Mathematics - 2014-12-11 (page 5 of 5)

Ted Shifrin

11:07 PM

@DanielF: You complied with Jasper and put up a new pic? You're incognito :P

Khallil

Hey, @Ted!

How've you been?

evinda

Hey @TedShifrin!!!
Martin told me that we can find the intersection multiplicity like that:
$I_P(x^5+x^4+y^2,x^6-x^5+y^2)\\
=I_P(x^5+x^4+y^2,(x^6-x^5+y^2)-(x^5+x^4+y^2))\\
=I_P(x^5+x^4+y^2,x^4 (x^2-2x-1))\\
=I_P(x^5+x^4+y^2,x^4)+I_P(x^5+x^4+y^2,x^2-2x-1)\\
=I_P(x^5+x^4+y^2,x^4)\\
=I_P(y^2,x^4)\\
=2 \cdot 4 \cdot I_P(y,x)\\
=8$

Could you explain me why we take the difference $(x^6-x^5+y^2)-(x^5+x^4+y^2)$ at the equality:
I_P(x^5+x^4+y^2,x^6-x^5+y^2)=I_P(x^5+x^4+y^2,(x^6-x^5+y^2)-(x^5+x^4+y^2))?

Ted Shifrin

hi Khallil ... Just begin to have my head above water. Too many homeworks to grade, exams to write, letters of recommendation to submit ...

Daniel Fischer

@TedShifrin Who says it's me on either pic?

Ted Shifrin

Hell if I know, @DanielF :)

Because this depends just on the ideal, @evinda, so you're free to do "row operations" on the generators, just so long as the two generators you have give the same ideal.

You healthy again, @DanielF?

I've been largely absent due to plethora of work ...

Daniel Fischer

11:13 PM

@TedShifrin Mostly. I'm still coughing a bit, but less every day.

Ted Shifrin

I hope you don't have a serious infection, @DanielF

evinda

@TedShifrin And how can we know if, after having done an operation, the two generators we have give the same ideal?

Ted Shifrin

I'm still recovering from having a basal cell carcinoma removed ... sore healing going on on my back where I can barely reach it :(

@evinda: Do you know about row operations on matrices? It's the same thing.

Daniel Fischer

@TedShifrin I don't think so.

Ted Shifrin

Have you seen someone who might have more authority than you, @DanielF?

evinda

11:15 PM

@TedShifrin Why is it the same thing? Could you explain it to me?

Ted Shifrin

When I had serious lung congestion, I needed antibiotics and more, @DanielF :(

Daniel Fischer

@TedShifrin Who could that be? I'm a free man.

Ted Shifrin

That's only what you think, @DanielF.

@evinda: Consider $\langle f,g\rangle$ and $\langle f, g+cf\rangle$. Prove those are the same ideals.

It seems like I'll never get back to being able to use ChatJax on my iPad :( More reason to quit chatting :D

Daniel Fischer

@TedShifrin Or quit the iPad.

Ted Shifrin

Well, I'm at my desk, now, but I guarantee that without the iPad I won't be here very much.

well, Kaj is furiously studying for his real analysis final exam, so we won't see him tonight :D

I'm still rewriting my probability final to make it shorter/easier .... :(

evinda

11:19 PM

@TedShifrin A ok.. Could we also take this difference: $x^6-x^5+y^2-x(x^5+x^4+y^2)$ ?

Pedro Tamaroff

@TedShifrin Heretics!.

@DanielFischer Hello Daniel.

Ted Shifrin

@evinda: You could, but he's trying to cancel out the $y^2$.

evinda

@TedShifrin So that we have only $x$?

Daniel Fischer

@PedroTamaroff Hello Pedro.

Ted Shifrin

It is somewhat presumptuous, @Pedro. Somehow, I don't think anyone will be stealing my name.

Right @evinda ... The other thing is, $x$ is not a unit, so you can't go backwards. Your two ideals aren't equal. One is contained in the other, but not the other way.

Khallil

11:22 PM

If we have a group $G$ and an element of $G$ called $g$, is $\langle g \rangle = \{ \dots, g^{-1}, 1, g, g^2, g^3, ... \}$ called the cyclic subgroup of $G$ with $g$ as it's generator?

