groups-rings-fields

[next obsidian]

Yeah idk this seems really hard to pick out some random example

I think anything in the realm of Z/nZ won't work

matrix rings aren't commutative

I don't think quotients of polynomial rings of fields seem very nice

Suddenly most of my examples for rings are gone lmao

[oblique river]

there are examples in quotient rings of polynomial rings

[next obsidian]

Huh

I tried dual numbers

[oblique river]

https://math.stackexchange.com/questions/1045673/two-elements-in-a-non-integral-domain-which-are-not-associates-but-generate-the

[next obsidian]

to introduce nilpotence

[oblique river]

this one feels weirder to me than the continuous function one

but it uses polynomials

[next obsidian]

This seems more like something I'd stumble upon

but also I literally never think about continuous functions for rings so

I should probably try to consider them more lol

[shy bluff]

I... dont' follow that example

so you have some ring K[x, y]/(xy^2)

Like what is ?

[next obsidian]

U is a polynomial of x and y

this is just a product of elements of the polynomial ring k[X,Y]

[shy bluff]

oh is U(x, y) just any unit in K[x, y]?

[next obsidian]

it came from the unit $u$ in the quotient

[shy bluff]

there was a unit in the quotient? Is this not the quotient by xy^2?

[next obsidian]

No

They start off by saying they use lowercase for the image of stuff in the quotient

they said assume that ux = x(1 + y) in the quotient

[shy bluff]

Ok

[next obsidian]

this turns into a statement U(X,Y)X - X(1 + Y) in (xy^2) in the polynomial ring

By taking representatives

[shy bluff]

What do you mean by taknig representatives

[next obsidian]

like

x, u are equivalence classes

you jsut take an element in that equivalence class

[shy bluff]

Oh ok

So the capitals are just elements that are in the equivalence classes?

[next obsidian]

you do this everytime you move from a quotient into statements in the lareger ring

yeh

then they do some stuff to prove something about U(X,Y) I think

to show that u is in a maximal ideal

so that it couldn't be a unit

even tho you started off by assuming it is

[shy bluff]

this is confusing

I shall go think about this

thanks

[next obsidian]

👍

[dawn kiln]

yoo where tf am i lmao

[next obsidian]

The algebra zone

[dawn kiln]

ah im p pissed rn

[shy bluff]

Ok so from that paper

With K[x, y, z]/(x - xyz), we get that x = xyz

And y and z aren't units?

[vestal snow]

You ever feel like cheating when you use a technique from a problem you've seen to solve the problem you're currently working on?

[solemn rain]

what

isnt this math

lmfao

[next obsidian]

You ever feel like cheating when you use a technique from a problem you've seen to solve the problem you're currently working on?
@vestal snow I feel the opposite

I feel like “IM LEARNING!”

[vestal snow]

Haha I guess I'm in the "glass is half empty" camp

[vestal snow]

I wonder what commutative algebra would be if we assumed the negation of Zorn's Lemma

Is there any text/paper about this?

Is there a typo here?

where it says $P_3, \cdots P_r$ are prime, maybe it should say $P_1, \cdots P_r$

[cloud walrus]

The formerly edible banana:

[shy bluff]

ok so from this,

why do they have x ~ xy?

THeir definition is that a ~ b if Ra = Rb

Oh is x ~ xy their way of writing (x) = (xy)

[next obsidian]

Is there a typo here?
@vestal snow no, you only need the ones past 1 and 2 to be prime

Usually anytime you use this all of them will be prime

But prime Avoidance is satisfied when the first two aren’t necessarily prime and I’m pretty sure you want to do something like that

[next obsidian]

@marble heath both groups are commutative

[marble heath]

That is the thing

if the set is of 4 elements

[next obsidian]

You can’t show they’re non-isomorphic by showing one is commutative and the other isn’t

Just write out multiplication tables

[marble heath]

Then both of them have to be conmutative if they are a group

[next obsidian]

Yes so you do it a different way

With 4 elements you can just write out multiplication charts until you find them

[marble heath]

I will be able to create just 4 charts

All of them are going to be conmutative

[next obsidian]

There’s other ways to show they’re not isomorphic

[marble heath]

So I have to find another thing

[next obsidian]

Look at the order of the elements

[marble heath]

But the order will be the same if the groups have the same properties

I think it is a hard ex hahaha

[next obsidian]

This is not hard

One of them has an element of order 4

The other doesn’t

Just find a group without any elements of order 4

With 4 elements

[marble heath]

Maybe I m wrong

But, if they are 4 elements

Then I cannot repeat one of them on a row or column

That is not going to asure me that the order will be the same for the 4 charts ?

[next obsidian]

You have to look at your charts

And follow repressed multiplication by the same element to figure out its order

If you do it you will discover a group with 1 element of order 4, 2 of order 2, and 1 of order 1

Versus one with 3 elements of order 2 and one of order 1

[marble heath]

You are right

I thought that the 4 possible groups had the same order

Could you please tell me what Groups and Rings are useful for ?

[fickle brook]

groups arise naturally any time you need to describe symmetry of any kind

[shy bluff]

Ok so I've been trying to figure this out, and by their definition, a ~ b is if Ra = Rb, which is the same as (a) = (b) right?

So then we'd have to have that Rx = Rxy

And I see that all of Rxy would be in Rx, but not necessarily the other way around?

[shy bluff]

Like in the earlier paragraph of this they have that

So I guess that I need to show somehow that q(x) = q(xy) right?

And in this quotient ring we have x = xyz, so 1 = yz?

I guess I'm asking what exactly does x and xy get sent to in this quotient ring

[smoky cypress]

Does the fact that R only has 1 finite degree field extension has something to do with the fact that every polynomial with real coefficients can be split into linear and quadratic terms?

[shy bluff]

so a UFD is also an integral domain, so there are no nonzero zero divisors, and every non zero element is cancellable under multiplication right? So if u and s are in a UFD R, then u/s, s/u both have meaning/exist?

[knotty mason]

cancellation yes, but division no

example Z: you can't do 2/3

[summer path]

I know the notation for "is a subgroup of" is $\leq$, but is there any notation for "is a Sylow p-subgroup of", or is it actually necessary to write out all of that (or does anyone have a nice short way of writing that)?

