Algebraic Geometry: Homework

Paris, Chennai

Affine and projective spaces

Let be the plane curve (i.e. is the zero-set of the polynomial ). Show that is isomorphic to the polynomial ring in one variable over .

Proof: We need to show that is isomorphic to . To do this, consider as a polynomial in adjointed with the solutions of . The solutions lie in . Alternatively, the kernel of the isomorphism is , by the first isomorphism theorem

Let be the plane curve . Show that is not isomorphic to the polynomial ring in one variable over .

Proof: contains an invertible element not in , and is therefore not a polynomial ring over

Let be the set . Show that is an affine variety of dimension . Find the generators for the ideal . Show that is isomorphic to a polynomial ring in one variable over . We say that is given by the parametric representation , and this curve is called the twisted cubic.


since the solutions of the two polynomials lie in . Generators of are therefore Futher, , and is hence an affine variety of dimension

Show that the Zariski topology on is not the product topology of the Zariski topologies on two copies of .

For homogenous ideal , show that the following conditions are equivalent:

  1. .
  2. either or the ideal .
  3. for some .


Suppose are maps to that are locally constant; i.e. for any , there is an open neighborhood of where the function is constant. Show that this is a sheaf.

Proof: In the axioms of a presheaf, let the restriction map be the identity map. The gluability axiom of a sheaf is satisfied trivially

Show that the pushforward is a sheaf, if is.

Proof: Suppose is a continuous function, and is a sheaf on . Then, , where is an open subset of . To show that this is a presheaf, let be the preimage of under . Then, we have a restriction map for the inclusion map . Further, let , where . We end up with the inclusion map , where . This satisfies the commutative diagram for the restriction map, and the gluability axiom for a sheaf, since does

Pushforward induces maps of stalks. Suppose is a continuous map and is a sheaf of sets, rings or -modules. If , describe the natural morphism of stalks .

Solution: The task is to describe the map , where . This maps the colimit of the sheaf of the preimage of to that of the subset ; in other words, what we have is the identity map

If is a ringed space, and is an -module, describe how for each , is an -module.

Solution: For to be an -module, the following diagram must commute, for :

Taking colimits for all open neighborhoods such that , we get the required result

Show that the sequence of presheaves is exact only if is, for all .

Suppose is a morphism of sheaves. Show that the presheaf kernel is in fact a sheaf. Further show that it satisfies the universal property of kernels.

Let be with classical topology. Let be the constant sheaf on associated to , the sheaf of holomorphic functions, and the presheaf of functions admitting a holomorphic logarithm. Describe the exact sequence of presheaves on :

where is a natural inclusion, and is given by . Show that is not a sheaf.

Prove that a section of a sheaf of sets is determined by its germs; i.e. the natural map is injective:

Show that is a closed subset of .

Prove that any choice of compatible germs for a sheaf of sets over is the image of a section of over .

Sheafification is a functor. Use the universal property to show that for any morphism of presheaves induces the natural map of sheaves. Hence, show that sheafification is a functor from presheaves on to sheaves on .

Proof: Using the universal property of sheafification, we get:

$$ \begin{xy} \xymatrix{ \mathscr{F}\ar[r]^{\text{sh}}\ar[dr]_g & \mathscr{F}^{\text{sh}}\ar[d]^f\ar@{.>}[dr]|{\exists!} & \ & \mathscr{G}\ar[r]^{\text{sh}} & \mathscr{G}^{\text{sh} } \end{xy} $$

completing our proof

Show that forms a sheaf using tautological restriction maps.

Describe the natural map of presheaves .

Show that the sheafification functor is left-adjoint to the forgetful functor from sheaves on to the presheaves on .

Show that describes as a quotient sheaf of . Find the open set in which this map is not surjective.

Morphisms of sheaves correspond to morphisms of sheaves on a base. Suppose is a base for topology of . A morphism of sheaves on the base is the collection of maps such that the following diagram commutes:

for all . Prove the following:

  1. A morphism of sheaves is determined by an induced morphims of sheaves on a base.
  2. A morphism of sheaves on a base gives a morphism of induced sheaves.

Suppose is an open cover, and we have sheaves on along with the isomorphisms , that agree on triple overlaps; i.e. on . Show that these sheaves can be glued together into a sheaf on , such that , and the isomorphisms over are the obvious ones. Thus, we can glue sheaves together using limited patching information.

Stalk of a kernel is the kernel of stalks. For all , show that:

Show that the cokernel of stalks is naturally isomorphic to the stalk of cokernels.

Suppose is a morphism of sheaves of abelian groups. Then, show that the image sheaf, is the sheafification of the image presheaf. Moreover, show that the stalk of an image is the image of a stalk.

Show that taking stalks of a sheaf of abelian groups is an exact functor. More precisely, if is a topological space and is a point, show that taking the stalk at defines the exact functor .

"Sections of U" is a left-exact functor. For topological space , and , if the sequence of sheaves is exact, then so is:

Moreover, show that if the sequence of sheaves is exact, then this is not:

Show that the pushforward is a left-exact functor.

Show that, if is a ringed space, then -modules form an abelian category.

