Let C be any category.
Definition 4.1. Suppose p is a
proper radical ideal of an object X.
Remark 4.2. (a) If f: Y -->
X is an arrow and p is a prime ideal then f-1(p)
is a prime ideal.
Denote by Spec(X) the set of prime ideals for an object X. Any arrow f: Y --> X induces a map Spec(f): Spec(X) --> Spec(Y) sending each prime ideal p of X to the prime ideal f-1(p) of Y.
If a is any ideal of X we define the subset V(a) Spec(X) to be the set of all prime ideals which contain a.
Proof. (a) If p Î
we have p a
b. Then a or b is in p as p is prime.
V(b). The other direction is trivial.
It follows from (4.3) that the subsets of the form V(a) are the closed subsets of a topology on Spec(X), which turns Spec(X) into a topological space.
Proof. If a is any ideal of Y, then we have f-1(V(a)) = V(f(a)). Thus Spec(f) is continuous.
It follows from (4.4) that Spec is a natural contravariant functor from C to Top.
Remark 4.6. (a) Any prime ideal is spatial
and any spatial ideal is radical by (3.5.a).
Proof. Suppose p is an irreducible spatial ideal of an object X and f: Y --> X is an arrow. We prove that f-1(p) is irreducible. First f-1(p) is proper and radical since it is an intersection of proper radical ideals. Suppose a and b are two ideals of Y such that a b f-1(p). Suppose q is a prime ideal of X containing p. Then f-1(q) is a prime ideal containing f-1(p). Thus a or b is in f-1(q). It follows that f*(a) or f*(b) is in q. Thus q contains f*(a) f*(b). Since p is an intersection of such q, we see that p contains f*(a) f*(b). Since p is irreducible, f*(a) or f*(b) must be contained in p. Hence a or b is in f-1(p). This proves that f-1(p) is irreducible. Thus p is prime.
Proof. Suppose W is an irreducible closed subset of Spec(X). Let p be the intersection of all the prime ideals in V. Then W = V(p). Suppose a and b are two ideals of X and p a b. Then W = V(p) V(a Ç b) = V(a) V(b). Since W is irreducible, we must have W V(a) or W V(b). But W V(a) means that any prime ideal in W contains a. Since p is the intersection of all the prime ideals in W, we have p a. Similarly W V(b) implies that p b. Thus we have p a or p b. It follows that p is irreducible. Since p is spatial, it is prime by (4.7). Clearly p is the generic point of W. This proves that Spec(X) is sober.
(b) We only need to verify that s(f(a b)) s(f(a)) s(f(b)). Suppose p is a prime ideal containing f(a b). Then f-1(p) contains a b. Since f- 1(p) is prime, it contains a or b. Then p contains f(a) or (f(b). Thus it contains s(f(a)) or s(f(b)). It follows that p contains s(f(a)) s(f(b)).
Corollary 4.11. (a) For each object X
the set Is(X) of spatial ideals of X
is a spatial frame.
Proof. (a) and (b) follow from (4.10).
Definition 4.12. (b) A category is spectral
if any non-terminal object has a prime ideal.
Proof. Suppose f: X --> T is an arrow and T is non-terminal. Then f*(1X) 0T because T has a prime ideal p and f- 1(p) is prime, which is proper. Thus the category is unitary by (2.9.b). Since each non-terminal object has a prime ideal which is radical and proper, the category is radical.
Proof. (a) Any proper ideal is contained in a proper 2-kernel
ker(f) of an arrow f: X --> Z, and ker(f)
is contained in the inverse image of any prime ideal of Z.
Proof. Suppose a is a proper radical ideal of X. For any 2-element r of X not in a let t: X --> Z be an arrow such that t(r) is invertible and t(a) is not. Let p be a prime ideal of Z containing t(a). Then f- 1(p) is a prime ideal containing a but not r. This shows that a is an intersection of prime ideals. The other direction follows from (4.6.a).