[cloud walrus]

Tubular Cat:

[woven obsidian]

I encountered this problem in my topology course and realized my group theory was really rusty

But how can I show that Z^2 and Z^3 are not isomorphic as groups?

[wind parrot]

Suppose $\phi\colon \mathbf Z^2 \to\mathbf Z^3$ is an iso. Consider the subgroups $H = 2\mathbf Z^2$ and $H' = \phi(H) = 2\mathbf Z^3$. This gives isomorphisms $(\mathbf Z/2\mathbf Z)^2 \cong \mathbf Z^2/H \cong \mathbf Z^3/H' \cong (\mathbf Z/2\mathbf Z)^3$. But the first has 4 elements and the latter has 8, so these can't be isomorphic.

[cloud walrus]

NielsK:

[timid hull]

For any two groups $G,H$ let $\text{Mor}(G,H)$ denote the set of group homomorphisms from $G$ to $H$ . Assume to the contrary that $\mathbb{Z}^2$ is isomorphic to $\mathbb{Z}^3$ via an isomorphism say $f:\mathbb{Z}^2\rightarrow \mathbb{Z}^3$ . Now consider the map $h:\text{Mor}(\mathbb{Z}^3,\mathbb{Z}/2\mathbb{Z})\rightarrow \text{Mor}(\mathbb{Z}^2,\mathbb{Z}/2\mathbb{Z})$ given by $h(g)=g\circ f$ since we assumed $Z^2$ is isomorphic to $Z^3$ so $h$ should be a bijection but clearly $|\text{Mor}(\mathbb{Z}^3,\mathbb{Z}/2\mathbb{Z})| > |\text{Mor}(\mathbb{Z}^2,\mathbb{Z}/2\mathbb{Z})|$ hence a contradiction .

[shy bluff]

Hrmmm

[cloud walrus]

WhyamIsohot?:

[next obsidian]

Does the fact that R only has 1 finite degree field extension has something to do with the fact that every polynomial with real coefficients can be split into linear and quadratic terms?
@smoky cypress you can do a proof which exploits this, and uses only this fact, along with a few others. So I’d say yes

[shy bluff]

Well gcd(a, b) | a always

So I can write a/gcd(a, b) right?

[next obsidian]

You really shouldn’t

Since / isn’t an operation, it’s supposed to mean you multiply by gcd(a,b)^-1 which might not exist

[shy bluff]

Hrm

[next obsidian]

But usually you’d just say let c be such that gcd(a,b)c = a

[shy bluff]

Also I still want to clarify some stuff about the queshion from yesterdy but I'm in bed right now lol

Oh

True

[next obsidian]

And I’m not even certain that such a c is unique

In an integral domain it will be

By cancellation

[shy bluff]

Well I'll ask more about that in a bit, I'm in bed right now and can't remember what I wrote down regarding that

Well in a UFD

[next obsidian]

I guess you can say like

“Anytime a | b, I will denote by b/a the unique element c such that ac = b”

[shy bluff]

Hrm

I see

[next obsidian]

But in general if you’re doing a/b and you don’t know everything has inverses people will be a little suspicious

[shy bluff]

Yea but like as somebody pointed out earlier, say 2/3 doesn't exist in Z

[next obsidian]

Yeah

So if you have x,y in Z and wrote x/y

I’d be like, woah woah woah, why is that in Z?

[shy bluff]

Well in a general UFD, if d = gcd(a, b) I can say that there exists a unique element c such that dc = a and some other unique e such that de= b?

[next obsidian]

Yup

Uniqueness is just

Suppose that you have c, d such that ca = b and da = b

Just use transitivity

ca = da

Then you can cancel to get c = d

Unless a is 0 lol

[shy bluff]

Yea

Wrapping my head around this is still a bit hard

Uh other question

So from the paper last night linked in the mathstack discussion, they had this

So I guess that it's saying that q(x) = q(xy) where q is the quotient function from K[x, y, z] to R

But why is that true?

[next obsidian]

I don’t think that’s true

[shy bluff]

I don't really get it

[next obsidian]

What does ~ mean

[shy bluff]

a~b means Ra = Rb

So same ideal generated?

But in the first screenshot above, does the bar not denote quotient?

[next obsidian]

Yeah I think so

I don’t see how you show that though uhh

[shy bluff]

Yea nor do I

[next obsidian]

Actually I do

[shy bluff]

They just claim it and I'm ????

[next obsidian]

So xy = x*y

And xy*z = xyz = x

Cuz quotient

So xy in (x)

And x in (xy)

[oblique river]

the more I think about it, the more this example seems really clever to me

the whole point with trying to find two elements which generate the same ideal is you need to find some a and b which are both multiples of each other

[shy bluff]

Like I see that xy in x

[oblique river]

the easiest thing you can do is take a = x and b = xy

[shy bluff]

But not that all of x is in xy?

[oblique river]

then b is clearly a multiple of a

[next obsidian]

Like I said

[oblique river]

but how do you get that a is a multiple of b?

[next obsidian]

xy*z = xyz

= x

[oblique river]

literally just adjoin another element to make it true

x = (xy)*z

[next obsidian]

Because you’re quotientinf by (x - xyz)

That just says x = xyz

[shy bluff]

I think there's something fundamental that I'm missing

[oblique river]

in the ring R, x is literally equal to xyz

[shy bluff]

Yes

[next obsidian]

The elements (xy)z in (xy)

[shy bluff]

That I get

[next obsidian]

But (xy)z = xyz = x

[oblique river]

x is a multiple of xy

and therefore x is in (xy)

because x is a multiple of xy

[shy bluff]

Ok x is in xy that I get

But why is xy in x?

[oblique river]

because

xy is a multiple of x

this is the whole point of the exercise -- we are trying to find two elements a and b in a ring which are multiples of each other

so that (a) = (b)

so in this ring, we're taking a = x and b = xy

[next obsidian]

Do you believe that x = xy * z

[shy bluff]

yes, x = xyz

because the quotient

[next obsidian]

Do you believe that xy*z in (xy)

[shy bluff]

yes

[next obsidian]

Then x in (xy)

[shy bluff]

OH

Ok

It's like a circle

x is in xy, xyz is in xy, and then x is xyz

[next obsidian]

Yes

This gives you equality of the two ideals

Now you have to show that they aren’t associate

the more I think about it, the more this example seems really clever to me
@oblique river yeah same

[shy bluff]

ok yea I think that I can figure out the second half now if I stare at it for a while

That first half confused me lol

what do they mean by setting y = z and x = 0 though?

