Definition 3.5.1. An analytic geometry is an analytic category A satisfying the following axioms: 
(Axiom 4) A is perfect (i.e. any intersection of strong subobjects exists). 
(Axiom 5) A is reducible (i.e. any non-initial object has a non-initial reduced strong subobject). 
(Axiom 6) A is locally disjunctable (i.e. any strong subobject is an intersection of disjunctable strong subobjects). 

Let A be an analytic geometry. 

Proposition 3.5.2. Any object has a radical. 

Proof. Since any intersection of strong subobject exist, the lattice R(X) of strong subobjects of an object X is complete, and the join of reduced strong subobjects is the reduced model red(X) of X by (3.1.6), and red(X) is unipotent by (3.1.8) because the category is reducible by (3.5.1.b). ––n 

Proposition 3.5.3. Suppose an object X is a join of two strong subobjects U and V. Then {U, V} is a unipotent cover on X. 

Proof. First consider the special case that U and V are disjunctable. We have U^c Ç V^c = (U Ú V)^c = X^c = 0 by (1.5.4), so {U, V} is a unipotent cover on X. 
Next we consider the general case. Suppose t: T ® X is a map that is disjoint with U and V. We prove that T is an initial map by contradiction. So we assume that T is non-initial. Replacing T by its reduced model red(T) if necessary we may assume that T is reduced. Since U is an intersection of disjunctable strong subobjects, there is a disjunctable strong subobject U₁ containing U such that t does not factor through U₁. Then t^-1(U₁) is a proper strong subobject of T. As T is reduced, t^-1(U₁) is not unipotent, thus there is a non-initial map s: W ® T which is disjoint with t^-1(U₁). The non-initial map t_°s is disjoint with U₁. Since t is disjoint with V, so is t_°s. Applying the same procedure to t_°s and V we can find a disjunctable strong subobject V₁ containing V and a non-initial map r: R ® W such that t_°s_°r is disjoint with V. We obtain a non-initial map t_°s_°r which is disjoint with U₁ and V₁, but this is absurd, as U₁ Ú V₁ = U Ú V = X implies that {U₁, V₁} is a unipotent cover on X, by the proceeding case. ––n 

Proposition 3.5.4. Suppose f: Y ® X is a map such that Y is reduced and X is the join of two strong subobjects U and V. Then Y is the join of f^-1(U) and f^-1(V). 

Proof. By (3.5.3) {U, V} is a unipotent cover on X, so {f^-1(U), f^-1(V)} is a unipotent cover on Y, and f^-1(U) Ú f^-1(V) is a unipotent strong subobject of Y. Since Y is reduced, we have f^-1(U) Ú f^-1(V) = Y as desired. ––n 

Proposition 3.5.5. Suppose U and V are two strong subobjects of an object X. Then red(U Ú V) = red(U) Ú r(V). 

Proof. Since A is reducible, by (3.1.8) the reduced model red(U Ú V) of U Ú V is the unipotent reduced subobject of U Ú V. Thus it suffices to prove that red(U) Ú red(V) is the unipotent reduced subobject of U Ú V. First red(U) Ú red(V) is reduced by (3.1.3.d). From (3.5.3) we know that {U, V} is a unipotent cover on U Ú V. If t: T ® U Ú V is a non-initial map, t is not disjoint with at least one of U and V, thus t is not disjoint with at least one of red(U) and red(V) by (3.1.9.b), so red(U) Ú red(V) is a unipotent subobject of U Ú V as required. ––n 

Remark 3.5.6. Denote by S(X) the set of reduced strong subobjects of X. 
(a) S(X) is closed under arbitrary join in the complete lattice R(X) of strong subobjects by (3.1.3)(d). Thus S(X) is a complete lattice. 
(b) The mapping red: R(X) ® S(X) is a right adjoint of the inclusion S(X) ® R(X). 
(c) The mapping red: R(X) ® S(X) preserves meets and finite joins by (a), (b) and (3.5.5). 
(d) If f: Y ® X is a map, we have a mapping red(f)^-1: S(X) ® S(Y) sending each reduced strong subobject U in X to the reduced model of f^-1(U). It has a left adjoint which coincides with the restriction of f⁺¹ on S(Y), denoted by red(f)⁺¹ . 

Proposition 3.5.7. (a) For any object X the dual S(X)^op of the lattice S(X) is a locale isomorphic to the locale F(X) of analytic sieves. 
(b) The functor Q sending each object X to S(X)^op and each map f: Y ® X to red(f)^-1 is equivalent to the analytic topology F_A on A. 

