Commutative Algebra 36

In this article, we will study the topology of Spec A when A is noetherian. For starters, let us consider irreducible topological spaces in greater detail.

Irreducible Spaces

Recall that an irreducible topological space is a non-empty space X satisfying any of the three equivalent conditions.

  • Any non-empty open subset of X is dense.
  • If U, V\subseteq X are non-empty open subsets then U\cap V\ne\emptyset.
  • If C, D\subseteq X are closed subsets whose union is X, then CX or DX.

Now let us go through some basic properties of irreducible topological spaces.

Proposition 1.

A non-empty open subset U of an irreducible space X is irreducible.

Proof

Let V, W\subseteq U be non-empty open subsets. Then they are non-empty open subsets of X also, so V\cap W \ne\emptyset. ♦

Proposition 2.

Let Y\subseteq X be a non-empty dense subset of X. If Y is irreducible so is X.

In other words, the closure of an irreducible subspace is irreducible.

Proof

For any subset A of a topological space X we let \mathrm{cl}_X(A) be the closure of A in X. We will use the topological fact: if A\subseteq Y \subseteq X then \mathrm{cl}_Y(A) = \mathrm{cl}_X(A) \cap Y.

Let U\subseteq X be a non-empty open subset. Since Y is dense in X, V := U\cap Y is a non-empty open subset of Y. We have \mathrm{cl}_Y(V) = \mathrm{cl}_X(V) \cap Y, but since Y is irreducible \mathrm{cl}_Y(V) = Y. Hence \mathrm{cl}_X(V) \supseteq Y. Since Y is dense in X we have \mathrm{cl}_X(V) = X. ♦

Exercise A

Decide if this statement is true: let Y\subseteq X be a non-empty dense subset of X. If X is irreducible, so is Y.

Proposition 3.

Let f:X\to Y be a continuous map. If X is irreducible, so is f(X).

Proof

Let U, V\subseteq f(X) be non-empty open subsets. Then f^{-1}(U), f^{-1}(V) are non-empty open subsets of X so f^{-1}(U\cap V) = f^{-1}(U) \cap f^{-1}(V) \ne \emptyset. Thus U\cap V \ne\emptyset. ♦

Proposition 4.

If X, Y are irreducible topological spaces, so is X\times Y.

Proof

Pick two non-empty open subsets of X\times Y; we wish to prove they intersect.

The open sets of the type U\times V (U\subseteq X, V\subseteq Y open) form a basis of X\times Y. Hence, we may assume the two open sets we picked were U_1 \times V_1 and U_2 \times V_2, where U_1, U_2 \subseteq X and V_1, V_2\subseteq Y are open. Since these are all non-empty,

U_1 \cap U_2 \ne \emptyset, V_1 \cap V_2 \ne \emptyset \implies (U_1 \times V_1) \cap (U_2 \times V_2) \ne\emptyset. ♦

Exercise B

Let k be an algebraically closed field. Prove that if V and W are irreducible affine k-varieties so is their product V\times W. [ Warning: this is not the product topology, as we noted earlier. ]

Prove that if A and B are finitely generated k-algebras which are integral domains, so is A\otimes_k B. Find a counter-example when k is not algebraically closed.

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Noetherian Topological Spaces

Following the idea of considering Spec A as the set of prime ideals of A, with closed sets corresponding to ideals, we define:

Definition.

A topological space is said to be noetherian if its collection of closed subsets forms a noetherian poset, when they are ordered by inclusion.

In other words, X is noetherian if and only if the following equivalent conditions hold.

  • Any non-empty collection of closed subsets of X has a minimal element.
  • If Y_1 \supseteq Y_2 \supseteq \ldots is a sequence of closed subsets of X then for some n we have Y_n = Y_{n+1} = \ldots.

Since the closed subsets of Spec A correspond bijectively to radical ideals, we have:

Proposition 5.

The spectrum of a noetherian ring is noetherian.

warningIt is possible for a non-noetherian ring to have a noetherian spectrum. For example if A = k[X_1, X_2, \ldots ]/(X_1^2, X_2^2, \ldots) then A has exactly one prime ideal but it is not noetherian.

Exercise C

Decide if each of the following claims is true.

  • A subspace of a noetherian topological space is noetherian.
  • A product of two noetherian topological spaces is noetherian.
  • If Y_1, Y_2 \subseteq X are noetherian subspaces, then so is Y_1 \cup Y_2.
  • If f:X\to Y is continuous and X is noetherian, then so is f(X).

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Irreducible Components

Definition.

An irreducible component of X is a non-empty maximal irreducible subset C\subseteq X.

We get the following.

