Notes from Hartshorne | artagnon.com

Modern algebraic geometry is built on the theory of schemes and cohomology. The subject is intricately linked with commutative algebra and homological algebra.

Affine varieties

Let be an algebraically closed field. Then affine variety over an -space, , is defined as the set of -tuples over the set . is called a point in , and s are called coordinates.

Let be a polynomial ring in -space over . Elements of are then functions from affine -space to , defined by , where . The zero-set of is defined as the set of simultaneous zeros of :

Note that if is an ideal, then .

The set is called an algebraic set if there exists such that . The union and intersection of any two algebraic sets is an algebraic set, as is the empty set, and the whole space.

Define the Zariski topology in by taking, as open subsets, the complement of algebraic sets. Considering the Zariski topology on the affine line , we note that every ideal is principal, so every algebraic set is the set of zeros of a single polynomial. Thus algebraic sets in are just the finite subsets and the whole space; in particular, this is not Hausdorff.

The nonempty subset of a topological space is irreducible if it cannot be expressed as the union, , of two proper closed sets. The empty set is not irreducible, but is.

An affine variety can then be expressed as an irreducible subset of . An open subset of an affine variety is called quasi-affine.

An ideal for is defined as:

So, we have the functions and , mapping subsets of to an algebraic set, and mapping the subsets of to ideals. Their properties can be summarized as follows:

If , then .
If , then .
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, a direct consequence of the Nullstellansatz.
, the closure of .

There is a one-to-one correspondence between the algebraic sets in and the radical ideals in , given by , and . Furthermore, the set is irreducible iff its ideal is a prime ideal. For example, is irreducible, since it corresponds to the zero ideal in .

If is an irreducible polynomial in , then the affine variety is defined as , an -dimensional hypersurface.

Let be an affine algebraic set. Then, the affine coordinate ring, , is defined to be . Furthermore, if is an affine variety, then is an integral domain, and is also a finitely generated -algebra.

A topological space is termed noetherian if it satisfies the descending chain condition. For instance, is noetherian. In this space, every subspace can be expressed as a union of irreducible subspaces, . If we also require that for , then is uniquely determined; they are termed the irreducible components of .

Every algebraic set of can be expressed uniquely as a union of varieties, with no single variety containing the other.

The dimension of an affine or quasi-affine variety is equal to the dimension of the topological space, and this is exactly the supremum integer in , the irreducible closed subsets of the topological space. As an example, dimension of is .

The height of a prime ideal is defined as the supermum of in the chain of distinct prime ideals . The Krull dimension of is the supremum of heights of the prime ideals in .

The dimension of an affine algebraic set, , is equal to the dimension of its affine coordinate ring .

Let be a field, and be an integral domain which is a finitely-generated -algebra. Then:

Dimension of is equal to the transcendence degree of the quotient field of over .
For any prime ideal , we have .

If is a quasi-affine variety, then . Let be a sequence of distinct closed irreducible sets in , be that of . is finite, so we can choose a maximal . In that case, must be a point , and the chain will also be maximal. Now, corresponds to the maximal ideal in affine coordinate ring of . s correspond to prime ideals contained in , so . Since is a point in affine space, . Hence, , and .

What follows is a bunch of results:

Krull's Haputidealsatz: Let be a noetherian ring, and be an element which is neither a zero-divisor nor unit. Then, every minimal prime ideal containing has height .
Every noetherian integral domain is a UFD iff every prime ideal of height is principal.
A variety has dimension iff it is the zero set of a single nonconstant irreducible polynomial in .

Projective varieties

We proceed as before, but in the projective space , the set of equivalence classes of -tuples, denoted , nonzero elements of with the equivalence relation , for all . An element of is called a point, and the -tuple in the equivalence class of is called set of homogenous coordinates for .

A graded ring is a ring together with the decomposition , the direct sum of abelian groups, such that for , . An element of is called a homogenous element of degree . Ideal is a homogenous ideal if . An ideal is said to be homogenous if it is generated by homogenous elements. The sum, product, intersection, and radical of homogenous ideals yields a homogenous ideal.

A polynomial cannot be used to define a polynomial in because of the nonuniqueness of homogenous coordinates. However, if is a homogenous polynomial of degree , then , so the property of being zero depends only on the equivalence class of . Thus, if is a set of homogenous elements, the zero-set can be defined as:

Since is noetherian, any set of homogenous elements has a finite subset such that .

Subset is an algebraic set if there exists a set of homogenous elements such that . The union of any two algebraic sets is an algebraic set, as is the intersection. The Zariski topology on is defined in the usual way.

A projective variety is the irreducible algebraic sets of along with the induced topology. An open subset of a projective variety is a quasi-projective variety. If is a subset, we define the projective ideal, to be the ideal . If is an algebraic set, we define the homogenous coordinate ring to be .