[next obsidian]

This is a polynomial

You’re just plugging in values

Y = Z means the same arbitrary value

[shy bluff]

oh

[next obsidian]

So this turns the polynomial of 3 variables into a polynomial of 1

[shy bluff]

didn't know you were allowe dto do that

[next obsidian]

I mean if you want to reason you can just plug in W for Y and Z

Then it’s a polynomial in only W

But that’s the same thing as just calling it Y

[elder valley]

i feel like this problem is too tricky for an intro algebra class

at least i wouldn't have been able to do it during my first algebra course without a hint of some kind or just looking up the answer

[shy bluff]

shrugs

[next obsidian]

Yeah this is kinda fucked haha

[shy bluff]

I tried googling some of the material and half of what shows up is like grad level stuff at other schools lol

Seems pretty arbitrary how they go and ste things up

[next obsidian]

Are you at like Stanford or something haha

Or Mich?

[shy bluff]

None of the above

waterloo

[elder valley]

o.O

[vestal snow]

I know that University of Toronto has pretty difficult intro sequences in Analysis and Algebra as weed out classes

I'm guessing Waterloo is the same

@next obsidian Would you recommend skipping some sections at the beginning from Matsumura?

[next obsidian]

Idk, I found them refreshing

They also cover some stuff you might not have seen before

Like section 2

[vestal snow]

They are much better written than A-M imo so I don't mind

[next obsidian]

has a proof that any projective module over a local ring is free

where as there's a much easier proof for finite projective

the general case used transfinite induction

and I thought it was mega cool

and I actually did a proof of the finite case for a midterm once, and knew it held more generally so seeing the proof was sick

[vestal snow]

I'll definitely have to check that out

Thanks for recommending this book to me

[next obsidian]

Yeah and it's just in the second section haha

I really like the book

The style of proof you'll see also is like

really clever with picking elements and doing element manipulation stuff

which I'm not a fan of normally, but I like how it's done in the book and it makes me want to get better at it

Also my general thoughts on it are that it isn't the best option for a first introduction to commutative algebra, but once you know a bit it's good and if you do the entire thing you're in a great spot to do anything in AG or ANT for the most part

[shy bluff]

Mmmm I've decided to wait until classes are in preson to take the next algebra course, galois theory, as it seems that online learning is not greate for me lol

[elder valley]

are you specializing in algebra?

[shy bluff]

no, but I'm interested in it

[next obsidian]

Galois theory is really cool, but if online learning isn't working no one can fault you

[shy bluff]

I don't know what I'm specializing in, for now I'm just treating this whole thing as a "hey iunno what I like but I think that this is interesting so imma go take a bunch of courses on it" kinda thingi

And perhaps retake this coursee in the future tbh

[elder valley]

when i struggle in a class usually ill try to sit on in the lectures the next time it's offered, but unregistered

[shy bluff]

Ah problem is that my school is doing all online again next term

[elder valley]

i guess you can take the lame courses online in the mean time xD

[shy bluff]

yea

Or at least ones that aren't as specialized?

I'm taking uh number theory, cryptography and combinatoricial enumeration instead lol

[elder valley]

sounds like a fun semester

[shy bluff]

Still a couple weeks left in this class though

this much left

[elder valley]

"the big theorem" lmao

[shy bluff]

It's this

[elder valley]

dunno about you all but i thought unique factorization was the most boring topic ever

[next obsidian]

Oh that’s a cool@one

[steep hull]

It becomes more interesting in the context of algebraic NT

[next obsidian]

It’s an exercise in Matsumura

I don’t know if it really qualifies to have that name tho tbh haha

[vital quail]

yeah same

jacobson just goes over it quick

[elder valley]

the name is nearly as suggestive as "fundamental theorem of ___"

[next obsidian]

Is it considered bad practice to introduce a UFD as an integral domain such that the latter condition holds, then make students show it's equivalent to unique factorization

lol

[old hollow]

Is Galois theory lit

[solemn rain]

yes

easy and fun

for me i think it was finally the moment where i turned

definition-thm-proof into ideas

that i can play with and so on

its cool

[shy bluff]

Ok so for this question I got it going in one direction, if I is prime and nonzero then as R is a PID, I is maximal, and so R/I is a field, and fields are PIDs, but I'm not really sure how to approach it from the other direction?

If R/I is a PID, something something

I feel like there's something to do with the quotient?

[vestal snow]

The "big theorem" that Liria posted looked familiar.

I literally saw it in Matsumura yesterday lol

[old hollow]

@solemn rain what are the prerequisites

[solemn rain]

group theory/field theory are must

and some knowledge of how polynomial rings work

if you knwo groups and rings most galois theory textbooks/sections will begin with fields

[vestal snow]

Can someone explain how we got the commutative diagram?

I don't know what theorem I'm supposed to look for in the appendix

[elder valley]

If R/I is a PID, something something
@shy bluff R/I is an integral domain if and only if I is a prime ideal

[shy bluff]

well that's the question lol

[elder valley]

😛

the pid part is not really needed

[shy bluff]

Well

How do you go the other dierciton? Where R/I is integral domain

[old hollow]

@solemn rain I’m a high schooler, I know some group/ring theory but not polynomial rings. Do you know of any books that might be good for me

[solemn rain]

check pinned in #book-recommendations

[old hollow]

Oh thanks

[solemn rain]

a 'guide' into abstract algebra textbooks

[elder valley]

just use the definitions @shy bluff

[shy bluff]

Integral domain has no zero divisors

[elder valley]

it's nothing tricky

[shy bluff]

I feel like there's a definition that I'm missing

So an integral domain is just that there are no zero divisors right?

[old hollow]

I assume GT means Galois theory? lol

[elder valley]

yeah, and commutative with 1

you want to prove I is prime, so take a,b in R such that a*b is in I

[shy bluff]

Then either a or b is in I

But what does that haev to do with R/I?

