Eisenbud-Harris
Posted on January 2, 2019Here 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.
;
(note
is a field);
(note
is not prime since
is not an integral domain);
;
;
- 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