Let
k be an
algebraically closed field (for example
C). We are going to make
k^{n} into a
topological space. In the case
of
k=C there is already a natural way to do this (since it
is a
metric space) but the topology we give here, called
the
Zariski topology is coarser
than the usual
topology, it has fewer open sets. The advantage of
this approach is that it works for all
k, not just
C.
Once we have this topology on
k^{n} then it becomes
possible to talk about
algebraic geometry.
We will be using the polynomial ring
k[x_{1},...,x_{n}].
If f(x)=f(x_{1},..., x_{n}) in
k[x_{1},...,x_{n}] is a polynomial
then we can evaluate it at a point
a=(a_{1},..a_{n}) in k^{n}
to give an element of
k denoted by f(a) or f(a_{1},...,a_{n}).
If S is a nonempty set of polynomials then define
Z(S) = { a in k^{n} : f(a)=0
for all f in S }
For example, if
n=2 and
S={xy}
then
Z(S)
is the set
{ (a,a) : a in
k }. This is just the
line
y=x. Note that to get an idea of what's going on you can
sketch the real points of of closed sets in
C^{2}
or
C^{3}.
Here's another example for
n=2
if we take
S={ xy } then this time
Z(S) is the union of the
xaxis
and the
yaxis. For some interesting examples take a look
at
plane algebraic curve. One final example. If we take
S={x,y}
then
Z(S) is a single point, namely the the origin
{(0,0)}
Definition Any subset of k^{n} of the form Z(S)
is called an affine algebraic variety.
In fact we are going to show that the sets of this form are the
closed sets for a topology on k^{n}
called the Zariski topology.
First notice that if I is the ideal of
k[x_{1},...,x_{n}] generated
by S then we have Z(I)=Z(S).
For since S<=I (S is a subset of I)
we obviously have Z(I)<=Z(S). Suppose that a is
in Z(S) and let f in I. Then
f=f_{1}g_{1} + ... + f_{m}g_{m}
for some
f_{i} in
S and polynomials
g_{i}.
Evidently then
f(a)=0 and we have
Z(I)=Z(S).
Since
k[x_{1},...,x_{n}] is
Noetherian
by
The Hilbert Basis Theorem we know that any
ideal has finitely
many generators. This means that in the definition of affine
varieties
set we only need to consider
finite sets of polynomials
S.
Theorem The affine algebraic varieties in k^{n}
are the closed sets of a topology.
Proof: We just have to check the axioms for a topology
 The empty set is a variety. For this is equals to
Z(k[x_{1},...,x_{n}])
as no point is a zero of the constant polynomial 1.

The whole set k^{n} is a variety. For this equals
Z({0}), by definition.

A finite union of closed sets is closed. By induction we just have to think
about two sets Z(S) and Z(T). As above we may suppose
that the subsets S,T are ideals.
Consider Z(ST). Since ST is a subset of S and
T we have Z(S) U Z(T) is a subset of Z(ST).
Let's show the reverse inclusion. Suppose that
a is in Z(ST) and that WLOG
it is not in Z(S).
Since a is killed by ST then it is killed
by every product fg with f in S and g
in T. But if f(a)g(a)=0 then either f(a)=0 or
g(a)=0. Since a is not in Z(S) there exists
f in S such that f(a) is nonzero. Thus
a is in Z(T), as required.

An arbitrary intersection of closed sets is closed.
For suppose that we have a family Z(S_{i}) of closed sets.
As usual we can assume that each S_{i} is an ideal.
Let S be the sum of the ideals S_{i}, itself
an ideal. Then I claim that Z(S) is the intersection of
the Z(S_{i}). Since each S_{i} is a subset
of S we see that Z(S) is contained in the intersection
of the Z(S_{i}). On the other hand, suppose that
a is in the intersection. Then a is killed by the polynomials
in each S_{i} so it is killed by any finite linear combination
of such polynomials, that is a is in Z(S).