[elder valley]

you want to conclude a or b is in I

[solemn rain]

GT means group theory

generally

[old hollow]

Oh

Thank you for the help

[shy bluff]

well ab is in I, so ab = 0 in R/I right?

[elder valley]

yup

[shy bluff]

Oh and because it's an integral domain, then neither a nor b can be zero divisors

So either a or b is in I?

[elder valley]

yeah that's it

[shy bluff]

Ohmright R is also an integral domain

[next obsidian]

Can someone explain how we got the commutative diagram?
@vestal snow construct it explicitly

You want to think about elements and where to send them and stuff, I forget exactly

At one point you map to like inverse images of stuff

[vestal snow]

Alright. I'll try what I did for the snake lemma

[next obsidian]

It isn’t quite that bad

So take basis element v_i in L_1

then look at f(v_i), as N -> M is surjective, take some w_i in N such that phi(w_i) = f(v_i)

Then map v_i to w_i

that makes the diagram commute for that square

you can do a similar kind of thing for the map L_2 -> K

[vestal snow]

Would you recommend doing all the exercises or is it fine to skip a few?

For example, one in section 2 requires a homological algebra

Is it fine if I skip it or should I go read the appendix?

[next obsidian]

I did all of them

if you can I'd say go for it

but I'd get familiar with homologial methods

1: they're powerful

2: I enjoy them

[vestal snow]

Got it

[next obsidian]

2: I enjoy them
@next obsidian Also by this I don't mean, "I like them so learn them reee", but me and a lot of other people I know found homological algebra fun so you might as well

[vestal snow]

Haha I'll keep that in mind

@next obsidian Are you sure that the map is well defined?

[next obsidian]

It doesn't matter

you pick arbitrary things in each fiber

Multiple such maps can work

[vestal snow]

I don't think I follow

[next obsidian]

So you chose things in the fiber of f(v_i)

to map v_i to

like you picked some w_i such that phi(w_i) = v_i

you might have multiple choices in such w_i

but it doesn't matter, you just pick one

[vestal snow]

Ahh gotcha

[next obsidian]

all that matters is existence

which is implied by exactness

for the other map

you do something similar, but you need to use the fact that whatever thing you want to map to is actually in the kernel

or something like that, just use exactness to know such a thing exists

[vestal snow]

Got it

[next obsidian]

Since the map K-> N isn't surjective

[vestal snow]

This way of defining the maps only works because L_1 is a free module right?

[next obsidian]

Yup

[vestal snow]

There is no linear relation between the generators, which is why we don't get contradictions

[next obsidian]

If it wasn't you'd have to check some relations are satisfied yup yup

[vestal snow]

S L I C K

Where did we use in the proof that M is of finite presentation?

[next obsidian]

L2 is finite

free

[vestal snow]

Oh right lol my bad

[next obsidian]

You can get a free module L2 there alwyas if M is finite

But to know it's finite is presentation bit

Also

This is the lemma I told you about

if you remember

[vestal snow]

I did!

[next obsidian]

Pretty neat haha

I was surprised to use it outside of like

the problems immediately following

which are concocted to be solved with it haha

[vestal snow]

*problem

[next obsidian]

I don't see finite presentation stuff talked about that much but Matsumura talked about it a bit so I was

[vestal snow]

Do you remember what the problem I asked about was?

[next obsidian]

Umm

Something about a surjection to a free module I think

[vestal snow]

I knew it had something to do with Kernel is finitely generated

Yes

[next obsidian]

For a finite module

show the kernel is finitely generated

[vestal snow]

If M -> A^n is a surjection and M is finite, then Kernel is also finite

[next obsidian]

Yup

I see

I think some of the problems in section 2 are best solved via snake lemma

I remember a few were kinda tricky

but I have solutions written up so

[vestal snow]

It's pretty neat seeing how free modules are analogous to polynomial rings

[next obsidian]

if you get stuck and don't want to look in the back lol

Ah

yeah so polynomial rings are free algebras

since the indeterminants form the like basis elements wrt algebraness

[vestal snow]

Yeah

Also the universal property

[next obsidian]

yup

same thing as for free modules but in the category of algebras

[vestal snow]

Ah yes category theory, the thing every non-math major asks about when I tell them I'm a math major

[next obsidian]

Do they actually

[vestal snow]

You'd be surprised

[next obsidian]

I usually just get "wow you like math?"

Or

"Wow I sucked at math, I was good until they introduced letters"

I havne't once had someone ask about category theory haha

maybe you're talking to computer scientists or something???

[vestal snow]

A lot think it has something to do with philosophy

[next obsidian]

lol

maybe only in the sense of trying to make it a foundations for math

[vestal snow]

Yeah

[next obsidian]

and the general philosophy of "we care how things map to things more"

[vestal snow]

I think someone once tried to tell me how I can make any equation true by using category theory

[next obsidian]

what uni do you go to if you don't mind sharing? You said you're going into your second year?

Lmao

[vestal snow]

and taking it to a universe where it is true

[next obsidian]

...

[vestal snow]

I'm going into my second year at University of Arizona

[next obsidian]

Oh I've heard some nice things about them

[vestal snow]

Also, the people I was talking about do not know anything about category theory

I'm pretty sure they saw a youtube video or something

[next obsidian]

They sound like cranks

plain and simple

[vestal snow]

Hah that might be it

[next obsidian]

I had someone once tell me about how like the odds can be in my favor if I could count the number of ways to buy different lottery tickets

and it was literally countable by an nCk

and I just told him right there how many combinations exist

and how like

no

you statistically lose massive money

it was a nightmare

and I was doing something like at a desk

[vestal snow]

That's what I like about math as opposed to other subjects

[next obsidian]

like a job sorta

so I couldn't leave

or tell him to piss off

;w;

[vestal snow]

It's really hard to bs your way out with mathematicians

[next obsidian]

yeah lol

If you have significant clout you can try and pull a Mochizuki thing but

in general no haha

[vestal snow]

Hah yeah

What if Mochizuki is on this server?

[next obsidian]

I'll dm him and ask him to explain

[vestal snow]

DeletedUser #2480

[next obsidian]

Are you taking your school's grad algebra sequence next year?

