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.
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 are non-empty open subsets then .
- If are closed subsets whose union is X, then C = X or D = X.
Now let us go through some basic properties of irreducible topological spaces.
A non-empty open subset U of an irreducible space X is irreducible.
Let be non-empty open subsets. Then they are non-empty open subsets of X also, so . ♦
Let 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.
For any subset A of a topological space X we let be the closure of A in X. We will use the topological fact: if then .
Let be a non-empty open subset. Since Y is dense in X, is a non-empty open subset of Y. We have , but since Y is irreducible . Hence . Since Y is dense in X we have . ♦
Decide if this statement is true: let be a non-empty dense subset of X. If X is irreducible, so is Y.
Let be a continuous map. If X is irreducible, so is .
Let be non-empty open subsets. Then are non-empty open subsets of X so . Thus . ♦
If X, Y are irreducible topological spaces, so is .
Pick two non-empty open subsets of ; we wish to prove they intersect.
The open sets of the type ( open) form a basis of . Hence, we may assume the two open sets we picked were and , where and are open. Since these are all non-empty,
Let k be an algebraically closed field. Prove that if V and W are irreducible affine k-varieties so is their product . [ 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 . 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:
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 is a sequence of closed subsets of X then for some n we have .
Since the closed subsets of Spec A correspond bijectively to radical ideals, we have:
The spectrum of a noetherian ring is noetherian.
It is possible for a non-noetherian ring to have a noetherian spectrum. For example if then A has exactly one prime ideal but it is not noetherian.
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 are noetherian subspaces, then so is .
- If is continuous and X is noetherian, then so is .
An irreducible component of X is a non-empty maximal irreducible subset .
We get the following.
Every non-empty irreducible subset is contained in an irreducible component.
Since the closure of an irreducible set in a space is irreducible, any irreducible component of X is closed in X.
Let be the collection of all irreducible subsets of X containing Y, ordered by inclusion. Note that so . It suffices to apply Zorn’s lemma to this set, so we need to show that any chain has an upper bound. Hence it suffices to show is irreducible.
Suppose are disjoint non-empty open subsets. Then and for some . Since is a chain we may assume without loss of generality that . This gives , non-empty disjoint open subsets of , which contradicts irreducibility of . ♦
Find the irreducible components of under (i) the usual topology, (ii) the cofinite topology (where the only closed subsets are the finite subsets and itself).
Components of Noetherian Space
Let be irreducible components of X. If is irreducible then for some i.
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
so by irreducibility of Y we have Y = D or Y = D’. In the former case we are done since . In the latter case so by induction hypothesis for some . ♦
A noetherian space X has only finitely many irreducible components . Furthermore, for any we have
The second statement follows from lemma 1. For the first statement, let be the collection of all closed subsets of X which are not finite unions of irreducible subsets. It suffices to show is empty.
If not has a minimal element C; since C is not irreducible we can write where are closed subsets of C. Since is minimal we have . Thus each of them is a finite union of irreducible subsets, and so is C, a contradiction. ♦
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:
A noetherian ring has only finitely many minimal primes.
Let be the variety defined by the equations
Solving gives us and hence or . The first case gives the z-axis while the second case gives a hyperbola for . Hence V has two irreducible components. The corresponding minimal primes of are:
To compute the nilradical of A, use the following facts (proof: exercise).
- The nilradical of a ring quotient is exactly .
- In a UFD B, for with prime factorization , the radical of the principal ideal (b) is .
Now A is isomorphic to . Since is a UFD, the radical of A is . Hence the intersection of all prime ideals of A is so