Mathematics - 2021-02-03 (page 2 of 3)

C.F.G

17:04

5

I am working on a problem (problem 10.19 (d)) in John M. Lee's Introduction to Smooth Manifold. Assume that $\pi$: $E$ $\rightarrow$ $M$ is a fiber bundle with model fiber $F$, I need to prove if $E$ is compact, then so are $M$ and $F$. Clearly, $M$ is compact and if we assume that $M$ is Hausdor...

fiber-bundles

Thorgott

well, we reason with them to get the axioms, but yeah

it's just so bizarre

BigSocks

17:23

why is $Spec(C_f) = Max(C_f) \cup \{(0)\}$? shouldn't $C_f$ have ideals generated by $2$ elements? So $Spec(C_f)$ will have principal ideals that can't be in $Max(C_f)$ other than the trivial one?

Astyx

What is $C_f$ ?

Mike Miller

@Thorgott traditional advice is to just ignore the definition until you really truly can't

BigSocks

$\overline{k}[x,y]/(f)$ for $f$ irreducible

Mike Miller

terrible notation

123

Hello Guys...

Astyx

17:25

Do you know what prime ideals of $\bar k[x,y]$ look like?

Thorgott

well, I can't ignore it before confirming the axioms, so I guess the advice is to ignore it after that

Astyx

hi 123

Mike Miller

No, you skip to the axioms

Astyx

...

Thorgott

it's just such a clean definition that there should be some better way of understanding it

Mike Miller

17:26

This is how it's presented in Milnor Stasheff for a reason

Axiomatically and then a definition later

BigSocks

@Astyx well they should be generated by less than $3$ polynomials I guess

Mike Miller

The definition just kills interest and causes confusion

@Thorgott I would proceed. In two weeks I would be glad to have extended discussions about the various defns.

BigSocks

I guess being generated by an irreducible polynomial is one of em. there's also $(0)$

I know the maximal ones for $C_f$ are generated by the images of $x-a$ and $y-b$ and that is weak Nullstellensatz

(and $(a,b)$ a solution of $f$)

Astyx

Ok let's do this first

let $p$ be a prime in k[x,y] with k alg. closed.

What kind of such primes can you think of?

BigSocks

17:41

I mean... $2$ is a prime in there I guess?

just constants that are primes

they work?

Astyx

2 is not a prime, it's a number

Be precise

BigSocks

you mean a prime ideal then

so $(2)$

Astyx

yes i do mean prime ideal

(2) is not a prime ideal: it's all of k[x,y]

BigSocks

oh bc you have $1/2$. forgot $k$ was alg. closed

so basically nothing like $(c)$ works at all

where $c \in k$

so you need some irreducible $f(x,y)$

and you take $(f)$

Thorgott

@MikeMiller sure, that would be much appreciated

Astyx

17:45

So that's some of the primes

(f) with f irreducible

Can you think of other primes?

BigSocks

$(0)$ but that is boring

Thorgott

I guess I have a very vague hunch on why this definition corresponds to pulling back from the complex Grassmannian, granted that its cohomology is generated by the Chern classes, but I'm far from familiar enough with this stuff to make this precise

Astyx

(0) is also a prime, although boring, it's important

Mike Miller

@Thorgott Right, because after all, H^*(BU(n); Z) ~ Z[c_1, ..., c_n] is not canonical if you're just saying it's a polynomial ring on a generator in each degree

Even in the universal case you need to somehow pin down particular classes

Playing with the universal case as much as you can you end up finding various recipes to pin down particular classes

Ted Shifrin

mumbles Schubert cycles, yum, yum (not sarcastically)

BigSocks

17:51

@Astyx ok well idk any more of them

I guess maybe stuff like $(x - a, y - b)$

$a, b \in k$

Astyx

You guess or you're sure?

Mike Miller

@TedShifrin I still don't get those.

BigSocks

well it's maximal for $C_f$ as long as $a,b$ is a solution for $f$, so here it's kind of the same, but with $f$ the zero polynomial, which I think means all $a$ and $b$ are solutions to it

Astyx

You can prove it directly

BigSocks

so, yeah I am pretty sure and if I am wrong I would have to review some things

Ted Shifrin

17:53

I can connect it up to the degeneracy loci for you sometime, @MikeM, if you're interested.

Astyx

What is a polynomial in $(x-a, y-b)$ ?

