Let k be an algebraically closed field (for example C). We are going to make kn 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 kn then it becomes possible to talk about algebraic geometry.

We will be using the polynomial ring k[x1,...,xn]. If f(x)=f(x1,..., xn) in k[x1,...,xn] is a polynomial then we can evaluate it at a point a=(a1,..an) in kn to give an element of k denoted by f(a) or f(a1,...,an). If S is a nonempty set of polynomials then define

Z(S) = { a in kn : f(a)=0 for all f in S }
For example, if n=2 and S={x-y} 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 C2 or C3. Here's another example for n=2 if we take S={ xy } then this time Z(S) is the union of the x-axis and the y-axis. 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 kn 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 kn called the Zariski topology.

First notice that if I is the ideal of k[x1,...,xn] 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=f1g1 + ... + fmgm
for some fi in S and polynomials gi. Evidently then f(a)=0 and we have Z(I)=Z(S). Since k[x1,...,xn] 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 kn are the closed sets of a topology.

Proof: We just have to check the axioms for a topology

  1. The empty set is a variety. For this is equals to Z(k[x1,...,xn]) as no point is a zero of the constant polynomial 1.
  2. The whole set kn is a variety. For this equals Z({0}), by definition.
  3. 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.
  4. An arbitrary intersection of closed sets is closed. For suppose that we have a family Z(Si) of closed sets. As usual we can assume that each Si is an ideal. Let S be the sum of the ideals Si, itself an ideal. Then I claim that Z(S) is the intersection of the Z(Si). Since each Si is a subset of S we see that Z(S) is contained in the intersection of the Z(Si). On the other hand, suppose that a is in the intersection. Then a is killed by the polynomials in each Si so it is killed by any finite linear combination of such polynomials, that is a is in Z(S).

The Zariski topology is weird. I don't pretend to understand noether's writeup, above, but I have read that, on the real numbers, R, the open sets that make up Zariski topology are all those sets whose complements in R are finite, together with the empty set.

We can see that it is a topology by considering its closed sets - all finite sets. Since the intersection of any number of finite sets is itself finite and the union of any finite number of finite sets is finite, the conditions for a topology are met.

Consider the sequence x1, x2 , ..., where xn = n for any natural number, n. This is just the sequence 1, 2, 3,... etc.

A sequence, x1, x2, ..., is said to converge on a point p, in a topological space, T, if and only if there's some term in the sequence beyond which all subsequent terms are members of all open sets in T of which p is a member, in other words:

for all open sets, V in T which contain p, there is some positive natural number N such that for all xj, j > N, xj is in V.
Suppose r is a real in the Zariski topology on R, and is a member of an open set, V.

Since the complement of V is finite, it has a largest member, v. Consequently, there is a natural number, N > v, which is greater than any member of V's complement. Since xn = n, any term of our sequence where n > N is not in the complement of V, and must therefore be in V, so without specifying r we've shown the sequence converges on r. Hence, the sequence converges on any real number, r, that we care to choose, with respect to the Zariski topology on the reals.

This is enough to show the Zariski topology on the reals is not a Hausdorff space (and therefore not a metric space), since in a Hausdorff space, any convergent sequence must converge on a single point.

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