Mathematics - 2014-11-17 (page 6 of 6)

Studentmath

23:00

@Ted Is it possible for any passenger to decide he wants to spend the rest of his lives in the elevator, and not get out on any floor?

Ted Shifrin

@Kaj: Does the criterion $f$ has a double root at $x=a$ iff $f(a)=f'(a)=0$ work in characteristic $p$?

No, I interpret it that he must get out on one of the floors $1,\dots,N$, @Studentmath.

Swapnil Tripathi

@KajHansen That was for me?

Kaj Hansen

For both of us @SwapnilTripathi, if you're referring to @TedShifrin's comment. But I'm just thinking out loud right now

evinda

@DanielFischer Can you explai me how we show in this way that $Im \phi \subset \mathbb{C}$ ?

Swapnil Tripathi

@KajHansen Haha. No, the message that this is trivial if the base field has characteristic 0. That gave me a shock!

Daniel Fischer

23:04

@evinda That follows directly from $\phi \colon \mathbb{C}[x,y] \to \mathbb{C}$. By definition that means the image of $\phi$ is contained in $\mathbb{C}$.

Studentmath

Thinking outloud, the chance of $k$ people stopping at the same floor is $1/N^k$

Ted Shifrin

@Zach: How did you do on your exam?

Kaj Hansen

We just showed that @SwapnilTripathi. For any field $F \subset K$, $K$ is separable if $char(F) = 0$.

Zach Saucier

@Ted eh, not great

I'm a bad test taker

Ted Shifrin

True, @Studentmath. Better to think about how many floors have no one getting off.

Kaj Hansen

23:05

So consider $F \subset E \subset K$ where each individual extension is separable.

Ted Shifrin

Ah, sorry, @Zach. I've had students like that ...

Studentmath

@Zach so am I :P I feel your pain

evinda

@DanielFischer Oh yes.. I got it.. But I still haven't understood how we conclude in tis way that $\mathbb{C} \subseteq Im \phi$. Could you explain it further to me?

Alizter

ouch. I had a bad hand cramp today.

Ted Shifrin

Be careful, @Kaj. You know $E/F$ Galois and $K/E$ Galois doesn't mean $K/F$ Galois :P

Swapnil Tripathi

23:06

@KajHansen: Oh, yes! It seems I suck pretty bad at applying stuff I already know!!

Kaj Hansen

Ugh, I need to look back at what a "normal" extension means @TedShifrin. One sec.

I didn't learn Galois extensions in this context.

Ted Shifrin

LOL, Galois = normal + separable

Kaj Hansen

Indeed. But you said Galois = $[K:F] = |Aut(K/F)|$ in your book, IIRC ;)

Mike Miller

@Ted Did you see the nonsense update to your integral question?

Ted Shifrin

I'm only doing separable stuff, @Kaj, so be careful.

Kaj Hansen

23:08

Or did you? You might've said Galois = splitting field of a polynomial

Ted Shifrin

I proved that as iff, @Kaj.

@Mike, which question?

Daniel Fischer

@evinda $\phi(c) = c$ for constant polynomials. Hence, for every $z\in\mathbb{C}$, we know a polynomial $p$ with $\phi(p) = z$, namely $p = z + 0\cdot x + 0\cdot y = z$.

Kaj Hansen

At any rate, we've already shown that all extensions of a char=0 field are separable, so what @SwapnilTripathi is now trying to prove is done except for the case of char>0.

Studentmath

Think I got it.. $N$ is given, right? Not distribued somehow?

Ted Shifrin

yes, @Studentmath, $N$ is fixed.

<--- wishes @Studentmath were in his class

Studentmath

23:10

@Ted don't be so hasty - I make tons of mistakes before I get things right :P

Twink

Hi, could somebody answer my question please? It has a +50 bounty math.stackexchange.com/questions/1019464/…

Ted Shifrin

So do I, @Studentmath.

Studentmath

Well, Human nature I guess.. anyhow just a sec

Kaj Hansen

Ugh, I have to go. I've been trying to multitask too much, and now I have a meeting.

Ted Shifrin

bubye, @Kaj.

Kaj Hansen

23:12

I look forward to looking at this further later tonight @Studentmath

Swapnil Tripathi

@KajHansen: Sure. Thank you so much for your help. :)

Ted Shifrin

wrong person, @Kaj

Studentmath

@Kaj I believe you meant to tag @Swapnil

Kaj Hansen

Oops, meant to ping SwapnilTripathi. Indeed I did.