Or did you take it last year

[vestal snow]

Yes I am

[next obsidian]

Ah, I did that my second year as well

[vestal snow]

I took the freshman sequence

[next obsidian]

I had so much fun

[vestal snow]

They didn't let me freshman year

[next obsidian]

You'll also be in a really good position probably haha

Rip

[vestal snow]

But it worked out for the best

[next obsidian]

I wasn't in a position to my freshman year so it worked out fine for me

I started doing like actual proof based math my first year so

It would've been rough to say the least

[vestal snow]

The professor who's teaching it next year also taught me in the undergraduate sequence

[next obsidian]

at my school algebra was "self-contained" so it's the only grad class you could theoretically do for the first take

but man it would be rough

[vestal snow]

And I got a research thing with him next semester

[next obsidian]

oh that's nice

I should try and do some sort of research

oof

[vestal snow]

If you're into Algebraic Geometry maybe you've heard of him

He's into arithmetic geometry though so idk

[next obsidian]

What's his name?

Hmm, yeah I'm not familiar. Granted I don't really know many names in any field honestly

[vestal snow]

I think Zeta knows him

[next obsidian]

That makes sense I think he went to Arizona

for grad school

And did ANT so probably brushed shoulders with an arithmetic geometer

[vestal snow]

Hold up a minute

There's a non trivial chance I might have met him at some point

[next obsidian]

I think he graduated more than a few years ago

since I think he's at a teaching position now

[vestal snow]

Ah right

[next obsidian]

But

I wanna say he's not doing like a tenure track standard thing

so maaaaybe he did it right after graduating

not sure though

That would be funny if you have though

he has a youtube channel you could try and listen to his voice

and see if you recognize it LMAO

[vestal snow]

Do you have the name?

Is it just zetamath?

[next obsidian]

I think it's just zetamath

Also, there's solution in Matsumura

but I also have done every one and texed them up till section 6 which I'm on rn

so I'm down to talk you through any of them, unless you end up passing me haha

in which case I won't have them done yet lol

[vestal snow]

Section 9 or Chapter 6?

[next obsidian]

section 6

the end of the second chapter

Uhh

The one about associated primes

[vestal snow]

Cool, we can work on it together!

[next obsidian]

I was stuck on the second exercise for a while

but it's just false

and the example given in the first exercise is a counterexample LOL

for context it's "is it true that..."

So it's not a fault of the exercise the statment is false haha

[vestal snow]

Answer: No

[next obsidian]

yeah

[vestal snow]

I did that on a midterm once

[next obsidian]

Lmao

I'm interested in problem 6.5

[vestal snow]

I got hit with the "Counterexample?"

[next obsidian]

it seems really interesting

Also 6.6

Is something I did already for a final

lol

it's also on wikipedia

Yeah honestly I don't see why people recommend this book more

It's rough for a first introduction I think

[vestal snow]

I mean

[next obsidian]

but I question really how good the advice of "Just do AM"

is

[vestal snow]

The exercises in A-M

felt like dragging my balls through glass

[next obsidian]

Haha, yeah

[vestal snow]

It took an average of 2 hours per problem

[next obsidian]

It's really RIP

Honestly the exercises were like

took a long ass time

or like

This was trivial wtf

or "I know this already"

But a lot of them were of the first category

But some of them also are like

fuck you I'm not gonna do some shit about Boolean lattices

[vestal snow]

I actually liked that one

[next obsidian]

I just didn't think it would ever be useful

[vestal snow]

I think I liked it because it helped me figure out stuff like Lattice Isomorphisms by myself

[next obsidian]

Ah

So I learned algebra in winter and spring of my first year

in a study group Shamrock started

because we weren't in major until fall

both of us were 1st years

like midway through the quarter is when we got accepted

Shamrock couldn't take algebra because he wasn't in major

and the sequence already started and he wanted to do grad algebra the next year

so he just made a study group

[vestal snow]

Stuff works differently in my university

[next obsidian]

Then him and me continued it this past year with us leading it like as the teacher I guess (the previous year it was our analysis TA)

So a lot of the basic ring and group stuff I got really solid on that second pass

the stuff I kinda was eh on the first time

[vestal snow]

You could usually just go into the math department and ask them you want to take whatever undergraduate course

and they just take an oral exam about set theory and some proofs

and basically you're good to go

That's what I did to get into the algebra sequence

[next obsidian]

oh lol

[vestal snow]

Also, my advisor allowed me to switch out diff eq and vector calc for two semesters of grad algebra

[next obsidian]

Lmao

I'm gonna be doing stuff like that

[vestal snow]

Him being the professor for the grad algebra course probably played a role

[next obsidian]

I think my TA had like complex analysis subbed in for diff EQ

Oh nice haha

[vestal snow]

Yeah

[next obsidian]

the math advisors at my school aren't teachers

the math dept kinda sucks like

you don't wanna work through them

you wanna email teacher directly

[vestal snow]

Mine aren't either

[next obsidian]

get permission

[vestal snow]

this was my faculty advisor

[next obsidian]

and then be like "pls register me"

does everyone get one at your school?

or are you just a baller

[vestal snow]

I don't think so

I had one for my first year because I basically went to every algebraists office and was like "please do research with me senpai"

[next obsidian]

lmao

[vestal snow]

One said that I can't do research but he's willing to be my faculty advisor

[next obsidian]

I see haha

Idk if our school even has faculty advisors

that reminds me tho I should email my algebra prof and ask what I should do for research stuff

and grad school I guess

I think we're on pretty friendly terms so

[vestal snow]

Tropical Geometry is something that was recommeded to me

[next obsidian]

I'll probably ask him to write a rec

Tropical Geometry

[vestal snow]

Because it's a fairly new subject

[next obsidian]

In terms of research I see

[vestal snow]

and if you're doing the research with a prof, there's a reasonable chance to get a publication

[next obsidian]

From what I heard it's like number theory related right?

