Math 631, Lecture 3

Notes by Bryden Cais

Sept. 11, 2002

As usual, let K be an algebraically closed field and k ⊂ K a subfield. Let X, Y be affine k varieties and let X(K) denote the K points of X. We say that X ⊂ Y if X(K) ⊂ Y (K) for all k algebras K. Denote by I(X) the ideal of all polynomials vanishing on X. Clearly,

I(X) = rad(I(X))

since if f^{N} vanishes on V so does f (since K is a field).

Definition 1. An ideal I ⊂ k[T] is called radical if rad(I) = I.

Observe that for any radical ideal I, the ring k[T]/I is reduced, that is, has no nilpotent elements. For any ideal I, define V (I) := Sol(I, K).

Theorem 1. There is a one to one correspondence between radical ideals I and algebraic k sets V given by

I ↦ V (I)
V ↦ I(V).

Proof. It is not difficult to see that V ↦ I(V) is an injective mapping of k varieties into the set of radical ideals: if V_{1} ≠ V_{2}, there exists x ∈ V_{1} not in V_{2}. If V_{i} is the zero set of {f_{1i}, …, f_{m(i)i}} for i = 1, 2, then there exists some n such that f_{n2}(x) ≠ 0. Consequently, f_{n2} ∉ I(V_{2}). Indeed, this argument shows that V (I(V)) = V for any variety V. Now Hilbert’s Nullstellensatz tells us that V (I) = V (J) if and only if rad(I) = rad(J), which shows that the reverse mapping I ↦ V (I) is injective since I is radical. Consequently I(V (I)) = I and we are done.

Theorem 2. Let K be algebraically closed. Then the maximal ideals of K[T_{1}, …, T_{n}] are in bijective correspondence with points of K^{n}.

Proof. Let M be a maximal ideal of K[T] and suppose that M ≠ (1). Then V (M) is nonempty (since K is algebraically closed), so pick x = (c_{1}, c_{2}, … c_{n}) ∈ V (M). Now let I = (T_{1} − c_{1}, …, T_{n} − c_{n}). Clearly I is a maximal ideal since K[T]/I_{x} is a field (each T_{i} is sent to c_{i} so the image is just K). Moreover, we see that x ∈ V (I) and that if f ∉ M then f does not vanish at x (if it did than every polynomial would vanish at x since M is maximal). Therefore, we have I ⊂ M and since I is maximal, I = M and we are done.

Theorem 2 is a very important result: it enables us to forget about the ambient space K^{n} and work only with the ring k[T_{1}, …, T_{n}] and its maximal ideals. We shall see later how to exploit this to the fullest extent possible.

We now define a topology on K^{n} called the k-Zariski topology by defining the closed sets to be the algebraic k sets. Clearly K^{n} and ∅ are closed. We must verify that arbitrary intersections and finite unions are closed.

1. Let S be a finite index set and let V_{s} for s ∈ S be algebraic k sets and let I_{s} = I(V_{s}). Define

I = ∏_{s∈S} I_{s}. Then we claim that

V (I) = ⋃_{s∈S} V_{s}.

To see this, observe that if x ∈ V_{s} for some s then every f ∈ I_{s} vanishes at x, so any finite sum of products of elements of all the I_{r} vanishes at x (since one of the factors must be in I_{s}); i.e. x ∈ V (I). Conversely, let x ∈ V (I) and suppose that x ∉ V_{s} for all s. Then for each s there exists f_{s} such that f_{s}(x) ≠ 0, whence f = ∏_{s∈S} f_{s}(x) ≠ 0, but f ∈ I, a contradiction.

2. Now let S be an arbitrary index set and let V_{s}, I_{s} be as above. Set I = ∑_{s∈S} I_{s}. Now we

claim that

⋂_{s∈S} V_{s} = V (I).

Let x ∈ V_{s} for all s. Then I_{S} vanishes at x for each s, so that I also vanishes at x; that is, x ∈ V (I). Conversely, since I_{s} ⊂ ∑_{s∈S} I_{s}, we see that V (I) ⊂ V (I_{s}) (since the V operator is inclusion reversing) and hence that V (I) ⊂ ⋂ V_{s}.

It is not difficult to see that in general, the Zariski topology is not Hausdorff: when n = 1 for example, the open sets are just the complements of finite sets, so that it is impossible to separate points.