That is surprisingly reductive for a proof strategy, I always thought for proofs related to integrals and functions, you cannot express it in terms of proving that some fundamental building blocks of it is bounded or finite, and must use a combination of absolute values, epsilon deltas and limits

In the end you will of course need to do the $\epsilon$ things, but the proof is by a construction and then verifying that hte construction works

I see

@s.harp You mean doing a construction and leaving the verification to the students/readers

Algebraic geometry is difficult to study.

so is anything else in grad school

^

Rang the DAAD Aussenstelle in London (for a scholarship) and the person on the other end of the phone picks up and simply says "Ja?"

I'm like "... r u the daad aussenstelle in london pls"

Algebraic geometry is particularly difficult tbh though

The difficulty is that, for example, I will never feel satisfied, and things get easily forgotten.

@MatheinBoulomenos What's the intuition behind the exactness of the direct limit functor on R-Mod, and failure of exactness for the inverse limit functor?

I can write down a proof but I don't quite "get" it.

Hi @Albas

@BalarkaSen not sure if I have any clear intuition for that. The only part that is not automatic is left-exactness and I feel that this is just "obvious enough" if you think about a direct limit as a union. Since the forgetful functor from R-Mod to Set preserves filtered colimits (but not general colimits!) and limits and reflects isomorphisms, you can reduce this question to a property of the category of Sets

actually for Sets, there's an iff: a category is filtered iff colimits over that category commute with finite limits

but that doesn't really make things more intuitive

this is a more conceptual proof that doesn't rely on computations, but rather on the fact that left adjoints preserve colimits mathoverflow.net/a/57124/117693

I meant set-valued colimits for the characterization of filtered categories

So quick question, given an arbitrary ring $R$, does the ideal generated by the union of all prime ideals in a ring $R$ equal $(1)$?

Ohh wait derp, no, consider any field $F$

daym, i'd help but don't know yet whats ring lol

Are the chess players still here?

@MatheinBoulomenos Thanks, that does shed some light

Filtered colimits commutes with finite limits in Set, R-Mod, etc

I have a question about Yoneda embedding.

Can we determine Hom(X,colim(Y i)) by Hom(X,Y_i) in some sense?

Well, in general, Yoneda embedding does not commute with colimits.

@Perturbative The union is the unit ideal if and only if it is not local.

discrete maths

+

@neraj How do you define $a\wedge b$?

The least upper bound of $a$ and $b$?

Try to prove that $(a\wedge b)\wedge c$ is the least upper bound of $a$, $b$, and $c$.

Note that $x\succeq a\wedge b$ iff $x\succeq a$ and $x\succeq b$

Hey folks, I just noticed that subspaces of V\in k-Vec make a modular lattice. Is there a nice general result for when this holds? like, for the substructure lattice of an object in any abelian category?

Why is $z^2 \sin{z}$ not analytic at $\infty$?

@user330477 consider the taylor series at $0$, this is also a Laurent series at a punctured neighborhood of $\infty$, using that $z^n=(\frac{1}{z})^{-n}$, so you get an essential singularity at $\infty$ from the classification of singularities by Laurent series

@MatheinBoulomenos We have not reached Laurent Series yet, but from what I can say, I think you want to say when viewed as a power series the limit does not exist.

that's one way of saying it, true I was thinking too complicated. The only entire functions which are analytic at $\infty$ are constant

and the only entire functions which have a pole at $\infty$ are polynomials

$\{v_k\} \subset H$ is a complete orthonrmal sequence in a Hilbert space. $\{x_k\}$ is an orthonormal sequence s.t $\sum ||v_i-x_i||^2 \lt \infty$. i want to show that if $x\in H$ satisfy $(x,x_k) = 0$ for all $k$ then $x=0$

there is a hint that $|(v_j-x_j,v_i)|=|(v_i-x_i,x_j)|$ which is obvious and that :

any ideas how to continue?

