Mathematics - 2014-11-20 (page 4 of 4)

@DanielFischer
I have to show that $I(A\ cup B)=I(A) \cap I(B)$

$$I(A)=\{ f \in K[x_1,x_2, \dots, x_n]:f(a)=0 \forall a \in A \}$$

where $a=(a_1, a_2, \dots, a_n) \in K^n$

Let $f \in I(A \cup B)$.
If $a \in A$, then $f(a)=0$
So, $f \in I(A)$
If $b \in B$, then $f(b)=0$

Then, why do we conclude that $f \in I(A) \cap I(B)$

I thought that that the above two conditions aren't satisfied simultaneously.. Am I wrong?

Jorge Fernández

that moment when you finish writing a solution and it was deleted two minutes ago

Alizter

10:28 PM

@Studentmath Why do electrons prefer to fill orbitals separately. They only pair if the have to?

Studentmath

@Alizter intutive explanation is that they carry negative..

thingy

So they don't like to be close, they want to be as far as possible

Alizter

So its like filling up a bus?

Everybody takes their own seat

Studentmath

Yep, if they can have more space, they will

Daniel Fischer

@evinda $f\in I(A\cup B)$ means $f$ vanishes on $A\cup B$. Then it vanishes on all subsets of $A\cup B$. In particular, $f$ vanishes on $A$, whence $f\in I(A)$, and $f$ vanishes on $B$, whence $f\in I(B)$. Together, that is $f\in I(A) \cap I(B)$.

@JorgeFernández Ah, that lovely feeling. How I detest it.

Studentmath

Hrm, can anyone guide me as slightly as possible as to why $H$ is normal in $G$, if $ind(H)=5$ and $o(G)$ is odd?

I just have no idea where to start.

Mike Miller

10:42 PM

what's g

evinda

@DanielFischer I understand!!! Thank you!!!!!!!!!!!

Studentmath

Oh, they are all groups @Mike

evinda

@DanielFischer And if we have $f(x)=0, \forall x \in A$ and $f(x)=0, \forall x \in B$, do we conclude that $f(x)=0 \forall x \in A \cap B$ or $f(x)=0 \forall x \in A \cup B$?

Daniel Fischer

@evinda We conclude $(\forall x \in A\cup B)(f(x) = 0)$. That implies $(\forall x\in A\cap B)(f(x)=0)$, but that is a weaker conclusion than each of the premises $(\forall x\in A)(f(x) = 0)$ and $(\forall x\in B)(f(x) = 0)$ separately.

And besides, we want to show $I(A)\cap I(B) \subset I(A\cup B)$ there, so $A\cup B$ is the set we're interested in.

Ted Shifrin

10:58 PM

hi @DanielF, @studentmath, @Mike

Studentmath

@Ted !

How's it going?

Daniel Fischer

@TedShifrin Hi.

skullpatrol

@TedShifrin Did you write the preface last with your books?

Ted Shifrin

yes @skull

3

skullpatrol

Hi btw :-)

Daniel Fischer

11:02 PM

@MikeMiller You might like this.

Possibly you too, @TedS.

Pedro Tamaroff

@TedShifrin Hello there.

Ted Shifrin

hi @Pedro

Pedro Tamaroff

So, I did finish your computation. It is quite nice how the differentials of such functions interact.

For example, if $z=\frac 12(w+w^{-1})$ then $dz=\frac 1 2(1-w^{-2})dw$, and $z^2-1=(\frac{w-w^{-1}}2)^2$.

Ted Shifrin

cool ...

Pedro Tamaroff

And if $z=\frac{w+1}{w-1}$, then $(z-1)^2(w-1)^2=4$, and $dz=\frac{-2}{(w-1)^2}$.

So transformations involving them are not so ugly.

Ted Shifrin

11:06 PM

Huh? These are unrelated.

Pedro Tamaroff

@TedShifrin Using Wolfram, I got that what maps to $|z|=2$ under the degree 2-rational map is a little ellipse around $0$.

@TedShifrin Ah?

Ted Shifrin

Yes, with or without the 1/2, this is classical stuff. See Ahlfors.

Pedro Tamaroff

@TedShifrin I haven't looked at it yet. It seems I should, if I want to have any hope in learning how to handle Möbius transformations...

Ted Shifrin

Quadratic ≠ Möbius ... I'm confused

Pedro Tamaroff

@TedShifrin Well, both came up in integrals in the problem sheet I was looking at. One was $$\int_{|z|=2} e^{z+z^{-1}}/(1-z^2)dz$$

Studentmath

11:08 PM

Ahlfors always reminds me of alfajores

Pedro Tamaroff

@Studentmath Alfajores.

Studentmath

Yeah, the first might be misread.

