ALGEBRAIC GEOMETRY

Alexei Skorobogatov

December 16, 2003

Contents

1 Basics of commutative algebra 3 1.1 Integral closure. Noetherian rings . . . . . . . . . . . . . . . . 3 1.2 Noether’s normalization and Nullstellensatz . . . . . . . . . . 7

2 Affine geometry 9 2.1 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 Category of affine varieties . . . . . . . . . . . . . . . . . . . . 10 2.3 Examples of rational varieties . . . . . . . . . . . . . . . . . . 14 2.4 Smooth and singular points . . . . . . . . . . . . . . . . . . . 17 2.5 Dimension. Application: Tsen’s theorem . . . . . . . . . . . . 19

3 Projective geometry 22 3.1 Projective varieties . . . . . . . . . . . . . . . . . . . . . . . . 22 3.2 Morphisms of projective varieties . . . . . . . . . . . . . . . . 24

4 Local geometry 26 4.1 Localization, local rings, DVR . . . . . . . . . . . . . . . . . . 26 4.2 Regular local rings . . . . . . . . . . . . . . . . . . . . . . . . 28 4.3 Geometric consequences of unique factorization in O_{P} . . . . . 29

5 Divisors 30 5.1 The Picard group . . . . . . . . . . . . . . . . . . . . . . . . . 30 5.2 Automorphisms of P_{k}^{n} and of A_{k}^{n} . . . . . . . . . . . . . . . . 33 5.3 The degree of the divisor of a rational function on a projective

curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 5.4 Bezout theorem for curves . . . . . . . . . . . . . . . . . . . . 36 5.5 Riemann–Roch theorem . . . . . . . . . . . . . . . . . . . . . 39 5.6 From algebraic curves to error correcting codes . . . . . . . . . 41 A More algebra 43

A.1 Krull’s intersection theorem . . . . . . . . . . . . . . . . . . . 43 A.2 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 A.3 The topological space Spec(R) . . . . . . . . . . . . . . . . . . 44

B More geometry 46 B.1 Finite morphisms . . . . . . . . . . . . . . . . . . . . . . . . . 46 B.2 Functoriality of Pic . . . . . . . . . . . . . . . . . . . . . . . . 48

Basic references

Algebra:

S. Lang. Algebra. Addison-Wesley, 1965. M. Reid. Undergraduate commutative algebra. Cambridge University Press, 1995. B.L. Van der Waerden. Algebra. (2 volumes) Springer-Verlag.

Geometry:

M. Reid. Undergraduate algebraic geometry. Cambridge University Press, 1988. I.R. Shafarevich. Basic algebraic geometry. (2 volumes) Springer-Verlag. 1994.

Further references

Advanced commutative algebra:

M.F. Atiyah and I.G. Macdonald. Introduction to commutative algebra. Addison-Wesley, 1969. H. Matsumura. Commutative ring theory. Cambridge University Press, 1986

O. Zariski and P. Samuel. Commutative algebra. (2 volumes) Van Nostrand, 1958.

Advanced algebraic geometry:

R. Hartshorne. Algebraic geometry. Springer-Verlag, 1977.

1 Basics of commutative algebra

Let k be a field. (Affine) algebraic geometry studies the solutions of systems of polynomial equations with coefficients in k. Instead of a set of polynomials it is better to consider the ideal of the polynomial ring k[X_{1}, …, X_{n}] generated by them. The subset of k^{n} consisting of common zeros of the polynomials of an ideal I ⊂ k[X_{1}, …, X_{n}] is called the set of zeros of I. Such subsets of k^{n} are called closed algebraic sets.

Hence our main object of study will be the polynomial ring k[X_{1}, …, X_{n}], its ideals, their sets of zeros in k^{n}, and the quotient rings of k[X_{1}, …, X_{n}]. Here is the list of principal facts that we prove in this chapter:

(1) Every ideal of k[X_{1}, …, X_{n}] is a finitely generated k[X_{1}, …, X_{n}]- module. (Hilbert’s basis theorem.)

(2) Every quotient ring of k[X_{1}, …, X_{n}] is of the following form: it contains a polynomial ring k[Y_{1}, …, Y_{m}] over which it is a finitely generated module. (Emmy Noether’s normalization lemma).

(3) If k is algebraically closed, then all maximal ideals of k[X_{1}, …, X_{n}] are of the form (X_{1} − a_{1}, …, X_{n} − a_{n}), a_{i} ∈ k, that is, consist of polynomials vanishing at a point (a_{1}, …, a_{n}) ∈ k^{n}.

(4) Let k be algebraically closed. If a polynomial f vanishes at all the zeros of an ideal of k[X_{1}, …, X_{n}], then some power f^{m} belongs to this ideal. (Hilbert’s Nullstellensatz.)

Fact (1) says that speaking about ideals and systems of (finitely many) polynomial equations is the same thing. The meaning of (2) will be made clear in the geometric part of the course (an important corollary of (2) is the fact that any algebraic variety is birationally equivalent to a hypersurface). Fact (3) speaks for itself. Finally, (4) implies that certain ideals of k[X_{1}, …, X_{n}] bijectively correspond to closed algebraic sets.

1.1 Integral closure. Noetherian rings

Most of the time we assume that k is an algebraically closed field. When k is not algebraically closed, \overline{k} denotes a separable closure of k (unique up isomorphism, see [Lang]). Let R be a commutative ring with a 1. We shall always consider ideals I ⊂ R different from R itself. Then the quotient ring R/I is also a ring with 1. By definition, a ring is an integral domain if it has no zero divisors.

If M is an R-module, then 1 ∈ R acts trivially on M. An R-module M is of finite type if it is generated by finitely many elements, that is, if there exist a_{1}, …, a_{n} ∈ M such that any x ∈ M can be written as x = r_{1}a_{1} + … + r_{n}a_{n} for some r_{i} ∈ R.

If a module A over a ring R also has a ring structure (compatible with that of R in the sense that the map R → A given by r ↦ r.1_{A} is a ring homomorphism), then A is called an R-algebra. An R-algebra A is of finite type (or finitely generated) if there exist a_{1}, …, a_{n} ∈ A such that any x ∈ A can be written as a polynomial in a_{1}, …, a_{n} with coefficients in R.

Prime and maximal ideals. An ideal I ⊂ R is called prime if the quotient ring R/I has no zero divisors. An ideal I is maximal if it is not contained in another ideal (different from R). Then the ring R/I has no non-zero ideal (otherwise its preimage in R would be an ideal containing I), hence every element x ∈ R/I, x ≠ 0, is invertible (since the principal ideal (x) must concide with R/I, and thus contain 1), in other words, R/I is a field. Since a field has no non-trivial ideals, the converse is also true, so that I ⊂ R is maximal iff R/I is a field.

Integral closure. We shall consider various finiteness conditions. Suppose we have an extension of integral domains A ⊂ B. An element x ∈ B is called integral over A if it satisfies a polynomial equation with coefficients in A and leading coefficient 1.

Proposition 1.1 The following conditions are equivalent:

(i) x ∈ B is integral over A,

(ii) A[x] is an A-module of finite type,

(iii) there exists an A-module M of finite type such that A ⊂ M ⊂ B and xM ⊂ M.

The proof of (i) ⇒ (ii) and (ii) ⇒ (iii) is direct. Suppose we know (iii). Let m_{1}, …, m_{n} be a system of generators of M. Then xm_{i} = ∑_{i=1}^{n} b_{ij}m_{j}, where b_{ij} ∈ A.

Recall that in any ring R we can do the following “determinant trick”. Let S be a matrix with entries in R. Let adj(M) be the matrix with entries in R given by

adj(M)_{ij} = (−1)^{i+j}det(M(j, i)),

where M(i, j) is M with i-row and j-th column removed. It is an exercise in linear algebra that the product adj(M) · M is the scalar matrix with det(M) on the diagonal.

We play this trick to the polynomial ring R = A[T]. For M we take the n × n-matrix Q(T) such that Q(T)_{ij} = T δ_{ij} − b_{ij}. Let f(T) = det(Q(T)) ∈ A[T] (this is the analogue of the characteristic polynomial of x). We have a matrix identity

adj(Q(T)) · Q(T) = diag(f(T)).

We consider this as an identity between matrices over the bigger ring B[T]. We are free to assign T any value in B. Substitute T = x ∈ B, and apply these matrices to the column vector (m_{1}, …, m_{n})^{t}. Then the left hand side is zero. Hence f(x)m_{i} = 0 for any i. Since the m_{i} generate M the whole module M, is annihilated by f(x) ∈ B. In particular, f(x).1 = 0, that is, f(x) = 0. Now note that f(T) has coefficients in A and leading coefficient 1. QED

Remark. This proof does not use the fact that A and B are integral. If we assume this we can remove the condition A ⊂ M in (iii).

Definitions. Let A ⊂ B be integral domains, then B is integral over A if its every element is integral over A. The set of elements of B which are integral over A is called the integral closure of A in B.

Important example. Let K be a number field, that is, a finite extension of Q. One defines the ring of integers O_{K} ⊂ K as the integral closure of Z in K.

Let us prove some basic properties of integral elements.

Proposition 1.2 (a) The integral closure is a ring.

(b) Suppose that B is integral over A, and is of finite type as an A-algebra. Then B is of finite type as an A-module.

(c) Suppose that C is integral over B, and B is integral over A, then C is integral over A.