@MatheinBoulomenos Your classification is essentially Lioville's Theorem. But, am not sure how to deal with the infinity case here. In particular, if we are to show that limit does not exists at infinity, the power series should hold for $\mathbb{C} \cup \{\infty\}$, but it only holds for $\mathbb{C}$.

@user330477 you can argue by Riemann's theorem on removable singularities: if $f$ is an entire function that has a removable singularity at infinity, then it is bounded in some neighborhood of $\infty$. Every neighborhood of infinity contains an open subset of the form $\{z \in \Bbb{C} \cup \{\infty\} \mid |z| > r \}$ for some fixed $r$. By continuity and compactness of $\{z \in \Bbb{C} \mid |z| \leq r\}$ you get that $f$ is bounded on $\Bbb C$ in that case, so $f$ is constant by Liouville

@MatheinBoulomenos can you take a look at my question?

@Liad I have no clue about your question

@MatheinBoulomenos Thank you for your comment. But I think there is a more elementary way to do it. I know this may sound stupid, but is it true that $\lim\limits_{z \rightarrow 0} z^2 \sin{z}$ does not exist?

@user330477 no that limit exists

@MatheinBoulomenos Sorry, I meant $\infty$ instead of $0$.

How do we show this?

If the limit exists, then for any sequence $(z_n)_{n \in \Bbb{N}}$ of complex numbers with $\lim\limits_{n \to\infty} |z_n| \to \infty$, we get that $\lim\limits_{n \to\infty} z_n\sin(z_n)$ exists. So you could try something like $z_n=n2\pi+\pi/2$

Hi @Mathei

Hi @Alessandro

Do you happen to have a ZF example of an affine scheme with infinitely many connected components? The last AG pset asked if an affine scheme can have infinitely many connected components and I (think I) found one, but made abundant use of choice for it

the spectrum of any boolean ring is totally disconnected

so take $\prod_{n \in \Bbb N} \Bbb{F}_2$

That's what I did, but you need choice

no

That spectrum is $\beta\Bbb N$

yes, but $\beta \Bbb{N}$ contains $\Bbb N$ even without invoking choice

so it's an infinite totally disconnected space

Oh of course

Principal ultrafilters correspond to kernels of the projections on the factors

right

And those are enough to have infinitely many components

I just overcomplicated it by arguing it is $\beta\Bbb N$ which is totally disconnected then :P

@user330477 for yet another overly complicated proof: let $f$ be an entire function with a removable singularity at infinity, then $f$ extends to a holomorphic function $\Bbb{C} \cup \{\infty\} \to \Bbb{C}$, but any such function is constant, since if it is not, the image is open by the open mapping theorem and also compact, thus closed but $\Bbb{C}$ doesn't contain any non-empty open compact subset, since the only clopen subsets are $\varnothing$ and $\Bbb{C}$ by connectedness

that argument proves that no compact riemann surface has nonconstant holomorphic functions in general

right

Hi @Balarka btw

Hi Mathein!

oh oh ... it's the algebra mob.

?? im offended

@Alessandro but I'm confused now: affine schemes are compact (even in ZF), but $\Bbb N$ in $\beta \Bbb N$ is discrete and ZF can't prove that $\beta \Bbb N \setminus \Bbb N$ is non-empty

@MatheinBoulomenos Thank you. Just a question of logic. Is it true that to show $z^2 \sin{z}$ is not analytic at $\infty$, it suffices to show that it's $\lim\limits_{|z| \rightarrow \infty} z^2 \sin{z}$ that doesn't exist. For if it is analytic, then $f(\infty)$ is not defined.

LOL, a @Balarka

@user330477: Analytic as a map to $\Bbb C$ or to $\Bbb C\cup\{\infty\}$?

(Sorta irrelevant in this case, but I still want to clarify that.)