Ted Shifrin

It's just a generalization of Poincaré-Hopf (for $c_n$), not surprisingly.

Mike Miller

I believe it.

And products in cohomology related somehow to products of RREF matrices I assume.

Ted Shifrin

A lot of this stuff I had to figure out for myself, although some is done nicely in Griffiths/Harris.

Mike Miller

Or something like this.

Ted Shifrin

17:55

Oh, I guess the Pieri formulas are combinatorial things with Dynkin somehow, but that's stuff I've never learned.

Mike Miller

Gotta write some exams. Take home, but hopefully with some fear of God instilled beforehand to make sure only one pair of eyes are on them.

Ted Shifrin

Good luck with that.

Do they have to sign a blood oath?

Mike Miller

An oath, but I can't confirm it's in blood while we are online.

Ted Shifrin

Hmm, always technical difficulties.

user10478

Does anyone know of a site that solves multivariable recurrence relations? Wolfram doesn't seem to like them.

Astyx

18:00

@BigSocks you still there?

BigSocks

yeah my bad had to deal with something irl

yeah they will be sums of products of those two generators with ring elements

(in $k[x,y]$)

Ted Shifrin

a @Balarka: The proof you were suggesting that the Weyl tensor vanishes identically is the last thing in that paper. So conformal flatness is the right condition (too bad Moishe didn't just say that in his answer).

Greetings, @Astyx and @BigS.

Astyx

Ok, but can you caracterize them in a meaningful manner?

BigSocks

hey there @TedShifrin

Astyx

Hello Ted

BigSocks

18:03

@Astyx well I guess they are polynomials that vanish at $(a,b)$

Astyx

Do more than guess

Is the set of polynomials that vanish at (a,b) a prime idieal?

BigSocks

ok well they have to be that because if you sub in $x \mapsto a, y \mapsto b$ you get sums and products of zero

Simone

Hello Ted!

Ted Shifrin

Salut, Simone.

Simone

I'm trying to solve an exercise in your book about multivariable calculus, here's how I did it:

3

Q: Do you find this proof convincing?

I must show that the function $f:\mathbb R^n \to \mathbb R$ defined as $ f(\mathbf x)=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2 $ for fixed $\mathbf a_1,...,\mathbf a_k \in\mathbb R^n$ has a global minimum. Here's my reasoning: The function $f$ is continuous. Pick $\delta \gt 0$, by def...

calculus analysis multivariable-calculus solution-verification

Astyx

18:06

@BigSocks Are you thinking about my question?

BigSocks

yes

writing some things down real quick

Simone

Like Servaes said I'm not done yet. But do you condone the approach?

Astyx

Do you know how to prove that something is a prime ideal?

BigSocks

yeah $fg \in P \Rightarrow [f \in P] \vee [g \in P]$

Ted Shifrin

@Simone: So the point is to find some compact set (say, a ball) so that all the values of $f$ outside that compact set are known to be larger than some value inside. Right?

Simone

18:07

Yes, that is the final step I'm missing

Astyx

Ok, so if f(a,b)g(a,b) =0, can we certify that g(a,b)=0 or f(a,b)=0 ?

Simone

I need that to show that y is indeed a member of the image set of f

BigSocks

@Astyx yeah bc they are each constants in $k$ and if they weren't zero, you'd have a zero divisor in an algebraically closed field

Ted Shifrin

Yeah, the way you're doing it is not so clear.

It's a very different approach from what I have suggested.

Well, you really want to know that eventually the minimum points are all the same.

Astyx

Right. Another way to see this is that this set is the kernel of $k[x,y]\to k$, $f\mapsto f(a,b)$

Ted Shifrin

18:10

It seems to me guidar (or whatever the name is) finesses the interesting part, which is getting a compact set with the property I said. Did he actually prove that?

Astyx

ie $\phi^{-1}((0))$ and $(0)$ is prime in k and $\phi$ is a ring homo, so the preimage of a prime ideal is prime

Ted Shifrin

He just says "you can see that."

BigSocks

And evaluation is a ring hom ok, yeah this makes sense

Astyx

However we haven't yet shown that $(x-a,y-b)$ is that set. To do that we also need to check that it isn't a proper ideal of the polynomials that vanish at (a,b)

Ted Shifrin

@Simone: You certainly need to use the quadratic nature of the function, as your proof won't work for just any function (you could have a function that decays as you go to infinity, for example). Did you ever do some actual estimates with the function?