Proof. (a) Let x, y ∈ B be integral over A. Consider the A-module generated by all the monomials x^{i}y^{j}, i, j ≥ 0. It is of finite type, and xy and x + y act on it.

(b) Suppose that B is generated by b_{1}, ... , b_{n} as an A-algebra, then B is generated by monomials b_{1}^{i_{1}} … b_{n}^{i_{n}} as an A-module. All higher powers of each of the b_{i}’s can be reduced to finitely many of its powers using a monic polynomial whose root is b_{i}. There remain finitely many monomials which generate B as an A-module.

(c) Let x ∈ C. Consider the A-subalgebra D ⊂ C generated by x and the coefficients b_{i} of a monic polynomial with coefficients in B, whose root is x. Then D is an A-module of finite type, as only finitely many monomials generate it (the b_{i} are integral, and the higher powers of x can be reduced to lower powers) Now use (iii) of the previous proposition. QED

Definition. A ring is integrally closed or normal if it is integrally closed in its field of fractions.

Examples. Z and Z[\frac{1+\sqrt{−3}}{2}] are integrally closed, but Z[\sqrt{−3}] is not. If k is a field then k[x] and k[x, y] are integrally closed, but k[x, y]/(y^{2}−x^{2}−x^{3}) is not.

Noetherian rings. Another important finiteness property of rings is given in the following definition.

Definition-Proposition. A ring R satisfying any of the following equivalent properties is called Noetherian:

(i) any chain of ideals I_{1} ⊂ I_{2} ⊂ I_{3} ⊂ … of R stabilizes (that is, there is an integer m such that I_{m} = I_{m+1} = I_{m+2} = …),

(ii) any set of ideals of R contains a maximal element,

(iii) any ideal of R is generated by finitely many elements, that is, is an R-module of finite type.

Proof. The equivalence of (i) and (ii) is completely formal.

(iii) ⇒ (i): Let I = ∑ I_{j}, then I is an ideal which is generated, say, by x_{1}, …, x_{n} as an R-module. Take m such that I_{m} contains all the x_{i}, then the chain stabilizes at I_{m}.

(ii) ⇒ (iii) is based on a trick called “Noetherian induction”. Suppose that I ⊂ R is an ideal which is not of finite type as an R-module. Consider the set of subideals of I which are of finite type as R-modules. This set is not empty: it contains 0. Now it has a maximal element J ≠ I. Take x ∈ I \ J, then the ideal J + (x) ⊂ I is strictly bigger than J, but is of finite type as an R-module. Contradiction. QED

Examples of Noetherian rings: Fields, principal ideals domains, the ring of integers in a number field.

An easy exercise: Quotients of a Noetherian ring are Noetherian.

Theorem 1.3 (Hilbert’s basis theorem) If R is Noetherian, then so are the polynomial ring R[T] and the formal power series ring R[[T]].

Sketch of proof. Let I ⊂ R[T] be an ideal. We associate to it a series of ideals in R:

A_{0} ⊂ A_{1} ⊂ A_{2} ⊂ …,

where A_{i} is generated but the leading coefficients of polynomials in I of degree i. Since R is Noetherian, this chain of ideals stabilizes, say, at A_{r}. Then we have a finite collection of polynomials whose leading coefficients generate A_{0}, ... , A_{r}. Then the ideal of R[T] generated by these polynomials is I. See Lang’s book for the proof of the other statement. QED

1.2 Noether’s normalization and Nullstellensatz

The elements r_{1}, …, r_{n} of a k-algebra R are algebraically independent (over k) if the only polynomial f(x_{1}, …, x_{n}) with coefficients in k such that f(r_{1}, …, r_{n}) = 0, is the zero polynomial.

Theorem 1.4 (E. Noether’s normalization lemma.) Let k be any field, and I ⊂ k[T_{1}, …, T_{n}] be an ideal, R = k[T_{1}, …, T_{n}]/I. There exist algebraically independent elements Y_{1}, …, Y_{m} ∈ R such that R is integral over k[Y_{1}, …, Y_{m}].

Proof. If I = 0 there is nothing to prove. Suppose we have a non-zero polynomial f ∈ I. Let d be a positive integer greater than deg(f). Let us choose new variables in the following tricky way:

X′_{2} = X_{2}−(X_{1})^{d}, X′_{3} = X_{3}−(X_{1})^{d^{2}}, X′_{4} = X_{4}−(X_{1})^{d^{3}}, …, X′_{n} = X_{n}−(X_{1})^{d^{n−1}}.

Substituting this into f we rewrite it as a linear combination of powers of X_{1} and a polynomial, say, g containing no pure powers of X_{1}. We observe that the pure powers of X_{1} are of the form i_{1} + di_{2} + d^{2}i_{3} + … + d^{n−1}i_{n}. Since d > i_{s} all these integers are different, hence there is no cancellation among the pure powers of X_{1}. At least one such power enters with a nonzero coefficient. On the other hand, any power of X_{1} in g is strictly less than the corresponding pure power. Therefore, we get a polynomial in X_{1} with coefficients in k[X′_{2}, …, X′_{n}] and leading coefficient in k. Normalizing this polynomial we conclude that X_{1} is integral over R_{1} = k[X′_{2}, …, X′_{n}]/I ∩ k[X′_{2}, …, X′_{n}]. Hence R is integral over R_{1}. We now play the same game with R_{1} instead of R, and obtain a subring R_{2} over which R_{1} is integral. By Property (c) of integral ring extensions R is also integral over R_{2}. We continue like that until we get a zero ideal, which means that the variables are algebraically independent. QED

Theorem 1.5 Let k be an algebraically closed field. All maximal ideals of k[X_{1}, …, X_{n}] are of the form (X_{1} − a_{1}, …, X_{n} − a_{n}), a_{i} ∈ k, that is, consist of polynomials vanishing at a point (a_{1}, …, a_{n}) ∈ k^{n}.

Proof. Any polynomial has a Taylor expansion at the point (a_{1}, …, a_{n}). The canonical map

k[X_{1}, …, X_{n}] ⟶ k[X_{1}, …, X_{n}]/(X_{1} − a_{1}, …, X_{n} − a_{n})

sends f to f(a_{1}, …, a_{n}), hence is surjective onto k. It follows that the ideal (X_{1} − a_{1}, …, X_{n} − a_{n}) is maximal.

Let M a maximal ideal (recall that M ≠ k[X_{1}, …, X_{n}]), then K = k[X_{1}, …, X_{n}]/M is a field containing k. By Noetherian normalization K is integral over its subring A = k[Y_{1}, …, Y_{m}]. But K is a field, and we now show that then A must also be a field, in which case k[Y_{1}, …, Y_{m}] = k (no variables at all), and hence K is integral over k. Indeed, let x ∈ A, then it is enough to show that x^{−1} ∈ K also belongs to A. Since x^{−1} ∈ K is integral over A it is subject to a polynomial relation (x^{−1})^{n} +a_{n−1}(x^{−1})^{n−1} +…+a_{1}x^{−1} +a_{0} = 0, for some a_{i} ∈ A. Multiplying this by x^{n−1} we express x^{−1} as a polynomial in x with coefficients in A, hence x^{−1} ∈ A.

The k-algebra of finite type K is integral over k, hence by Proposition 1.2 (b) K is a k-module (= vector space over k) of finite type (= of finite dimension). Thus the field K is an algebraic extension of k. Since k is algebraically closed, we must have k = K. Now let a_{i} ∈ k be the image of X_{i} under the map k[X_{1}, …, X_{n}] → k = k[X_{1}, …, X_{n}]/M. Then M contains the maximal ideal (X_{1} − a_{1}, …, X_{n} − a_{n}), hence coincides with it. QED

Remark. When k is not supposed to be algebraically closed, this proof shows that the quotient by a maximal ideal of k[X_{1}, …, X_{n}] is a finite extension of k.

Corollary 1.6 Let k be an algebraically closed field. If the polynomials of an ideal I ⊂ k[X_{1}, …, X_{n}] have no common zeros in k^{n}, then I = k[X_{1}, …, X_{n}].

Proof. Assume I ≠ k[X_{1}, …, X_{n}]. Hilbert’s basis theorem says that k[X_{1}, …, X_{n}] is Noetherian. Then I is contained in a maximal ideal, since the set of ideals that contain I has a maximal element, by (ii) of Definition-Proposition above. Therefore I ⊂ (X_{1} − a_{1}, …, X_{n} − a_{n}), for some a_{i} ∈ k, since all the maximal ideals are of this form by the previous result. But then all the polynomials of I vanish at the point (a_{1}, …, a_{n}), which is a contradiction. QED

Theorem 1.7 (Nullstellensatz.) Let k be an algebraically closed field. If a polynomial f vanishes at all the zeros of an ideal I ⊂ k[X_{1}, …, X_{n}], then f^{m} ∈ I for some positive integer m.

Proof. We know that I is generated by finitely many polynomials, say, I = (g_{1}, …, g_{r}). Let T be a new variable. Consider the ideal J ⊂ k[T, X_{1}, …, X_{n}] generated by g_{1}, …, g_{r} and T f − 1. We observe that these polynomials have no common zero. The previous corollary implies that J = k[T, X_{1}, …, X_{n}], in particular, J contains 1. Then there exist polynomials p, p_{1}, …, p_{r} in variables T, X_{1}, …, X_{n} such that

1 = p(T f − 1) + p_{1}g_{1} + … + p_{r}g_{r}.

