4.3 Immersions
O of stalks is surjective.
_{Y,y}
X from the associated geometric space X
of _{red}X to X is an immersion.
(c) Suppose f: X ® Y
is an immersion of local ringed spaces, then the induced morphism f:
_{red}X ® _{red}Y
of the associated geometric spaces is an immersion.
_{red}(d) Any effective morphism in the category of ringed (resp. local ringed, resp. geometric) spaces is an immersion. This is obvious for ringed and local ringed spaces. Suppose U is a subset of a geometric space,
then the effective morphism U ®
_{red}X is the composition U ®
_{red}U ® X where U ®
X is effective in LSp. Since U ®
_{red}U and U ® X are immersions,
U ® _{red}X is an
immersion by (a).
^{#}
= h_{z}^{#}: O ®
_{X,x}O for any _{Z,z}z ®
Z, where x = f(z) = g(z). Since
f is an immersion, f_{x}^{#}: O
® _{Y,f(x)}O is surjective,
hence _{X,x}g_{z}^{#}f_{x}^{#}
= h_{z}^{#}f_{x}^{#} implies
that g_{z}^{#} = h_{z}^{#}
as desired. This proof also works for local ringed spaces. For geometric
spaces we may replace the stalks by the residue fields of the stalks.
(a) q: X × ®
_{S} YY is an immersion.
(b) If X, Y are local ringed spaces over a local ringed spaces S,
then the fibre product (X ×
_{S} Y, p, q) of X and Y over
S in the site of ringed spaces is the fibre product as local ringed spaces.(c) Any immersion in RSp (resp. LSp, resp.
GSp) is universally bicontinuous. (See 5.2.)
^{#}: O
® _{Y,x}O is an isomorphism.
Since _{X,x}f is an immersion, f_{x}^{#} is surjective,
it remains to prove that f_{x}^{#} is injective.
Suppose s is a regular section of an open neighborhood f(x)
such that f_{x}^{#}(s) is zero at x.
Shrinking U if necessary we may assume that f_{x}^{#}(s)
= 0 over U. Thus for any y Î
U, s is in the kernel of _{y}f_{x}^{#},
hence a non-unit. Since Y is a geometrically reduced local ringed
space, this means that s = 0. Thus f_{x}^{#}
is injective.
Y be the effective inclusion. Then
is a geometric space. The induced morphism UX ®
is an isomorphism by (4.3.6).
U
Y is effective above U in the category of geometric spaces.
Since by assumption X is a geometric space, f factors through
a unique surjective immersion X ®
, hence an isomorphism by (4.3.6).
This proves that U_{red}f is effective in the category of geometric spaces.
Recall that a
g is bicontinuous, and for each s Î
S, the composition (f)_{g}(s)g_{s}^{#}
= g_{s}^{#}f_{g}(s)^{#}:
O ® _{S,s}O
® _{X,g(s})O is the identity
map of _{S,s}O. Hence _{S,s}g_{s}^{#} is
surjective for any s Î S. Thus any section
g is an immersion. In particular, any diagonal morphism of a morphism
is an immersion (5.1.3). Since any regular monomorphism is the result of
making a base extension of a diagonal morphism (see (5.1.10c), and immersions
are stable under base extensions, it follows that any regular monomorphism
is an immersion. Since an immersion is universally bicontinuous, any regular
monomorphism in these sites are universally bicontinuous.
Combining (4.3.8) and (4.3.9) we obtain the following
O Ä_{x}_{k}O given by _{x}a ®
a Ä 1 and a
® 1 Ä
a respectively are different since for any a Î
O- _{x } the elements ka Ä
1 and 1 Ä a
are linearly independent over k.
(c) Assume O has only one prime ideal.
Then _{x}x = spot O, _{x}w = spot O
Ä_{x}_{k}O, and _{x}p^{#} = i, q^{#}
= j. Applying (a) and (b) we see that f is an isomorphism.
(d) Suppose O has more than one prime
ideals. Moduling a minimal prime ideal _{x}p if necessary we may assume
that O is an integral domain (the composition _{x}spot
O/_{x}p ®
spot O ®
_{x}spot k is a monomorphism since the morphism spot O
® _{x}spot k
and the immersion spot O/_{x}p ®
spot O are monomorphisms). If _{x}k is algebraically
closed, then O Ä_{x}_{k}O is an integral domain, and because _{x}O
is a localization of _{w}O Ä_{x}_{k}O, we have _{x}O Ê
_{w}O Ä_{x}_{k}O. Since _{x}O is not a filed, _{x}f
is not an isomorphism, then by (b) the maps i, j: O
® _{x}O
Ä_{x}_{k}O are different. Composing with the inclusion map _{x}O
Ä_{x}_{k}O ®
_{x}O we obtain different maps _{w}p^{#}, q^{#}:
O ®
_{x}O, contradicting to (a). This proves that _{w}O
can not have infinite prime ideals under the assumption that _{x}k is
algebraically closed. We have thus proved the proposition for algebraically
closed k.
(e) It remains to consider the case that k is
not algebraically closed. By (c) we may assume x is not a geometric
point. Suppose y' is a geometric point over y whose field
is an algebraic closure of k. Then x = _{y}'x
×_{y}y' ®
y' is a monomorphism and the local ring of x
is faithful flat over _{y}'O since _{x}y' ®
y is so. This implies that x is not a geometric
point, which is impossible according to the above (c) and (d). This finishes
the proof.
_{y}'
Y be the canonical immersion sending spot k
® _{y} to y, where
k_{y} is the residue field of the local ring O_{y} of y.
Then the projection X_{y} = X ×_{Y} Spec k_{y}Spec k
_{y}
is an isomorphism.
Spec k is a monomorphism, thus we may apply (4.3.11).
_{y} |