Tensor product of -modules. Then:

  1. Suppose is a sheaf of rings on . Then, what is the tensor product of two -modules?
  2. Show that the tensor product of a stalk is the stalk of a tensor product.

Schemes I

Show that a map of differentiable manifolds with induces a morphism of stalks . Show that . In other words, if you pull back a function that vanishes at , you get a function that vanishes at .

Describe , where is very small. is called the ring of dual numbers. This is the first example of a non-zero function which is not determined by its values at points; its value at all points is .

Describe .

Deduce that for an algebraically closed field, the only maximal ideals are , from the Nullstellansatz.

If is a map of rings, and is a prime ideal in , prove that is a prime ideal in .

Consider the map sending . The "source" can be interpreted as the x-line, and the "target" can be interpreted as the y-line. This can be pictured as the parabola in the xy-plane mapping to the y-line. Interpret the corresponding maps of rings as given by . Verify that the preimage, or fiber above are the points .

Suppose is a field, and are given. Let be the ring morphism defined by .

  1. Show that induces a map for ideals such that .
  2. Show that the map of (a) sends the point to:

Check that the -axis is contained in .

Solution: Since is one solution of , the -axis is indeed contained in .

Prove that .

Proof: Given that , we need to show that . The ideal generated by vanishing points is a set of functions vanishing at those points, and this is contained in the same prime ideal

is closed for all . Check this is a topology:

  1. Show that both and are open sets of .
  2. If is a collection of ideals, show that . Hence, union of any collection of open sets is open.
  3. Show that . Hence, intersection of any finite number of open sets is open.

Solution: Let be the open cover of : since arbitary unions of open sets is open, is open in . It is also closed, because it is the whole space. Similarly, is both open and closed in , finishing the proof for (a).

By showing that closed sets pull back to closed sets, show that is a continous map. Interpret as a contravariant functor .

Radicals commute with finite intersections. Prove that, for ideals , .

Suppose are an ideal and multiplicative subset respectively.

  1. Show that is a closed subset of , and that is an open subset of . Further, show that, for arbitary , need not be open or closed.
  2. Show that and are the subspace topology induced by inclusion in .

Suppose is an ideal. Show that vanishes on iff ; i.e. for some .

Proof: is the set of points which are contained in , a prime ideal in . It remains to be shown that , given that . This is obvious from the fact that is prime

Show that distinguished open sets form a base for the (Zariski) topology.

Solution: Since forms a topology, it remains to be shown that, given a subset , the complement of is . From the definitions of and , this is clear

Suppose as runs over some index set . Show that iff , or equivalently, there are all but finitely many zero, such that .

Show that if is an infinite union of distinguished open sets, , then in fact, it is a union of finitely many number of these; i.e. for subset , .

Show that .

Show that iff for some , and is an invertible element of .

Show that iff .

For , describe homeomorhpism for which each is mapped onto distinguished open set of . Thus, as topological spaces.

  1. Show that, in an irreducible topological space , any nonempty open set is dense.
  2. If is a topological space, and is an irreducible subset, show that is also irreducible in .

Solution: To prove (a), let be this nonempty open set, and . Then, , leading to a contradiction since is irreducible. To prove (b), take closure of , leading to . Since , and , we get

If is an integral domain, show that is irreducible.

Solution: has no nilpotents other than . Hence, .

Show that an irreducible topological space is connected.

Solution: Since is irreducible, , where are proper closed subsets. Since all open subsets of irreducible spaces are dense, . Taking closure and complement, we get ; hence, for nonempty open subsets , we get ; this is exactly the condition for separability, as required by the definition of connectedness

Prove that is quasicompact.

Proof: Let be an open cover, with . Now,

Hence, and .

Since , there exists some finite subset , such that . Then, for , . Hence, , and

  1. If is any topological space which is a finite union of quasicompact spaces, show tht is quasicompact.
  2. Show that every closed subset of a quasicompact space is quasicompact.

  1. If is a field, and a finitely-generated -algebra, show that the closed points of are dense, by showing that if , and is a non-empty distinguished open subset of , then contains a closed point of .
  2. Show that if is a -algebra that is not finitely generated, then the closed poitns of need not be dense.

If show that is a specialization of iff . Hence, show that .

Verify that is a generic point for .

Suppose is a generic point for the closed subset . Show that it is "near every point of " (every neighborhood of that contains ), and not "near every point not contained in ".

Show that every point is contained in an irreducible component of .

Show that is Noetherian, and in the classical topology is not.

Show that a ring is Noetherian iff every ideal is finitely-generated.

If is Noetherian, show that is a Noetherian topological space. Describe when is a non-Noetherian topological space.

Show that .

Solution: We know that , since is the set of functions vanishing on , and we have defined this to be closed on a Zariski topology. Now, it remains to be shown that .

Prove that, if is an ideal, . To gain some intuition about this statement, consider the ideal ; operating on it with V, we get y-axis as the variety; operating on the variety with I, we get the ideal , which is the radical of .