Note that this is an identity in variables T, X_{1}, …, X_{n}. Thus we can specialize the variables anyway we like. For example, we can set T = 1/f. Multiplying both sides by an appropriate power of f we get an identity between polynomials in variables X_{1}, …, X_{n}, which gives that some power of f belongs to I = (g_{1}, …, g_{r}). QED

2 Affine geometry

2.1 Zariski topology

Let k be any field. Let us prove some easy facts about closed algebraic sets. If X ⊂ k^{n} we denote by I(X) ⊂ k[X_{1}, …, X_{n}] the ideal consisting of polynomials vanishing at all the points of X. We denote by Z(J) the set of zeros of an ideal J ⊂ k[X_{1}, …, X_{n}]. It is a tautology that X ⊂ Z(I(X)) and J ⊂ I(Z(J)). If X is a closed algebraic set, then X = Z(I(X)) (if X = Z(J), then I(Z(J)) ⊃ J, hence Z(I(Z(J))) ⊂ Z(J)).

Exercise. Show that if X ⊂ A_{k}^{n} and Y ⊂ A_{k}^{m} are closed subsets, then X × Y ⊂ A_{k}^{n+m} is a closed subset.

It is clear that the function J ↦ Z(J) reverses inclusions; associates the empty set to the whole ring, and the whole affine space k^{n} to the zero ideal; sends the sum of (any number of) ideals to the intersection of corresponding closed sets; and sends the intersection I_{1} ∩ I_{2} to Z(I_{1}) ∪ Z(I_{2}) (a part of the last property is not completely obvious: if P ∉ Z(I_{1}) ∪ Z(I_{2}), then f(P) ≠ 0 for some f ∈ I_{1} and g(P) ≠ 0 for some g ∈ I_{2}, but then (fg)(P) ≠ 0, whereas fg ∈ I_{1} ∩ I_{2}).

Because of these properties we can think of closed algebraic sets as the closed sets for some topology on k^{n} (any intersections and finite unions are again closed, as are the empty set and the whole space). This topology is called Zariski topology. In the case when k = C or k = R we can compare it with the usual topology on C^{n} where closed sets are the zeros of continuous functions. Any Zariski closed set is also closed for the usual topology but not vice versa. Hence the Zariski topology is weaker. Another feature is that any open subset of k^{n} is dense (its closure is the whole k^{n}).

Definition. A closed algebraic subset X ⊂ k^{n} is irreducible if there is no decomposition X = X_{1} ∪ X_{2}, where X_{1} ≠ X and X_{2} ≠ X are closed algebraic sets.

Proposition 2.1 A closed algebraic subset X ⊂ k^{n} is irreducible iff I(X) is a prime ideal. Any closed set has a unique decomposition into a finite union of irreducible subsets X = ∪_{i}X_{i} such that X_{i} ⊄ X_{j} for i ≠ j (these X_{i}’s are called the irreducible components of X).

Proof. Let us prove the first statement. If we have X = X_{1} ∪ X_{2}, then since X = Z(I(X)) for any algebraic set, I(X) is a proper subset of I(X_{i}). If f_{i} ∈ I(X_{i}) \ I(X), then f_{1}f_{2} ∈ I(X), hence I(X) is not a prime ideal. Conversely, if I(X) is not prime, we can find two polynomials f_{1} and f_{2} not in I(X) such that f_{1}f_{2} ∈ I(X), and define I_{i} = (I(X), f_{i}), X_{i} = Z(I_{i}). There exists a point P in X such that f_{1}(P) ≠ 0, hence P ∉ X_{1} which implies X_{1} ≠ X. Similarly we have X_{2} ≠ X. Therefore X = X_{1} ∪ X_{2} is not irreducible.

Let us prove the second statement. If X is not irreducible, we have some decomposition X = X_{1} ∪ X_{2}, and then continue for X_{1} and X_{2}. At some point we must stop because the chain of ideals I(X) ⊂ I(X_{1}) ⊂ … stabilizes somewhere (the ring k[X_{1}, …, X_{n}] being Noetherian). If ∪_{i}X_{i} = ∪_{j}Y_{j} are two decompositions into irreducible subsets, then X_{i} = ∪_{j}(X_{i} ∩ Y_{j}), and hence X_{i} = X_{i} ∩ Y_{j} for some j (since X_{i} is irreducible). For the analogous reason we have Y_{j} = Y_{j} ∩ X_{i′} for some i′. Then X_{i} ⊂ X_{i′}, hence i = i′. It follows that X_{i} = Y_{j}. Hence the two decompositions differ only in order. QED

Up till now we did not use Hilbert’s Nullstellensatz. Let k be an algebraically closed field. Let us call an ideal I ⊂ k[X_{1}, …, X_{n}] radical if f^{m} ∈ I implies f ∈ I. A corollary of Hilbert’s Nullstellensatz is that radical ideals bijectively (via operations I and Z) correspond to closed algebraic sets. The most important class of radical ideals are prime ideals. Again, by Hilbert’s Nullstellensatz, these bijectively correspond to irreducible closed algebraic sets. A particular case of prime ideals are maximal ideals, they correspond to points of k^{n}.

Zero sets of irreducible polynomials of k[X_{1}, …, X_{n}] are called irreducible hypersurfaces.

2.2 Category of affine varieties

An affine variety is a closed irreducible algebraic subset of k^{n} for some n. The variety k^{n} will be also denoted A_{k}^{n}, and called the affine space of dimension n.

Let X ⊂ A_{k}^{n} be an affine variety. Let J = I(X) be the corresponding prime ideal. Let us denote k[X] := k[X_{1}, …, X_{n}]/J. Then k[X] is an integral k-algebra of finite type: k[X] contains no zero divisors. k[X] is called the coordinate ring of X. The fraction field of k[X] is denoted by k(X), and is called the function field of X. Its elements are called rational functions as opposed to the elements of k[X] which are called regular functions.

The function field k(X) is an important object defined by X. Two affine varieties X and Y are called birationally equivalent if k(X) = k(Y). A variety X is called rational if k(X) is a purely transcendental extension of k, that is, k(X) = k(T_{1}, …, T_{l}). In other words, X is rational if and only if X is birationally equivalent to the affine space. It is a classical, and often a difficult problem of algebraic geometry to determine whether or not two given varieties are birationally equivalent.

Affine varieties form a category, where a morphism X → Y, X ⊂ A_{k}^{n}, Y ⊂ A_{k}^{m}, is given by a function representable by m polynomials in n variables. The varieties X and Y are called isomorphic if there are morphisms f : X → Y and g : Y → X such that fg and gf are identities.

Proposition 2.2 Let X ⊂ A_{k}^{n} and Y ⊂ A_{k}^{m} be affine algebraic varieties.

(a) A morphism f : X → Y defines a homomorphism of k-algebras f^{∗} : k[Y] → k[X] via the composition of polynomials.

(b) Any homomorphism of k-algebras ϕ : k[Y] → k[X] is of the form ϕ = f^{∗} for a unique morphism f : X → Y.

(c) f : X → Y is an isomorphism of affine varieties if and only if f^{∗} : k[Y] → k[X] is an isomorphism of k-algebras.

Proof. (a) follows from the fact that the composition of polynomials is a polynomial.

(b) Let x_{1}, …, x_{n} be the coordinates on X, and t_{1}, …, t_{m} be the coordinates on Y. Let Φ be the composition of the following homomorphisms of k-algebras

k[t_{1}, …, t_{m}] ⟶ k[Y] = k[t_{1}, …, t_{m}]/I(Y) ⟶ k[X] = k[x_{1}, …, x_{n}]/I(X).

Let f_{i} = Φ(t_{i}), i = 1, …, m. The polynomial map f = (f_{1}, …, f_{m}) maps X to A_{k}^{m}. Let F (t_{1}, …, t_{m}) be a polynomial. Since we consider homomorphisms of k-algebras we have

F (f_{1}, …, f_{m}) = F (Φ(t_{1}), …, Φ(t_{m})) = Φ(F (t_{1}, …, t_{m})).

If F ∈ I(Y), then Φ(F) = 0. Hence all the polynomials from I(Y) vanish on f(X), that is, f(X) ⊂ Z(I(Y)) = Y.

Finally, f^{∗} = ϕ since these homomorphisms take the same values on the generators t_{i} of the k-algebra k[Y].

(c) follows from (a) and (b). QED

Examples. (1) Hypersurfaces in A_{k}^{n} bijectively correspond to principal ideals in k[x_{1}, …, x_{n}] (via the usual operations Z and I).

(2) Square matrices of size n are parametrized by the points of A_{k}^{n^{2}}. The multiplication of matrices is a morphism A_{k}^{n^{2}} ×A_{k}^{n^{2}} → A_{k}^{n^{2}}. The determinant is a morphism A_{k}^{n^{2}} → A_{k}^{1}.

The closed subset given by det(M) = 1 is the special linear group SL(n). Note that the inverse is an isomorphism SL(n) → SL(n). The group operations are morphisms – when that is the case then the group is called an algebraic group. SL(n) ⊂ A_{k}^{n^{2}} is a hypersurface of degree n.

(3) The orthogonal group O(n) ⊂ A_{k}^{n^{2}} given by the conditions M ·M^{t} = I is another example of a closed subset which is an algebraic group. It is defined by n^{2} quadratic polynomials.