@user330477 yes, but I think a proper way to phrase the question is to say that $z^2 \sin(z)$ doesn't have a removable singularity at infinity (or even that is has an essential singularity, if you want the stronger result)

$z^2 \sin(z)$ does have an essential singularity at infinity, as a map to $\Bbb P^1$.

Even as a map to $\Bbb C$, it does.

Look at the Laurent series for $f(1/z)$.

oops yeah that's what I meant

Now, $z^2\tan(z)$ is more interesting :P

@TedShifrin Nothing is given in the question. But I assume it has to be $\Bbb C\cup\{\infty\}$. The actual question was to find the zeroes of $z^2 \sin(z)$, find their order and then show that $z^2 \sin(z)$ is not analytic at $\infty$.

I am always careful and extend my functions to maps of $\Bbb P^1$ whenever I have to deal with infinity.

@user330477: So, right, it would have to have a removable singularity at $\infty$ to be analytic. But as the question is posed, it really doesn't make sense, as they haven't defined the function at $\infty$.

@TedShifrin Easier, you can go $z \to \infty$ in two different directions and get different limits. If it was a pole or a removable singularity that would not be possible.

I'm not convinced that the proof that affine schemes are compact goes through in ZF @Mathei

Right? If I ask you if $f(x) = x^2 \sin (1/x)$ is differentiable at $0$, your answer is that $0$ isn't even in the domain of the function.

@Alessandro there's no Tychonoff or anything

@TedShifrin But in any case, it takes a value in $\Bbb C\cup\{\infty\}$, whereas $\lim\limits_{|z| \rightarrow \infty} z^2 \sin{z}$ that doesn't exist.

@Balarka: Easier, it's not going to have a limit even restricting to $\Bbb R$.

The analytic continuation of $\log(z+2)$ is apparently not compact. I'm not sure why.

Or do you need choice to show that it's sufficient to check compactness on a basis?

that might be it

@TedShifrin Haha, touch\'e.

Hi, I am reading measure-integral from ETH: people.math.ethz.ch/~kowalski/measure-integral.pdf and on page 3 they make a statement about multiple integral, which I do not understand. "Riemann’s definition encounters very serious difficulties here because the lack of a natural ordering of the plane makes it difficult to find suitable analogues of the subdivisions used for one-variable integration" (about R^3). What do they mean?

@Oskar: I don't even know what that means.

@isquared-KeepitReal: It's a totally stupid statement.

@Ted: Analytic continuation?

The analytic continuation of $\log(z+2)$ does not have a compact Riemann surface as it's domain, I suppose, is what he's saying.

@OskarTegby i have seen that in 3b1v video XD

@isquared-KeepitReal: Really? Which one?

Can you even prove that $I(V(a))$ is nonemtpty in ZF?

@BalarkaSen: Exactly. Mea culpa. I should've phrased it more carefully.

@isquared-KeepitReal: I can tell you how to take a subdivision of a rectangle easily enough. So it is truly a stupid statement. Look at any decent text on multivariable analysis and it will show you.

"Every commutative ring with 1 has a prime ideal" is independent of ZF

@OskarTegby i mean the analytic continuation part, which I thought Ted was referring to

@isquared-KeepitReal: I think Ted very well knows what analytic continuation is.

To me, an analytic continuation is a function. A function is never compact. @Oskar

So they're talking about the Riemann surface, of course.

We're talking about Riemann surfaces here.

I should've mentioned that. I'm sorry.

I don't think the $+2$ is relevant in the least.

yeah thats super random lol

Indeed. A translation shouldn't make any difference.

The issue is for you to see that there are countably infinitely many points in the Riemann surface for almost every $z$.

Currently reading this paper

core.ac.uk/download/pdf/82385648.pdf

Topology and voting theory

hi DogAteMy

Wouldn't've thought the two were connected

Hi

Oh, there was a whole summer thing on that last summer for mathematicians, DogAteMy

Was there?