Astyx

18:12

One way to see this is to compute k[x,y]/(x-a, y-b) and see that this is iso to k

Thorgott

it should just be the reverse triangle inequality

Astyx

Which implies that (x-a, y-b) is maximal, hence prime (and hence equal to the set of polynomials that vanish at (a,b))

Simone

@TedShifrin Well later I find the actual point constructively by finding the critical point with the derivative

BigSocks

Ok yeah so far so good

Ted Shifrin

Yes, but maybe the critical point is not the global minimum. Such things happen.

I think you have to show that $f$ is strictly increasing outside the ball of radius $n\delta$ for $n$ large enough.

Astyx

18:13

Ok. Now we're established that (f) with f irreducible, (0), and (x-a, y-b) are prime ideals of k[x,y]. I claim all prime ideals of k[x,y] are of this form

Do you see a way to check this?

Simone

@TedShifrin indeed, This is why I used the approach with the closed balls: to use the minimum value theorem

BigSocks

Not really :/

Ted Shifrin

But you do not know that there is a global minimum by your argument.

You must show, as I said, that outside a (suitably large compact set) the function is everywhere large.

BigSocks

Not immediately

Astyx

I'm not asking for an argument, but for a line of thought that could potentially prove the claim

Ted Shifrin

18:15

quotes Monty Python for Astyx: "I've come for an argument."

Astyx

No you haven't

Ted Shifrin

"Yes I have."

Astyx

hehe

Simone

@TedShifrin why not? if the sequence of minima converges to a point in the image set of f and that point is in the image set then that same point is the global minimum.

BigSocks

Well we know there aren't any larger ones since $(x-a, y-b)$ is maximal. So we need to show that any other prime ideal that isn't maximal is like $(f)$ with $f$ irreducible

Astyx

18:17

Yes, or the contrapositive

BigSocks

and if you extend it by something that $f$ doesn't divide, it will always be maximal

Ted Shifrin

@Simone: If you never have to use the nature of the function, I am convinced the proof cannot be correct.

Simone

@Astyx @BigSocks I'm sorry I'm polluting the chat a bit

Astyx

That every prime ideal that isn't principal is of the form (x-a, y-b)

Ted Shifrin

I can give you an everywhere positive function on $\Bbb R^2$ that has no global minimum.

BigSocks

18:18

please do not worry @Simone

my eyes have good filters

Astyx

Don't worry about it Simone, we are too

I don't own the place (and even if I did I wouldn't mind)

Thorgott

Astyx and Big discussing why dim k[x,y]=2?

Astyx

More like finding prime ideals of k[x,y], but essentially yeah

BigSocks

it was in relation to $Spec(C_f) = Max(C_f) \cup {(0)}$

Astyx

@BigSocks do you agree that every principal prime ideal is of the form (f) with f irreducible?

Thorgott

18:20

yeah that's just saying dim C_f=1 + C_f domain

BigSocks

and me thinking that the LHS might have some stuff that's like $(f)$ that might not be in the RHS

Ted Shifrin

@Simone: You literally never used anything about the function $f$. That makes the proof very suspicious.

BigSocks

@Thorgott right, but doesn't it have max ideals that are $(x-a, y-b)$ by weak Nullstellensatz?

ah hold up

Thorgott

that's not the weak Nullstellensatz, it's a direction observation

the weak Nullstellensatz is the converse

Ted Shifrin

There's nothing to stop the points $\mathbf x_n$ going off to infinity and the value $y$ being an inf, not a min.

Thorgott

18:21

but yes, those are maximal ideals alright

that's not an issue, though

BigSocks

@Astyx I am not sure if I could see any other alternative

I thought about it a bit, but yeah, it seems to be a very general form

@Thorgott I just know it as the equivalence so I guess I could believe you

Astyx

You not succesfully constructing a counterexample isn't a proof however

BigSocks

@Astyx yeah true

Thorgott

don't believe me, prove it

BigSocks

I mean, I have seen the proof of the equivalence. what you refer to is a naming convention. that's what I would believe

Simone

18:24

@TedShifrin Well I agree that the proof isn't complete yet. I need to show that with a closed ball that contains all the $\mathbf a_i$s, for all $\mathbf x$ outside that ball, I have that the distance is greater.