Exercise. Show that O(n) is not irreducible. (Hint: what are the possible values of det(M)?)

Zariski topology on A_{k}^{n} induces a topology on a variety X ⊂ A_{k}^{n}. An open subset U ⊂ X is an intersection of X with an open set of A_{k}^{n}. Such sets are called a quasi-affine varieties. An example of a quasi-affine variety is the general linear group GL(n) (square matrices with non-zero determinant). Later we’ll see that GL(n) is isomorphic to an affine variety (see the end of this subsection).

Definition. A rational function f ∈ k(X) is called regular at a point P of X if f = g/h, where g, h ∈ k[X] and h(P) ≠ 0. A function is regular on an open set U ⊂ X if it is regular at every point of U.

The ring of regular functions on an open subset U ⊂ X is denoted by k[U]. Since k[X] ⊂ k[U] ⊂ k(X) the fraction field of k[U] if k(X).

To a rational function f ∈ k(X) one associates “the ideal of denominators” D_{f} ⊂ k[X] consisting of regular functions h such that hf ∈ k[X] (it is clearly an ideal!). The set of all points P where f is regular is X \Z(D_{f}). Indeed, we can write f = g/h, g, h ∈ k[X], h(P) ≠ 0, if and only if P ∉ Z(D_{f}).

An immediate corollary of the Nullstellensatz says that if I ⊂ k[X] is an ideal, and f ∈ k[X] vanishes at all the common zeros of I in X, then f^{s} ∈ I for some s > 0. (Apply Theorem 1.7 to the pre-image of I in k[x_{1}, …, x_{n}] under the natural surjective map.) We’ll often use the Nullstellensatz in this form.

Lemma 2.3 Let X be an affine variety. The subset of k(X) consisting of functions regular at all the points of X is k[X]. A function is regular on the open subset given by h ≠ 0, for h ∈ k[X], if and only if f ∈ k[X][h^{−1}], in other words, if f = g/h^{s} for some g ∈ k[X] and s > 0.

Proof. Let f be such a function. Then Z(D_{f}) = ∅. By Corollary 1.6 D_{f} must be the whole ring, hence contains 1, hence f ∈ k[X]. This proves the first statement. To prove the second statement we note that Z(D_{f}) is contained in the closed set given by h = 0. By Nullstellensatz if h vanishes on Z(D_{f}), then a power of h is in D_{f}. QED

One defines rational maps using rational functions instead of regular ones. Rational maps are not everywhere defined, that is, are not functions! A rational map is called dominant if its image (=the image of the set of points where the map is actually defined, that is, is regular) is dense, that is, not contained in a smaller subvariety. The following proposition is proved along the same lines as Proposition 2.2.

Proposition 2.4 (a) A dominant rational map f : X − − > Y defines a homomorphism of k-algebras f^{∗} : k(Y) → k(X).

(b) Any homomorphism of k-algebras ϕ : k(Y) → k(X) is of the form ϕ = f^{∗} for a unique dominant rational map f : X − − > Y.

We need the condition that f is dominant, as otherwise the composition of rational functions is not always defined. (On substituting the f_{i} for the coordinates we may have to divide by zero! But this can’t happen if the map is dominant, as the f_{i} then do not satisfy a non-trivial polynomial condition.)

Finally, a morphism of quasi-affine varieties is an everywhere defined rational map. An isomorphism is a morphism that has an inverse. For example, the map M ↦ M^{−1} is a morphism GL(n) → GL(n) (and in fact, an isomorphism). On the affine space of square matrices A_{k}^{n^{2}} this is just a rational map.

A rational map which has an inverse (at the same time left and right) is called a birational map or a birational equivalence. For example, the map M ↦ M^{−1} is a birational equivalence of A_{k}^{n^{2}} with itself.

Proposition 2.5 The following conditions are equivalent: a rational map f : X − − > Y is

(i) birational,

(ii) f is dominant, and f^{∗} : k(Y) → k(X) is an isomorphism of kalgebras,

(iii) there exist open sets in X and Y such that f defines an isomorphism between them.

Proof. The only implication which is not immediate is (i) ⇒ (iii). But this is also easy, see [Reid, UAG], (5.8).

Theorem 2.6 Any affine variety is birationally equivalent to a hypersurface.

Proof. By Noether’s normalization there exist algebraically independent elements y_{1}, …, y_{m} ∈ k[X] such that k[X] is integral over k[y_{1}, …, y_{m}]. Then the field extension k(y_{1}, …, y_{m}) ⊂ k(X) is finite. In fact, one can arrange that this extension is separable (this is automatic if the characteristic of k is 0, see [Reid, UAG], (3.16) for the case of finite characteristic). Any separable finite extension can be obtained by adding just one element (primitive element theorem), say, y. Let F (t) be an irreducible polynomial with coefficients in k(y_{1}, …, y_{m}) such that F (y) = 0. By fiddling with the coefficients can assume that the coefficients of F (t) are in k[y_{1}, …, y_{m}], and that F is irreducible as a polynomial in y_{1}, …, y_{m}, t. Then k(X) is the fraction field of k[y_{1}, …, y_{m}, t]/(F (t)). Let V ⊂ A_{k}^{m+1} be the hypersuface given by F = 0. Then k(V) = k(X). QED

The following lemma allows us to consider only affine open neighbourhoods.

Proposition 2.7 Every open neighbourhood of a point of an affine variety contains a neighbourhood isomorphic to an affine variety (and not just quasiaffine).

Proof. It is enough to show how to remove zeros of polynomials. Let X ⊂ A_{k}^{n} be a closed set given by the polynomial equations f_{1} = … = f_{m} = 0 in variables T_{1}, …, T_{n}, and let g be another polynomial, g(P) ≠ 0. Let X_{0} ⊂ X be given by g ≠ 0. Consider the closed subset X_{1} ⊂ A_{k}^{n+1} with coordinates T_{0}, T_{1}, …, T_{n} given by f_{1} = … = f_{m} = T_{0}g − 1 = 0. Then the projection (T_{0}, T_{1}, …, T_{n}) ↦ (T_{1}, …, T_{n}) defines an isomorphism of X_{1} with X_{0}. QED

Exercise. Show that GL(n) is isomorphic to a closed subset of an affine space.

2.3 Examples of rational varieties

Let’s do some concrete algebraic geometry.

Let X ⊂ A_{k}^{n} be a hypersurface given by f(X_{1}, …, X_{n}) = 0. We shall always assume that no linear change of coordinates X_{1}, …, X_{n} reduces f to a polynomial in n − 1 variables, and that no linear change of coordinates reduces f to a homogeneous polynomial. Then X is called non-conical. We shall only consider non-conical hypersurfaces. We call the degree of X the degree of f.

In this section we examine some examples of rational hypersurfaces. By definition X ⊂ A_{k}^{n} is rational if there is a birational map A_{k}^{m} − − > X. Such a map is given by rational functions in m variables. The meaning of rationality is that the points of X can be parametrized by rational functions such that this parametrization is an isomorphism on a non-empty open set.

Remark. We observe that a rational variety over any infinite field k has infinitely many k-points. Indeed, it contains an dense open subset isomorphic to a dense open subset of A_{k}^{n}, and the latter contains infinitely many k-points (easy exercise).

(1) Quadrics (degree 2). If k is not algebraically closed there may be no k-point on X, e.g. x^{2} + y^{2} + 1 = 0 over k = R. Then X is not rational.

The following proposition is a generalization of the classical parametrization of the conic x^{2} + y^{2} = 1 by rational functions x = (t^{2} − 1)/(t^{2} + 1), y = 2t/(t^{2} + 1).

Proposition. A non-conical quadric with a k-point is rational.

Proof. The idea is to exploit the classical stereographic projection. We can assume that the k-point is N = (0, 0, …, 0) (N stands for the North Pole ...). Then we can write

f = Q(x_{1}, …, x_{n}) + L(x_{1}, …, x_{n}),

where the homogeneous polynomials L and Q have degrees 1 and 2, respectively (there is no constant term). Chose a hyperplane H not passing through N, say the one given by x_{1} = 1. The coordinates on H are x_{2}, …, x_{n}. Let L be the line passing through N and the point (1, x_{2}, …, x_{n}). Consider L ∩ X. The line L is the set (t, tx_{2}, …, tx_{n}), t ∈ \overline{k}, hence the \overline{k}-points of L ∩ X correspond to the roots of the following equation in t:

t^{2}Q(1, x_{2}, …, x_{n}) + tL(1, x_{2}, …, x_{n}) = 0.

The root t = 0 is the point N, but it is the other root t = −L/Q that is interesting to us. Define the rational map ϕ : A_{k}^{n−1} − − > X by sending (x_{2}, …, x_{n}) to “the other” (residual) intersection point:

ϕ(x_{2}, …, x_{n}) = −\frac{L(1, x_{2}, …, x_{n})}{Q(1, x_{2}, …, x_{n})}(1, x_{2}, …, x_{n}).

The image is obviously in X. The inverse map sends (x_{1}, x_{2}, …, x_{n}) to (x_{2}/x_{1}, …, x_{n}/x_{1}). (Check that this is indeed the inverse to ϕ, both right and left.) QED

(2) A cubic surface. Assume that the characteristic of k is not 3, and that k contains a non-trivial cubic root ρ of 1. Consider the hypersurface X ⊂ A_{k}^{3} given by the equation

x_{1}^{3} + x_{2}^{3} + x_{3}^{2} = 1.