Proof: Let

Introduce a new variable , so that

such that . Then,

Multiply both sides by to obtain:

This identity holds if , and hence , as required

Show that and give a bijection between irreducible closed sets of and prime ideals of . From this conclude that, in , there is a bijection between points of and irreducible closed subsets of , where a point determines an irreducible closed set by taking closure. Hence, any irreducible closed set has exactly one generic point: irreducible closed subset can be written as .

If is any ring, show that the minimal prime ideals are in bijective correspondence with irreducible components of . In particular, is irreducible iff has only one minimal prime ideal.

What are the minimal prime ideals of where is a field?

Schemes II

Show that is an isomorphism.

Describe a bijection between maps between and rings .

Suppose . Show that there is a natural isomorphism .

If is a scheme and any open set, then show that is also a scheme.

If is a scheme, show that affine open sets form a base for the Zariski topology.

  1. Show that a disjoint union of a finite number of affine schemes is also an affine scheme.
  2. Show that a disjoint union of an infinite number of affine schemes is not an affine scheme.

Proof: For part (b), notice that never holds if is infinite, as the of any ring is quasicompact.

Show that the sheaf of at point is the local ring .

Show that it is possible to glue an arbitarary collection of schemes together. Given:

  1. Schemes .
  2. Open subschemes such that .
  3. Isomorphisms , with identity.

Such that the cocycle condition is satisfied. That is, isomorphisms agree on triple intersections; i.e. . The cocycle condition ensures that are inverses. Show that there is a unique scheme along with open subsets respecting the gluing data in the obvious sense.

Show that affine line with doubled origin is not affine.

Solution: Let us compute the ring of global sections.

In plane with doubled origin, , describe two open affine subsets whose intersection is not an affine subset.

Show that the only functions on are constants; i.e. , and hence is not affine if .

Projective schemes

Consider with projective coordinates . Define a scheme that would be interpreted as in .

Solution: The affine open subset corresponding to is ; i.e. . The affine open subset corresponding to and can be defined similarly.

Consider the projective scheme with coordinates . Given the homogenous polynomials , define the scheme cut out by in .

Suppose is homogenous.

  1. Give a bijection between prime ideals and homogenous prime ideals of .
  2. Interpret the set of prime ideals of as a subset of .

Properties of schemes

Show that, if is a quasicompact scheme, then each point has a closed point in its closure. Show that every nonempty closed subset of contains a closed point of . In particular, every nonempty quasicompact scheme has a closed point.

Show that a scheme is quasicompact and quasiseparated iff can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.

Show that all projective -schemes are quasicompact and quasiseparated.

Show that, if is a reduced ring, so are , , and .

Suppose is quasicompact, and vanishes at all points of , then show that for some . Show also that, this may fail if is not quasicompact.

Show that a scheme is integral iff it is irreducible and reduced.

Show that locally Noetherian schemes are quasiseparated.

Show that a Noetherian scheme has a finite number of irreducible components, and a finite number of connected components, each a finite union of irreducible components.

Show that a Noetherian scheme is integral iff is nonempty, connected, and all stalks are integral domains. Thus, under "good conditions", integrality is a union of local (stalks are integral) and global (connected) conditions.

Normality and factoriality

Show that UFDs are integrally closed. Hence, factorial schemes are normal, and if is a UFD, is normal.

Show that , , and are normal.

Suppose is a UFD with invertible, such that is irreducible in . Show that has no repeated prime factors iff is normal.

UFD-ness is not affine-local. Let denote the homogenous degree of the ring . In other words, if it consists of quotients where has pure degree . Show that distinguished open sets cover . Show that are UFDs.

Associated points and primes

Suppose is an integral domain. Show that the generic point is the only associated point of .

Show that associated primes/points behave well with respect to localization: associated points/primes of are precisely those points/primes of that lie in .

Suppose is an -module. Show that if is maximal among all proper ideals that are annihilators of , then is prime, and . Thus, if is Noetherian, then is nonempty. Quite generally, proper ideals that are maximal with respect to some property have an uncanny tendency to be prime.

Suppose is a Noetherian ring, and is a finitely generated -module. Show that associated primes of satisfy:

  1. Show that every associated point is the generic point of an irreducible component of , for some .
  2. If , show that the closure of is that set of associated points at which has nonzero germ.

Morphisms of schemes

-modules push forward. Given a morphism of ringed spaces , show that a sheaf pushforward induces a functor .

Given a morphism of ringed spaces , with , show that there is a map of stalks .

Suppose is a morphism of rings. Define to be a morphism of ringed spaces as follows. The map of topological spaces is defined in the usual way. To describe a morphism of sheaves on , it suffices to describe a morphism of sheaves on the distinguished base of . On , we define:

by . Show that this describes a morphism of sheaves on a distinguished base.

  1. Show that is a locally ringed space.
  2. Show that the morphism of ringed spaces is defined by a ringed morphism is a locally ringed morphism.

Show that a morphism of schemes is a morphism of ringed spaces that looks locally like morphisms of affine schemes. Put precisely, if is an affine open subset of and is an affine open subset of , and , then the induced morphism of ringed spaces is a morphism of affine schemes. Show that it suffices to check on a set where form an open cover of and form an open cover of .

Show that a category of rings, and opposite category of affine schemes are equivalent.

Show that given by:

is a morphism of schemes.