## 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.

# 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.

It 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 $f(X)$.

# 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).

# 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. ♦

# 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

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

This entry was posted in Advanced Algebra and tagged , , , , . Bookmark the permalink.