BigSocks

@Astyx hmm

Astyx

How could you prove that?

BigSocks

well if it was just a PID you could mod out and show you get a field

but that won't work

Ted Shifrin

How do you even know there is a closed ball containing all the $\mathbf a_i$'s?

I can give you continuous functions $f$ where $\|\mathbf a_i\|\to\infty$.

Astyx

Any principal ideal is of the form (a) for some element a (by definition). What if a isn't irreducible?

Ted Shifrin

18:26

My suggestion is to forget your proof and do what I suggested :)

Simone

@TedShifrin ahahm ok ;)

BigSocks

Then it is a product of irreducibles. I think you can find them to be linear since $k$ alg. closed

Simone

thanks

BigSocks

or rather somehow factor them over $k$, is what I mean

Astyx

That's too complicated

What does it meant hat a is not irreducible?

BigSocks

18:27

oh ok

Alessandro Codenotti

Ted's argument tells you that this minimum is also unique, but to establish existence of a global minimum it's enough to note that your function is nonnegative and check whether it happens to be 0 somewhere

Simone

@AlessandroCodenotti I remarked upon this fact :P

Thorgott

@TedShifrin huh? it's just finitely many points, no?

Astyx

What doe sit mean for an element to be irreducible?

Ted Shifrin

Simone has used nothing about the function at hand (other than positivity). How do I know the minima on the expanding balls don't go off to infinity?

Of course, if you prove that the points stabilize, then there's nothing more to do.

Simone

18:33

@TedShifrin because they are nonicreasing

Ted Shifrin

No, I said stabilize. We want to know that $\mathbf a_i = \mathbf a_N$ for all $i\ge N$.

You cannot hope to prove anything for "all" functions $f$, which is what you're doing here.

You literally never used the explicit formula for $f$, other than to observe that its values are positive.

Simone

@TedShifrin I agree, I need to show that $y=f(\mathbf x)$ for some $\mathbf x \in \mathbb R^n$

Ted Shifrin

I'm curious to see you do that.

Without proving what I said.

Simone

By saying that the points stabilize you mean that there is a ball big enough to contain them all correct?

Ted Shifrin

I said what I meant. The sequence will be eventually constant.

I need to leave for a few hours. So you have plenty of time :)

Simone

18:42

I'm confused, the sequence of minima will be constant I agree. By $\mathbf a_i$ you mean the minima? because the terms are used differently in the problem

@TedShifrin Ok... tata :D

Ted Shifrin

Oh, darn. I wrote $\mathbf x_i$ first, and then something you said made me switch.

BTW, there's a completely elementary proof of this just using high school algebra. One needs NO calculus and NO maximum value theorem.

Simone

Ok... thanks Ted

BigSocks

19:15

@Astyx hey, @Astyx, my bad, a friend showed up at my house unannounced and I ran out

Astyx

ok

BigSocks

an element is irreducible when it's not the product of 2 irreducibles

so if you write it as a product it's a unit times an irreducible

Astyx

So x^3 is irreducible?

BigSocks

2 or more*

Astyx

Ok. So more generally when it's not the product of two non units

BigSocks

19:18

ok yeah, not the product of 2 non-units is fair I think

Astyx

So if a is not irreducible, what do we get

BigSocks

$a$ can be written as $a = nm$ where $n,m$ are nonunits (in $k[x,y]$)

Astyx

What are the non units of k[x,y] ?

BigSocks

I think it's the irreducible polynomials (it should be for this to work), but I don't have a proof offhand

Astyx

What is the definition of a unit?

BigSocks

19:21

$g$ is a unit if there exists an $h$ such that $gh = 1$

so nonunits are $g$'s such that $\neg \exists h [gh = 1]$

Astyx

So do you think x^3 is a unit in k[x,y]

BigSocks

no you can't multiply by $1/x^3$

Astyx

Ok, follwing this line of thought, I ask again: what are units in k[x,y] ?

Thorgott

I disagree with this definition

BigSocks

units seem like elements you can divide by, so I guess just constants

Astyx

19:24

nonzero constants, but yes

Thorgott

an irreducible should be a non-zero non-unit that cannot be written as product of two non-units

BigSocks

right nonzero

Thorgott

non-zero or not may just be convention, but you definitely don't wanna call units irreducible (at least I don't)

BigSocks

so nonunits are the things that are not nonzero constants

so nonconstant polynomials are actually the nonunits. so irreducibles are just one kind of nonunit

Astyx

Ah yes, Thorgott is right

We don't want units to be considered irreducibles

BigSocks

19:27

oh yeah and we were allowing that

Astyx

So coming back to (a), if a is not irreducible, what can we say about a?