Pedro Tamaroff

@DanielFischer Let's try to see in how many seconds you solve this: let $\mathcal A$ be a family in $\mathcal O(\Bbb C)$ that is bounded at a point (wlog $0$) and such that the family of derivatives $\mathcal A'$ is normal. Then the family is normal. Give an example where the family of derivatives is normal, but $\mathcal A$ is not.

@Khallil Yes.

Ted Shifrin

Child's play, @Pedro: Cauchy Integral Formula for the first.

Pedro Tamaroff

@TedShifrin I solved it without Cauchy.

Only the standard estimate is necessary.

evinda

@TedShifrin I understand!!! Thanks a lot!!!!!

Ted Shifrin

You're welcome, @evinda.

Khallil

11:23 PM

It's true that $\langle g \rangle = \{ \dots, g^{-1}, 1, g, g^2, g^3, \dots \} = \{ 1, g, g^2, g^3, \dots \}$, right @Pedro?

evinda

@TedShifrin And something else...
Let the algebraic curve $f(x_0, x_1, x_2) \in K[x_0, x_1, x_2]$. The inflection points are the non-singular points of the curve that are the intersection points with the hessian.

If we have the curve $x^3+y^3+z^3=0$ the hessian is equal to $216 \cdot x \cdot y \cdot z$. How can we find the non-singular points of the curve that are the intersection points with the hessian?

Pedro Tamaroff

@Khallil No.

Take $\Bbb Z$, $g=1$.

Ted Shifrin

Remember that the Hessian needs an equation @evinda. What's the equation?

Pedro Tamaroff

@Khallil It is true iff $g$ has finite order.

Ted Shifrin

What is "the standard estimate"? @Pedro To me, Cauchy is it.

evinda

11:24 PM

@TedShifrin How can we find this equation? :/

Daniel Fischer

@PedroTamaroff Which sense of "normal"? Is $f_n \to \infty$ allowed or not?

Khallil

Oh, sorry! Yep, I forgot to mention that part. Thanks, @Pedro! ^_^

Ted Shifrin

You need to set your expression equal to $0$, @evinda !

Ah @DanielF ... I hate that normal on the Riemann sphere stuff :D

evinda

@TedShifrin So do I have to solve the system $216 \cdot x \cdot y \cdot z=0$ and $x^3+y^3+z^3=0$?

Ted Shifrin

Yes, @evinda.

Pedro Tamaroff

11:26 PM

@DanielFischer Normal means the family is relatively compact in the compact convergence metric in $\Bbb C$. No $\infty$ allowed. So every sequence has a subsequence that converges uniformly in every compact of $\Bbb C$.

evinda

@TedShifrin But then we have three variables but two equations.. So is one variable a free one?

Daniel Fischer

@PedroTamaroff Okay, first subsequence such that $f_{n_k}(0)$ converges. Next subsequence such that $f_{n_{k_m}}'$ converges compactly.

Ted Shifrin

Yes, @evinda. Isn't this all happening in the projective plane? Remember homogeneous coordinates.

Daniel Fischer

For the second one, @Pedro, take constants.

Ted Shifrin

@DanielF: Why should we do @Pedro's homework? :D

He's smart enough to do it himself.

Pedro Tamaroff

11:28 PM

@DanielFischer What I'd do is look at $B(0,M)$ for every $M$, since that is convex, so we can join $0$ to any $z$ with a straight line $\gamma_z$ in there and use $$f_n(z)-f_n(0)=\int_{\gamma_z}f'_n(w)dw$$

evinda

@TedShifrin Yes, we want to find the inflection points of the curve in $\mathbb{P}^2(\mathbb{C})$.. How can we then solve the system? :/

Mary Star

@DanielFischer Do you mean that $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^{\gcd(n,m)}}$$ ??

Pedro Tamaroff

@TedShifrin I already solved this!

Ted Shifrin

Answer that for yourself, @evinda.

Pedro Tamaroff

It was a midterm question.

Daniel Fischer

11:29 PM

@TedShifrin We have to make good with our future moderator.

Ted Shifrin

LOL, I knew I'd get you with that, @Pedro :)

No, @DanielF, I can just quit here just like I do at the university.

Daniel Fischer

@MaryStar Yes.

Ted Shifrin

Besides, the future moderator might want a letter of recommendation from me :P

Chris's sis

@TedShifrin I was scared to death when I first discovered some similar thing a couple of years ago, but happily the doctor said it was just a verruca vulgaris with a weird appearance (on my back).