It contains two skew lines (non-parallel with empty intersection)

L_{0} = \{(1, x, −x)\}, x ∈ \overline{k}, L_{1} = \{(y, −ρy, 1)\}, y ∈ \overline{k}.

Let us write P_{x} = (1, x, −x), Q_{y} = (y, −ρy, 1). Let L_{x,y} be the line passing through P_{x} and Q_{y},

L_{x,y} = \{tP_{x} + (1 − t)Q_{y}\}, t ∈ \overline{k}.

We now contruct a rational map ϕ : A_{k}^{2}−− > X by sending (x, y) to the third (“residual”) point of intersection of L_{x,y} with X. More precisely, substituting tP_{x} + (1 − t)Q_{y} into the equation of X we get a cubic polynomial in t with coefficients in k(x, y). It has two obvious roots t = 0 (the point Q_{y}) and t = 1 (the point P_{x}). After checking that its degree is exactly 3 (this is enough to check for a particular choice of x and y, say x = 1, y = 0) we can write it as c(x, y)t(t − 1)(t − λ(x, y)), where λ(x, y) ∈ k(x, y). Set

ϕ(x, y) = λ(x, y)P_{x} + (1 − λ(x, y))Q_{y}.

By construction this is rational map A_{k}^{2} − − > X.

Exercise. Find λ(x, y), then check that ϕ is a birational equivalence. There is a geometric construction of the inverse map. Let R be a point

of X \ (L_{0} ∪ L_{1}). Let Π_{i} be the plane spanned by R and L_{i}. Then for R in a non-empty Zariski open subset of X the intersection Π_{0} ∩ L_{1} is a single point on L_{1}, call it Q. Similarly, Π_{1} ∩ L_{0} is a point on L_{0}, call it P. We have a well-defined rational map

f : X \ (L_{0} ∪ L_{1}) − −− > A_{k}^{2}, f(R) = (P, Q).

Check that f is inverse to ϕ.

(3) Conic bundles over the affine line. Let k be algebraically closed. Let

F (x_{0}, x_{1}, x_{2}) = ∑_{i_{j}≥0, i_{0}+i_{1}+i_{2}≤2} f_{i_{0},i_{1},i_{2}}(t)x_{0}^{i_{0}}x_{1}^{i_{1}}x_{2}^{i_{2}}

be a rank 3 quadratic form over the field K = k(t). By Tsen’s theorem (see Theorem 2.15 below) the conic C ⊂ A_{K}^{2} given by F (1, x_{1}, x_{2}) = 0 has a Kpoint. But we know from (1) that a non-conical quadric with a rational point is rational. Hence C is rational over K, in other words, K(C) = K(y) = k(t, y) is a purely transcendental extension of K, and hence of k as well.

Multiplying the coefficients of F by a common multiple, we can assume that they are actually in k[t]. Hence the equation F (t; 1, x_{1}, x_{2}) = 0 defines a surface X ⊂ A_{k}^{3}. If we assign to t a value in k we get a plain conic. Thus X is a pencil (1-parameter family) of conics, or a conic bundle. We thus proved that any conic bundle over the affine line over an algebraically closed field is rational.

Exercise. Let X be a smooth cubic surface, and L ⊂ X be a straight line. Planes Π passing through L form a 1-dimensional family. Show that for almost all planes Π the intersection Π ∩ X is the union of L and a conic. Deduce that X is birationally equivalent to a pencil of conics. Now (3) gives another proof of rationality of X.

2.4 Smooth and singular points

Let X ⊂ A_{k}^{n} be an affine variety, and suppose that the ideal of X is generated by polynomials f_{1}, …, f_{m}. Let P = (a_{1}, …, a_{n}) be a point of X. We define the partial derivatives ∂f_{i}/∂T_{j} as partial derivatives of a polynomial (in a purely algebraic way, this works over any field). Then we get some constants ∂f_{i}/∂T_{j}(P) ∈ k. Recall that an affine subspace of A_{k}^{n} is a translation of a vector subspace.

Definition. The affine subspace of k^{n} given by the system of linear equations in T_{1}, …, T_{n}

∑_{j=1}^{n} ∂f_{i}/∂T_{j}(P)(T_{j} − a_{j}) = 0, i = 1, …, m,

is called the tangent space to X at P = (a_{1}, …, a_{n}), as is denoted by T_{X,P}.

Exercise. Let m = 1. Show that T_{X,P} is the union of lines passing through P such that the restriction of f to this line is a polynomial in one variable with a multiple root at P.

All we need to know to compute T_{X,P} are partial derivatives of the equations definiting X at P. Thus it makes sense to define T_{U,P}, where U is an open subset of X containing P, by the same formula, so that T_{U,P} = T_{X,P}.

We have a function from X to non-negative integers given by dim(T_{X,P}).

Lemma 2.8 (Upper semi-continuity) The subset of X given by the condition dim(T_{X,P}) ≥ s is a closed subset of X.

Proof. The condition dim(T_{X,P}) ≥ s is equivalent to the condition that the rank of the matrix (∂f_{i}/∂T_{j}(P)) is at most n − s, which is equivalent to the vanishing of the determinants of all its square submatrices of size n − s + 1. But these determinants are polynomials in a_{1}, …, a_{n}, which implies our statement. QED

It follows that the subset of X consisting of points where dim(T_{X,P}) takes its minimal value is open. It is non-empty, and X is irreducible, hence this subset is dense. The points in this subset are called smooth or non-singular, and all the other points are called singular.

Exercises. When is the curve X ⊂ A_{k}^{2} given by x^{a} + y^{b}, where a and b are positive integers, non-singular at (0, 0)?

Show that any hypersurface defined by a homogeneous polynomial of degree at least 2 is singular at the origin.

Proposition 2.9 Let m_{P} = (T_{1} − a_{1}, …, T_{n} − a_{n}) ∈ k[X] be the maximal ideal of a point P in the coordinate ring of X. Then the tangent space T_{X,P} (considered as a vector space with origin at P) is canonically dual to the quotient ring m_{P} /m_{P}^{2} .

Proof. We first do an exercise in linear algebra. Let V be a vector space, V^{∗} its dual space (that is, the space of linear forms V → k), and S a subspace of V^{∗}. Consider the subspace

W = {v ∈ V |f(v) = 0, for any f ∈ S}.

Its dual space W^{∗} can be identified with V^{∗}/S (the restrictions of linear functions on V). Since W is canonically isomorphic to (W^{∗})^{∗} we conclude that W is canonically isomorphic to (V^{∗}/S)^{∗}.

After a translation in k^{n} we can assume without loss of generality that a_{i} = 0, i = 1, …, n. Let V = k^{n}. The coordinates T_{i} form a basis of the dual space V^{∗}. Let S ⊂ V^{∗} be the subspace of linear terms of functions from I(X), that is, linear combinations of ∑_{j=1}^{n} ∂f_{i}/∂T_{j}(P)T_{j}. The k-vector space m_{P} /m_{P}^{2} consists of linear combinations of T_{i}’s modulo linear terms of functions from I(X). This means that m_{P} /m_{P}^{2} = V^{∗}/S. On the other hand, by definition T_{X,P} ⊂ V is the space of zeros of the functions from S. Now the exercise in linear algebra above gives the required isomorphism. QED

The proposition implies that the tangent space to X at P is an intrinsic invariant of X, in the sense that it only depends on the isomorphism class of X, and not on the particular embedding.

Exercise. Prove that computing m_{P} /m_{P}^{2} for X and for an open subset U ⊂ X containing P gives the same thing. (The resulting vector spaces are canonically isomorphic.)

Let p : A_{k}^{n} → A_{k}^{m} be a collection of polynomials p_{1}, …, p_{m} in n variables that defines a morphism p : X → Y. The m × n-matrix of partial derivatives J = (∂p_{i}/∂x_{j}) is called the Jacobian matrix. Using the chain rule for partial derivatives one easily checks that J defines a linear map p_{∗} : T_{X,P} → T_{Y,Q}, where Q = p(P). The construction of p_{∗} from p is functorial in the sense that if q : Y → Z is a morphism, then (qp)_{∗} = q_{∗}p_{∗}. In particular, if p is an isomorphism, then so is p_{∗}. This is a practical way to check that a given map is not an isomorphism at a given point.

2.5 Dimension. Application: Tsen’s theorem

Definition. Let X be an affine variety over a field k. The transcendence degree of k(X) over k is called the dimension of X. The dimension of a closed affine set is defined as the maximum of the dimensions of its irreducible components.

For example, dim(A_{k}^{n}) = n since k(A_{k}^{n}) is the purely transcendental field extension k(x_{1}, …, x_{n}).

Proposition 2.10 Let X ⊂ A_{k}^{n} be a variety, Y ⊂ X a subvariety, Y ≠ X. Then dim(Y) < dim(X).