Pedro Tamaroff

@TedShifrin The other was $$\int_{|z|=2}\log\frac{z+1}{z-1}dz$$

Alizter

@PedroTamaroff I count 3 poles on the first.

Pedro Tamaroff

@Alizter Well, no.

Alizter

11:10 PM

@PedroTamaroff Why not?

Pedro Tamaroff

Two poles. The other is not a pole.

Alizter

What is the other?

Daniel Fischer

Essential.

Pedro Tamaroff

An essential singularity.

Alizter

Ah ok. Thank you.

So what happens with that then?

Pedro Tamaroff

11:11 PM

That integral actually vanishes.

Alizter

So how would you evaluate it?

Does the residue theorem still hold for essential singularities?

I guess no?

Daniel Fischer

@Alizter It holds for all isolated singularities.

Just, computing the residue in an essential singularity can be a bitch.

Pedro Tamaroff

@Alizter First, use $w=z^{-1}$ to get it equals $$\int_{|z|=1/2}\frac{e^{z+z^{-1}}}{1-z^2}dz$$ so now you've avoided two singularities.

Alizter

Oh cool.

Pedro Tamaroff

Now there is a biholomorphic mapping from the punctured unit disk to the complex plane slit along $(-\infty,-1]\cup [1,\infty)$, which is $w=\frac 1 2(z+z^{-1})$.

This gives up to a constant multiplier the integral $$\int_{\gamma}\frac{e^{2w}}{1-w^2}dw$$ and this is holomorphic inside the domain in question.

Daniel Fischer

11:20 PM

@PedroTamaroff Hold on, one is simply connected, the other isn't.

Pedro Tamaroff

Here $\gamma$ is a path inside the unit disc, so we're safe.

@DanielFischer checks notes

SHOOT. I mean

The upper half plane.

FUUUUUUUUUUUUUU

That thing sends the upper half plane biholomorphically onto $\Bbb E^{\times}$.

Well, the other way around.

Still, no singularities there. YAY.

Alizter

Wait. What does $\gamma$ look like?

Pedro Tamaroff

I have no idea.

But it is irrelevant.

Alizter

Ok

Pedro Tamaroff

It is some closed countour inside $\{\Im z>0\}$.

Studentmath

11:23 PM

What does this falls into? Complex analysis?

Alizter

Yes.

@PedroTamaroff So no singularities therefore that thing is zero?

@PedroTamaroff Wait. Does this not mean that essential residue was just zero?

Studentmath

Blach, really no idea where to start. Will go back to the videos..

Pedro Tamaroff

@DanielFischer @DanielFischer Sorry, the last time I solved that integral I used that same thing gives a biholomorphism from $\Bbb E^\times$ to $\Bbb C\smallsetminus [-1,1]$, so I didn't even bother changing the circle. =)

Alizter

@Studentmath Starting what?

Pedro Tamaroff

@Studentmath What are yer talking about?

Studentmath

11:33 PM

I want to show that if o(G) is odd and ind(H) is 5, then H is normal subgroup of G

I read the chapter twice and it's the second time I watch the videos now.. nothing seems somehow related

Pedro Tamaroff

@Studentmath OK, so $|G|$ is odd, and $H$ is a subgroup of index $5$, yes?

Studentmath

@Pedro yes

I thus know $H$ has five different disjoint classes in $G$. If that's the right word in English.

Pedro Tamaroff

@Studentmath Have you tried anything? Usually, one would want to invoke group actions here. Since $H$ has index $5$, we can make $G$ act on the coset space $G/H=\{H,g_1H,\ldots,g_4H\}$ giving a morphism of groups $\eta :G\to S_5$.

Studentmath

facepalm

Pedro Tamaroff

What's going on?

Studentmath

11:36 PM

I didn't think about quantative groups..

That's not the right term.

Alizter

@PedroTamaroff Can you respond to my earlier mutterings?

Pedro Tamaroff

Symmetric.

@Alizter ?

@Alizter Yes.

and the second?

Looks like it too.

Interesting.

11:38 PM

Thanks @Pedro, no idea why I was stuck on showing it via definitions of normal groups themselves

Alizter

I guess this also means that the two poles cancel each other as well.

How nice.

Pedro Tamaroff

@Studentmath I am not sure if the above will work.

Did you figure it out?

Studentmath

@Pedro I think so, I think that if $gH\neq Hg$ I can show there is a subgroup\element of order 15, and the morphism will show there isn't one

Pedro Tamaroff

Note that $G/\ker\eta$ injects into $S_5$. $S_5$ has order $5\times 8\times 3$, and $G/\ker \eta$ has odd size, moreover, it divides $5\times 8\times 3$, so it divides $5\times 3$.

Studentmath

Wait, isn't $o(S_5)=5!=25$?

Alizter

11:43 PM

5!=120

Studentmath

Erm

No.

Alizter

:)

Studentmath

I am too tired.