Algebraic Geometry

For simplicity, we will assume \(k\) is algebraically closed from now on. This is not really needed for many of the statements that follow, but it will be tedious to keep track of which ones really do, and in some cases where it is not needed, the proofs are harder without that assumption.

Recall that a topology on a set \(X\) is a specified collection of subsets, known as the open subsets, such that (1) \(X\) is an open subset of \(X\text{,}\) (2) \(\emptyset\) is an open subset of \(X\text{,}\) (3) the intersection of any two open subsets it open, and (4) an arbitrary union of open subsets is open. Such a specified collection of open subsets makes \(X\) into a topological space. In any topological space, a closed subset is any subset that is the complement of an open one. One may equivalently describe a topology as a collection of specified “closed subsets” such that (1) \(X\) is an closed subset of \(X\text{,}\) (2) \(\emptyset\) is an closed subset of \(X\text{,}\) (3) the intersection of an arbitrary collection of closed subsets is closed, and (4) the union of two closed subsets is closed. By what we have proved above about affine varieties, we may define:

Definition 1. The Zariski topology on \(\mathbb{A}^n_k\) is the topology in which a subset is closed if and only if it is an affine variety (i.e., cut out by a collection of polynomial equations). Given a variety \(X\) in \(\mathbb{A}^n_k\text{,}\) the Zariski topology on \(X\) is the subspace topology — since \(X\) is closed in \(\mathbb{A}^n_k\text{,}\) this means that a subset \(C\) of \(X\) is closed if and only if \(C\) is an affine subvariety of \(X\) (i.e., an affine variety in \(\mathbb{A}^n_k\) that is contained in \(X\)).

So, a subset \(U\) of \(\mathbb{A}^n_k\) is open iff \(U = \mathbb{A}^n_k \setminus V(J)\) for some ideal \(J\) (which we may take to be radical), and more generally a subset of an affine variety \(X\) is open iff it has the form \(X \setminus V(J)\) for some ideal \(J\) in \(A(X)\text{.}\)

Remark 1. The Zariski topology on \(\mathbb{A}^n_k\) is not Hausdorff (provided \(n > 0\)). Indeed, the intersection of any two non-empty open sets is non-empty. This is equivalent to the assertion that the union of any two proper closed subset is proper. Using the Nullstellensatz, this is equivalent to the assertion that if \(I\) and \(J\) are any two non-zero ideals of \(k[x_1, \dots, x_n]\) then \(I \cap J\) is non-zero, and that holds since \(k[x_1, \dots, x_n]\) is a domain.

Proposition 1. Every affine variety \(X\) is (quasi-)compact for the Zariski topology — that is, every open cover of \(X\) by Zariski open subsets admits a finite subcover or, equivalently, if \(\{C_\alpha\}\) is a collection of closed subsets of \(X\) such that \(\bigcap_\alpha C_\alpha= \emptyset\) then \(C_{\alpha_1} \cap \cdots \cap C_{\alpha_m} = \emptyset\) for some finite list \(\alpha_1, \dots, \alpha_m\) of indices.

Proof. We may translate the first assertion into a statement about ideals in the ring \(R = A(X)\text{:}\) We need to prove that if \(\{I_\alpha\}\) is a collection of ideals such that \(\sum_{\alpha} I_\alpha= R\) then \(I_{\alpha_1} + \cdots + I_{\alpha_m} = R\) for some finite sub-collection. This holds since \(\sum_{\alpha} I_\alpha= R\) iff \(1 \in \sum_{\alpha} I_\alpha\) iff \(1 \in I_{\alpha_1} + \cdots + I_{\alpha_m}\) for some \(\alpha_1, \dots, \alpha_m\text{.}\)