Proposition 6.

Every non-empty irreducible subset Y\subseteq X is contained in an irreducible component.

Note

Since the closure of an irreducible set in a space is irreducible, any irreducible component of X is closed in X.

Proof

Let \Sigma be the collection of all irreducible subsets of X containing Y, ordered by inclusion. Note that Y\in\Sigma so \Sigma \ne \emptyset. It suffices to apply Zorn’s lemma to this set, so we need to show that any chain \Sigma'\subseteq \Sigma has an upper bound. Hence it suffices to show C := \cup_{Z\in \Sigma'} Z is irreducible.

Suppose U, V \subseteq C are disjoint non-empty open subsets. Then U\cap Z_1 \ne \emptyset and V\cap Z_2 \ne\emptyset for some Z_1, Z_2 \in \Sigma'. Since \Sigma' is a chain we may assume without loss of generality that Z_1 \subseteq Z_2. This gives U \cap Z_2, V \cap Z_2\ne\emptyset, non-empty disjoint open subsets of Z_2, which contradicts irreducibility of Z_2. ♦

Easy Exercise

Find the irreducible components of \mathbb R under (i) the usual topology, (ii) the cofinite topology (where the only closed subsets are the finite subsets and \mathbb R itself).

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Components of Noetherian Space

Lemma 1.

Let C_1, \ldots, C_k \subseteq X be irreducible components of X. If Y \subseteq C_1 \cup \ldots \cup C_k is irreducible then Y\subseteq C_i for some i.

Proof

Apply induction on k; when k = 1 there is nothing to prove so assume k > 1. Since irreducible components are closed, Y is the union of closed subsets

D := Y \cap C_k, \quad D' := Y \cap (C_1 \cup \ldots \cup C_{k-1})

so by irreducibility of Y we have YD or YD’. In the former case we are done since Y \subseteq C_k. In the latter case Y \subseteq C_1 \cup \ldots \cup C_{k-1} so by induction hypothesis Y \subseteq C_i for some 1\le i \le k-1. ♦

Proposition 7.

A noetherian space X has only finitely many irreducible components C_1, \ldots, C_k. Furthermore, for any 1\le i \le k we have

C_i \not\subseteq \cup_{j\ne i} C_j.

Proof

The second statement follows from lemma 1. For the first statement, let \Sigma be the collection of all closed subsets of X which are not finite unions of irreducible subsets. It suffices to show \Sigma is empty.

If not \Sigma has a minimal element C; since C is not irreducible we can write C = C_1 \cup C_2 where C_1, C_2 \subsetneq C are closed subsets of C. Since C\in\Sigma is minimal we have C_1, C_2 \not\in\Sigma. Thus each of them is a finite union of irreducible subsets, and so is C, a contradiction. ♦

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Geometric Interpretation

Let us interpret the above for X = Spec A. Recall (proposition 6 here) that the irreducible closed subsets of X correspond to prime ideals of A. Thus the irreducible components correspond to minimal primes of A and we have again shown that every prime ideal of A contains a minimal prime. Since a noetherian space has only finitely many irreducible components we have:

Proposition 8.

A noetherian ring has only finitely many minimal primes.

Example

Let V\subset \mathbb A^3(\mathbb C) be the variety defined by the equations

f(X, Y, Z) = X^2 - YZ, \quad g(X, Y, Z) = X^3 - Y.

Solving gives us XYZ = Y and hence Y=0 or XZ=1. The first case gives the z-axis X=Y=0 while the second case gives a hyperbola (X, Y, Z) = (X, X^3, \frac 1 X) for X\in \mathbb C - \{0\}. Hence V has two irreducible components. The corresponding minimal primes of A = \mathbb C[X, Y, Z]/(X^2 - YZ, X^3 - Y) are:

\mathfrak p_1 = (X, Y), \quad \mathfrak p_2 = (XZ - 1, Y - X^3).

To compute the nilradical of A, use the following facts (proof: exercise).

  • The nilradical of a ring quotient B/\mathfrak b is exactly r(\mathfrak b)/\mathfrak b.
  • In a UFD B, for b\in B with prime factorization b = \prod_{i=1}^r \pi_i^{e_i}, the radical of the principal ideal (b) is (\prod_{i=1}^r \pi_i).

Now A is isomorphic to \mathbb C[X, Z]/(X^2 - X^3 Z). Since \mathbb C[X, Z] is a UFD, the radical of A is (X(1-XZ))/(X^2 - X^3 Z). Hence the intersection of all prime ideals of A is (X(1-XZ)) so

\mathfrak p_1 \cap \mathfrak p_2 = (X(1-XZ)).

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