@Alessandro okay I'm convinced you can't prove the compactness in ZF

Yeah, training mathematicians to be qualified to advise legislatures on how to undo all the unfair gerrymandering

@MatheinBoulomenos I mean, this is a proof that you can't prove compactness in ZF

Ahh

@Alessandro good point

Right, that's Moon Duchin's thing

Is it not compact because it doesn't contain some part of its boundary?

I still don't know where exactly the standard proof of compactness fails in ZF

commutative algebra in ZF is something I don't want to do

as well as other people I assume

(I only mention Moon 'cause I met her at MathCamp)

Fair enough, I might ask that on main though

Doesn't seem related to the paper though

The paper seems to be a paper on algebraic topology

You tell me how you're defining this thing, @Oskar.

Another interesting (to me at least) question is if "every affine scheme is compact" is equivalent to some more well known statement independent of ZF

I forget who taught the thing, DogAteMy. Someone at Tufts, I think.

hi , i think that $k[x,y]/(x^2+y^2-1)$ is isomorphic to $k[x,y]/(xy-1)$ , im not sure how to prove it though.. any hints?

@Alessandro I'd say it implies the ultrafilter lemma at least

"every affine scheme is compact" feels like it should imply at least BPI

@Liad: Why do you think that?

I find it funny how we wrote two distinct but equivalent statements :P

@TedShifrin because k is algebraically closed

Why is that relevant, and why does it tell you?

@TedShifrin am i wrong?

it tells me that there is a root for $-1 $

So how can you rewrite $x^2+y^2-1$?

Ahh

$x^2+y^2+i^2$

Not useful

That's not going to help.

Factor $x^2+y^2$

rip akiva u gave it away

Sorry I thought he got it when he mentioned that it has $\sqrt{-1}$

When he said that, I realized how to do it

DogAteMy is good at giving things away. So is Mathein. I haven't trained them well enough.

@MatheinBoulomenos I'm writing a question on main, do you mind if I use your argument here?

@Alessandro it's not obvious though, at least to me. Any filter $\mathcal{F}$ on a set $I$ defines an ideal in $R:=\prod_{i \in I} \Bbb{F}_2$ via setting $I(\mathcal{F})=\{(x_i)_{i \in I} \in R \mid \forall i \in I: x_i=0 \Leftrightarrow i \in \mathcal{F}\}$, now just looking at $R/I(\mathcal{F})$ doesn't really help since $R/I(\mathcal{F})$ might just have empty spectrum

its thanksgiving at the USA isnt it ?@TedShifrin :P

LOL, it was.

Are you preparing to give thanks?

@Alessandro no, feel free to use it anywhere you want

after i finish this question ^^

@MatheinBoulomenos Wait what are you arguing here? What is not obvious

That was last Thursday.

Well, DogAteMy told you what to do, so do it.

Whose DogAteMy

Akiva?

Yes

weird nickname O_o

@Alessandro it's not obvious that compactness of affine schemes implies the ultrafilter principle

what I wrote above is the standard proof that every non-zero ring has a prime ideal implies the ultrafilter principle

@MatheinBoulomenos You want $\{i\in I\mid x_i=0\}$ to be in $\mathcal F$

the problem is that the empty space is compact

@Alessandro oops, right

Ah, I see what you're saying

I guessed BPI just because this is the statement that makes prime ideals work well, but it was 100% a feeling

I guess that it's not compact as $\log 0$ isn't defined.

Nor is $\log\infty$.

But there are zillions of reasons it cannot be compact.

Yeah.

I guess that answers my question as I now have not one, but a zillion answers to it.

you certainly get that every infinite set admits a non-principal ultrafilter

That should suffice, @Oskar.

Haha! Indeed.

@MatheinBoulomenos are you familiar with right shift operator?