I’ll just sketch the proof of the second assertion: Let \(U\) be an open subset of an affine variety \(X\text{,}\) say \(U = X \setminus V_X(J)\) for an ideal \(J\text{.}\) We have \(J = \langle g_1, \dots, g_m \rangle\) (Hilbert Basis Theorem), and so \(U = \bigcup_{i=1}^m X \setminus V(g_i)\text{.}\) Since this is a finite union, it suffices to show each \(X \setminus V(g_i)\) is quasi-compact. So, without loss, we may assume \(U = X \setminus V(g)\) for some \(g\text{.}\) The points of \(U\) are in bijective correspondence with the maximal ideals of the ring \(A(X)[1/g]\) and the closed subsets are given by ideals of this ring. So, the same proof as in the first part applies. ◻

In the previous proof I used some notation not yet introduced:

Definition 1. Given an affine variety \(X\) and element \(f \in A(X)\) of its coordinate ring, let \(V_X(f) = \{P \in X \mid f(P) = 0\}\text{.}\) Equivalently, if \(f = \tilde{f} + I(X)\) for \(\tilde{f} \in k[x_1, \dots, x_n]\text{,}\) then \(V_X(f) = V(I(X) + \langle \tilde{f} \rangle)\text{.}\) Define \(V_X(f_1, \dots, f_c)\) for \(f_1, \dots, f_c \in A(X)\) in the evident way.

Proposition 1. Every affine variety \(X\) is noetherian — by definition, this means the poset of its closed subsets satisfies the dcc.

Proof. This holds since \(A(X)\) has the acc on (radical) ideals. ◻

Example 1. \(\mathbb{A}^1_k\) has dcc on closed subsets since the only proper closed subsets of it finite sets, and clearly every finite set has dcc on closed subsets.

Example 1. Starting with a surface such as \(X = V(z - x^2 - y^2)\) (a paraboloid), a proper closed subset of \(X\) is a curve on \(X\) or a point on \(X\) or a finite union of such. Each (irreducible) curve on \(X\) has as its only proper closed subsets finite subsets. Accepting these facts (on faith) it is clear \(X\) has dcc on closed subsets.

Definition 1. An affine variety is irreducible if it cannot be written non-trivially as a union of closed subsets — that is, \(X\) is irreducible provided whenever \(X = Y \cup Z\) with \(Y\) and \(Z\) affine subvarieties of \(X\text{,}\) we have \(Y = X\) or \(Z = X\text{.}\)

Example 1. A single plane in three-space is irreducible, but the union of a plane and a line (not contained in the plane) is reducible.

Example 1. A surface \(X = V(f(x,y,z))\) in three-space \(\mathbb{A}^3_k\) is irreducible if and only if \(f\) is irreducible or a power of an irreducible. This holds in general: The hypersurface \(X = V(f(x_1, \dots, x_n))\) in \(\mathbb{A}^n_k\) is irreducible if and only if \(f\) is irreducible or a power of an irreducible.

Exercise 1. Prove that I just asserted: Assume \(f \in k[x_1, \dots, x_n]\) is a polynomial without repeated factors (so that \(\langle f \rangle\) is a radical ideal). Prove \(X = V(f)\) is irreducible if and only if \(f\) is irreducible.

Exercise 1. Show an affine variety \(X\) is irreducible iff every non-empty open subset \(U\) of \(X\) is dense in \(X\) (i.e., the intersection of \(U\) with any other non-empty open subset of \(X\) is non-empty).

Proposition 1. An affine variety \(X\) is irreducible if and only if \(A(X)\) is an integral domain. More generally, the collection of irreducible affine subvarieties of a given affine variety is in bijective correspondence with the collection of primes ideals in its coordinate ring.

Proof. If \(X\) is not irreducible, then \(X = Y \cup Z\) for proper closed subsets \(Y\) and \(Z\) of \(X\text{.}\) It follows that \(I(X) \subsetneq I(Y)\text{,}\) \(I(X) \subsetneq I(Z)\text{,}\) and \(I(X) = I(Y) \cap I(Z)\text{.}\) So pick \(g \in I(Y) \setminus I(X)\) and \(g \in I(Z) \setminus I(X)\text{.}\) Then \(fg \in I(X)\text{,}\) proving \(I(X)\) is not prime.

