Last update: 4 June 2013
Let be a ringed space. An (note the capital is a sheaf of abelian groups such that, for each open set in the group carries a structure of an these structures being compatible with the restriction homomorphisms: explicitly, if are open sets in then the restriction is compatible with the restriction that is to say, if and then Then each stalk has a natural structure, defined as follows: if say is the image of the image of for some sufficiently small open neighbourhood of then is the image of in
In particular, itself is an
Most of the concepts of module theory have their counterparts for Modules: -
(i) An homomorphism is a sheaf homomorphism (i.e. a family of homomorphisms commuting with the restrictions) such that each is an homomorphism. Then each is an homomorphism.
(ii) Sub-Modules. A subsheaf of an is a sub-Module of if, for each open set in is a of Then each is a of and the embedding is an homomorphism. In particular, a sub-Module of is called an Ideal (with a capital I).
(iii) Quotient Modules. Let be an a sub-Module of For each open set in form with the induced restriction homomorphisms, is a presheaf, but not necessarily a sheaf. So we form the sheaf associated with this presheaf: this is the quotient Module Since is exact, we have
(iv) Kernel. Let be an homomorphism. For each open set in let be the kernel of Then is a sheaf called the kernel of Clearly is an We have for all
(v) Image. For each open set in we can form which is a submodule of is a presheaf (not necessarily a sheaf). Let be the sheaf associated with this presheaf. Then is a subsheaf of called the image of Again by the exactness of we have Also is isomorphic to the quotient where is the kernel of
(vi) Cokernel. The cokernel of is We have the formulas
The class of is an abelian category. Exact sequences are defined in the usual way.
Lemma (7.1). A sequence is exact if and only if is exact for all
Proof. | |
is exact for all for all is exact for all |
Lemma (7.2). The "section functor" is left exact: if is exact, then is exact. This follows from (iii) above.
(vii) Direct sum. Let be any family of Their direct sum is the sheaf If each is equal to we write for the direct sum. In particular, if is finite and has elements, we write for the direct sum of copies of
(viii) Tensor product. If are their tensor product is defined to be the sheaf associated with the presheaf Since commutes with we have This tensor product has all the usual properties: it is commutative, associative, distributive over and is right exact in each variable (look at the stalks and use (7.1)). Also
(ix) Global Hom. is the group of all homomorphisms It has a natural structure: if and define by
(x) Sheaf The presheaf is easily checked to be a sheaf, denoted by Thus has a natural structure. Both Hom and are left exact in each variable (contravariant in the 1st variable, covariant in the 2nd). We have
Let Then is represented by say which gives rise to a homomorphism i.e. an element of Hence we have homomorphism
which in general is neither injective nor surjective (but see (7.9)).
(xi) Direct image. Let be a morphism of ringed spaces. If is an (thus a sheaf on we define its direct image which is an (thus a sheaf on as follows: for each open set in is an hence an via the homomorphism
is a left-exact functor from to For the section functor is left exact by (7.2). Hence if is an exact sequence of then is exact for each open hence is exact.
In particular, if is the ringed space consisting of a single point and the ring then Thus Thus is a 'relativization' of the section functor
If is an a homomorphism gives rise to i.e. to a global section of Conversely, given we may reconstruct if is open in and then Hence we have a one-one correspondence between homomorphisms and global sections of hence between homomorphisms and families of global sections of where is any index set. is an epimorphism if and only if each is generated (as an by the (for is an epimorphism if and only if each is an epimorphism, by (7.1)).
is said to be quasi-coherent if each has an open neighbourhood such that is the cokernel of a homomorphism where the index sets are of arbitrary cardinal (and depend on Clearly itself is quasi-coherent as an
Thus is quasi-coherent if and only if is locally generated by its sections and if the 'sheaf of relations' is locally generated by its sections.
An is of finite type if each has an open neighbourhood such that is generated by a finite set of sections of over i.e. if there exists an epimorphism for some integer If are of finite type, then so are and (the latter because is right exact). If is of finite type and is a homomorphic image of then is of finite type.
is said to be coherent if
(i) | is of finite type; |
(ii) | for each open set in and each homomorphism (n a positive integer), is of finite type. |
Clearly a coherent sheaf is quasi-coherent. All these properties (quasi-coherence, finite type, coherence) are local with respect to the base-space
We shall use the following notation. If is an open set in the phrase (over shall mean Similarly for diagrams of sheaves and homomorphisms.