[vestal snow]

Vaguely

[next obsidian]

Idek if anyone at my uni does tropical geometry

[vestal snow]

It's more combinatorics/optimization related I would say

No one at my school does either

I asked my professor if he would be willing to do it

and he said yes

[next obsidian]

Oh wow the wikipedia has only really recent citations

dang that's chad

Will you do this research with me

that they don't already do

[vestal snow]

you should try asking your algebra prof

I mean

[next obsidian]

I think he's waaaay too busy to do it

[vestal snow]

I know that this research won't be his main focus

[next obsidian]

I asked if he could do a reading course

and he said he was just too busy

He's like a fairly well-known algebraic geometer so rip

[vestal snow]

And he himself said that he doesn't really care what the research is about because what he really wants me to learn is how to do research in math

[next obsidian]

sounds like a nice guy

[vestal snow]

Worth a shot imo

[next obsidian]

I might give it a shot haha

[vestal snow]

Definitely

[next obsidian]

I'd ideally like

want to try my hand at something in AG

but like

RIP

[vestal snow]

Tropical Geometry is like

[next obsidian]

he told shamrock he should just do all of Hartshorne and try to do research in his senior year

and we're in a similar boat I think

same year, know a lot of the same stuff

[vestal snow]

Classical AG but replace + with min and * with +

[next obsidian]

Hmm

time to learn about varieties I guess

☠️

[vestal snow]

All of Hartshorne

Rip

[next obsidian]

He said it like

in his second quarter of second year tho

so the idea was he could finish in 2 years

[vestal snow]

Oh that's doable

[next obsidian]

Yeah

[vestal snow]

I hope so

[next obsidian]

still massively hard for someone like

doing full workload for their degree tho

And now I'm tryna finish it by the end of next year so

I should get on that

[vestal snow]

Online courses

[next obsidian]

Yeah

[vestal snow]

I just do all my gen eds online

[next obsidian]

Really actually I'm taking some AG stuff next year

I struggle to do gen ed and math at the same time

messes up my workflow

[vestal snow]

What i do is this

[next obsidian]

I do really well with the 1 assignment every week on same day

So this past year I was taking only grad math and same next year

Then my senior year is like do gen eds and separate math stuff

not in classes

[vestal snow]

1. Find an online class that fulfils the requirements
2. Grind out the entire course over a weekend
3. Profit

[next obsidian]

lmao

Idk if any of them have it setup so you can just do it all in 1 go like that

if there is I would totally do that though

[vestal snow]

You have to find those

Usually iCourses are like that

[next obsidian]

what's an iCourse?

are you not taking it through your university?

Also, I have a lot of gen eds from community college

[vestal snow]

Asyncronous online

[next obsidian]

so that really isn't a problem for me

[vestal snow]

No they are through my uni

[next obsidian]

Huh

Honestly tho my senior year idk what I would take

I'll take real analysis

but at the end of next year I'll have

[vestal snow]

Graduate?

[next obsidian]

I thought about it

but I still do need gen eds

but in terms of math

I'd have a lot of AG, Algebra, Manifolds, Complex Analysis all at graduate level

I guess I could try like combinatroics

and reals

Do french

[vestal snow]

Damn that's a lot of courses

[next obsidian]

and I only need a few more out of major credits

[vestal snow]

I usually do 2 math classes/semester

[next obsidian]

Yeah haha, 2 years straight I'd only take grad math courses so

Yeah I'm rocking 3 a quarter this year

and 4 next year

but that's fine cuyz

2 are AG and like sort of linked

[vestal snow]

Here's my plan:

[next obsidian]

so It's sorta just 1 I think

[vestal snow]

Sophomore: Grad Algebra and UG Real Analysis
Junior: ANT and Complex, both grad
Senior: AG and Topic in Algebra, both grad

[next obsidian]

Maybe we should go to #discussion

[vestal snow]

yeah

[plucky flicker]

Hello world! I have a question here: let the ideal generated by f_1,...,f_s be J. So, the marked equality is clear for me, but what happens when we take the right hand side to be I(V(J))? The equality is still true, right?

[drifting plover]

The marked equality isn’t always true

The left hand side has to be a radical ideal for it to be true

If that’s the case, then yeah, I(V(J)) = J as well

[plucky flicker]

okay, so the extra condition is that J has to be radical, right?

ahh, okay

thank you 🙂

[plucky flicker]

under what condition is an ideal its own radical ideal?

[next obsidian]

When if x^n in I that x in I

Equivalently if R/I is reduced (only nilpotent is 0)

[vestal snow]

How do we get $I$ is generated by $(u_1,\cdots , u_n, v_1y,\cdots , v_my$?

[cloud walrus]

The formerly edible banana:

[vestal snow]

Nvm figured it out

[next obsidian]

Yeah you just have to bash it out

[shy bluff]

for this question, what excatly is Z[alpha]? Is it just a + balpha such that alpha is purely imaginary?

And what exactly is |x|^2? Is that like the magnitude of it?

[supple marsh]

presumably yes

[shy bluff]

Can't I just be like "As N(x) defined as |x|^2, we get a value that is greater than or equal to 0"?

[scarlet estuary]

whats the question

[woven delta]

|n+im|^2 = n^2 + m^2

[shy bluff]

Yea

[supple marsh]

show that N(xy) = N(x)N(y) and rewrite x | y.

[shy bluff]

x | y is the same as saying y = sx for some s, then we can show that N(x) | N(y) if there exists t such that N(y) = tN(x), so you basically want to go and show that s maps to t?

[elder valley]

well you can define t to be N(s) but you still have to show the equation N(y)=N(s)N(x) holds

not sure if that's what you were asking

[shy bluff]

Yea no I figured it out

I think I just typed it funny here

[shy bluff]

Wait question, @woven delta, I emailed the prof to clarify but he hasn't gotten back to me yet, but uh in general, for Z[alpha], would you say that |a + b(alpha)| = sqrt(a^2 + b^2alpha^2)?

[woven delta]

So for Z[sqrt(d)], the norm is N(a+bsqrt(d)) = a^2+db^2

If d <= 0, then this gives you a norm which is always nonnegative

[shy bluff]

hrm

Because I'm getting this

I think I messed up my expansion somewher?

[woven delta]

Yeah you did

First off what is alpha^2

I'm assuming alpha is sqrt(d)

[shy bluff]

An integer?

Yea e

The question is uh

like Iget that the two of them differ by one term

Which is like bleh so close yet so far

so iunno what I'm doing wrong

[woven delta]

Which term for they differ by?

The middle one?

