5. Solvable Categories

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).
(b) For any arrow f: X --> Z and any two radical ideals a and b of X, we have 
r(f(a))  r(f(b)) = r(f(a  b)).

Proof. Note that (a) is a special case of (b) with f = 1X
(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( b)).
It suffices to see that if g: Z --> W is an arrow and g(f(a)  f(b)) is invertible, then g(f(a  b)) is invertible. But g(f(a)  f(b)) is invertible implies that g(f(a)) = (gf)(a) and g(f(b)) = gf(b) are invertible, i.e. (gf)(a  b) = g(f(a  b)) is invertible.   

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)) = 1X  1X = 1X . Applying (3.8) this means that f( b) is invertible.   

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

Proposition 5.5. (a) r: I(X) --> Ir(X) preserves joins and finite meets in a solvable radical category. 
(b) fr: Ir(Y) --> Ir(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) Ir(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 Ir(X) is a complete lattice, we verify the infinite distributive law for Ir(X). Suppose a and {bi}are radical ideals of X. We only need to verify that (ri bi)  ri ( bi), where r is the join operation in Ir(X), because the other direction is trivial. Consider a morphism f: Y --> X such that fr( (ri bi)) is invertible. Then fr(a) and fr( (ribi)) are invertible. Applying (5.5.b) we see that 

fr(ri bi)ri fr( bi)ri( fr(a)  fr(bi))ri fr(bi) = fr( (ri bi)
is invertible. Since ri ( bi) is reduced, this implies that (ri bi)  ri ( bi). 
(b) follows from (a) and (5.5.b).  

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 = 1X = f*(b) + m. Thus (tf)*(a) + t*(m) = 1Z = (tf)*(b) + f*(m), i.e. (tf)*(a) = 1Z = (tf)*(b). Hence (tf)*(a  b) = 1Z 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 (1X, 1X). 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.  
 

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