@ninjahatori zft

zft means?

it's a right shift of yes

LOL

lmao

slow clap

zum Feier teuer

(I know that doesn't work)

Hi chat

hi Semiclassic

Tangentially related but I'm somewhat annoyed at the terminology "quasicompact" because some people assume that "compact" implies Hausdorff

@Ted lol

Trying to figure out how to phrase a linear algebra question

@Alessandro I assume "topology" implies Hausdorff

I was about to put your comment as an answer, and then I saw your comment, @Balarka.

That guy frequently writes very obtuse answers ...

Hi @Semiclassical

@BalarkaSen You're not going to like AG

To assume topologies are always Hausdorff is very (za)risky.

Zariski topology is not a topology, it's a philosophy 8)

It's the Zariski philosophy

I’ve got, to my misery, a specific 6-by-14 matrix M. I want to consider this as a linear transformation on the 13-simplex

The resulting image will be some polytope in R^6

@TedShifrin (x+iy)(x-iy)

@Semiclassic: What do you mean by the 13-simplex? The face $\sum x_i = 1$ in the first "octant"?

@Liad: Right. Proceed.

@Fargle "In Absentia" makes my skin crawl, yet I go back to listening to it from start to end every time.

@Fargle: Did you notice my reply to your group of group emails?

@TedShifrin Yep

@TedShifrin so we want to show $k[x,y]/( (x+iy)(x-iy)-1) $ is isomorphic to $k[x,y]/(xy-1)$

@TedShifrin Teehee. And yes---I was very tired, so I was for some reason trying to come up with a hom by thinking about a normal subgroup first, like a big old goober.

@Semiclassic: So it's determined as the convex hull of where the standard basis vectors map ... or something.

@Liad: Yes. Maybe you should use different letters in one of 'em.

Yeah

@TedShifrin sure, maybe the right one will be $k[z,t]/(zt-1)$

OK.

So now what?

@BalarkaSen Lots of great musical moments on that album.

both of the form $ab-1 $ @TedShifrin ^^

So can you write down an isomorphism?

I’m not quite done, though. I’m only interested in the subspace of the image where the first three components are equal to 1

The subgroup you were looking at certainly suggests an obvious group homomorphism, @Fargle. The proof goes slightly differently, but you'll get where you want.

@Fargle Yeah but the contrast between the lyrical material and the music is so unsettling.

maybe $z\to x+iy$ and $t\to x-iy$ @TedShifrin

@Semiclassic: That doesn't sound like a subspace.

hmm

So write it down carefully, @Liad.

@TedShifrin Yeah. I was struggling to find a homomorphism. Kind of still am. Let me keep thinking about it

Subset, then

it looks right, why the "carefully"? i missed something?

Silly goose. You have an obvious homomorphism if you think about what you're talking about, @Fargle.

You're working with quotient rings, @Liad, so you always have to be a bit careful.

Oh---not with my subgroup as a kernel? So just $x \mapsto x^2$?

Right.

Note my comment above about "goes slightly differently."

the kernal of the homomorphism i wrote is $(zt-1)$ , right ?@TedShifrin

I think Wittgenstein would have a lot to say about me reading your messages.

I haven't read Wittgenstein, so that one is lost on me.

Whereof thereof speaketh silence

@Liad: That makes no sense at all. What are the domain and range of your map?

This is probably where my lack of recent linear algebra is kicking my butt

i mean it is $0$ in $k[z,t]/(zt-1)$ which is what i want @TedShifrin

Roughly, the impossibility of interpreting language perfectly because our use of it is lensed by our experience

Aha.

its $\overline{zt-1}$ if you want me to be more precise

@Liad: I'm not going to talk about it any more.

Geometrically, what I’m doing is taking a particular 3D cross section of a 6D polytope

And I imagine that what I get should be another 3D polytope

I think?

(Very confident reasoning, I know)

I posted it as a question on main @Mathei

You intersect the faces of the original polytope with your 3D-plane, so that sounds like it should be a polytope, unless some degeneracy occurs.

Yeah

@Alessandro: Refrain from starting questions with "Every algebraic geometry course ..." Certainly that's very false.