[shy bluff]

Yea

They differ by 4adbc

[woven delta]

I see

[shy bluff]

I fuckede up somewhere but idk where

[woven delta]

Yeah one sec let me check your work

[shy bluff]

oke

Thank

[woven delta]

Oh I think you are defining the norm wrong

[shy bluff]

Oh

[woven delta]

No it's correct

I agree with your first computation

Let's see if I agree with your second one

[shy bluff]

hrmmmm

[woven delta]

Wait actually

Okay so the norm is supposed to be $N(a+b\sqrt(d)) = a^2 - b^2 d$

[cloud walrus]

Liquid:

[shy bluff]

why? How?

[woven delta]

Which is why it is always positive when d is negative

[shy bluff]

oh wait ok I see

Yea because d is purely imaginary right?

[woven delta]

So when d is -1 it works out to N(a+bi) = a^2+b^2

sqrt(d) is purely imaginary

[shy bluff]

Yea

[woven delta]

Yeah

So yeah you need that negative

[steep hull]

It needs to be a map from Q(sqrt(d)) to Q.

[woven delta]

Sorry I got confused

Yeah

Lol you can define the norm for finite field extensions of Q

[shy bluff]

pardon?

[woven delta]

In terms of the galois group of a galois extension

And that's where the - comes from

[shy bluff]

oh

Ok, so I was using the wrong definition of norm right?

[steep hull]

Yes, these indeed always coincide for Galois extensions (you will have repeated conjugates for non-primitive elements though)

[woven delta]

Because the norm of an element x of Q(sqrt(d)) is the product over the elements of the galois group of Q(sqrt(d))/Q of sigma(x)

[shy bluff]

I .... We have'nt learnt that

[woven delta]

That's fine

Sorry I just thought it's important to know it comes from somewhere

I wonder if there's an easier way to come up with the norm

[shy bluff]

But ok so I should do it again but this time with N(x) = N(a + balpha) = a^2 - b^2alpha^2?

[woven delta]

Yeah

[shy bluff]

Ok

[steep hull]

For the case of non-Galois extensions (really extensions in general), the norm is the product of the images of alpha over all the embeddings L into an algebraic closure that fix K (of which there are always [L:K] of them for separable extensions by the primitive element theorem).

[woven delta]

Yeah

[steep hull]

Sorry for the digression @shy bluff

[shy bluff]

It's ok

So what you just said is how you define this stuff yea?

I just haven't learnt enough yet to understand it

[steep hull]

There’s a deeper theory behind the definition

[shy bluff]

o okie

[woven delta]

Basically the only automorphisms of Q(sqrt(d)) are the identity and congugation. The norm is if you apply both to an element and then multiply those 2 numbers together

[shy bluff]

oke I got it now

THank you!

How do you show that a ring satisfies the ascending chain condition for principal ideals?

Like how do you show that there is no infinite strictly increasing sequence of principal ideals?

Actually what does it mean to have an infinite strictly increasing sequence of principal ideals?

[woven delta]

So you should figure this out by looking at examples

And proving that certain classes of rings satisfy the ascending chain condition

[shy bluff]

Yea like the notes say that PIDs satisfy it

but there are no examples unfortunately

[woven delta]

A nice non PID that satisfies it is Z[x]

Try proving that

Also there's a good theorem that makes the ascending chain condition make sense

Prove that a ring satisfies the ascending chain condition iff every ideal is finitely generated

Oh for principle ideals

Lol

[shy bluff]

Yea for principal ideals

[woven delta]

I thought you just meant the ascending chain condition

Okay, so any ring that satisfies the ascending chain condition satisfies the ascending chain condition for principle ideals

So that's an example

[shy bluff]

Like my understanding is basically that

• principal ideals are just sets generated by a single element that are closed under addition and multiplication?
• If it stabilizes then there exists an ideal that contains all other ideals?

[woven delta]

So stabilizes means that your ascending chain is eventually constant

[shy bluff]

what does that mean?

[woven delta]

So call your ascending chain A_n

[shy bluff]

Wait another question, what do they mean by "ascend"

[woven delta]

It means that you have a sequence of ideals with
$A_0 \subseteq A_1 \subseteq A_2 \subseteq ... \subseteq A_n \subseteq ...$

[cloud walrus]

Liquid:

[woven delta]

That's what ascending means

[shy bluff]

So each ideal is contained within another ideal

[woven delta]

Each ideal is contained within all of the ideals of higher number label

Each ideal is labeled by a natural number

[shy bluff]

Sure

[woven delta]

Eventually constant means that all of the ideals labeled with a number greater than some number k is the same ideal

[shy bluff]

Oh so it's saying that eventually you get a big enough ideal that you can't go any bigger?

[woven delta]

Yeah

[shy bluff]

Err that any bigger ideal will just be the same thing

[woven delta]

Like it means that you can't have a strictly increasing chain of principle ideals

[shy bluff]

Oh so that's what the (A_n) = (A_k) thing is about?

[woven delta]

Yeah

Did you guys talk about the ascending chain condition in general?

Or just for principle ideals

[shy bluff]

just for principal I think

[woven delta]

Oh okay

[shy bluff]

I'm mildly lost right now because his lectures have been increasingly disjointed as we near the end of the term and he realizes that he went too slow at some point and is now rsuhing to complete

[woven delta]

That sucks

[shy bluff]

Yea I think we only did principal ideals

[woven delta]

But yeah the ascending chain condition on principle ideals makes sense if you think about divisibility

[shy bluff]

hrmmm what do you mean?

Also seperate question, just from proving these two bit scan I say that Z[alpha] is a euclidean domain? As we've shown that there's a function N: R -> N u {0}

[next obsidian]

I don’t know what exactly you’ve shown, but you have to still show the Euclidean algorithm works with respect to that norm

I don’t think it follows directly from the divisibility thing at the end of (a)

Each ideal is labeled by a natural number
@woven delta isn’t ascending chain condition for arbitrary length chains, not just countable?

I feel like I once had issues because of this

I think I was trying to show Ascending chain <==> finitely generated ideals but doing it the wrong way and countability threw a wrench in it

[shy bluff]

hrm

Ok

What exactly does it mean to show that the euclidean algorthim works?