Proof. Let tr.deg._{k}k(Y) = m, and choose u_{1}, …, u_{m} ∈ k[X] such that their images in k[Y] are algebraically independent. Then u_{1}, …, u_{m} are algebraically independent in k(X). In particular, tr.deg._{k}k(X) ≥ m. For contradiction assume that tr.deg._{k}k(X) = m. Since Y ≠ X the ideal I(Y) ⊂ k[X] is non-zero. Any u ∈ I(Y), u ≠ 0, must be algebraically dependent on u_{1}, …, u_{m}. Thus there exists a polynomial F (t, t_{1}, …, t_{m}) = ∑_{i} a_{i}(t_{1}, …, t_{m})t^{i} such that F (u, u_{1}, …, u_{m}) is zero in k[X]. We can assume that the a_{i} are polynomials, and that F is irreducible in all the variables t, t_{1}, …, t_{m}. In particular, the constant term a_{0}(t_{1}, …, t_{m}) is not the zero polynomial. Since the image of u in k[Y] is zero, the image of a_{0}(u_{1}, …, u_{m}) is zero in k[Y]. This gives an algebraic relation between the images of u_{1}, …, u_{m}, contrary to our choice of u_{1}, …, u_{m}. This contradiction proves that tr.deg._{k}k(X) < m. QED

Corollary 2.11 A subvariety X ⊂ A_{k}^{n} has dimension n − 1 if and only if X is a hypersurface. Then I(X) is a principal ideal of the polynomial ring k[T_{1}, …, T_{n}].

Proof. It remains to show that dim(X) = n − 1 implies that X is a hypersurface. Since the dimensions are different we have X ≠ A_{k}^{n}. Let F be a non-zero element in I(X). We write F as a product of irreducible factors F = F_{1} … F_{m}. Then X = ∪(X ∩ {F_{i} = 0}). Since X is irreducible we have X = X ∩ {F_{j} = 0} for some j, hence F_{i} ∈ I(X). Replace F by F_{j}. Then X is contained in the irreducible hypersurface given by F = 0. Since the dimensions are equal, these varieties coincide by Proposition 2.10. QED

Theorem 2.12 The dimension of the tangent space at a smooth point of X equals dim(X).

Proof. We know that X is birationally equivalent to a hypersurface V ⊂ k^{m+1} given by some (non-constant) polynomial F = 0. Since k(X) = k(V) we conclude that dim(X) = dim(V).

The dimension of the tangent space to X at any smooth point is the same, and can be computed in any non-empty open subset of X. Wee can arrange that the birational map X − − > V is an isomorphism on this open set. This reduces the whole computation to V.

By Proposition 2.10 dim(V) ≤ m. On the other hand, the transcendence degree of k(V) is at least m. Indeed, F depends on at least one variable, say x_{1}. Then the images of x_{2}, …, x_{m+1} in k[V] are algebraically independent (as otherwise we would have a polynomial G(x_{2}, …, x_{m+1}) in the principal ideal (F) which is impossible since F depends on x_{1} whereas G does not). Therefore, dim(V) = m.

On the other hand, the dimension of T_{V,P} for P in a certain dense open subset of V is m: if all the partial derivatives of F vanish everywhere on V, they must belong to the principal ideal (F) by Hilbert’s Nullstellensatz. Since they have degrees less than the degree of F, they must be zero polynomials. If the characteristic of k is zero, this implies that F is a constant, which is a contradiction. If the characteristic is p, then F is a p-th power, which contradicts the irreducibility of F. QED

We quote the following result without proof (see Ch. 1 of [Shafarevich] or Ch. 11 of [Atiyah-McDonald]).

Theorem 2.13 Let k be an algebraically closed field. Let X ⊂ A_{k}^{n} be a variety, F be a polynomial taking both zero and non-zero values on X. Then dim(X ∩ {F = 0}) = n − 1.

Corollary 2.14 Let F_{1}, …, F_{m}, m ≤ n, be homogeneous polynomials in n variables. Then the closed affine set X given by F_{1} = … = F_{m} = 0 has dimension at least n − m. In particular, X is not empty.

Proof. Note that the all-zero point is in X. Thus at each successive intersection the dimension drops at most by one. QED

The natural place of this statement is in intersection theory of subvarieties of the projective space. To demonstrate its force we now deduce just one corollary.

Recall that k is algebraically closed. Let

F (x_{0}, x_{1}, x_{2}) = ∑_{i_{j}≥0, i_{0}+i_{1}+i_{2}≤2} f_{i_{0},i_{1},i_{2}}(t)x_{0}^{i_{0}}x_{1}^{i_{1}}x_{2}^{i_{2}}

be a homogeneous polynomial of degree 2 with coefficients in k[t].

Theorem 2.15 (Tsen) There exist non-zero polynomials p_{0}(t), p_{1}(t), p_{2}(t) with coefficients in k such that x_{0} = p_{0}(t), x_{1} = p_{1}(t), x_{2} = p_{2}(t) is a solution of the equation F (x_{0}, x_{1}, x_{2}) = 0.

Proof. Let m be a positive integer. Polynomials of degree m form an m + 1-dimensional vector space over k. Hence the dimension of the vector space of coefficients of p_{0}(t), p_{1}(t), p_{2}(t) is 3m + 3. We observe that the conditions on these coefficients that must be satisfied in order for x_{0} = p_{0}(t), x_{1} = p_{1}(t), x_{2} = p_{2}(t) to be a solution, are given by homogeneous (quadratic) polynomials. Let us compute the number of these conditions.

Let ℓ be the maximum of the degrees of the f_{i_{0},i_{1},i_{2}}(t). Suppose that the degree of p_{j}(t), j = 1, 2, 3, is at most m. Then the degree of F (p_{0}(t), p_{1}(t), p_{2}(t)) is at most ℓ + 2m. This means that the coefficients of the p_{j}(t) must satisfy ℓ + 2m + 1 homogeneous polynomial conditions (1 must be added to provide for the zero constant term). For large m we have 3m + 3 > ℓ + 2m + 1, hence the number of variables exceeds the number of equations. By Corollary 2.14 the dimension of the closed affine set of polynomials that are solutions, is positive. We conclude that there exist such non-zero polynomials. QED

A more general approach to dimension. Let R be a ring. The Krull dimension of R is defined as the supremum of all integers n such that there exists a chain I_{0} ⊂ I_{1} ⊂ … ⊂ I_{n} of distinct prime ideals of R. For example, the Krull dimension of a field is 0. The dimension of Z and, more generaly, of the ring of integers in a number field is 1. The dimension of k[T] is also 1. The dimension of k[X, Y] is 2, etc.

In fact all given definitions of dimension are equivalent.

Theorem 2.16 Let X be an affine variety, then the Krull dimension of k[X] equals the transcendence degree of k(X).

This can be deduced from Theorem 2.13.

3 Projective geometry

3.1 Projective varieties

Below is the list of main notions, and the outline of differences with the affine case.

The projective space P_{k}^{n} is the set of equivalence classes of points of A_{k}^{n+1}\ {(0, …, 0)}, where two points are equivalent if they differ by a common non-zero multiple. The equivalence class of (x_{0}, x_{1}, …, x_{n}) is denoted by (x_{0} : x_{1} : …: x_{n}).

The zeros of homogeneous polynomials (also called forms) are closed projective sets. One defines open sets as their complements. This gives rise to Zariski topology on P_{k}^{n}. The condition T_{i} ≠ 0 defines an open subset of P_{k}^{n} isomorphic to the affine space A_{k}^{n} with coordinates T_{0}/T_{i}, …, T_{n}/T_{i}. We get n + 1 affine spaces which provide an open covering of P_{k}^{n}.

Let f(T_{1}, …, T_{n}) be a polynomial of degree d. It can be written as the sum f = f_{0} + … + f_{d}, where f_{i} is a form of degree i. The homogenization of f is the form of degree d in n + 1 variables given by

F (T_{0}, T_{1}, …, T_{n}) = F = T_{0}^{d}f_{0} + T_{0}^{d−1}f_{1} + … + f_{d}.

If X ⊂ A_{k}^{n} is a closed affine set, then associating to polynomials in the ideal of X their homogenizations defines the projective closure of X.

There are serious reasons for working with projective varieties:

(1) The set of complex-valued points of P^{n} is compact in the usual complex topology (e.g. P^{1} over C is just the Riemann sphere),

(2) classifications are simpler (e.g. quadratic forms are classified only by their rank),

(3) intersection theory is simpler (any two curves in the plane, or more generally, any n hypersurfaces in P_{k}^{n} have a common point).

An ideal J ⊂ k[T_{0}, …, T_{n}] is called homogeneous if whenever f ∈ J we also have f_{i} ∈ J, where f_{i} is the homogeneous part of f of degree i. Homogeneous ideals are generated by homogeneous polynomials.

The affine cone of a closed projective set X is the set of points in A_{k}^{n+1} given by the same (homogeneous) equations as X. To a closed projective set X one associates the ideal I(X) of all polynomials in k[T_{0}, …, T_{n}] vanishing on the affine cone of X. The ideal I(X) is clearly homogeneous. To any homogeneous ideal J one associates its set of zeros Z(J) ⊂ P_{k}^{n}. This always gives a non-empty set unless J = 1 (empty affine cone) or J = (T_{0}, …, T_{n}) (the affine cone consists of the zero point). Hence the projective variant of Nullstellensatz, which immediately follows from the affine Nullstellensatz, reads as follows.

Theorem 3.1 (Projective Nullstellensatz) If J is a homogeneous ideal, then

(1) Z(J) = ∅ iff the radical of J contains the ideal (T_{0}, …, T_{n}) (the maximal ideal of the zero point in A_{k}^{n+1}),

(2) if Z(J) ≠ ∅, then I(Z(J)) is the radical of J.

