5. Solvable Categories
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.
(ra)
(rb) = (ra
b).f: X --> Z and any two radical ideals a
and b of X, we have
(rf(a))
(rf(b)) = (rf(a
b)).
(a) We only need to verify that (ra)
(rb) Í (ra
b). It suffices to see that if f: X --> Z is an
arrow and f((ra)
(rb)) is invertible, then f(a
b) is invertible. But f((ra)
(rb)) is invertible implies that f((ra))
and f((rb)) are invertible, i.e. f(a)
and f(b) are invertible, which is equivalent to that f(a
b) is invertible by Axiom A_{0} .
(b) Applying (a) we only need to verify that (rf(a))
(rf(b)) = (rf(a)
(f(b)) Í (rf(a
b)).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.
f(a
b)) = (rf(a))
(rf(b)) = 1_{X}
1_{X} = 1
. Applying (3.8) this means that _{X}f(a
b) is invertible.
Let 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 (rf(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 preserves join and _{r}f
preserves meet. Applying (3.6) and (5.3)
we obtain
^{-1}
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.
X) is a frame for any object
X.
(b) I may be viewed as a functor from C to the category
of frames.
X) is a complete
lattice, we verify the infinite distributive law for I(^{r}X).
Suppose a and {b}are radical ideals of
_{i}X. We only need to verify that a
(^{r}_{i}
b)_{i} (^{r}_{i}
a b),
where _{i} is
the join operation in ^{r}I(^{r}X), because the
other direction is trivial. Consider a morphism f: Y --> X
such that f(_{r}a
(^{r}_{i}
b)) is invertible. Then _{i}f(_{r}a)
and f(_{r} (^{r}_{i}b))
are invertible. Applying (5.5.b) we see that
_{i}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}(^{r}_{i}
a b)
is reduced, this implies that _{i}a
(^{r}_{i}
b)_{i} (^{r}_{i}
a b).
_{i}(b) follows from (a) and (5.5.b).
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 (rf(a
b)) (rp)
= p. By (5.3.b) we have (rf(a))
(rf(b)) = (rf(a
b)) p.
But p is prime, thus (rf(a)) or (rf(b))
is in p. Thus f(^{-1}(rf(a))
or f(^{-1}(rf(b)) is in f(^{-1}p).
But a f(^{-1}(rf(a))
and b f(^{-1}(rf(b)).
Thus a or b is in f(^{-1}p) .
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
by Axiom A_{Z}_{0}. This means that a
b is not in f(^{-1}m). Thus f(^{-1}m)
is prime.
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). Then (_{X},
1_{X}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. |