I read that as "electrostatistics" and died

lmao

LOL

@Thorgott electrostatistics is what you do when something electronic fails and you tape it together crudely and say "that'll probably be fine"

I took the copybook for the electrostatics course at my school, but never actually read it

Are you studying plasmas or something ?

ehh i dont know what those are

I used to

Question: If you have a non-Noetherian ring, and you localize at the union of all finitely generated ideals, is the resultant ring Noetherian?

if you localize at the union of all finitely generated ideals, you get the zero ring, as $0$ becomes a unit in the localization

so the answer is yes, but probably not in a way you expected

Huh? What is the multiplicative set?

He means complement of all the finitely generated ideals surely

Sorry, that was a result of the unfortunate syntax of $R_\mathfrak{P}$ meaning $R$ localized at $R\setminus\mathfrak{P}$

ah

So, the complement, yes

You need ideals to be prime for this to be multiplicative no ?

why is that multiplicative

so you mean the complement of all f.g. prime ideals? otherwise I don't see how this yields a multipicatively closed subset

Yes, yes, the complement of the union of finitely generated prime ideals

The union of finitely generated prime ideals isn't a prime ideal for all I know

(Technically speaking, you just need the complement of the union of all ideals maximal with respect to being finitely generated, but yeah)

Isn't $(x_1, \cdots, x_n) \subset k[x_1, \cdots]$ a prime ideal?

it isn't, but it's multiplicative

its complement*

Oh ok

Therefore localizing at the intersection of the complement of all finitely generated prime ideals doesn't seem to do anything in this case

in fact, saturated multiplicative <=> complement is union of prime ideals

Since every element lies in one of these guys

@MikeMiller if you invert all the $(x_1, \cdots, x_n)$ you get a field right

So, you'd wind up with the same ring, you think, Mike?

saturated as in you can't add more without getting the full thing ?

union of all f.g. primes is all polynomials without constant term for the reason Mike gives

Saturated means that if $fg\in S$, then $f,g\in S$

oh complement

yeah so the intersection is empty lol

what intersection

Alright, then that route is dead (I'm trying to construct a Noetherian ring which has infinite Krull dimension)

the complement of the union of all f.g. primes are the polynomials with nonzero constant term

ah wait no I'm being stupid

everything is in a finitely generated prime ideal like mike said, so the localization does nothing

Yeah, I also get tripped up by polynomials with nonzero constants being possibly primes themselves

@Rithaniel Does that exist ?

yeah Nagata leaf

@Rithaniel Nagata intensifies

Yeah, but I'm trying to see if I can come up with one myself

I honestly don't remember the example

But it's ok

But my thoughts are a little scatterbrained, as per the usual

maybe I'm dense, but why is everything in a f.g. prime ideal

Commutative algebra is scatterbrained, so it makes sense

because every polynomial is a polynomial on finitely many elements so is in $(x_1, \cdots, x_n)$ for some $n$ which is a prime ideal

i think

polynomials are hard

no

well, $1+x_1$ is not in such an ideal

$(x_1,..,x_n)$ only contains those with zero constant term

lol

yeah

@BalarkaSen noted

so it does something

and I'm not sure which polynomials with nonzero constant term are in f.g. prime ideals

but $1+x_1$ is in a f.g. prime ideal nonetheless

@MikeMiller you started this

i have nothing to do with this revolutionary movement

if $f$ is a polynomial, take an irreducible factor $g$ of $f$, then $f$ is in $(g)$ which is prime. $k[x_1, \dots]$ is factorial so irreducible elements are prime

i am turning myself over

this works for any non-constant polynomial

ah yeah, makes sense

so localization does do nothing

You can also argue that the units $U$ are a multiplicatively closed set, so given any non-unit polynomial $f$, you can find an ideal $I_f$ where $f\in I_f$ and $I_f$ is maximal with respect to $I_f\bigcap U=\emptyset$, and this $I_f$ is necessarily prime.

(This stuff is in my brain, but sometimes I just don't manage to put it all together)

yeah, but that isn't necessarily f.g.

Ah, good point

Ah, I get the Nagata example. Basically you take $\mathbb{F}[\{x_i:i\in\mathbb{N}\}]$ and localize it at $$\bigg(\bigcup\limits_{i\in\mathbb{N}}\big(\{x_j:i^2<j\leq(i+1)^2\}\big)\bigg)^c$$Then you have a bunch of "islands" where ideals are only comparable to other ideals in that island, but there are islands of arbitrarily large size (In my precise example, one island of size $2n+1$ for every $n\in\mathbb{N}$).

ooo spooky

Yeah, it's pretty neat

I don't know if I can make a less contrived example, but maybe I can come up with something using this as inspiration

@TedShifrin For the sphere.

But also Euclidean.

Did you ever define precisely what your map was that you want to know what $S^2/(x \sim f(x))$ is

"For the sphere" sounds like it could be a battle cry in a science fiction novel

@Rithaniel note that Nagata uses a rather obscure criterion to show that his example is Noetherian

what do you call curves in a manifold that leave the metric invariant? The example I know is, invariant hyperbola, which preserve the quantity $\Delta s^2 = \Delta x^2 - c \Delta t^2$

yeah

Was just reading about that

ah nice, I had that as an exercise in my commutative algebra lecture

did everyone attend church today?

What is the definition of a singularity in complex analysis? I found the classification of singularities in my textbook, but I couldn't find a definition.

it's a point such that the function is holomorphic in a punctured neighborhood around it

Hmm...wouldn't that mean all points are singularities?

hey all, I have a problem with an indefinite integral of dx/sqrt(x^2-1). I don't understand why my substitution t^2=x^2-1 isn't working. Then dt=dx, we have dt/sqrt(t^2) = dt/t = ln|t| = ln(sqrt(x^2-1)). I'm probably making some elementary error but it's late;_; (place integral notation as necessary)

Isn't $f(z) = z$ analytic on $B(z_0,r) \setminus \{z_0\}$ for any $z_0$?

point outside of the domain*

Ah, okay.

(alright, tdt=xdx, feel very stupid;_;)

hehe

$h(z) = w^2$ has a zero at $w=0$ of order $2$, right?

Also, does $\tan z$ have a zero at $\infty$?

mordekaiser es numero uno

I don't think it has a zero at $\infty$ because $\tan(1/w) = \frac{\sin (1/w)}{\cos (1/w)}$ is not defined at $w=0$.

i'm never

never solo queue

stoopid brazil stop solo queue

Hey chat

I think I’ve probably already said a bit about this issue, but once again I’m here

I’m taking Advanced Linear Algebra and just finished General Linear Operator Theory - everything concerning minimal and characteristic polynomial, primary and cyclic decompositions, Jordan and Rational forms

linear algebra is absolutely everywhere

I wanted something more algebraic. Took a look at the connectedness of GL(n, C) but I don’t know a lot of applications of this fact and if it’s even relevant

So my question is: what are some (pure) mathematical applications of this specific part of linear algebra?

if you wanted to be more original, try representation theory

BRAZILLLL

yo

hello edward, do you love Brazil

err

I don't know anything about Brazil except that portuguese is spoken there

or some variant thereof

VERY NICE !

are you doing borat?

i haven't watched the new borat film yet

please no spoilers

@EdwardEvans yeah, Brazilian portuguese

@JoeShmo what can I talk about it, in this context?

I don't know representation theory enough to talk about it, but the idea is to approximate operators by linear operator as far as I know