## Algebraic Geometry, Zariski Topology

### Zariski Topology

Zariski
(biography)
enjoyed turning algebraic structures into topological spaces.
For instance,
the prime ideals in an arbitrary ring,
even a noncomutative ring,
can act as points in a topological space
known as spec R.
Here is a different zariski topology, placed on the vector space Ln.
Let L/K be a field extension, and let Ln be a vector space
associated with the variables x1 through xn.
A region R in Ln is an algebraic set if it is S′ for some set of functions S.
In the last section we showed such a region is closed.
That is, R′′ = R.
Conversely, assume R is closed.
Now R′′ = R, and if S = R′, then R = S′, and R is an algebraic set.
In other words, R is closed iff it is an algebraic set.

So far we have not specified the universal set of functions.
It is customary to use polynomials over L.
At other times we will restrict attention to polynomials with coefficients in K.
Regardless of the universal set,
R is closed iff it is an algebraic set.
Let's see if this definition makes topological sense.

The function 1 shows the empty set is closed, and the function 0 shows all of Ln is closed.
(I'll assume the universal set is a ring, hence it contains 0 and 1.)

Consider the intersection of arbitrarily many closed regions.
Let S hold all the corresponding functions, for all the regions,
and the intersection becomes an algebraic set.
In other words, the intersection of closed sets is closed.

To take the union of R1 and R2,
multiply all the functions in S1 by all the functions in S2.
The resulting functions vanish on R1∪R2,
hence the finite union of closed regions is closed.
We have a valid topology.

Any point in Kn is closed via the equations x1 = c1,
x2 = c2, x3 = c3, etc.
If K = L then the space is half hausdorff.

Take the finite union of points,
and any finite subset of Kn is closed.

In L1,
which is a pid,
S′ is the zeros of the gcd of the polynomials in S.
If L = K,
a region in L1 is closed iff it is finite (or the entire space).
This is the cofinite topology.

### Complex Space

If **C** is the field of complex numbers, we already have a topology for **C**n, based on the distance metric.
The zariski topology is consistent with the metric topology, though it is weaker.
Let R be a region in **C**n that is closed under the zariski topology.
Thus the functions of S vanish on R.
Our universe of functions could be those functions that are complex analytic, or any subring thereof, such as the polynomials.
Let p be a point not in R, and evaluate each function at p.
Some function f in S is nonzero at p, and since f is continuous, it is nonzero throughout a neighborhood about p.
This open neighborhood lies outside of R, hence R complement is open, and R is closed.

Conversely, let R be an infinite set of isolated points in **C**.
We showed above that **C**1 has the cofinite topology,
hence R is metric closed, but not zariski closed.

### Valuation Metrics

Let K be a field with a valuation metric,
such as the p-adic numbers,
or the fraction field of any pid, or the completion thereof.
The key to the earlier proof is the continuity of the functions in S.
Since addition and multiplication are continuous,
polynomials are continuous,
and within this context the zariski topology is consistent with the metric topology, but weaker,
as shown by the powers of p in K1,
an infinite set of isolated points that is
metric closed, but not zariski closed.
Closed sets correspond to ideals, and a smaller closed set comes from a larger ideal.
We already said the ring of polynomials
is noetherian.
This makes Kn a noetherian space.
It is also compact,
as is every subspace thereof, be it open, closed, or otherwise.