Proof. (a) Since any strong mono u is an intersection of disjunctable monos, the sieve Øu is analytic by (3.1.10). We obtain a order-reversing map S(X) ® F(X) sending each reduced strong subobjects u to Øu. It is injective by (3.1.9.c) and surjective by (3.1.10). Thus S(X)^op is isomorphic to the locale F(X). 
(b) Consider any map f: Y ® X. If u: U ® X is a reduced strong mono we have f^-1(Øu) = Øf^-1(u) = Ør(f^-1(u)) = Ø(r(f)^-1(u)). This shows that the isomorphism in (a) is functorial, thus the functor Q is equivalent to the framed topology F_A. ––n 

Proposition 3.5.8. The following are equivalent for a non-initial object X. 
(a) Any two non-initial fractions are disjoint with each other. 
(b) X is quasi-primary. 
(c) The reduced model of X is integral. 

Proof. Clearly (a) implies (b). 
Suppose X is a quasi-primary. We prove that red(X) is integral. Since red(X) is reduced, by (3.2.6) it suffices to prove that red(X) is irreducible. Assume red(X) is the join of two proper strong subobjects U and V; then U and V are not unipotent subobject of X. Since A is locally disjunctable, we can find two non-initial analytic monos H and G which are disjoint with U and V respectively. Then H Ç G is non-initial because X is quasi-primary. Thus there is a non-initial map t: T ® X in H Ç G, which is disjoint with U and V. But by (3.5.3) U, V is a unipotent cover on red(X); since red(X) is a unipotent subobject of X, U, V is a unipotent cover on X. We obtain a contradiction. This shows that X is integral. This shows that (b) implies (c). 
Finally we prove that (c) implies (a). Suppose red(X) is integral. Consider two non-initial fractions U and V of X. Since red(X) is unipotent, U Ç red(X) and V Ç red(X) are noninitial fractions of red(X). Since red(X) is integral, therefore irreducible by (3.2.6), (U Ç red(X)) Ç (V Ç red(X)) = U Ç V Ç red(X) is non-initial. It follows that U Ç V is non-initial. Thus (a) holds as desired. ––n 

Proposition 3.5.9. Any singular mono in a strict analytic geometry is analytic. 

Proof.. We only need to prove that any singular mono u: U ® X is coflat. Since the pullback of any singular mono is a singular mono, it suffices to prove that any singular mono is precoflat. By (3.1.10 ) u has a analytic cover u_i: U_i ® U, where each U_i is an analytic subobject of X. Suppose t: T ® X is an epic map. Let s: S ® U be the pullback of t along u, and let s_i: S_i ® U_i be the pullback of t along u_i. Since each u_i is analytic and t is epic, each s_i is epic. Suppose s factors through a strong subobject V of U. Then each s_i factors through the strong subobject V Ç U_i of U_i. But s_i is epic implies that V Ç U_i = U_i. Thus V contains the analytic cover U_i. Since by assumption A is strict, U is the colimit of analytic subobjects {U_i Ç U_j}, so {U_i} is not contained in any proper subobject of U. Thus we have U = V. Therefore t is epic by (1.1.3.e). This shows that u is precoflat. ––n 

Proposition 3.5.10. Suppose X = U Ú V is the join of two strong subobjects and U Ç V = 0 in a strict analytic geometry. Then the induced map U + V ® X is an isomorphism. 

Proof. We know from (3.5.3) that (U, V) is a unipotent cover on X. First we show that U and V are analytic mono. Suppose V is an intersection of disjunctable strong subobjects {V_i} and let U_i = (V_i)^c. According to (3.1.10) we have ØØ{U_i} = ØV, so {U_i} together with V form a unipotent cover on X. Since U Ú V = X and each U_i is coflat, we have 

Proposition 3.5.11. (Categorical form of Chinese Remainder Theorem) Suppose U₁, U₂, ..., U_n are strong subobjects of an object X in a strict analytic geometry, such that U_i, U_j are disjoint for all i ¹ j, then the induced map S U_i ® Ú U_i (where S U_i is the sum of {U_i}) is an isomorphism. 

Proof. This follows from (3.5.10) by induction on n. ––n 
 
The above theorem is inspired in turn by the following General form of the Chinese remainder theorem given in [Eisenbud 1995, p.79]: 

Let R be a ring, and let Q₁, ..., Q_n be ideals of R such that Q_i + Q_j = R for all i ¹ j. Then R/(Ç_iQ_i) is isomrophic to Õ_iR/Q_i.