Suppose \(I(X)\) is not prime. Then there are polynomials \(f\) and \(g\) neither of which is in \(I(X)\) and yet \(fg \in I(X)\text{.}\) Set \(J = I(X) + \langle f \rangle\) and \(L = I(X) + \langle g \rangle\) and \(Y = V(J)\) and \(Z = V(L)\text{.}\) Then \(Y \subsetneq X\text{,}\) \(Z \subseteq X\) and \(Y \cup Z = V(I) \cup V(L) = V(I \cdot L) = V(X)\) since \(I \cdot L = I(X)\text{.}\)

For the second assertion, we already know there is a bijective correspondence between affine subvarieties of an affine variety \(V\) and radical ideals of \(A(V)\text{,}\) given by \(W \mapsto I(W)/I(V)\text{.}\) The assertion holds since \(I(W)/I(V)\) is prime in \(A(V)\) if and only if \(I(W)\) is prime in \(k[x_1, \dots, x_n]\text{.}\) ◻

Corollary 1. Every affine variety \(X\) is a finite union of irreducible closed subsets, \(X = X_1 \cup \cdots \cup X_m\text{,}\) such that \(X_i \not\subseteq X_j\) for all \(i \ne j\text{.}\) Moreover, the list of such subsets is unique up to ordering, and are known as the irreducible components of \(X\text{.}\)

Proof. This holds since the ring \(A(X)\) is noetherian and hence has a finite number of minimal primes \({\mathfrak p}_1 \dots, {\mathfrak p}_m\) — we take \(X_i = V({\mathfrak p}_i)\text{.}\) ◻

Remark 1. In general, if \(X\) is any noetherian topological space (i.e., a space with dcc for closed subsets), then the conclusion of the previous Corollary holds. The existence is given as follows: If \(X\) is irreducible, then we may take \(m = 1\) and \(X_1 = X\text{.}\) If not, then \(X = X_1 \cup Y_1\) for two proper closed subsets \(X_1\) and \(Y_1\) of \(X\text{.}\) If both \(X_1\) and \(Y_1\) are irreducible, we are done; otherwise we can decompose at least one of them — without loss say it is \(Y_1\) that decomposes. Then \(Y_1 = X_2 \cup Y_2\) with \(X_2\) and \(Y_2\) closed a proper in \(Y_1\text{.}\) Continuing in the fashion gives a strictly descending chain \(X \supsetneq Y_1 \subsetneq Y_2 \subsetneq\) of closed subsets. By dcc, this process must stop after a finite number of steps, and thus \(X = X_1 \cup \cdots \cup X_m\) for some irreducible closed subsets \(X_1, \dots, X_m\text{.}\) If \(X_i \subseteq X_j\) for some \(i \ne j\text{,}\) delete \(X_i\) from the list.

I’ll skip a proof of uniqueness.

Example 1. Consider \(X = V(xy, xz) = V(x) \cap V(y,z)\) in \(\mathbb{A}^3_k\text{.}\) That is \(X\) is the union of a line and a plane, specifically the \(x\)-axis and the \(yz\)-plane. Then \(X\) has two irreducible components, the line and the plane.

Example 1. For \(f \in k[x_1, \dots, x_n]\text{,}\) the irreducible components of \(V(f)\) are \(V(g_1), \dots, V(g_m)\) where \(f = \prod_i g_i^{e_i}\) for \(e_i \geq 1\) and \(g_i\) irreducible.

One particular kind of Zariski open subset is particularly importent:

Exercise 1. Let \(X\) be any affine variety.

Show \(D_X(f) \cap D_X(g) = D_X(fg)\text{;}\) in particular, the intersection of any two distinguished open subsets of \(X\) is again distinguished.

Show that the collection of distinguished open subsets of \(X\) form a “basis” for the Zariski topology on \(X\text{.}\) Recall that a collection of subsets \(\mathcal B\) for a set \(T\) is a basis for a topology if:

And the topology generated by such a basis is the collection of subsets of \(T\) that are arbitrary unions of members of \(\mathcal B\text{.}\) So, you are being asked to show the above two properties hold for the collection of distinguished open sets and that every Zariski open set is a union of distinguished ones.