the kth cylcotomic polynomial. Is this the information you needed?

[formal ermine]

the algebraic closure of Fp uses the same addition and multiplication as Fp

[static yew]

Maybe

[static yew]

But then x^q = q and frobenius is an identity transform, no?

[coral spindle]

No

The Frobenius endomorphism is not invertible in the algebraic closure of F_p

Wait hold on, that's wrong

Sorry, I mean to say that it is not of finite order

[static yew]

Yeah that's what I read

It's an infinite union

But this is where I'm hung up

Elliptic curves over prime field Fp just means computing the x and y coordinates modulo p

But that doesnt hold got higher powers of p, does it?

[coral spindle]

I don't know what the confusion is exactly

If you're in a different field, you use the addition and multiplication defined in that field

The field, as it were, is packaged with this information.

[coral spindle]

Now as it turns out this definition above comes with the information you need, if you know how to read it

Is that what the confusion is?

[static yew]

Right this is where I have trouble

The algorithm description doesn't define the field

It just says it's F_q-bar

[coral spindle]

Yes, because typically we know what F_q bar is

Writing down what it is in an explicit way is a bit tricky.

[static yew]

I have a mathematical description of a prescriptive algorithm

It works fine for proving it's valid, cuz all fields of equal size are isomorphic, so the choice of field is arbitrary

[coral spindle]

Here's a rough explanation: every element of F_p bar lives in some F_p^n for some n>0. The addition and multiplication within these individually is the same as normal. We think of F_p being contained in F_p^2, being contained in F_p^3 etc. which is what allows us to cross-pollinate the addition and multiplication.

[coral spindle]

cuz all fields of equal size are isomorphic
All finite fields of equal size are isomorphic

We certainly cannot substitute F_p bar for Q.

(despite their being the same size)

[static yew]

Wait we can't?
Interesting

Not relevant though I'm working in finite fields

[slim kayak]

Mfw Q and Q^2 are isomorphic

[coral spindle]

You're not working in finite fields. You've mentioned F_p bar

[coral spindle]

He said fields

Anyway

F_p bar is an infinite field.

[static yew]

Not isomorphic to Q, got it

[slim kayak]

Ye sry

[coral spindle]

He STILL said fields lol

An example is Q and A, fwiw

A is the field of algebraic numbers

[static yew]

Isn't Q a field?

[coral spindle]

Q^2 is not.

[delicate orchid]

sure, Q^2 isn't th- STUF

wfoejih;jio;grfe

[static yew]

Smh

[slim kayak]

Yeah but fields have no proper product operation

[coral spindle]

I aim to steal all your thunder. it's ALL mine

[delicate orchid]

_ _

[static yew]

Okay let's get off the infinites, I'm an engineer

Everything is finite

[delicate orchid]

F_p bar is infinite you can't get around this point

[static yew]

R and C are birds

Then maybe f_p bar doesn't exist but a finite version exists that's good enough for my needs

[south patrol]

isn't engineering built on R lol

and C

[static yew]

Sanity check: Groups of order n

All subgroups are of orders that are factors of n, right

[delicate orchid]

yes

and f_p bar exists as much as f_p does

that's a bad notion to have

[formal ermine]

not factors, divisors

[delicate orchid]

what

oh I see

[ivory trail]

R exists, C exists, F_p bar exists

[delicate orchid]

yeah, combinations of factors

[delicate orchid]

R does NOT exist

the other two do

[ivory trail]

oh my bad

[formal ermine]

yeah who even believes in the axiom of choice??? amiright???

[delicate orchid]

https://tenor.com/view/black-panther-black-panther-black-panther-marvel-marvel-gif-25406660

[static yew]

Real engineering is mostly built on a horrific union of a subset of the rationals, a single element that has a+e=a for most a, two absorptive elements, and set of a few million even more absorptive elements

[ivory trail]

you need choice to show the algebraic closure is unique in general iirc

[static yew]

Anything that leads to Banach-Tarski is sus

I dont actually need the full closure I thini

[ivory trail]

yeah but no one says that a 32 bit int is Z mod 2^32

[static yew]

I think all I need is a field that can contain x^(q^2)

[static yew]

I do if I'm explaining wraparound

[slim kayak]

Axiom of determinacy enjoyer

[ivory trail]

to be clear you don't need choice to construct the reals

[south patrol]