Swapnil Tripathi

@TedShifrin: You're up next for helping me out? ;)

Ted Shifrin

23:15

I haven't helped you, @Swapnil :)

Studentmath

Don't think it matters, actually? Or does it?

Swapnil Tripathi

@TedShifrin: So will you now? :P

Ted Shifrin

No, in a moment of panic, I thought it should, but it doesn't, @Studentmath ... so I removed my comment :p

nope, @Swapnil. I'm going to dinner.

Hippalectryon

Eating infants again ? ;)

Swapnil Tripathi

@TedShifrin Haha. See you. :)

Ted Shifrin

23:18

De quoi parles-tu, @Hippa?

Hippalectryon

@TedShifrin chat.stackexchange.com/transcript/36?m=18635781#18635781

Ted Shifrin

ah, @Hippa, your memory is so much better than mine ... worthlessly so, but so.

Hippalectryon

@TedShifrin :D

I can at least remember the first 8 Bell numbers here

Ted Shifrin

well, wait 'til you're my age :D

Hippalectryon

:O long time

Ted Shifrin

23:20

ok, bubye, all, for now

Swapnil Tripathi

@TedShifrin bye. :)

Hippalectryon

Bye

Swapnil Tripathi

@hippa

Hippalectryon

@SwapnilTripathi ?

Swapnil Tripathi

do you get pinged when someone writes half of your name! was just checking. new here. ;)

Hippalectryon

23:22

:-)

Swapnil Tripathi

You got pinged. Say yes atleast! And lol, nice profile. :D

@hippa

Hippalectryon

Yes

:D

@SwapnilTripathi Have you seen the second half of my desc ?

Swapnil Tripathi

So what area of mathematics do you like?

What does that mean?

desc?

Hippalectryon

Well I'm in some kind of equivalent of college (I'm french), and I like everything :D

@SwapnilTripathi Description. On my profile.

Studentmath

@Ted $N-N*e^{-10/N}$?

Swapnil Tripathi

23:26

@hippa Of the photo? :P

*oh

Hippalectryon

@SwapnilTripathi :-))

@SwapnilTripathi You can edit your messages if you have made a mistake

Just press the 'up' arrow in the empty chatbox

Swapnil Tripathi

Oh. Thanks. Yeah got it! @hippa

@Hippa. I am from India and doing my Master's.

Hippalectryon

Ok :)

Swapnil Tripathi

Though here. They don't teach us much maths. I've seen people on SE who have done their bachelors and have a much more diverse knowledge of subjects.

Studentmath

@Ted using expectation of the conditional expecatation where I set $Y=y$

Swapnil Tripathi

23:29

Kinda saddens me.

@hippa

Hippalectryon

@SwapnilTripathi SE isn't a good example :) lots of exceptional people here

Swapnil Tripathi

You mean to say every second one of them? ;)

Studentmath

I think it's right... I can't find the question on my edition though, now to wait for Ted to return from dinner..

evinda

@DanielFischer I understand!!!! Thank you!!!! I also want to show that the radical of the ideal $I=<X=x^5,y^3>$ of the ring $\mathbb{C}[x,y]$ is $Rad(I)=<x,y>$.

So far, I have shown that $<x,y> \subset Rad(I)$.

It remains to show that $Rad(I) \subseteq <x,y>$, right? But,how can we show this?

Hippalectryon

@SwapnilTripathi I am but a banana here :)

Swapnil Tripathi

23:32

@hippa SE will make us better soon then. :D

Hippalectryon

I hope so :)

Daniel Fischer

@evinda What is the characterising property of polynomials not belonging to $\langle x,y\rangle$?

Studentmath

@Ted Alternatively you could view the distribution of the number of floors where you do stop as $Bin(N, (\frac{N-1}{N})^Y$, and then N-E[Bin], and run expectation over that for $Y=y$ from 0 to $\infty$

Mike Miller

@Ted The question about integrating $w_2 \cap c_1$. If you interpret his comment the obvious way his question is now worthless...

evinda

@DanielFischer If they don't belong to $\langle x,y\rangle$, and since we are working at $\mathbb{C}[x,y]$, the polynomials have to be constants.. Or am I wrong?

Mike Miller

23:35

If you like to think about rings, here's a question about fixing an incorrect question of Matsumura's.