We want to show that every projective -space has a covering by the affine -space, and hence, every projective (or quasi-projective) variety has a covering by an affine (or quasi-affine) variety. If is a linear homogenous polynomial, then the zero set of is called a hyperplane; we denote the zero-set of to be . Let be open subsets. Then, is covered by , since if is a point, then atleast one is nonzero, and hence . We define mapping , such that if and , where is the point defined by:

with omitted. Note that is well-defined since is independent of the choice of homogenous coordinates.

is a homeomorphism of with its induced topology to its Zariski topology. is clearly bijective, so it will be sufficient to show that closed sets of are identified with the closed sets of by . We can assume that and write for , for ; then we have . Let , be a map , the set of homogenous elements of to , be a map from to . Set and . Let , and let be its closure in . This is an algebraic set so , for some subset . Let , so . Checking the converse as well yields the result that both and are closed maps, so is a homeomorphism.

Morphisms

A function is regular at point , if for open neighborhood , and , is nowhere zero on , and . is regular on if it is regular on every point of .

A regular function is continuous when is identified with in its Zariski topology. It suffices to show that of a closed set is closed. A closed set of is a finite set of points, so it's enough to show that , for all . Subset of a topological space is closed iff it can be covered by open subset , such that is closed in for every . Let be an open set on which can be expressed as , , nowhere zero in . So, . But iff . Hence, , and is closed in . A similar argument holds for a quasi-projective variety.

If are regular functions in variety , and in some subset , then everywhere.

If are two varieties (affine, projective, quasi-affine, or quasi-projective), and a morphism, such that for each open subset , and every regular function , the function is regular. This defines the isomorphism and its inverse in the category of varieties. Isomorphisms are necessarily bijective and bicontinuous.

Let be a variety, the ring of regular functions on . If is a point of , we define local ring of on , , as the ring of germs of regular functions of near . An element of is the pair such that is an open subset of near , and is a regular function on . We identify two such pairs and if in .

is indeed a local ring; its maximal ideal is the set of germs of regular functions that vanish at . If , then is regular on some neighborhood of . The residue field .

Elements of function field of variety are the equivalence classes of the pair , where is an open neighborhood of , and is a regular function on .

is a field that can be made into a ring by defining addition and multiplication. Set so that is a regular function on . Then, is the inverse of .

By restricting functions, we obtain the natual map which is injective. We normally treat as subrings of . If we replace by an isomorphic variety, we obtain isomorphic rings, and say that are invariants of the variety .

Let be the affine coordinate ring, and be the homogenous coordinate ring (as in projective spaces). Then, , but is not invariant; it depends on the embedding of in the projective space.

Let be an affine variety, and the affine coordinate ring. Then:

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For each point , let be the set of functions vanishing at . Then, there is a one-to-one correspondence between the points of and the maximal ideals of , given by .
For each , , and .
is isomorphic to the quotient field of , and hence is a finitely generated extension field of , with transcendence degree = .

To prove (a), notice that every polynomial is regular on , and hence ; thus, the kernel of the injective homomorphism is just .

To prove (b), notice that there is a one-to-one correspondence between points of and maximal ideals of , passing to the quotient . Identifying with regular functions on , the maximal ideal corresponding to is .

To prove (c), notice that, for every point , there is a natural map which is injective; it is also surjective by definition of a regular function. Hence, . Now, . Since , .

From (c), it follows that the quotient field of is isomorphic to the quotient field of , and this is equal to , because any rational function is in some . Since is a finitely generated -algebra, is a finitely generated field extension of . Furthermore, its transcendence degree equals . This proves (d).

Let be an open set defined by the equations . Then, is an isomorphism of varieties.

Let us introduce some notation. If is a graded ring, and is a homogenous prime ideal in , then denotes the subring of elements of degree in the localization of with respect to multiplicative subset consisting of homogenous elements of . Notice that has a natural grading given by , where is homogenous in , and . is a local ring given by . In particular, if is a domain, then for , we obtain a field . Similarly, if is homogenous, denotes a subring with elements of degree in the localized ring .

Let be a projective variety with homogenous coordinate ring . Then:

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For any point , let be the ideal generated by the set of homogenous such that . Then, .
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Let be the open set , and let . is isomorphic to , and is an affine variety. There is a natural isomorphism of the affine coordinate ring with the localization of the homogenous coordinate ring of . First, make an isomorphism from to leaving out ; this isomorphism sends to , so passing to the quotient, we obtain .

To prove (b), let be any point, and choose so that . Then, . where is the maximal ideal of corresponding to . Since , , and localization is transitive, we obtain .

To prove (c), notice that , which is equal to , which is the quotient field of . But by , this is isomorphic to .

To prove (a), let be a global regular function. For each , is regular on , so . But we know that , so can be written as , where is homogenous of degree . Thinking of as subrings of the quotient field of , we get . Now, choose . Then, is spanned as a -vector space as monomials of degree in , and in any such mononomial atleast one occurs as with power . Thus, and . This shows that a subring of is contained in , which is a finitely-generated -module. Since is noetherian, is a finitely-generated -module, and therefore, is integral over . This means that there are elements such that:

Now, , so and is algebraic over . But is algebraically closed, so , completing the proof.