What are absoprtive elements

[ivory trail]

@ all readers

[formal ermine]

you missed the joke

[static yew]

Also a 32 bit signed int is Z/2^32Z minus 2^31

[south patrol]

Isn't Banach Tarski more just a warning sign that we need to define measurability lol

I've never seen the issue with it

[ivory trail]

we have some people here who may not be able to detect the joke

[static yew]

I am 100% certain we had this convo yesterday

[formal ermine]

that's their problem, respectfully 👍

[formal ermine]

and the day before that

and the day before that

and the day before that

and the day before that

[coral spindle]

This is the Theorem of Lagrange

Nb the converse is not true

[formal ermine]

only if abelian

[delicate orchid]

A_4 has no subgroup of order 6

[coral spindle]

Whoops internet moment

[delicate orchid]

hahahah beat you to it this time

[chilly ocean]

illuminator goes crazy arc

[static yew]

Ngl the effects of BT make me think of the proof that 1=2 that leverages division by zero

[south patrol]

Which one

lol

[south patrol]

Really

[static yew]

Wait does that mean the contents of this channel form a cyclic group

[slim kayak]

Crazy? I was crazy once. They put me in a rubber room. A rubber room with rats. Rats make me crazy.
Crazy? I was crazy once

[south patrol]

A_4 also has no subgroups of order 48

[formal ermine]

Crazy? I was crazy once. They put me in a rubber room. A rubber room with rats. Rats make me crazy.
Crazy? I was crazy once

[static yew]

What is A?

[formal ermine]

A_4 also has no subgroups of order 1024

[south patrol]

a capital letter

[delicate orchid]

no not really

[formal ermine]

alternating group

[south patrol]

Oh okay wew thanks

[formal ermine]

kernel of sgn or some shit like that

[delicate orchid]

group of even permutations on a set of size n

[formal ermine]

idk I don't believe in A_n, it doesn't exist

[south patrol]

unique subgroup of index 2 of S_n

[formal ermine]

group theorists keep making these groups up

what's next?? tits group??? oh wait...

[south patrol]

Imma be bullied for that typo

[static yew]

If I have an l torsion subgroup of a finite group of order n, then gcd(l,n)=l, right?

[slim kayak]

S_n doesn't consent to be abelianized

[formal ermine]

what's an l torsion subgroup

[static yew]

There is no axiom of consent

[ivory trail]

Lagrange's theorem

[formal ermine]

torsion group of order l?

[static yew]

That's a yes, right?

[ivory trail]

your language is a bit weird but yeah

[formal ermine]

I'm feeling left out

[ivory trail]

oh wait

it might be something else

lmao

[formal ermine]

oh I don't know elliptic curves yet

[ivory trail]

like the l-primary part

[slim kayak]

Now is your opportunity

[static yew]

Yeah I have no formal education at this level it's all self taught

I skipped over vector spaces and went straight to polynomial rings

[formal ermine]

I have so much stuff to learn at the moment...

[next obsidian]

What the FUCK is going on

[slim kayak]

Moar

[formal ermine]

p adic analysis... toric geometry... algebraic number theory... diff geo...

[next obsidian]

[static yew]

We can learn together!

[formal ermine]

[ivory trail]

that's the subgroup annihilated by l

[formal ermine]

do you know what a fan is

[delicate orchid]

check out my cool ring

[ivory trail]

those elements who go to 0 when multiplied by l

[south patrol]

Fake

[next obsidian]

That’s not a ring, that’s some rep theory shit

[delicate orchid]

it's also a ring

[next obsidian]

I know what those [] mean…..

[south patrol]

I have a lot to learn rn

Pain

[formal ermine]

hey @delicate orchid

[south patrol]

Fun times

[static yew]

Point multiplied, right?

[formal ermine]

what is a monoid algebra??? it sounds like rep theory

[delicate orchid]

I've learnt a lot and now I'm going back through it all and computing examples as I go

[formal ermine]

is it??