Pedro Tamaroff

@DanielFischer Hehe, I took $\mathcal A=\{z-\nu,\nu\in\Bbb Z\}$.

Ted Shifrin

11:30 PM

oh, glad it was nothing, @Chris'ssis. To be honest, after cancer that was almost death, the most benign skin cancer didn't bother me at all :P It's gone. It's just taking its time to heal :)

Same equivalence class, @Pedro :D

Pedro Tamaroff

@TedShifrin Equivalence class of what?

Ted Shifrin

functions :)

Pedro Tamaroff

I don't get it.

Chris's sis

@TedShifrin Great! Get well soon! :-)

Daniel Fischer

@PedroTamaroff $\{ f + c\}$ for some entire $f$ and an unbounded set of constants $c$.

Ted Shifrin

11:32 PM

Thanks for sparing me the trouble, @DanielF.

Pedro Tamaroff

@DanielFischer Ah. OK.

evinda

@TedShifrin From $216 x \cdot y \cdot z=0$ we get $x=0$ or $y=0$ or $z=0$.
From the other equation($x^3+y^3+z^3=0$), when $x=0$ then $z=-y$, when $y=0$ then $z=-x$ and when $z=0$ then $x=-y$. So are the inflection points of the form $(x,y,z)=(-y,y,-y)=y(-1,1,-1)$? Or am I wrong?

Pedro Tamaroff

But what's the equivalence relation here?

Ted Shifrin

You're wrong @evinda. You lost track of your $0$'s.

LOL @Pedro. You insist on taking me literally.

$f\equiv g \iff f-g = c$ for some constant $c$.

evinda

@TedShifrin Where have I done something wrong? :/

Ted Shifrin

11:35 PM

No, @Pedro, I got that wrong. The hell with it :P

First of all, @evinda, what field are you working over?

Mary Star

@DanielFischer After having said that "$x^{p^{n-m}-1} = 1$, i.e. $x\in \mathbb{F}_{p^{n-m}}$" what do we have to say to prove that $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^{\gcd(n,m)}}$$ ??

cxseven

if a continuous map from $f:S^n \rightarrow S^n$ has a degree other than 1 or -1, must it be noninjective?

evinda

@TedShifrin Since we have to find the flexes of the curve in $\mathbb{P}^2(\mathbb{C})$ isn't the field $\mathbb{C}$ ?

Ted Shifrin

Certainly true for a smooth map, @cxseven.

cxseven

you mean differentiable? yeah

what about just continuous

Ted Shifrin

11:37 PM

Oh, ok, over $\Bbb C$, how do you solve $x^3=-1$, @evinda?

cxseven

i can rule out an injective map having degree 0

Ted Shifrin

Still true for continuous, @cxseven, but I'm cheating and using smooth stuff to prove it (smooth degree counts numbers of preimages, with sign). I don't know how much you know.

Daniel Fischer

@MaryStar After seeing $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}}$$ for $n > m$, iterating yields $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{\gcd (m,n)}}.$$ But if $k\mid m$, then $\mathbb{F}_{p^k} \subset \mathbb{F}_{p^m}$, which yields the reverse inclusion.

evinda

@TedShifrin Isn't the solution $e^{\frac{\pi \cdot i \cdot k}{3}}$ ?

Ted Shifrin

$2\pi i$, @evinda, yes.

cxseven

11:39 PM

oh i guess you take a sequence of points that map to the same value, pass to a convergent subsequence, then argue convergence

Ted Shifrin

So your solutions need a $0$ in one slot, and then ... ?

cxseven

actually i think there are some catches to that argument

Mike Miller

@Pedro I commented. There's no reasonable thing to call a "generalized Riemann mapping theorem" in higher dimensions.

Not even in some super weak sense.

Ted Shifrin

un-hello @Mike

like what, @cxseven?

Daniel Fischer

@cxseven Since $S^n$ is compact, an injective continuous $f\colon S^n \to S^n$ is a homeomorphism.

Mike Miller

11:40 PM

Hi Ted... did you say hello earlier?

Ted Shifrin

Oh, @Daniel brings out invariance of domain. :)

Oh, no, not even that.

I'm rusty.

yields the floor and goes off to cook dinner

Mike Miller

@DanielFischer Which still isn't enough yet, since who's to say it's surjective? (Alexander duality does, for what it's worth.)

