### Eisenbud-Harris

Posted on January 2, 2019

Here are my exercise solutions and notes for the beginning of Eisenbud and Harris’s The Geometry of Schemes.

Let be a ring and a prime ideal. Recall that we define , i.e., is the localization of with respect to the multiplicative set

1-1.

1. ;
2. (note is a field);
3. (note is not prime since is not an integral domain);
4. ;
5. ;
6. Recall the prime ideals of are precisely the quotients of prime ideals Thus

I-2. Note that sends an element of to Thus takes the value in and the value in

I-3.a. This follows since the map defines an isomorphism

I-3.b. Let The maximal ideals of are for all points Thus the maximal ideals of are the images of these ideals under the natural map and correspond the points of Moreover, for a point the map is an isomorphism Therefore given a polynomial we have where is the maximal ideal corresponding to the point

I-4.a. The points of are the ideals generated by irreducible polynomials, which in are the polynomials Thus the closed points of are precisely the maximal ideals, i.e., the points corresponding to “actual points” in For a point to be open we would have some ideal such that for all and ; this is clearly impossible, since would then have an infinite number of distinct factors.

I-4.b. The localization is a DVR with uniformizing parameter Thus it has prime ideals and , i.e., The closed sets are precisely

I-5.a. Let be a sheaf on We may pick and to be arbitrary sets. Then is the set consisting of a unique for each distinct and such that and If we take , then and the closed sets are , i.e., every subset is closed. Thus the topology is discrete as desired.

I-5.b. If we take then has the desired topology (see exercise 1-4).

1-6. Let be a vector bundle over a topological space Let be the sheaf of sections of on and let be the sheaf of continuous functions on Then is a -module for each open subset

I-8.a. We will first show that is continuous. Suppose that is open. Then But if then and for all So we in fact have So is open and is continuous.

Now we will show that is a continuous section of over By the definition of we have that is the identity on ; under and projects It suffices to show is open for each open and since form a basis. Observe that

I-8.b. For each pick an open set , , in containing Observe that since is the identity on , we must have Note that is an open cover of Moreover, if , then So Thus by the sheaf axiom, there exists such that So is in fact defined by

Note. So is in fact isomorphic to the sheaf of germs of continuous sections of

I-9. Suppose that is injective for each open Let and such that Then there exists an open containing and and But then since is injective. So If is also surjective for each , then is surjective for each

Conversely, suppose that is injective for each Suppose that is open and such that Then for all So for all and So is injective.

It remains to show that if is bijective for each , then is also surjective for each Let Since each is surjective for each there exists an open containing and such that Observe that for each we have But since is injective, Thus by the sheaf axiom there exists such that for each , and hence

I-13. Suppose that is an open cover of an open set , and for each we have with the collection of sections satisfying the property that for every We must show that there is a unique element such that Note that

Let be such that Define a basic open cover of For each we define a section by picking some such that and defining Observe that this is well defined as if is also contained in another set in the cover the requirement that ensures that Moreover, if then for any basic open set a similar argument shows that

Since was a -sheaf, and was a cover of , our argument above actually defines a section for every basic open set (not just those in ) and these sections satisfy the condition that for any two basic open sets the sections and restrict to the same section on any basic open set in their intersection. Therefore is a well defined element of and by construction It is clear that this construction was unique and completely determined by the sections