Projective variety is an irreducible closed projective set. (The definition of irreducible is the same as in the affine case.) Quasi-projective varieties are dense open subsets of projective varieties. Rational functions on a projective variety X are fractions of forms of equal degree \frac{F}{G}, where G ∉ I(X), modulo natural equivalence: \frac{F}{G} = \frac{F_{1}}{G_{1}} if F G_{1} − F_{1}G ∈ I(X). The rational function \frac{F}{G} is called regular at a point P ∈ X if G(P) ≠ 0. The field of rational functions on X is again denoted by k(X), but it is not the field of fractions of the ring of regular functions on X! Indeed, the only regular functions on P_{k}^{1} are constants^{1}: k[P_{k}^{1}] = k (easy exercise, in fact every regular function on A_{k}^{1} ⊂ P_{k}^{1} is a polynomial, and every non-constant polynomial has a pole at infinity).

A rational map f : X − − > P_{k}^{n} is (a not necessarily everywhere defined function) given by (F_{0}, …, F_{n}), where F_{i} ∈ k(X)^{∗}, defined up to an overall multiple from k(X)^{∗}. A rational map f is regular at P ∈ X if there exists a representative (F_{0}, …, F_{n}), such that all the F_{i}’s are regular at P, and (F_{0}(P), …, F_{n}(P)) ≠ (0, …, 0). A morphism is an everywhere regular rational map.

Examples of projective varieties, rational maps and morphisms (a) Rational normal curves. This is a map f : P_{k}^{1} → C ⊂ P^{n} given by

f : (X : Y) ↦ (X^{n} : X^{n−1}Y : …: XY^{n−1} : Y^{n}).

One checks that f is a morphism, whose image is given by equations T_{0}T_{2} = T_{1}^{2}, T_{1}T_{3} = T_{2}^{2}, and so on. The inverse map is given by

g : (T_{0} : …: T_{n}) ↦ (T_{0} : T_{1}) = (T_{1} : T_{2}) = … = (T_{n−1} : T_{n})

which is everywhere defined (check!). Hence f is an isomorphism of P_{k}^{1} with a closed subvariety C ⊂ P^{n}, called the rational normal curve of degree n. For n = 2 one recovers the rational parametrization of the conic.

1The same is true for any projective variety, see page 20.

(b) The Veronese embedding of P_{k}^{n}. This is a natural generalization of the previous example, where one considers all monomials of degree d. This defines an isomorphism of P_{k}^{n} with a closed subvariety of P_{k}^{N}, where N = C_{n}^{n+d}. For d = 2 and n = 2 one gets the Veronese surface in P_{k}^{5}.

(c) Any quadric in Q ⊂ P_{k}^{3} is isomorphic to P_{k}^{1} × P_{k}^{1}. (Think of two families of P_{k}^{1}’s on a quadric). The stereographic projection from a point P ∈ Q defines a birational map Q − − > P_{k}^{2} which is not a morphism (not regular at P). Neither is the inverse map a morphism (two lines of Q passing through P are contracted to points).

(d) The Segre embedding P_{k}^{n} × P_{k}^{m} ⊂ P_{k}^{nm+n+m}. This map associates to two vectors their tensor product. The map is well defined everywhere, and is a bijection with the image. The image can be interpreted as the set of non-zero matrices of rank one. In particular, when n = m = 1 we get a quadric in P_{k}^{3}.

The Segre embedding can be used to define the structure of a projective variety on P_{k}^{n}×P_{k}^{m}. As a consequence we realize the product of two projective varieties as a closed subset of some projective space.

(e) Elliptic curve. These are smooth plane cubic curves with a k-point. It can be proved that such a curve is isomorphic to a projective curve y^{2}z = x^{3} + axz^{2} + bz^{3}. The map (x : y : z) → (x : −y : z) is a non-trivial automorphism with exactly four fixed points (there are three obvious fixed points with y = 0, z = 1, and also (0 : 1 : 0)). This shows that the curve is not P_{k}^{1}, as any element of P GL(2) that fixes three different point is an identity. This follows from the fact that Aut(P_{k}^{1}) =PGL(2), see Proposition 5.4 below.

3.2 Morphisms of projective varieties

Examples show that if f : X → Y is a morphism of affine varieties, then f(X) ⊂ Y need not be a closed subset. A standard example is X ⊂ A_{k}^{2} given by xy = 1, mapped to A_{k}^{1} by the morphism (x, y) ↦ x. It is another pleasant feature of projective varieties that the image of a projective variety under a morphism is always closed! We start with a lemma.

Lemma 3.2 Let X and Y be quasi-projective varieties. The graph Γ_{f} of any morphism f : X → Y is closed in X × Y.

Proof. It is enough to consider the case Y = P_{k}^{n}. Consider the morphism (f, Id) : X ×P_{k}^{n} → P_{k}^{n} ×P_{k}^{n}, and let Δ ∈ P_{k}^{n} ×P_{k}^{n} be the diagonal (the graph of the identity map). Then Γ_{f} = (f, Id)^{−1}(Δ). It is clear that the preimage of a closed subset is closed. Thus it is enough to prove that Δ ⊂ P_{k}^{n} × P_{k}^{n} is closed. But Δ is given by T_{i}X_{j} = T_{j}X_{i} for all i and j, and hence is closed. QED

Theorem 3.3 Let X be a projective variety, and Y be a quasi-projective variety. Then the projection to the second factor p : X ×Y → Y maps closed subsets to closed subsets.

Corollary 3.4 Let f : X → Y be a morphism, where X is a projective variety. Then f(X) is closed in Y.

Proof. Apply the theorem to p(Γ_{f}) = f(X). QED

Proof of the theorem. We can assume that X = P_{k}^{n}. Next, the closedness can be checked locally in a small affine neighbourhood of every point. Thus we can assume that Y is a closed subset of A_{k}^{m}. But then Y can be replaced by A_{k}^{m}. All in all, we see that the general statement follows from the statement for the projection P_{k}^{n} × A_{k}^{m} → A_{k}^{m}. Let us prove it.

Let a closed subset in P_{k}^{n} × A_{k}^{m} be given by equations

g_{i}(T_{0}, …, T_{n}; Y_{1}, …, Y_{m}) = 0, i = 1, …, s,

where the g_{i}’s are homogeneous polynomials in variables T_{0}, …, T_{n} whose coefficients are polynomials in variables Y_{1}, …, Y_{m}. We must prove that the set U of y = (y_{1}, …, y_{m}) ∈ k^{m} such that the ideal J_{y} = (g_{i}(T_{0}, …, T_{n}; y_{1}, …, y_{m})) has no zeros in P_{k}^{n}, is open. By the projective Nullstellensatz, Z(J_{y}) = ∅ iff all T_{i}’s are contained in the radical of J_{y}, that is, T_{i}^{l_{i}} ∈ J_{y} for some l_{i}. Let l = l_{0}+…+l_{n}, then in any monomial of degree l at least one variable T_{i} enters in the power greater or equal to l_{i}. Hence J_{y} contains the ideal I_{l} generated by all monomials of degree l. Let U_{l} be the set of y = (y_{1}, …, y_{m}) ∈ k^{m} such that J_{y} ⊃ I_{l}. Then U is the union of all U_{l}’s, l = 1, 2, …, hence it is enough to prove that each U_{l} ⊂ A_{k}^{m} is open.

If one can represent a monomial as a linear combination of homogeneous polynomials, then the coefficients can be chosen to be homogeneous polynomials. Let d_{i} be the degree of g_{i}(T_{0}, …, T_{n}; y_{1}, …, y_{m}). Then y ∈ U_{l} iff the products of the g_{i}(T_{0}, …, T_{n}; y_{1}, …, y_{m}) with all monomials of degree l − d_{i} span the vector space of forms of degree l. Equivalently, the corresponding matrix has maximal rank, which is the condition on the non-vanishing of the determinants of its square submatrices of maximal size. This cleary describes an open subset. (Which may well be empty, but it does not matter!) QED

This result hints at the following definition.

Definition. A morphism f : X → Y of quasi-projective varieties is proper if it is a composition of a closed embedding X ↪ P_{k}^{n} × Y and the projection P_{k}^{n} × Y → Y.

It follows from the previous theorem that f(X) ⊂ Y is closed provided f is proper. It is clear that if f is proper, and P ∈ Y, then f^{−1}(P) is a projective variety.

Proposition 3.5 Any regular function on a projective variety is constant. Any morphism of a projective variety X to an affine variety sends X to a point.

Proof. The second statement clearly follows from the first one. A regular function gives rise to a morphism f : X → P_{k}^{1} whose image does not contain the point at infinity. Hence f(X) ≠ P_{k}^{1} is a union of finitely many points. Since X is irreducible, f(X) is just one point. QED

4 Local geometry

4.1 Localization, local rings, DVR

Let R be a ring. A subset S ⊂ R is called multiplicative if it is closed under multiplication and contains 1. The localization S^{−1}R of R with respect to S is defined as the set of formal fractions \frac{a}{b}, with a ∈ R and b ∈ S, up to the equivalence relation: \frac{a}{b} = \frac{a_{1}}{b_{1}} iff (ab_{1} − a_{1}b)s = 0 for some s ∈ S. When R has no zero divisors, the natural map a ↦ \frac{a}{1} is an injective homomorphism of rings, so that we can think of R as a subset of S^{−1}R. Then S^{−1}R is simply the fractions with “restricted denominators”. Note that if I ⊂ R is an ideal, then S^{−1}I is an ideal in S^{−1}R.

Exercise. A localization of a Noetherian ring is Noetherian.