Lemma (7.3). If is a subsheaf of and is of finite type and is coherent, then is coherent.
Proof. | |
Let be the embedding. If we have (over then (over but is coherent, hence is of finite type. |
Lemma (7.4). Let be If we have a diagram
with the row exact, then there exists an homomorphism (over a (smaller) neighbourhood of such that
Proof. | |
The map defines sections belonging to some open neighbourhood of Explicitly, maps into and is the image of the generator of Since is an epimorphism, there exist such that Each is represented by say agrees with at hence in some open neighbourhood of say Let then the define (over and we have (over hence |
Theorem (7.5). If is an exact sequence of on and if any two of are coherent, then so is the third.
Proof. | |
(1) coherent. By (7.3) it is enough to show that is of finite type. Let Since is of finite type we have an epimorphism (over some neighbourhood of Since is coherent, the kernel of is of finite type, hence we have an exact sequence Hence a commutative diagram with exact rows: We wish to define such that and show that is an epimorphism (over Since so we can define to be To show that is an epimorphism, let consider the corresponding diagram of stalks over and verify that is an epimorphism by diagram-chasing. Hence by (7.1) is an epimorphism and therefore is of finite type. (2) coherent. is of finite type, hence so is Let and let be a homomorphism (over an open neighbourhood of By (7.4) we can lift to (over a smaller open neighbourhood of so that is of finite type, hence we have say (over some open neighbourhood of Hence we have the following diagram: in which the rows are exact and the bottom row is split: Define then the diagram is commutative. Since is coherent, the kernel of is of finite type and we can therefore enlarge the diagram: Verify that the right-hand column is exact, e.g. by considering the corresponding diagram of stalks over a point Hence is coherent. (3) coherent. Since and are of finite type we have with epimorphisms; hence as in (2) we can define Since are epi, so is (by the 5 lemma). Hence is of finite type. Now let be a homomorphism (over some open neighbourhood of we have to show that is of finite type. Since is coherent we have an exact sequence of the form (over some open neighbourhood of hence a diagram here we have hence so we can define (over so that Now is coherent, hence the kernel of is of finite type, hence we can enlarge the diagram: Here the first column (as well as the top row) is exact, and we verify (e.g. by diagram-chasing in the stalks) that the second column is exact. Hence is of finite type and therefore is coherent. |
Corollary (7.6). and are coherent if and only if is coherent.
Proof. | |
If are coherent, the exact sequence shows that is coherent. If is coherent then is of finite type because it is a homomorphic image of it is also a subsheaf of hence coherent by (7.3). |
Corollary (7.7). If is a homomorphism of coherent then the kernel, image and cokernel of are all coherent.
Proof. | |
is a homomorphic image of hence is of finite type; it is also a sub-Module of hence coherent by (7.3). Now apply (7.5) to the exact sequences |
Corollary (7.8). If is an exact sequence in which all but are coherent, then is coherent.
Proof. | |
From (7.7) and the exact sequence |
Proposition (7.9). If are coherent then and are coherent.
Proof. | |
Consider Let since is coherent there is an exact sequence. hence, as tensoring with is right exact and an exact sequence since is coherent, so are by (7.6), hence is coherent by (7.7) and the fact that coherence is a local property. For operate on The argument is similar. |
Proposition (7.10). If are and is coherent, the mapping
is an isomorphism.
Proof. | |
From we have exact, hence by the left exactness of and we have exact sequences Since the second and third vertical arrows are isomorphisms, hence so is the first. |
If itself is coherent as an we shall say that is a coherent sheaf of rings.
Proposition (7.11). Let be a coherent sheaf of rings and let be an Then is coherent if and only if it is locally finitely presented, i.e. for each there is an exact sequence over some neighbourhood of
Proof. | |
If is coherent it is locally finitely presented (whether is coherent or not). Conversely, if is coherent, so are and by (7.6), hence so is by (7.7) (since coherence is a local property). |
This is a typed excerpt of the book "Algebraic Geometry: Introduction to Schemes - I.G. Macdonald".