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Solutions to Hartshorne's Algebraic Geometry/Separated and Proper Morphisms

< Solutions to Hartshorne's Algebraic Geometry

The reference for this section is EGA II.5, EGA II.6, EGA II.7. For the discrete vaulation ring questions at the end see Samula and Zariski's Commutative Algebra II.


Exercise II.4.1Edit

Let   be the finite morphism. Finite implies finite type so we only need to show that   is universally closed and separated.

  is separated. We want to show that   is a closed immersion. To check that a morphism is a closed immersion it is enough to check for each element of an open cover of the target. Let   be an open affine cover of  . The pull-back of   along each   is   where  . The ring homomorphism corresponding to these morphisms of affine schemes is surjective, and so they are all closed immersions according to Exercise II.2.18(c).

  is universally closed. The proof of Exercise II.3.13(d) goes through to show that finite morphisms are stable under base change (in fact, the proof becomes easier). Secondly, we know that finite morphisms are closed (Exercise II.3.5) and therefore finite morphisms are universally closed.

Exercise II.4.2Edit

Let   be the dense open subset of   on which   and   agree. Consider the pullback square(s):


Since   is separated, the lower horizontal morphism is a closed immersion. Closed immersions are stable under base extension (Exercise II.3.11) and so   is also a closed immersion. Now since   and   agree on  , the image of   in   is contained in the diagonal and so the pullback is, again   (at least topologically. But this means that   factors through  , whose image is a closed subset of  . Since   is dense, this means that  . Since   is a closed immersion, the morphism of sheaves   is surjective. Consider an open affine   of  . Restricted to  , the morphism   continues to be a closed immersion and so   is an affine scheme, homeomorphic to  , determined by an ideal  . Since   is a homeomorphism,   is contained in the nilradical. But   is reduced and so  . Hence,   and therefore  .

  1. Consider the case where  , the affine line with nilpotents at the origin, and consider the two morphisms  , one the identity and the other defined by  , i.e. killing the nilpotents at the origin. These agree on the complement of the origin which is a dense open subset but the sheaf morphism disagrees at the origin.
  2. Consider the affine line with two origins, and let   and   be the two open inclusions of the regular affine line. They agree on the complement of the origin but send the origin two different places.

Exercise II.4.3Edit

Consider the pullback square


Since   is separated over   the diagonal is a closed immersion. Closed immersions are stable under change of base (Exercise II.3.11(a)) and so   is a closed immersion. But   is affine since all of   are. So   is a closed immersion into an affine scheme and so   itself is affine (Exercise II.3.11(b)).

For an example when   is not separated consider the affine plane with two origins   and the two copies   of the usually affine plane inside it as open affines. The intersection of   and   is   which is not affine.

Exercise II.4.4Edit

Since   is proper and   separated it follows from Corollary II.4.8e that   is proper. Proper morphisms are closed and so   is closed.

  is finite type. This follows from it being a closed subscheme of a scheme   of finite type over   (Exercise II.3.13(a) and (c)).

  is separated. This follows from the change of base square and the fact that closed immersions are preserved under change of base.

f(Z) \ar[d]^\Delta \ar[r] & Y \ar[d]^\Delta \\
f(Z) \times_S f(Z) \ar[r] & Y \times_S Y 

  is universally closed. Let   be some other morphism and consider the following diagram

T \times_S Z \ar[r] \ar[d]^{f'} & Z \ar[d]^f \\
T \times_S f(Z) \ar[r] \ar[d]^{s'} & f(Z) \ar[d]^s \\
T \ar[r] & S

Our first task will be to show that   is surjective. Suppose   is a point with residue field  . Following it horizontally we obtain a point   with residue field   and this lifts to a point   with residue field  . Let   be a field containing both   and  . The inclusions   give morphisms   and   which agree on   and therefore lift to a morphism   giving a point in the preimage of  . So   is sujective.

Now suppose that   is a closd subset of  . Its vertical preimage   is a closed subset of   and since   is universally closed the image   in   is closed. As   is surjective,   and so  . Hence,   is closed in  .

Exercise II.4.5Edit

  1. Let   be the valuation ring of a valuation on  . Having center on some point   is equivalent to an inclusion   (such that  ) which is equivalent to a diagonal morphism in the diagram
Spec\  K \ar[r] \ar[d] & X \ar[d] \\ 
Spec\  R \ar[r] \ar[ur] & Spec\  k

But by the valuative criterion for separability this diagonal morphism (if it exists) is unique. Therefore, the center, if it exists, is unique.

  1. Same argument as the previous part.
  1. The argument for the two cases is the same so we will prove: Suppose that every valuation ring   of   has a unique center in  , then   is proper. This is clearly true for integral  -schemes of finite type of dimension zero. Suppose that it is true for integral  -schemes of dimension less than   and that   is an integral  -scheme of dimension  . We will use the valuative criteria. Suppose that we have a diagram
Spec\  L \ar[r] \ar[d] & X \ar[d] \\ 
Spec\  S \ar[r] & Spec\  k

with   a valuation ring of function field  . If the image of the unique point of   is not the generic point of   then let   be the closure of its image with the reduced structure. We have a diagram

Spec\  L \ar[r] \ar[d] & Z \ar[r] & X \ar[d] \\ 
Spec\  S \ar[r] & Spec\  k \ar@{=}[r] & Spec\  k

The scheme   is an integral  -scheme of dimension less than   and so the square on the lest admits a lifting, which gives a lifting for the outside rectangle. Moreover, as closed immersions are proper, any lifting of the outside rectangle factors uniquely through   by the valuative criteria and so the lifting is unique.

Now suppose that the image of the point of   is the generic point of  . Then we have a tower of field extensions   and the valuation on   induces a valuation on  . We then have the following diagram.

