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).
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 Uc 
Vc = (U 
V)c = Xc = 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 ts is disjoint with U1. Since
t is disjoint with V, so is ts. Applying the same
procedure to ts and V we can find a disjunctable strong subobject
V1 containing V and a non-initial map r:
R --> W such that tsr is disjoint with V. We obtain a
non-initial map tsr which is disjoint with U1
and V1, but this is absurd, as U1
V1 = U V = X
implies that {U1, V1} is a unipotent cover
on X, by the proceeding case.
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)
f-1(V) is a unipotent strong subobject of Y.
Since Y is reduced, we have f-1(U)
f-1(V) = Y as desired.
Proposition 3.5.5. Suppose U
and V are two strong subobjects of an object X. Then red(U
V) = red(U)
r(V).
Proof. Since A is reducible, by (3.1.8)
the reduced model red(U
V) of U V is the
unipotent reduced subobject of U
V. Thus it suffices to prove that red(U) Ú
red(V) is the unipotent reduced subobject of U
V. First red(U)
red(V) is reduced by (3.1.3.d).
From (3.5.3) we know that {U, V} is a unipotent
cover on U V. If t:
T ® U
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
V as required.
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 
(X)
of analytic sieves.
(b) The functor Red sending each object X to S(X)op
and each map f: Y --> X to red(f)-1
is equivalent to the analytic topology A
on A.
Proof. (a) Since any strong mono u is an intersection
of disjunctable monos, the sieve 
u
is analytic by (3.1.10). We obtain a order-reversing
map S(X) --> (X)
sending each reduced strong subobjects u to u.
It is injective by (3.1.9.c) and surjective
by (3.1.10). Thus S(X)op
is isomorphic to the locale (X).
(b) Consider any map f: Y --> X. If u: U -->
X is a reduced strong mono we have f-1(u)
= f-1(u)
= r(f-1(u))
= (r(f)-1(u)).
This shows that the isomorphism in (a) is functorial, thus the functor
Red is equivalent to the framed topology A.
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 G is non-initial
because X is quasi-primary. Thus there is a non-initial map t:
T --> X in H 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
red(X) and V
red(X) are noninitial fractions of red(X). Since
red(X) is integral, therefore irreducible by (3.2.6),
(U red(X))
(V red(X))
= U V
red(X) is non-initial. It follows that U
V is non-initial. Thus (a) holds as desired.
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 ui: Ui
--> U, where each Ui 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 si:
Si --> Ui be the pullback of t along ui.
Since each ui is analytic and t is epic, each
si is epic. Suppose s factors through a strong
subobject V of U. Then each si factors
through the strong subobject V
Ui of Ui. But si is
epic implies that V Ui
= Ui. Thus V contains the analytic cover Ui.
Since by assumption A is strict, U is
the colimit of analytic subobjects {Ui
Uj}, so {Ui} 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.
Proposition 3.5.10. Suppose X =
U V is the join of two
strong subobjects and U
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 {Vi} and let Ui = (Vi)c.
According to (3.1.10) we have {Ui}
= V, so {Ui}
together with V form a unipotent cover on X. Since U
V = X and each Ui is coflat, we have
X = Ui (U
V) = (Ui
U) (Ui
V) = Ui
U. 
{Ui}.
Thus {t-1(Ui)} is an analytic cover
on T contained in t-1(U). Since the analytic
geometry is strict, there is no proper subobject of T containing
each t-1(Ui). This implies that T
= t-1(U). Thus t factors through U.
It follows that U = Vc. Similarly V = Uc.
We have proved that U and V are analytic subobject. Thus
(U, V) is an analytic cover on X. As the analytic geometry
is strict, X is the smallest subobject containing U and V.
But by (1.3.9) U + V --> X is a
mono which is the smallest subobject containing U and V.
This shows that U + V --> X is an isomorphism.
Proposition 3.5.11. (Categorical form
of Chinese Remainder Theorem) Suppose U1, U2,
..., Un are strong subobjects of an object X in a
strict analytic geometry,
such that Ui, Uj are disjoint for all i 
j, then the induced map 
Ui --> Ui
(where Ui
is the sum of {Ui}) is an isomorphism.
Proof. This follows from (3.5.10) by induction
on 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, ..., Qn
be ideals of R such that Qi + Qj
= R for all i
j. Then R/(iQi)
is isomrophic to 
iR/Qi.