Daniel Fischer

@evinda They need not be constants, $1+x \notin \langle x,y\rangle$.

evinda

@DanielFischer I see.. But, how can I find a property, that must be satisfied? :/

Daniel Fischer

@evinda If you write down a generic polynomial in $x,y$, which part of it is not divisible by either $x$ or $y$?

Hippalectryon

What's a good way to show that any positive symmetric matrix $A$ can be written $A=^t MM$, $M$ invertible ?

Mike Miller

By positive matrix you mean all its eigenvalues are positive?

Hippalectryon

23:40

In $\mathcal{S}_n^+$

Swapnil Tripathi

Bye everyone! See you tomorrow, maybe. :)

Hippalectryon

Matrix such that $^tXAX>0,\forall X$

@SwapnilTripathi See you :)

Karl Kronenfeld

I got your email @PedroTamaroff I will have time to read it through on Wednesday or Thursday

evinda

@DanielFischer Do you mean a polynomial of the form $\sum_{i,j=0}^n x^i y^{j}$ ?

Mike Miller

OK, good. That's equivalent to my claim.

@KarlKronenfeld Did you see my Matsumura question?

Hippalectryon

23:41

@MikeMiller Oh true

Mike Miller

Anyway, no, I can't help you. :)

Karl Kronenfeld

@MikeMiller Yes. Currently thinking about it.

Can't recall if I drew a similar conclusion.

Mike Miller

I'm really offended by the close vote that just fell. Did he even read my question?

Hippalectryon

@MikeMiller Was that for me or @KarlKronenfeld ?

Mike Miller

You, @Hippalectryon

Don't recall how the proof goes.

Hippalectryon

23:44

Ah :/

Daniel Fischer

@evinda There are coefficients missing.

evinda

@DanielFischer So, it should be of the form $\sum_{i,j=0}^n a_{ij}x^i y^{j}$, right? It is not divisible by either $x$ or $y$, if the constant term $a_{00}$ is non-zero, right? How can we use this fact?

Karl Kronenfeld

@MikeMiller In your example $X$ is a zero divisor.

Bleh.. Of course, that's your point isn't it.

(Derp)

Daniel Fischer

@evinda All terms except the constant term are divisible by $x$ or $y$. Hence $p\notin \langle x,y\rangle$ if and only if it has a nonzero constant term. Or, if and only if $p(0,0) \neq 0$.

Mike Miller

@Karl The commenter is a moron.

evinda

23:48

@DanielFischer I understand!!! And how can we use this, in order to show that $Rad(I) \subset <x,y>$ ? :/

Twink

Why "This question does not appear to be about math within the scope defined in the help center"? math.stackexchange.com/questions/1026829/…

Karl Kronenfeld

@MikeMiller He is. It still stands that your example is problematic. Indeed $\langle X, Y\rangle$ intersects $S$. Take for example $X+Y$, a non zero-divisor.

Daniel Fischer

@evinda If $p(0,0) \neq 0$, can $(p^k)(0,0) = p(0,0)^k$ be $0$?

Hippalectryon

@MikeMiller math.stackexchange.com/a/1026838/150347

Mike Miller

@Karl That's pretty upsetting.

I'm definitely cross about this.

evinda

23:54

@DanielFischer Why do we take at the beginning polynomials that do not belong to $\langle x,y\rangle$ ?

Mike Miller

I don't know how I missed this. Ugffffh

Daniel Fischer

@evinda We want to show $\operatorname{Rad}(I) \subset \langle x,y\rangle$. So we want to show $p \notin \langle x,y\rangle \implies p \notin \operatorname{Rad}(I)$.

Karl Kronenfeld

A more abstract way to see that a non-minimal prime $\mathfrak p$ intersects $S$ is to note that $S$ is the complement of the minimal primes. If $\mathfrak p$ were a union of minimal primes, then use prime avoidance to derive a contradiction.

Twink

@DanielFischer Thank you so much for your answer, it explains everything I didn't understand very clearly

Karl Kronenfeld

@MikeMiller Well, it's a nontriviality that zero-divisors of a Noetherian ring occur only in minimal primes.

Mike Miller

23:57

Ok, that's what I needed. I hate that that guy was right.

Twink

I need to wait 5 more minutes to give you the bounty

Daniel Fischer

@Twink You're welcome.

Mike Miller

Yes, I know. But you just provided a proof.

Karl Kronenfeld

@MikeMiller Fortunately or unfortunately, he turns out to be an asshole who is pretty damn knowledgeable about ring theory.

@MikeMiller I don't claim to truly understand prime avoidance. :)

so it's still nontrivial in my mind

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