4.3 Immersions  Definition 4.3.1. A morphism f: Y ® X of ringed space (resp. local ringed spaces) is called an immersion if f is bicontinuous, and for any x Î Y, the induced map fx#: OX,f(y) ® OY,y of stalks is surjective.  Remark 4.3.2. A morphism f is an immersion if and only if f is bicontinuous and the underlying morphism f(f): f(Y) ® f(X) of ringed sets (resp. local ringed sets) is an immersion.  Remark 4.3.3. (a) The composite of two immersions is an immersion.  (b) If X is a local ringed space, then the canonical morphism Xred d X from the associated geometric space Xred of X to X is an immersion.  (c) Suppose f: X ® Y is an immersion of local ringed spaces, then the induced morphism fred: Xred ® Yred of the associated geometric spaces is an immersion.  (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 Ured ® X is the composition Ured ® U ® X where U ® X is effective in LSp. Since Ured ® U and U ® X are immersions, Ured ® X is an immersion by (a).  Proposition 4.3.4. An immersion f: X ® Y of ringed spaces (resp. local ringed spaces, resp. geometric spaces) is a monomorphism in RSp (resp. LSp, resp. GSp).  Proof. Suppose f: X ® Y is an immersion of ringed spaces. Suppose g, h: Z ® X are two morphisms of ringed spaces such that gf = hf. Since f is bicontinuous, the underlying map of g and h must be the same, i.e. g(z) = h(z) for any z Î Z. Therefore it suffices to prove that gz# = hz#: OX,x ® OZ,z for any z ® Z, where x = f(z) = g(z). Since f is an immersion, fx#: OY,f(x) ® OX,x is surjective, hence gz#fx# = hz#fx# implies that gz# = hz# 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.  Proposition 4.3.5. Suppose f: X ® S, g: Y ® S are morphisms of ringed spaces. Let (X ×S Y, p, q) be the fibre product of X and Y over S in the site of ringed spaces. Suppose f is an immersion. Then   (a) q: X ×S Y ® Y 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.)  Proof. (a) is a consequence of (4.2.12.b and d).  (b) also follows from (4.2.12.d).  (c) comes from (a) and the fact that immersions are bicontinuous morphisms.  Proposition 4.3.6. Suppose f: X d Y is a surjective immersion of local ringed spaces and Y is a geometric spaces. Then X is a geometric space and f is an isomorphism.  Proof. Since f is a homeomorphism, we may identify X with Y. We only need to prove that for any x Î X, the induced map fx#: OY,x ® OX,x is an isomorphism. Since f is an immersion, fx# is surjective, it remains to prove that fx# is injective. Suppose s is a regular section of an open neighborhood f(x) such that fx#(s) is zero at x. Shrinking U if necessary we may assume that fx#(s) = 0 over U. Thus for any y Î U, sy is in the kernel of fx#, hence a non-unit. Since Y is a geometrically reduced local ringed space, this means that s = 0. Thus fx# is injective.  Corollary 4.3.7. Suppose f: X ® Y is an open immersion of local ringed spaces and Y is a geometric space. Then X is a geometric space and f is an open effective morphism.  Proof. Let U = f(X) and let i: U ® Y be the effective inclusion. Then U is a geometric space. The induced morphism X ® U is an isomorphism by (4.3.6).  Proposition 4.3.8. A morphism f: X ® Y in the site of geometric spaces is an immersion if and only if it is effective.   Proof. One direction has been noticed (4.3.3.d). Suppose f is an immersion. Write f(X) = U. The inclusion morphism Ured ® 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 ® Ured, hence an isomorphism by (4.3.6). This proves that f is effective in the category of geometric spaces.  Recall that a regular monomorphism in a category C is a monomorphism which is the kernel (i.e., equalizer) of some pair of morphisms (see (5.1) for the properties of regular monomorphisms).  Proposition 4.3.9. Any regular monomorphism in RSp, LSp, GSp, RSet, LSet, GSet, RPot, LPot, ASch, GASch is an immersion (hence is universally bicontinuous).  Proof. Suppose g: S ® X is a section of a morphism f: X ® S in any of these sites. Then we have fg = 1X, which implies that g is bicontinuous, and for each s Î S, the composition (fg(s)gs)# = gs#fg(s)#: OS,s ® OX,g(s) ® OS,s is the identity map of OS,s. Hence gs# 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  Corollary 4.3.10. Any regular monomorphism in GSp is effective.  Proposition 4.3.11. Suppose f: X ® y is a monomorphism in LSp where y is a geometric point. Then X = {x} is a geometric point and f is an isomorphism.   Proof. Suppose y = spot k for a field k. Since LSp is separable, f is injective. Thus X is a local ringed point x. We make the following observations:  (a) Since f: x ® y is monomorphism, the canonical projections p, q: x ×y x ® x are isomorphisms and p = q. Thus w = x ×y x is a local ringed point.   (b) If f: x ® y is not an isomorphism, then the canonical maps i, j: Ox ® Ox Äk Ox given by a ® a Ä 1 and a ® 1 Ä a respectively are different since for any a Î Ox - k the elements a Ä 1 and 1 Ä a are linearly independent over k.   (c) Assume Ox has only one prime ideal. Then x = spot Ox, w = spot Ox Äk Ox, and p# = i, q# = j. Applying (a) and (b) we see that f is an isomorphism.  (d) Suppose Ox has more than one prime ideals. Moduling a minimal prime ideal p if necessary we may assume that Ox is an integral domain (the composition spot Ox/p ® spot Ox ® spot k is a monomorphism since the morphism spot Ox ® spot k and the immersion spot Ox/p ® spot Ox are monomorphisms). If k is algebraically closed, then Ox Äk Ox is an integral domain, and because Ow is a localization of Ox Äk Ox, we have Ow Ê Ox Äk Ox. Since Ox is not a filed, f is not an isomorphism, then by (b) the maps i, j: Ox ® Ox Äk Ox are different. Composing with the inclusion map Ox Äk Ox ® Ow we obtain different maps p#, q#: Ox ® Ow, contradicting to (a). This proves that Ox can not have infinite prime ideals under the assumption that 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 xy' = x ×y y' ® y' is a monomorphism and the local ring of xy' is faithful flat over Ox since y' ® y is so. This implies that xy' is not a geometric point, which is impossible according to the above (c) and (d). This finishes the proof.   Corollary 4.3.12. Suppose f: X ® Y is a monomorphism of local ringed spaces. For any y Î Y, let Spec ky = spot ky ® Y be the canonical immersion sending spot ky to y, where ky is the residue field of the local ring Oy of y. Then the projection Xy = X ×Y Spec ky ® Spec ky is an isomorphism.   Proof. Since f is a monomorphism, the projection Xy ® Spec ky is a monomorphism, thus we may apply (4.3.11).      [Next Section][Content][References][Notations][Home]