Spec\  L \ar[r] \ar[d] & Spec\  K \ar[r] & X \ar[d] \\ 
Spec\  S \ar[r] & Spec\  R \ar[r] & Spec\  k

By assumption the valuation ring   has a unique center   on   and so there is a unique extension of the diagram above

Spec\  L \ar[r] \ar[d] & Spec\  K \ar[r] & Spec\  \mathcal{O}_{X,x} \ar[r] & X \ar[d] \\ 
Spec\  S \ar[r] & Spec\  R \ar[rr] \ar[ur] && Spec\  k

Hence, there is a unique lifting of our original square. By the valuative criteria, the scheme   is then proper.

  1. Suppose that there is some   such that  . Consider the image  . Since   is algebraically closed,   is transcendental over   and so   is a polynomial ring. Consider the localization  . This is a local ring contained in   and therefore there is a valuation ring   that dominates it. Since   we see that  .

Now since   is proper, there exists a unique dashed morphism in the diagram on the left.

Spec\  K \ar[r] \ar[d] & X \ar[d] && K & \Gamma(X, \mathcal{O}_X) \ar[l] \ar@{-->}[dl] \\
Spec\  R \ar[r] \ar@{-->}[ur] & Spec\  k && R \ar[u] & k \ar[l] \ar[u]

Taking global sections gives the diagram on the right which implies that   and so  . But   and so   This gives a contradiction since  .

Exercise II.4.6Edit

Since   and   are affine varieties, by definition they are integral and so   comes from a ring homomorphism   where   and   are integral. Let  . Then for valuation ring   of   that contains   we have a commutative diagram

Spec\  K \ar[r] \ar[d] &  X \ar[d] \\
Spec\  R \ar[r] \ar@{-->}[ur]^{\exists !} & Y

Since   is proper, the dashed arrow exists (uniquely, but we don't need this). From Theorem II.4.11A the integral closure of   in   is the intersection of all valuation rings of   which contain  . As the dashed morphism exists for any valuation ring   containing   so it follows that   is contained in the integral closure of   in  . Hence every element of   is integral over  , and this together with the hypothesis that   is of finite type implies that   is finite.

Exercise II.4.7Edit

Exercise II.4.8Edit

  • Let   and   be the morphisms. The morphism   is a composition of base changes of   and   as follows:

\tbd{\mathfrak{m}arginpar{Should really check that the all the claims made about pullbacks in here are true.}}

& X \ar[dd] \\
X \times X' \ar[ur] \ar[dd]  \\
& Y \\
Y \times X' \ar[ur] \ar[dd] \ar[dr] \\
& X' \ar[dd] \\
Y \times Y' \ar[dr] \\
& Y'

Therefore   has property  .

  • Same argument as above but we should also note that since   is separated the diagonal morphism   is a closed embedding and therefore satsifies  .
& Y \ar[dd] \\
X \ar[ur] \ar[dd]  \\
& Y\times_Z Y \\
X \times_Z Y \ar[ur] \ar[dd] \ar[dr] \\
& X \ar[dd] \\
Y \ar[dr] \\
& Z

  • Consider the factorization
X_{red} \ar@/^/[drr]^{id} \ar@/_/[ddr]_{f_{red}} \ar[dr]^{\Gamma_{f_{red}}} \\
& Y_{red} \times_Y X_{red} \ar[r] \ar[d] & X_{red} \ar[d] \\
& Y_{red} \ar[r] & Y

The morphism   is a composition of a closed immersion and a morphism with property   and therefore it has property  . Therefore the vertical morphism out of the fibre product is a base change of a morphism with property   and therefore, itself has property  . To se that   has property   it therefore remains only to see that the graph   has property   for then   will be a composition of morphisms with property  . To see this, recall that the graph is following base change

X_{red} \ar[r] \ar[d]^\Gamma & Y_{red} \ar[d]^\Delta \\
X_{red} \times_Y Y_{red} \ar[r] & Y_{red} \times_Y Y_{red}

But   and   and so   is a closed immersion. Hence,   is a base change of a morphism with property  .

Exercise II.4.9Edit

Let   be two projective morphisms. This gives rise to a commutative diagram

X \ar[r]^{f'} \ar[dr]_f & \mathbb{P}^r \times Y \ar[d] \ar[r]^{id \times g'} & \mathbb{P}^r \times \mathbb{P}^s \times Z \ar[d]  \\
& Y \ar[r]^{g'} \ar[dr]_g & \mathbb{P}^s \times Z \ar[d] \\
& & Z }

where   and   (and therefore  ) are closed immersions. Now using the Segre embedding the projection   factors as
So since the Segre embedding is a closed immersion then we are done since we have found a closed immersion   which factors  .

Exercise II.4.10Edit

Chow's Lemma is in EGA II.5.6.

Exercise II.4.11Edit

See Samula and Zariski's Commutative Algebra II.

Suppose that  . Then define:
The ring   is a discrete noetherian local domain with maximal ideal   and quotient field  . By induction then, we can reduce to the case when   is a finite field extension of  . Now consider a set of generators   of   such that  \mathfrak{m}arginpar{does such a set always exist?} (if   is principal wait for the next step). We claim that the ideal   is not the unit ideal in  . If it were then there would be some polynomial   of degree, say  , in the   such that  . Let   be the degree 0 part of   and   be the higher degree part. Since   the element   has an inverse, say  . Now with this in mind, our equality   implies that   which then implies that  . Since   is made up of terms of degree higher than zero, the element   which implies that   contradicting our assumption. So   is not the unit ideal in  . Now let   be a minimal prime ideal of  , and consider the localization  .

Exercise II.4.12Edit

See Samuel and Zariski's Commutative Algebra II.