I.1.4

11/30/2020

It looks like this time we've got a topological exercise rather than an algebraic one? Actually, the whole "point" of
the Nullstellensatz appears to be a duality between algebraic sets (rings) and topological spaces (A^{n}), so can
someone explain to me why this subject is called "algebraic geometry" rather than "algebraic topology"? I mean,
I know the latter name is already reserved for another branch. But where's the "geometry" in all
this? Is it because the spaces we're looking at are the zero sets of polynomials? That's geometry?

Lol, whatever. Wait. WAIT. I guess this exercise demonstrates exactly what I was talking about? We're comparing
the "topological" definition of A^{2} (i.e. using the product topology) with the "geometric" definition of A^{2} (i.e. using
the Zariski topology–the zero sets of polynomials). Am I right here, or am I just flailing around like an
idiot.

Well, anyway, here we go. The Zariski topology of A^{1} has already been expained in the book:

I.e. the open sets of A^{1} fall into the following classes

1) ∅

2) complements of finite sets

3) A^{1}

So a base for the product topology of A^{2} is (S an arbitrary complement of a finite set):

1) ∅

2) S × A^{1}

3) A^{1} × S

4) S_{1} × S_{2}

5) A^{2}

Well, "complements of finite sets" are annoying to think about. Let's make a closed set base instead, by complementing all of these.

We get the following closed set base for the product topology of A^{2} (F an arbitrary finite set):

1) A^{2} (the whole space)

2) A^{1} × F (finitely many horizontal lines)

3) F × A^{1} (finitely many vertical lines)

4) F_{1} × F_{2} ("a finite set of points")

5) ∅ (the empty set)

A closed set base generates the topology by taking intersections of the base elements. Clearly (using this theorem) intersecting any of these classes of sets gives us
another set in the class. So the base above is actually a complete description of the topology (wtf? kind of suspicious
but ok).

So, obviously Y = Z(y^{2} -x^{2} - 1) (the unit circle) is closed in the Zariski topology of A^{2} but isn't a closed set in the
product topology. So the spaces are not homeomorphic. (Though I believe the Zariski topology is finer than the
product topology? Don't quote me on that though lol)