OK, I'm outta here for an hour. Have webcam "class" to teach :P

Bye.

Enjoy!

Goodbye to you all I'm off to celebrate.

Hi @Balarka!

hi @MatheinBoulomenos

Hi @LeakyNun

@MatheinBoulomenos does every uncountable subset of $\Bbb R$ contain a limit point of itself?

yeah

how would you prove it?

$\Bbb R$ is $\sigma$-compact and for metric spaces, compact is equivalent to limit point compact

lol @BalarkaSen

As a humble member of the pre-trash subset of this site's users, I just wanted to step in and quickly ask, whenever I'm working with bilinear forms, I'm always going to be working with transposes and not inverse matrices when I'm diagonalizing, right?

@JakeS yes

Many thanks!

@JakeS Since the matrix $Q$ represents the quadratic form $\vec v \mapsto \vec v^\top Q \vec v$

changing a basis is sending $\vec v$ to $B \vec v$ for some invertible $B$

so now we have $\vec v \mapsto \vec v^\top B^\top Q B \vec v$

so the new quadratic form is $B^\top Q B$

which is why we use transpose instead of inverse

(the above is meant to be understood to be correct up to ordering)

I understand! Thank you.

@MatheinBoulomenos could you elaborate?

or you're saying that $[-n,n] \cap S$ must be uncountable for some $n \in \Bbb Z$?

basically, yeah

and then use the equivalence between compact and limit point compact here?

it's enough that is it infinite

the same argument works in any $\sigma$-compact metric space

cool

it's enough that $[-n,n]$ is infinite for some $n$, but of course you even get uncountable

how does being singular have anything to do with not being integrally closed?

also you only get a limit point?

how do you get the limit point to be inside $S$?

a set without limit points is closed

so the intersection with a compact subset will be closed inside that compact subset

thus compact

I should've written the whole argument

hmm

but why bother with such basic topology questions?

It's just an exercise. =)

But why bother with exercise?

Because exercise is good for health. =)

@MatheinBoulomenos I don't know

or you argue that a set without limit points is discrete and $\Bbb R$ can't have an uncountable discrete subset

@MatheinBoulomenos What do you specialise in?

@WillHunting algebraic number theory

@MatheinBoulomenos what do you specialize in?

Huh?

Is this a joke?

no, it's not a joke

Oh, I don't understand why he asked the same question I did.

If this is not a joke, what is happening here?

maybe it's about -ize vs. -ise

I have no idea

Maybe it is a joke after all, lol.

But I assure you the joke is not my intention.

it means, which sub-division of algebraic number theory?

Oooooh.

This is quite funny.

I've not yet decided on that yet. I've been doing some Langlands stuff lately, but I also want to look into Iwasawa theory at some point

you managed the butcher both names

:P

What about waterlands?

@Daminark sniped?

Apparently

I typo alot

You make typos when you get excited.

maybe

WhAt AbOuT WaTeRlAnDs?

As for computations you mentioned earlier, if $F$ is some homogeneous irreducible polynomial of degree $d$, $V(F)$ has degree $d$, as it ought

hmm...

@LeakyNun it's a difficult theorem in commutative algebra that a regular local ring is a UFD (thus integrally closed)

:o

well, difficult as in I don't find the proof particularly easy, but it's in Eisenbud

what does a cubic with singularity at infinity look like?

The idea is that you can compute $h_{V(F)}(l)$ for $l > d$, then $k[x_0,\ldots,x_n]_{(l)}$ has dimension $\binom{n+l}{n}$, and then $(F)_{(l)}$ is just $F$ homogeneous polynomials of degree $l-d$, so that has dimension $\binom{n+l-d}{n}$

what does the subscript mean anyway

the homogeneous part?

Yup

Or

Specifically choosing the degree to be that guy

So $h_{V(F)}(l) = \binom{l+n}{n} - \binom{n+l-d}{n} = \frac{1}{n!}(l+1)\ldots (l+n) - \frac{1}{n!}(l-d+1) \ldots (l-d+n)$, so this polynomial has degree $n-1$, so that's the dimension of $V(F)$

I don't know what I'm asking. That's exactly what elliptic curves look like.

And the coefficient of $l^{n-1}$ is $\frac{1}{n!}((1+2+\ldots + n) - ((1-d) + \ldots + (n-d))) = \frac{1}{n!}(\frac{n(n-1)}{2} - \frac{n(n-1)}{2} + nd) = \frac{d}{(n-1)!}$

So the degree is that guy times $(n-1)!$, which is $d$

@LeakyNun ?? elliptic curves are smooth

oh

then what do they look like?

Donuts

Mmmmm.

That sounds tongue in cheek but it turns out that elliptic curve are topologically tori

So... have been thinking about leaky's double rational question:

in $\Bbb C^2$

(hence the group operation)

(or something)

I think the increasing sequence $\{q_i^-\}$ will diverge, since the sequence of open neighbourhoods will shrink towards $q_- \to "-\infty_+"$ which is not in the set, thus that sequence diverges

(I need to reread Poncelet's Theorm)

But otherwise, every point should be accessible from some net in $\Bbb{Q}^+ \cup \Bbb{Q}^-$ so it should be dense in $2\cdot \Bbb{R}$ and also $\Bbb{R}$