Definition 5.1. A unitary category is called solvable if the following axiom is satisfied: 
Axiom A₀: If f: Y --> X is an arrow and a and b are two ideals of Y such that f(a) and f(b) are invertible then f(a  b) is invertible. 

Proposition 5.2. Any spectral category is solvable. 

Proof. Since any spectral category is unitary by (4.13), we only need to verify Axiom A₀. Suppose a and b are two ideals of an object Y and f: Y --> X is an arrow such that f(a) and f(b) are invertible. If f(a  b) is non-invertible then it is contained in a prime ideal p by (4.14.a). Then a  b is contained in f^-1(p). Since f^-1(p) is prime, it contains a or b, which means that p contains f(a) and f(b). But this is absurd as p is proper while f(a) and f(b) are invertible. This implies that we must have f(a  b) is invertible. Thus the category is solvable. 

Proposition 5.3. Let C be a solvable radical category. Then 
(a) For any two ideals a and b of an object X we have 

r(a)  r(b) = r(a  b).

r(f(a))  r(f(b)) = r(f(a  b)).

Proof. Note that (a) is a special case of (b) with f = 1_X. 
(a) We only need to verify that r(a)  r(b) Í r(a  b). It suffices to see that if f: X --> Z is an arrow and f(r(a)  r(b)) is invertible, then f(a  b) is invertible. But f(r(a)  r(b)) is invertible implies that f(r(a)) and f(r(b)) are invertible, i.e. f(a) and f(b) are invertible, which is equivalent to that f(a  b) is invertible by Axiom A₀ . 
(b) Applying (a) we only need to verify that 

r(f(a))  r(f(b)) = r(f(a)  (f(b)) Í r(f(a  b)).

  f

 

  f

 

 

Proposition 5.4. Suppose C is a radical category. Then Axiom A₀ is equivalent to (5.3.b). 

Proof. We only need to verify that (5.3.b) implies Axiom A₀ because the other direction is given by (5.3). Suppose f: Y --> X is an arrow and a and b are two ideals of Y such that f(a) and f(b) are invertible. Then (5.3.b) implies that r(f(a  b)) = r(f(a))  r(f(b)) = 1_X  1_X = 1_X . Applying (3.8) this means that f(a  b) is invertible.   

Let C be a solvable radical category. For any object X consider the set I^r(X) of radical ideals of X. Suppose f: Y --> X is an arrow. Then f induces a mapping f_r: I^r(Y) --> I^r(X) sending each radical ideal a of Y to the radical ideal r(f(a)) of X, which is the left adjoint of the mapping f^-1: I^r(X) --> I^r(Y) sending each radical ideal b of Y to the radical ideal f^-1(b) of Y. Thus f_r preserves join and f^-1 preserves meet. Applying (3.6) and (5.3) we obtain 

Proposition 5.5. (a) r: I(X) --> I^r(X) preserves joins and finite meets in a solvable radical category. 
(b) f_r: I^r(Y) --> I^r(X) preserves joins and finite meets for any arrow f: Y --> X in a solvable radical category. 

Proposition 5.6. Suppose C is a solvable radical category. Then 
(a) I^r(X) is a frame for any object X. 
(b) I may be viewed as a functor from C to the category of frames. 

Proof. (a) Since I^r(X) is a complete lattice, we verify the infinite distributive law for I^r(X). Suppose a and {b_i}are radical ideals of X. We only need to verify that a  (^r_i b_i)  ^r_i (a  b_i), where ^r is the join operation in I^r(X), because the other direction is trivial. Consider a morphism f: Y --> X such that f_r(a  (^r_i b_i)) is invertible. Then f_r(a) and f_r( (^r_ib_i)) are invertible. Applying (5.5.b) we see that 

f_r(^r_i a  b_i) = ^r_i f_r(a  b_i) = ^r_i( f_r(a)  f_r(b_i)) = ^r_i f_r(b_i) = f_r( (^r_i b_i)

a 

a 

  ^r_i

a 

Proposition 5.7. Any irreducible ideal p of an object in a solvable radical category is prime. 

Proof. Since p is a proper radical ideal of X, f^-1(p) is a proper radical ideal of Y. Now consider two ideals a and b of Y such that a  b  f^-1(p). Then f(a  b)  p. Since p is radical we have r(f(a  b))  r(p) = p. By (5.3.b) we have r(f(a))  r(f(b)) = r(f(a  b))  p. But p is prime, thus r(f(a)) or r(f(b)) is in p. Thus f^-1(r(f(a)) or f^-1(r(f(b)) is in f^-1(p). But a  f^-1(r(f(a)) and b  f^-1(r(f(b)). Thus a or b is in f^-1(p) .  

Proposition 5.8. Any maximal ideal in a solvable category is prime. 

Proof. Suppose m is a maximal ideal of an object X. Then there is an arrow t: X --> Z such that m = ker(t). Suppose f: Y --> X is an arrow and a, b are two ideals of Y not contained in the proper ideal f^-1(m). Then f_*(a) + m = 1_X = f_*(b) + m. Thus (tf)_*(a) + t_*(m) = 1_Z = (tf)_*(b) + f_*(m), i.e. (tf)_*(a) = 1_Z = (tf)_*(b). Hence (tf)_*(a  b) = 1_Z by Axiom A₀. This means that a  b is not in f^-1(m). Thus f^-1(m) is prime.  

Corollary 5.9. A solvable category is spectral if any non-terminal object has an irreducible (or maximal) ideal.  

Proposition 5.10. Any simple object in a solvable category is integral. 

Proof. This follows from (5.8). 

Definition 5.11. An simple and integral object is called a field . 

Proposition 5.12. (a) A category is spectral if for any non-terminal object there is an arrow to a integral object (or field). 
(b) A solvable category is spectral if any non-terminal object is the domain of an arrow to a simple object. 

Proof. (a) If the condition is satisfied then any non-terminal object has a prime ideal because the 2-kernel of any arrow to a integral object is prime. 
(b) This follows from (a) and (5.10).  

Proposition 5.13. Suppose C is a category in which any pair of parallel arrows has a coequalizer and any such coequalizer has a kernel pair. Suppose T is any object and (X, t) is an object in T/C. 
(a) The natural projection F: T/C --> C induces a natural bijection from I((X, t)) to I(X). 
(b) If C is unitary (resp. radical, resp. solvable, resp. spectral), then T/C is also unitary (resp. radical, resp. solvable, resp. spectral). 

Proof. (a) If a is an ideal of (X, t) denote by F_*(a) the ideal of X generated by all the 2-elements (F(r), F(s)) where (r, s) is any 2-element in a. We obtain a map F_*: I((X, t)) --> I(X), which is clearly injective. Consider any ideal b of X. For any 2-element (r, s) in b let (r', s'): Z --> X be the kernel pair of the coequalizer of (r, s). Denote by d: X --> Z be the diagonal arrow induced by the identities (1_X, 1_X). Then (Z, dt) is an object of T/C and (r', s') may be viewed as a 2-element of (X, t). Denote by a' the ideal generated by all such (r', s') . It is easy to see that F_*(a') = b. This shows that F_* is a natural bijection. 
(b) follows from (a) immediately.