Daniel Fischer

@TedShifrin I think we need invariance of domain for surjectivity.

Mike Miller

or that

Daniel Fischer

For generic compact spaces, a continuous injective self-map could be a homeomorphism to a proper subspace.

Ted Shifrin

11:42 PM

Oh, good, so I'm not a total twit. :)

Mike Miller

Mine's a teeny bit overkill, I suppose.

Ted Shifrin

Yes, @Mike. But some would say smooth theory is overkill, too. Not I.

Mary Star

@DanielFischer Do we have to prove that $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}}$$ ??

evinda

@TedShifrin So, is the solution $2\pi i$ or $e^{\frac{\pi \cdot i \cdot k}{3}}$?

Also, do we want that one coordinate is equal to $0$, so that the solution belongs to $\mathbb{P}^2(\mathbb{C})$?

Mike Miller

@Ted Actually, I was being silly... it's of course homotopic to a smooth map, so we can use that and move on with our lives.

Ted Shifrin

11:43 PM

sigh @evinda.

Daniel Fischer

@MaryStar Yes, but if you look above, we did prove that.

Ted Shifrin

I asked @cxseven what his objection to my using smooth was, but he got quiet.

evinda

@TedShifrin Why? :(

anon

@MaryStar that n-m should be gcd(n,m)

Ted Shifrin

@evinda: (1) You told me that for $xyz=0$, one of $x$, $y$, $z$ had to be $0$.

(2) I told you that $e^{\pi i k/3}$ was wrong. You need $e^{2\pi ik/3}$.

Mike Miller

11:45 PM

@Ted Presumably the objection was "I'm not using smooth maps", which is reasonable lest you know that every continuous map is homotopic to a smooth one and that degree is a homotopy invariant.

Daniel Fischer

@MikeMiller Yup, the ball and the polycylinder are biholomorphically equivalent only in dimension $1$, for example.

@anon $n-m$ is the first step to $\gcd(n,m)$.

Ted Shifrin

Without even mentioning homotopy, I can use the fact that I can arbitrarily closely approximate a continuous function by a smooth one ... and the preimages of a point will stay a bounded distance away ...

Ooh, I mentioned ball and polydisk to @Mike months ago.

@anon :)

anon

@DanielFischer oh, it's a containment and you have n>m, I see.

Mike Miller

@Ted Bur that doesn't help, since degree can't be defined that way for continuous maps. If one wants to define it for continuous functions either we define it as the degree of a homotopic smooth map or we do it homologically/homotopic ally...

evinda

@TedShifrin A ok, I am sorry!!! Then we will have the equation $a^3=-b^3$, where $a,b \in \{x,y,z\}$, right? But how can we find now the flexes?

anon

11:47 PM

@Mike I voted to reopen the one you closed.

Ted Shifrin

Right, @Mike. My view was smooth degree $\ne\pm 1$ implies non-injective, so, by approximation of a continuous function, it must still be true for continuous degree.

Mary Star

@DanielFischer Oh yes!! But how can we conclude that $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{\gcd (m,n)}}$$ ??

Ted Shifrin

@evinda: You need to sit down quietly and work this out.

Mike Miller

@DanielFischer I was being even more general with one might call a generalized RMT: a contractible open subset of $\Bbb R^n$ needn't be homeomorphic to $\Bbb R^n$. (Of course, this also prevents the much harder question of viholomorphicity, but the poly disc and cylinder are a better counterezample here.)

Ted Shifrin

OK, I'm going to cook.

Mike Miller

11:49 PM

@anon You're going to need to be more specific...

@TedShifrin Oh, I see.

anon

the QxQ->Q one

Daniel Fischer

@MikeMiller You mean you closed more than one?

Mike Miller

@anon I agree with the commenter that he's essentially asking for an expository paper; I am not convinced by the argument that broad questions are good because they allow one to say many different things. But if five others disagree, so be it.

@DanielFischer Don't tell on me.

evinda

@TedShifrin What is the property that should be satisfied so that a point $\in \mathbb{P}^2(\mathbb{C})$ ?

cxseven

@TedShifrin sorry I got called away for a few minutes

the argument you proposed could work, i just need to think about the fact that "passing to a smooth map" doesn't necessarily preserve injectivity

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