I'd every algebraic geometry which defines affine schemes proves that they are compact

I'm not going to argue.

The intuition makes sense, but intuition isn’t sufficient

I rephrased it :P

@Semiclassic: It seems to me you want to make sure that slicing plane meets all the defining hyperplanes in something of the same dimension (or not at all).

Sound right

@Alessandro: It might not hurt to say that $\beta\Bbb N$ is the Stone-Cech compactification ... or at least I assume it is. I don't know that everyone would recognize that notation.

What I ultimately want is to be able to visualize the 3D polytope

(Assuming it is indeed a polytope of course)

So I guess you should find where the 3D slice meets the edges from the origin to the various vertices ?

I'm not thinking about this very carefully.

Hmm, I think I see what you mean

@TedShifrin Another fair point

Hey everyone!

Hi @Daminark

How's everything going?

uni is pretty stressfull, my talk went well, but I'm considering dropping a course I'm just taking for fun.

And yourself?

Getting toward the end of first quarter, so I've got grad apps going on which are annoying, and classes are really piling up now

But next quarter will hopefully be a lot more relaxed

math.stackexchange.com/questions/3015011/about-direct-sum @MatheinBoulomenos

Please help regarding this

@ninjahatori I'm not good at functional analysis

OK fine thanks

@LeakyNun any idea about above problem

Don't just go and start pinging random people about a question. Post it and if people are interested they're interested

Sorry for that

Hey everyone!

The laplacian of $\ln(1+\frac{x^2}{y^2})$ is $\frac2{y^2}$

Strange that the input depends only on the ratio between $x$ and $y$ and the output depends only on $y$

Did you delete the question? @Ninja

Hi @Perturbative

The Laplacian of $\ln(\frac2{x^2})$ is $\frac2{x^2}$. The Laplacian of $\ln(\frac2{x^2}+\frac2{y^2})$ is $\frac2{x^2}+\frac2{y^2}$. The Laplacian of $\ln(\frac8{1-x^2-y^2})$ is $\frac8{1-x^2-y^2}$.

I'm trying to see if there's a pattern for when $\nabla\ln f=f$ and I'm not seeing one

Or if there's a way to combine two solutions in some way

@AkivaWeinberger one trivial observation is that you can multiply a solution of $\nabla \ln(f) = f$ with a solution of $\nabla \ln(g) = 0$ and get another solution for $\nabla \ln(f)=f$

Hey @AlessandroCodenotti, how's things going?

So I'm reading through a proof that $\operatorname{Spec}(A)$ is compact, and there's one step (the highlighted one) in the following proof that I don't quite understand

How exactly can we conclude that $\sum_{i \in I} (f_i) = (1)$?

@MatheinBoulomenos What do you mean? If $\nabla\ln(f)=f$ and $\nabla\ln(g)=0$ then $\nabla\ln(fg)=f\ne fg$.

So $fg$ is not a solution.

@Akiva right, that was nonsense

$\nabla\ln(x^2+y^2)=0$

Maybe I should look at this in polar coordinates

tomorrow

Good night

Okay so the only approach I can think of at the moment, is the following, since no prime ideal contains all the ideals $(f_i)$, letting $M$ be a maximal ideal of $A$, since $M$ is a prime ideal of $A$ there exists a $j \in I$ such that $(f_j) \not\subseteq M$ so there exists an $x \in (f_j)$ such that $x \not\in M$. Now if I can show that $\sum_{i \in I}(f_j) \supseteq M$ then I'd be done, but I don't know if that's possible to show

Expanding this out is gonna be a fun time

I saw the following: Fair enough, I've been doing mostly AG lately but variety is important

What does AG stand for?

Nice coincidence, we were talking about compactness of spectra earlier in chat @Perturbative

@Bob algebraic geometry

thanks

I am wondering something

when most people do math, do they first do it with pencil and paper then translate it to LaTex

or do they go to LaTex right away?