[woven delta]

Hmm can you try to reconstruct where it fails?

[shy bluff]

Like we have this

Err wait no

[woven delta]

@next obsidian

[shy bluff]

And this as our example?

[next obsidian]

It means that for any p,q

There exists some a such that

p = aq + r

And N(r) < N(q)

It’s just division with remainder

I think this is for non-zero q or you have to be careful if N(q) = 0 or whatever

[shy bluff]

Yea non zero q

[next obsidian]

Also Liquid, I don’t actually remember

Anyway you have to show that the norm you construct

Works like that

There’s a lot of conventions here, but if all you needed was A norm

Then everything would be a Euclidean domain taking the norm N(x) = 0 when x = 0, else N(x) = 1

The presence of the norm isn’t what makes the domain a Euclidean domain, it’s that using that norm you can do the Euclidean algorithm / division with remainder

[shy bluff]

I see

How do you rprove that that's possible though?

[next obsidian]

¯_(ツ)_/¯ idk what your norm is or where

[shy bluff]

Like in general

[next obsidian]

But like you probably saw this done for Z[i]

[woven delta]

Lol

[next obsidian]

Idk

[shy bluff]

An example that we have is this

[next obsidian]

You try hard

And come up with an idea

[shy bluff]

Which seems to be done as if they're just descrbiing that you go and add it a bunch of times

[next obsidian]

Idk

[woven delta]

They are talking about Z[sqrt(-d)]

[next obsidian]

I don’t really have a lot of practice showing it

[shy bluff]

hrm

Ok

[next obsidian]

I think I’ve only shown two things are Euclidean domains

Z[i] and a polynomial ring over a field

[woven delta]

Usually with these problems I just stare at it and eventually I figure it out lol

Idk offhand how I do it

[next obsidian]

Yeah I’m with Liquid on this one

[shy bluff]

Can I just be like "As Z[alpha] is a ring, it's closed under addition and multiplication, so we can go and repeatedly take multiples of x until we have some number s of x's such that N(sx) > N(y), but then we reduce s by 1 and taket he difference between y and (s-1)x to be r"?

idk if that makse sense

[next obsidian]

I mean

If it’s true

[woven delta]

I mean

[next obsidian]

Then yeah

But you’d have to convince yourself and others it is

[woven delta]

You need to figure out something specific to this ring

[shy bluff]

I see

[woven delta]

Or these class of rings

I'll think about it later liria

This is actually a thing I should know for my Qual lol

[shy bluff]

Ehhh I think I canfigure it out lol

[next obsidian]

So my thought is

Do you remember when i outlined the proof of why Z[i] is a Euclidean domain?

You basically estimate the quotient in C

By the closest element of Z[i]

And the distance you go is < 1 (from the actual quotient and the thing you estimate it by)

So when you multiply back through by the thing you want to divide by

You get the necessary N(r) < N(q)

I think the idea is the same??

Just instead of a nice 1 x 1 square grid

Z[alpha] makes a different lattice

[shy bluff]

Yea

I get what you mean?

[oblique river]

hi all, just curious, what are you trying to prove?

[next obsidian]

Wait are you even trying to prove it’s a Euclidean domain?

[oblique river]

what is alpha?

[next obsidian]

¯_(ツ)_/¯

[oblique river]

just sqrt of a negative?

those aren't all euclidean

[next obsidian]

I think it’s sqrt(-d)

Apparently

Yeah

That’s what I thought so

[oblique river]

(which is why I'm trying to figure out what you're tryign to prove)

[next obsidian]

Yeah I only spoke about it because she asked if we can say it’s a Euclidean domain so I said you’d have to show the Euclidean algorithm works

[oblique river]

you could have just said "it's not" :P

(in general, at least)

[shy bluff]

Alpha is just some purely imaginary number

Uhh I'll send the question again in a minute, I'm cooking

[next obsidian]

you could have just said "it's not" :P
@oblique river at the time I didn’t know what alpha was either

[shy bluff]

@oblique river part b over here

[kindred rivet]

What book is this liria

[shy bluff]

Uhhh ostensibly dunnit and foote but he's yet to refer to it

dummit*

[oblique river]

oh, yeah that is just going to follow from part a

if you have an ascending chain of principal ideals, look at the norms of each generator in Z

what can you say about that sequence of integers?

you'll also need the fact that the only things with norm 1 are units in the ring Z[alpha]

[shy bluff]

I don't really know how to do that and I was just going to prove that it's euclidean because that's ... easier to w rap my head around?

[oblique river]

well

it's not euclidean

[shy bluff]

oh

it's not?

rip

[oblique river]

no

[shy bluff]

[oblique river]

did someone here tell you it was?

[shy bluff]

ok uh what do youmean by look at the norms of each generator in Z

[oblique river]

i'm going to go beat them up

[shy bluff]

No

That was my own conclusion

[oblique river]

oh okay

[shy bluff]

Because I saw that we had N

And I thought that we could make the conditions for euclidean true

[oblique river]

just because it's called N doesn't mean it satisfies the euclidean condition

[shy bluff]

But I guess not

[oblique river]

it's just a norm, not a euclidean norm

[shy bluff]

Ok

[oblique river]

if you ahve an ascending chain of principal ideals (a1) subset (a2) subset (a3) subset (a4) ...

look at the norms of the a_i

those are integers

[shy bluff]

how do you know that

[oblique river]

know what

[shy bluff]

Well fo rone thing, where are you getting a_1, a_2.. a_n from, and why do you know that they're integer

How do you know that*

[oblique river]

they're principal ideals

so they have a generator

the a_i aren't integers

their norms are

[shy bluff]

hrm

By "they have a generator" you mean that a_1 itself is the generator right

[oblique river]

yes

you have a chain of principal ideals

[shy bluff]

Yes

[oblique river]

each of those has a generator

[shy bluff]

Following so far

[oblique river]

great

now the fact that (a1) is a subset of (a2)

what does that tell you about how a1 and a2 are related

[shy bluff]

All of a1 is contained in a2, which means that all multiples of a1 are in a2?

[oblique river]

yes

but multiples of a1 aren't helpful here

if (a) is contained in (b)

[shy bluff]

Oh

[oblique river]

what does that tell you abotu how a and b are related