Mathematics

12:00 AM

Even if it was strictly dominated by $M$, the closure would be dominated by 'at most $M$, and then I could show its interior is empty just the same - isn't it so?

Pedro Tamaroff

Yes.

Studentmath

Alright, I think I get it better now :) Thanks @Pedro!

evinda

@PedroTamaroff Do we have to suppose that $V$ is irreducible? If so, then , there won't be $V_1, V_2 \subset V$, such that $V=V_1 \cup V_2$.

But how can we conclude that $I(V)$ is a prime ideal?

Pedro Tamaroff

@evinda Suppose that $I $ is not a prime ideal. Then there are $x,y\notin I$ with $xy\in I$.

Consider $J_1=I+(x)$ and $J_2=I+(y)$.

Then $J_1J_2\subseteq I$ but $I\subsetneq J_1,J_2$.

evinda

@PedroTamaroff Why do we consider $J_1=I+(x)$ and $J_2=I+(y)$? :/

Pedro Tamaroff

12:06 AM

@evinda Well, because now we have $V(I)=V(J_1)\cup V(J_2)$.

evinda

@PedroTamaroff Could you explain me how we conclude that $V(I)=V(J_1)\cup V(J_2)$? :/

Pedro Tamaroff

@evinda $I\subseteq J\implies V(J)\subseteq V(I)$

evinda

@PedroTamaroff What does $I+(x)$ mean? That we add a multiple of I to a multiple of $x$ ?

Pedro Tamaroff

@evinda Do you know what a principal ideal is? Do you know what is $I+J$ when $I,J$ are two ideals?

evinda

12:24 AM

@PedroTamaroff A principal ideal is an ideal that is generated by a single element of R, $\langle h(x)\rangle$.

$a \in I+J \Rightarrow a=x \cdot i+y \cdot j$

We have that $V(I+J)=V(I)\cap V(J)$.

Am I right?

Pedro Tamaroff

@evinda Yes.

But if you know that you shouldn't be asking this.

evinda

@PedroTamaroff $a \in J_1 J_2 \Rightarrow a=m \cdot n $, where $m \in J_1$ and $n \in J_2$

How does this imply that $a \in I$?

So that we have $J_1 J_2 \subseteq I$.

evinda

1:22 AM

We assume that $V$ is irreducible. Suppose $I(V)$ is not a prime ideal. Then there exist two polynomials $f, g \in k[x_1, x_2, \cdots, x_n]$ such that $f \notin I(V), g \notin I(V)$ but $fg \in I(V).$ Set $V_1 := \{ x \in V | f(x) = 0\}$ and $V_2 := \{ x \in V | g(x) = 0\}.$ Then both $V_1$ and $V_2$ are non-empty and proper closed subsets of $V$ with $V = V_1 \cup V_2.$

Could you explain me why $V=V_1 \cup V_2$ ?

Pedro Tamaroff

@evinda Since $V_1,V_2\subseteq V$, $V_1\cup V_2\subseteq V$. Now pick $x\in V$. Then $g(x)f(x)=0$, so either $g(x)=0$ or $f(x)=0$. This means $x\in V_2$ or $x\in V_1$. Hence $V=V_1\cup V_2$:

evinda

@PedroTamaroff And why do we know that both $V_1= \{ x \in V | f(x)=0 \}$ and $V_2=\{ x \in V | g(x)=0\}$ are non-empty and proper closed subsets of $V$ ?

Pedro Tamaroff

@evinda Think.

evinda

@PedroTamaroff I thought that it is either f(x)=0 or g(x)=0.. Why are both of the sets non-empty?

Pedro Tamaroff

@evinda Think.

Pedro Tamaroff

4:42 AM

@TedShifrin

Do you know what the Poincaré sphere is?

Mike Miller

5:10 AM

@Pedro Whatcha wanna know about it

Pedro Tamaroff

@MikeMiller What the hell that is.

Kaj Hansen

Wiki says it could be two different things? en.wikipedia.org/wiki/Poincar%C3%A9_sphere

Mike Miller

@Pedro It's a 3-manifold with the same homology as the 3-sphere. It's famous for being a counterexample to Poincaré's original formulation of his hypothesis - "every homology 3-sphere is homeomorphic to the 3-sphere". Which it's not.

There are, in fact, lots of homology 3-spheres. It was just the first.

Pedro Tamaroff

@MikeMiller But how is it defined?

And why is it not contractible although it is acyclic?

Mike Miller

Huh? Link me to what you're talking about. If something has the same homology as a 3-sphere it's definitely not acyclic.

There are a bunch of ways to define it and I don't remember any of them.

Pedro Tamaroff

5:16 AM

@MikeMiller Sorry, I meant punctured.

I wanted an example of an acyclic space that is not contractible. The punctured Poincaré sphere is apparently an example of this.

Mike Miller

Link to me to what you're talking about. Also, what do you mean by acyclic.

Pedro Tamaroff

@MikeMiller All the homology groups vanish. As in acyclic complex.

"For example, if one removes a single point from a manifold M which is a homology sphere, one gets such a space. The homotopy groups of an acyclic space X do not vanish in general, because the fundamental group need not be trivial. For example, the punctured Poincaré sphere is an acyclic, 3-dimensional manifold which is not contractible."

Mike Miller

Oh. Is that standard terminology?

Pedro Tamaroff

@MikeMiller Yes.

Karl Kronenfeld

yeah

Pedro Tamaroff

5:18 AM

Literally, it means it has no friggin cycles.

Mike Miller

For spaces, not for complexes.

Pedro Tamaroff

Well, yeah.

Mike Miller

I know it's common in homological algebra.

Pedro Tamaroff

Oh.

Mike Miller

Puncturing a 3-manifold induces an isomorphism on fundamental groups. You can check that the homology groups are trivial by Meyer-Vietoris.

Pedro Tamaroff

5:20 AM

I don't know all that stuff Mike.

Mike Miller

What did you expect?

Pedro Tamaroff

I don't know.

Mike Miller

When you take algtop you'll know them. The first fact is Van Kampen's theorem, the second is an important (standard) result you'll learn.

The point is that the punctured Poincare sphere has fundamental group $A_5$, and the first homology group is the abelianization of the fundamental group.

Pedro Tamaroff

I read that in Rotman's book, yes.

Mike Miller

And $A_5$ is perfect.

Pedro Tamaroff

5:24 AM

Yes.

Such a nice group.

Mike Miller

5:46 AM

I think the perfect group is $SL_2(\Bbb R)$.

Pedro Tamaroff