BigSocks

$a$ is either a unit or a reducible nonconstant polynomial

Astyx

What if a is a unit

BigSocks

then $(a) = (1)$

Astyx

=k[x,y]

BigSocks

19:28

mhm

Astyx

What if a is a reducible polynomial?

Sha Vuklia

Guys, does someone know why $\Omega^1_{K(X)}$ would be given by this equivalence class? Clearly the map $d\colon K(X)\to\Omega^1_{K(X)}$ should send some $(U,f)$ to $d_U f$, where $d_U\colon \mathcal O_X(U)\to\Omega^1_{\mathcal O_X(U)}\to\Omega^1_X(U)$, but how do we construct a map $\phi$ from $\Omega^1_{K(X)}$ given some derivation $D\colon K(X)\to M$, where $M$ is a $K(X)$-vector space, such that $\phi d=D$

BigSocks

then $(a) = (\Pi g_i)$ where the $g_i$ are irreducibles

so nonconstant

Astyx

too complicated

BigSocks

so... $(a) = (cf) = (f)$ where $f$ irreducible and $c \in k$?

Astyx

19:31

a is reducible, not irreducible

BigSocks

well if you could factor out a $c$ it was reducible, but I guess the more interesting case is if you could factor the polynomial without just factoring out a constant

Astyx

$c\in k$ is a unit (if it's nonzero)

BigSocks

yeah

Astyx

So if a = cf, then a is irreducible

BigSocks

right of course, I slipped up

so yeah $a = gf$ at least, where $g, f$ irreducibles

so 2 nonconstant polynomials

Astyx

19:35

not irreducible, nonconstant

(or zero if a is zero, but that case is trivial)

BigSocks

right, they could be reducible. they were just nonunits

again, slipped up

Astyx

ok, so are $g\in (a)$ ? $f\in (a)$ ? $gf\in (a)$ ?

BigSocks

$a \in (a)$, so $gf \in (a)$. if $g$ or $f$ were in $(a)$ it would be prime

moreover, $(a) = (g)$ if $g \in (a)$

since you could just multiply by $f \in k[x,y]$ to get $a$ again

Astyx

But is g in a ?

BigSocks

well no because $f$ not a unit so you can't get $f^{-1} \in k[x,y]$ to write $a f^{-1}= g$

so, $(a)$ is not prime...

whenever $a$ is reducible

so $(a)$ is prime only if $a$ is irreducible

Astyx

19:42

Again, your failure to prove something doesn't prove the contrary

BigSocks

what did I fail to prove

Astyx

One correct argument is that if $g\in (a)$, g = au for some element u, then $auf = a$, so f is invertible (because k[x,y] is a domain), which contradicts our hypothesis

BigSocks

I mean I kinda wrote that

I just wrote it in regular english so it's covered in $\neg$

Astyx

Well, your logic was flawed, because your argument (as I understand it) is "Since I cannot invert f, I can't write $af^{-1} = g$ which would be my usual way of proving $g\in(a)$, so $g\notin (a)$"

Ok, so now we've proven that if a prime is principal, it's either (0) or (f) for some irreducible f

Right?

BigSocks

hmm ok I'll think more about that. and yeah

Astyx

19:48

What remains to be done?

BigSocks

I am not sure- did we show that all the maximal ones were like $(x-a, y-b)$?

Astyx

No

BigSocks

So "if a prime is not principal it's maximal" I guess?

once we have that

Astyx

That's the right idea: we're going to prove that a non principal prime ideal is of the form (x-a, y-b)

So let p be a non principal prime ideal. Show that you can find two elements f,g in p with no common irreducible factors

BigSocks

could be some more...

Astyx

19:55

Right, but you're on the right track

BigSocks

ok well since $p$ nonprincipal there are at least 2 generators.

Astyx

Don't think about generators yet

Pick a nonzero element f. Can you find an element $g\in p$ such that $g\notin (f)$ ?

BigSocks

hmm

Transcript for

$Mathematics$ Mathematics