[delicate orchid]

wtf is a monoid algebra

[formal ermine]

ok nvm lol

[ivory trail]

l is an integer, actually a prime as stated

[agile burrow]

monoidal gebra

[delicate orchid]

do you mean like

monoid objects in R-mod or something

aka algebras

[static yew]

Right so its P+P+P... l times is 0

[formal ermine]

nono

[agile burrow]

group algebra but monoid

[delicate orchid]

oh right, duh

[agile burrow]

oh i was just guessing

[formal ermine]

given a cone lattice

you can define the monoid algebra

[delicate orchid]

yeah ok so it sounds like a group algebra

[static yew]

Right so its P+P+P... l times is 0 (point at infinity) right?

[delicate orchid]

or at least some kind of algebra on the poset structure of the lattice

[formal ermine]

like if $(\omega, M)$ is your cone lattice then $$\bC[\omega \cap M] = \bigoplus_{u \in \omega \cap M} \bC\chi^u$$ is your monoid algebra

[delicate orchid]

not sure what the operation would be

[formal ermine]

yeah

[formal ermine]

the multiplication is just $\chi^u \cdot \chi^{u'} = \chi^{u + u'}$

[formal ermine]

we are doing things at an incredibly fast pace in lecture and my prof has now started not even giving proofs for theorems and propositions because they would be too complex for one course

[delicate orchid]

ok so it's just the free module generated by the elements of the monoid endowed with a linear extenstion to the module operation

or as I like to call it

F(M)

[delicate orchid]

that's a sign he should slow down

what course is it

[formal ermine]

newton okounkov theory

[static yew]

Okokok theory?

[formal ermine]

I like to call it NOT

[delicate orchid]

never heard of this SHIT

[formal ermine]

it's some relatively new theory

[delicate orchid]

dude you're 16 why the fuck are you learning new theories

learn real analysis you pointed skill set mf

[slim kayak]

That's a really specific course

[formal ermine]

so far we've done algebraic geometry, convex geometry, and now we're doing toric geometry

[slim kayak]

You sure it isn't just the current topic in the course?

[ivory trail]

do you know schemes

[formal ermine]

I know the definition of a scheme yes, haven't dived into any properties or anything like that yet tho lol

started learning diff geo

[formal ermine]

15

me

[static yew]

Notation question

Is "F_q[x,y]/(y+x+b)" a polynomial ring modulo y+x+b?

[ivory trail]

you don't actually need schemes for toric geometry

[formal ermine]

yeah

today was the first time I used schemes

[ivory trail]

just the gluing part

[formal ermine]

to understand why it's called "ramified"

like in algebraic number theory

apparently it comes from geometry...

[formal ermine]

yes

[delicate orchid]

don't ping me I'm listening to text to speech reddit compilations

[ivory trail]

ramified means branched

[static yew]

Use a Ponzi scheme to make money off of toric geometry

[formal ermine]

@delicate orchid hey I got a question for you

@delicate orchid ok so

@delicate orchid what do you call $\bC[M]$ for $M = \Hom(N, \bZ)$

[static yew]

How TF do I do modulo a multivariate polynomial

[delicate orchid]

@formal ermine what is the @formal ermine problem my dear friend @formal ermine please let me know

[formal ermine]

N is a free Z mod

[delicate orchid]

see this is why you need to learn ring theory properly

this is just a quotient R[x_1, ..., x_n]/(f)

[formal ermine]

k[x,y] = k[x][y]

[ivory trail]

look up quotient ring

[slim kayak]

It's a set cosets of ideals which forms a ring, so yeah, do your ring theory

[delicate orchid]

dunno

[coral spindle]

I cannot stress enough that you should read a book

[formal ermine]

the definition is clear to me but my prof never gave it a name

[ivory trail]

or we could do the radical approach of learning the universal property and only using that LMAO

[formal ermine]

you don't already do that?

[agile burrow]

does it need a name

[formal ermine]

yes

[agile burrow]

Fred

[coral spindle]

I call it Robert

[formal ermine]

do I just read out "C M"

[coral spindle]

fuck you walter

[formal ermine]

like if I'm talking to someone

what would I say

[coral spindle]

You'd say C M

yeah

I'd say that

[delicate orchid]

I should really know what Hom(F_n, Z) is

oh is it just S_n lol

[formal ermine]

is it bad that I just view algebras as vector spaces with haha funny vector product

[coral spindle]

Should be

[ivory trail]

people don't really do this in intro algebra classes

[delicate orchid]

yeah cool

ok that didn't take long

[coral spindle]

Well something like that anyway

ZS_n

[delicate orchid]

yeah

[ivory trail]

somewhat

[delicate orchid]

just pick one of the generators of F_n

[coral spindle]

ye

universal property that bad boi

u get elements of Z^n

oh shit

Z^n it is

[delicate orchid]

I just thought of matrices ngl

[agile burrow]

just pick n-tuple

[delicate orchid]

yur

ok thanks for the help my best friends

[coral spindle]

Wait wew

my best friends

am I not your bestie, bestie?

