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 Ç 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 Ç b) is non-invertible then it is contained in a prime ideal p by (4.14.a). Then a Ç 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 Ç b) is invertible. Thus the category is solvable. – ––n 

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 

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

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 Ç b)) = r(f(a)) Ç r(f(b)) = 1_X Ç 1_X = 1_X . Applying (3.8) this means that f(a Ç b) is invertible.  ––n 

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. ––n 

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 Ç (Ú^r_i b_i) Í Ú^r_i (a Ç b_i), where Ú^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 Ç (Ú^r_i b_i)) is invertible. Then f_r(a) and f_r(Ç (Ú^r_ib_i)) are invertible. Applying (5.5.b) we see that 

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 Ç b Í f^-1(p). Then f(a Ç b) Í p. Since p is radical we have r(f(a Ç b)) Í r(p) = p. By (5.3.b) we have r(f(a)) Ç r(f(b)) = r(f(a Ç b)) Í 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 Í f^-1(r(f(a)) and b Í f^-1(r(f(b)). Thus a or b is in f^-1(p) . n 

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 Ç b) = 1_Z by Axiom A₀. This means that a Ç b is not in f^-1(m). Thus f^-1(m) is prime. n 

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

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

Proof. This follows from (5.8). n 

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). n 

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. n