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. Proposition 4.3. (a) If a and b
are two ideals, then V(a
b) = V(a)
V(b).
Proof. (a) If p Î
V(a b)
we have p a
b. Then a or b is in p as p is prime.
Thus V(a
b) V(a)
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. Proposition 4.4. Suppose f: Y > X is an arrow. Then Spec(f): Spec(X) > Spec(Y) is a continuous map. 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. Definition 4.5. An ideal is called spatial if it is an intersection of prime ideals. Remark 4.6. (a) Any prime ideal is spatial
and any spatial ideal is radical by (3.5.a).
Proposition 4.7. Any irreducible spatial ideal is prime. 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. Proposition 4.8. Spec(X) is a sober space. 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. Definition 4.9. The intersection s(a) of all the prime ideals containing a set a of 2elements of X is called the spatial radical of a. Proposition 4.10. (a) Suppose {a_{i}} is a set of ideals of X. Then for any ideal a of X we have (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 I^{s}(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 nonterminal object has a prime ideal.
Proposition 4.13. Any spectral category is radical. Proof. Suppose f: X > T is an arrow and T is nonterminal. Then f_{*}(1_{X}) 0_{T} 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 nonterminal object has a prime ideal which is radical and proper, the category is radical. Proposition 4.14. Suppose C is
a spectral category.
Proof. (a) Any proper ideal is contained in a proper 2kernel
ker(f) of an arrow f: X > Z, and ker(f)
is contained in the inverse image of any prime ideal of Z.
Proposition 4.15. An ideal of an object in a spectral category is radical if and only if it is spatial. Proof. Suppose a is a proper radical ideal of X. For any 2element 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). Corollary 4.16. (a) In a spectral category
we have r(a) = s(a) for any ideal
a.
Proof. (a) follows from (4.15).