[delicate orchid]

Hom(Z, Z^n) is way easier...

[static yew]

If I have y^2-x^3-ax-b

Are the ideals x and y?

[delicate orchid]

you want to quotient this I imagine

you'd quotient by (y^3-x^3-ax-b) < this is the ideal generated by that polynomial

[coral spindle]

Ofc Hom(F_n, Z) is 'the same' as Hom(Z^n, Z) in some way that I don't wish to make precise, because Abelianisation.

[formal ermine]

as what structure is $\bZ^2 \iso (\bZ^2)^\times$

[coral spindle]

?

[delicate orchid]

none lol?

order argument

[formal ermine]

the fuck

[coral spindle]

I am going to repeat that you should read a book. Books tell you exactly what ideals are.

[delicate orchid]

(Z^2)^\times is just (\pm 1, \pm 1) is it not

[coral spindle]

Nobody here wants to teach you a whole undergrad AA course

[formal ermine]

wait loooooool I'm not sure if he means multiplicative group or dual space

[static yew]

I fixed my typo (twice) its y^2

Is it still the same if the high exponents differ?

[delicate orchid]

WHO USES TIMES FOR DUAL SPACE

[formal ermine]

yeah like bruh

[delicate orchid]

if it's an \ast it's dual space you muppet

[formal ermine]

he uses \vee for another dual thing

[delicate orchid]

of course it is?

[ivory trail]

go consult wikipedia

[delicate orchid]

no, go consult a book

[ivory trail]

consult wikipedia so you know you need to consult a book

[static yew]

I've read lots of books, but most books dont cover this specific topic

If you have a specific suggestion I'm all ears

[delicate orchid]

learning maths from wikipedia is a REALLY bad idea if you're not already well versed in the surrounding field

[static yew]

I am consulting wikipedia

[delicate orchid]

I can't think of any AA books that don't cover quotient rings/ideals

artin absolutely does

[coral spindle]

This is just wrong, sorry. As wew says, any intro book will cover ideals.

[slim kayak]

I am pretty sure any abstract algebra book will tell you how to understand forming the factor ring with an arbitrary ideal

[delicate orchid]

unless it's a pure group theory book for whatever reason lol

[coral spindle]

If you really need a suggestion, I personally learned from Fraleigh

But people here prefer Artin and others. I haven't read Artin.

[ivory trail]

artin has arrows

so it's good

[formal ermine]

artin 💪

[static yew]

Arrows?

[open sluice]

commutative diagrams i think

[slim kayak]

Abstract nonsense

[open sluice]

o

[ivory trail]

category theory and universal properties

[delicate orchid]

OHHHH YEAHHHH

[agile burrow]

grp cohomology

[open sluice]

must be a pain to code these diagrams

[delicate orchid]

co this homo nerd

[open sluice]

i struggle with straight lines, imagine curves

[delicate orchid]

just use quiver lo

[slim kayak]

Which is it written out

God gave us the complex category for a reason

[delicate orchid]

alg topologists writing "fold" for a map and expecting me to take their field seriously

[formal ermine]

so true king

[open sluice]

i like the nuclear explosion in the middle

[ivory trail]

https://q.uiver.app/ this actually

[open sluice]

WTF

[delicate orchid]

yeah lol

mf thinks we do it by hand lolololol

[static yew]

The set of polynomials with coefficients in F2 modulo an irreducible polynomial looks like a quotient ring

Did I miss something?

[ivory trail]

does your book seriously have the (n-1)th suspension of S^0

is this how algebraic topologists are

[open sluice]

i should have figured, these need to be advertised more

[delicate orchid]

we explained this yesterday lol, all field extenstions are quotient rings

[coral spindle]

quiver and tikzcd is pretty good but I've heard that some people like to draw their diagrams more precisely to get marginally better results

[delicate orchid]

"modulo" is a quotient

working in R mod I is the same as working in R/I

[ivory trail]

"modulo J" means " if J was 0"

and then we have some properties of 0: 0 + 0 = 0 and r0 = 0

so that's where ideals come from

of course you need to prove that this works

[static yew]

Okay so all I need to understand is how to compute y^5 mod x^3+y^2+3

[coral spindle]

You just did

[ivory trail]

if x^3+y^2+3=0, then y^2 = -(x^3 + 3)

[static yew]

I did?

[coral spindle]

Just because we have something we mod out by doesn't mean there's a nice representation of it

For example we can consider the group R modulo Q

[static yew]

Wait hrmmmm

[coral spindle]

Can you 'compute' pi mod Q?

No ofc not

that's meaningless

it is just pi mod Q

[ivory trail]

so y^5 mod x^3+y^2+3 is just y(y^2)^2 = y(-(x^3+3))^2 mod x^3+y^2+3

[coral spindle]

Now you're confusing modulo for computing a nice representative modulo

[ivory trail]

have fun with that expression

[static yew]

Since I'm dealing with prime fields, neither pi nor Q exist

[coral spindle]

You are missing the point

In arithmetic modulo 5, for example we really think of 5 mod 5 being equal to 0 mod 5, which is equal to 15 mod 5.

The way we write it doesn't matter.

Now as it happens we can always find an integer 0 <= x < 5 such that n mod 5 = x mod 5

and this is the representative

There is no general way of choosing representatives when we mod out by something.

The point is: mod is not a function!

[wraith cargo]

ofc there is
It's the one you choose

[static yew]

Yup

Formally "congruent"

With mod x^2 + x + 1

X^0 = 1
x^1 = x
X^2 = x + 1

[coral spindle]

So when you said this... the reason I said "you just did" is that there's not anything to compute yet

[static yew]

If I'm in F_p, it's the value in the range 0..p-1

[coral spindle]

Yes ofc

this is exactly what I said just above

[ivory trail]

the word "congruent" is a special word you use so you don't have to say "mod J" over and over

[coral spindle]

You need to think carefully first about what representatives in K[x,y] you want to choose

[ivory trail]

even then people don't care that much

[coral spindle]

then you can think about computing the representative

[ivory trail]

the coefficients are in F_p, the powers can be whatever

y^5 is already written with coefficients that are canonical representatives of the classes of F_p

if you want to reduce the power a lot, that's different

[static yew]

From my perspective, with univariate, its trivial. Reduction modulo the polynomial is polynomial long division and take the remainder

[ivory trail]

and it's not always helpful

[coral spindle]

Ahufhsakjfshkafhjk

THis is what I'm saying

[ivory trail]

it depends on what you are modding out by

[coral spindle]

You're doing polynomial long division because that gives you a representative: a unique nice thing in the ring for every element in the quotient ring.

[static yew]

The stuff I have experience with is modulo a univariate polynimial

Yes

Is there another common way that I don't know of?

[coral spindle]

Now you should see that the property that makes it unique is its degree: when you compute the remainder, it is of minimal degree such that it represents the thing you see

[static yew]

Yes. That I understand

[coral spindle]

Now you should also see that you can't do the same with bivariate polynomials. Firstly the division algorithm doesn't work the same way unfortunately, but also because the degree doesn't work in the same way.

[static yew]

Ugh.

[coral spindle]

So if you want to have nice representatives, you're going to have to think about what they ought to be

Then we can talk about 'computing' y^2 mod that polynomial

Here's a really grody one. Chew on F_2[x,y]/(xy - 1) for a bit

There are nice representatives you can choose but you will have to ponder it for a while

[static yew]

I can't because I don't understand what that ring is

[coral spindle]

Well I've suggested books

[chilly ocean]

grody

[coral spindle]

who'd you call grody

[delicate orchid]

it's all polynomials in x, y but we set xy-1 = 0

[chilly ocean]

joe

[next obsidian]

That ring is super nice

[coral spindle]

Shut your trap you are literally a chair

[ivory trail]

don't be hyperbolic, it's just nice

[next obsidian]

So true

[static yew]

Ugh I gotta get back to work

[chilly ocean]

call her laurent the way she invert my indeterminate