Examples. (1) If S = R \ {0}, then S^{−1}R is just the field of fractions.

(2) Let P ⊂ R be a prime ideal, then S = R \ P is a multiplicative system. Then S^{−1}R is denoted R_{P} and is called the localization of R at P. The ring R_{P} has a very important property: S^{−1}P is its only maximal ideal (every element not in S^{−1}P is by definition invertible, hence S^{−1}P contains all other ideals). Such rings have a name.

Definition. Rings with just one maximal ideal are called local rings.

Examples: (a) rational numbers such that p does not divide the denominator,

(b) p-adic integers Z_{p},

(c) rational functions in one variable over a field k such that the denominator does not vanish at 0,

(d) formal power series k[[T]],

(e) rational functions in two variables such that the denominator does not vanish at (0, 0).

In all these examples, except the last one, the maximal ideal is principal. Such ring forms the simplest class of local rings.

Definition. A local ring whose maximal ideal is principal is called a discrete valuation ring (DVR).

Definition. Any generator of the maximal ideal of a DVR is called a uniformizer or a local parameter.

Lemma 4.1 Let R be a Noetherian integral domain, t ∈ R \ R^{∗}. Then ∩_{i=1}^{∞}(t^{i}) = 0.

Proof. ([Reid, UCA], 8.3) For contradiction let x ≠ 0 be contained in (t^{i}), for any i ≥ 1. We write x = t^{i}x_{i}, then (x) ⊂ (x_{1}) ⊂ (x_{2}) ⊂ … is an ascending chain of ideals. Then (x_{i+1}) = (x_{i}) = (tx_{i+1}) for some i. Hence x_{i+1} = tax_{i+1}, but this implies t ∈ R^{∗} since x_{i+1} ≠ 0 and R is an integral domain. Contradiction. QED

Proposition 4.2 Let R be a DVR with maximal ideal m = (π) and the field of fractions K. Then

(1) there exists a discrete valuation on K defined by R, that is, a homomorphism v : K^{∗} → Z such that v(x + y) ≥ min{v(x), v(y)}, and R \ {0} = {x ∈ K^{∗}|v(x) ≥ 0}, m \ {0} = {x ∈ K^{∗}|v(x) ≥ 1}.

(2) All non-zero ideals of R are principal ideals m^{i} = (π^{i}), i ≥ 1.

Proof. (1) By the previous lemma every x ∈ R, x ≠ 0, is in m^{i} \ m^{i+1} for some i ≥ 0. Then x = π^{i}u, where u ∈ R must be a unit. We also write y = π^{j}u′ with u′ ∈ R^{∗}. Then xy = π^{i+j}uu′, hence v(xy) = v(x) + v(y). Suppose that i ≤ j, then x + y = π^{i}(u + π^{j−i}u′), hence v(x + y) ≥ v(x) = min{v(x), v(y)}. We now can extend v to K^{∗} by the formula v(x/y) = v(x) − v(y). The remaining properties are clear.

(2) Let s be the infinum of v on the ideal I ⊂ R, then there exists x ∈ I such that v(x) = s. Then m^{s} = (x) ⊂ I. On the other hand, v(I \ {0}) ⊂ {s, s + 1, … }, and m^{s} \ {0} = {x ∈ K^{∗}|v(x) ≥ s}, hence I ⊂ m^{s}. All in all we have I = m^{s}. QED

Observe that (1) implies that R^{∗} = {x ∈ K^{∗}|v(x) = 0}.

A DVR, like any other PID, is a UFD (note my excellent style).

Exercise. Prove that any DVR is normal (=integrally closed in its field of fractions).

The local ring of a subvariety. Let X ⊂ P_{k}^{n} be a variety, and Y ⊂ X a subvariety. We define the local ring O_{Y} of X at Y as the subring of k(X) consisting of the rational functions that are regular on an open set with a non-trivial intersection with Y. It is easy to check the ring axioms. An ideal m_{Y} ⊂ O_{Y} consisting of functions vanishing on Y is maximal since O_{Y} /m_{Y} is the field k(Y). Any function in O_{Y} \ m_{Y} is invertible in O_{Y}, so that m_{Y} is the unique maximal ideal. Thus O_{Y} is a local ring.

Let U ⊂ X be an affine open subset of X. Then O_{Y} is the localization of k[U] at the prime ideal I(Y ∩ U) (i.e. we divide by the functions that do not vanish on Y).

4.2 Regular local rings

Definition 4.3 A Noetherian local ring R with maximal ideal m and residue field k is regular if the Krull dimension of R is dim_{k}(m/m^{2}).

Key example. It follows from Proposition 2.9, Theorem 2.12 and Theorem 2.16 that if P is a smooth point of X, then O_{P} is a regular local ring.

A theorem of Auslander and Buchsbaum says that a regular local ring is a UFD ([Matsumura], 20.3). A very important corollary is

Theorem 4.4 The local ring of a smooth point of an algebraic variety is a UFD.

See Appendix A.2 for a sketch of proof of this theorem.

Exercise. Let X be a curve in P_{k}^{2}, and P ∈ X a smooth point. Here is a low level proof that O_{P} is a DVR (and hence also a UFD). The question being local we can assume that X ⊂ A_{k}^{2}, P = (0, 0). To fix ideas suppose that T_{X,P} is the line y = 0. Then X is given by

y + ∑_{i+j≥2} a_{ij}x^{i}y^{j} = 0.

The maximal ideal m_{P} ⊂ O_{P} is m_{P} = (x, y). All we need to do is to show that it is principal. I claim that m_{P} = (x). Indeed, on X we have

y = −x\frac{∑_{i+j≥2,i≥1} a_{ij}x^{i−1}y^{j}}{1 + ∑_{j≥2} a_{0j}y^{j−1}} ,

and the fraction is regular at P, that is, belongs to O_{P}. Hence y ∈ (x).

4.3 Geometric consequences of unique factorization in O_{P}

The following statement is a generalization of Corollary 2.11.

Theorem 4.5 In a sufficiently small neighbourhood of a smooth point any subvariety of codimension 1 can be given by one equation.

Proof. Let P be a smooth point of X contained in a subvariety Y ⊂ X of codimension 1. Replacing X by a neighbourhood of P we can assume that X is affine with coordinate ring k[X]. Choose a non-zero element f ∈ I(Y) ⊂ k[X]. Now O_{P} ⊂ k(X) is a UFD by Theorem 4.4, so that f is a product of prime factors f = f_{1} … f_{r}, f_{i} ∈ O_{P}. By further shrinking X we can assume that f_{i} ∈ k[X], i = 1, …, r. We have

Y = (Y ∩ {f_{1} = 0}) ∪ … ∪ (Y ∩ {f_{r} = 0}).

Y is irreducible hence Y = Y ∩ {f_{i} = 0} for some i. Replacing f by this f_{i} we can assume without loss of generality that f is a prime element of O_{P}.

Consider the closed subset Z = Z(f) ⊂ X. By Theorem 2.13 the codimension of Z is 1. Now Proposition 2.10 implies that Y is an irreducible component of Z, so that we can write Z = Y ∪ Y′, where Y′ is the union of irreducible components of Z other than Y. If P ∉ Y′ we replace X by a small affine neighbourhood of P that does not intersect with Y′. Then Y = Z(f). Let us show that f is an equation of Y, that is, the ideal I(Y) is generated by f. (Recall that we shrinked X a several times by now, so that f does not have to be a global equation of Y in the original X). Indeed, let g ∈ I(Y) = I(Z(f)). By the Nullstellensatz there exists a positive integer m such that g^{m} is divisible by f in k[X]. Then the same thing is also true in O_{P}. Since O_{P} is a UFD we conclude that g is divisible by f.

It remains to exclude the possibility that P ∈ Y′. Choose h ∈ I(Y) \ I(Y′), h′ ∈ I(Y′) \ I(Y), then hh′ ∈ I(Z) whereas h, h′ ∉ I(Z). By the Nullstellensatz there exists a positive integer n such that (hh′)^{n} is divisible by f in k[X]. Then the same thing is also true in O_{P}. Since O_{P} is a UFD we conclude that h or h′ is divisible by f, hence vanishes on Z. Contradiction. QED

Corollary 4.6 Let X be a smooth variety. Then the local ring of a subvariety Y ⊂ X of codimension 1 is a DVR.

Proof. The maximal ideal m_{Y} ⊂ O_{Y} is generated by any local equation of Y in X, which exists by the previous theorem. Hence m_{Y} is principal which means that O_{Y} is a DVR. QED

Recall that a rational map is regular on a non-empty open set. For smooth projective varieties there is a better result:

Corollary 4.7 A rational map of smooth projective varieties is regular away from a closed subset of codimension at least 2.

Proof. Without loss of generality we assume that the target space is P_{k}^{n}. So let f : X − − > P_{k}^{n} be a rational map. There exist a dense open set U ⊂ X such that f restricted to U is regular, and U is maximal with this property. Suppose that Z ⊂ X is a subvariety of codimension 1 contained in X \ U. Write f = (f_{0}, …, f_{n}), where f_{i} ∈ k(X). Multiplying all the f_{i} by a common multiple does not change f. We can choose this common multiple so that all the f_{i} are in the DVR O_{Z} and have no common factor. Then at least one f_{i} is not divisible by the generator of the maximal ideal of O_{Z}, hence is non-zero at some point of Z. Therefore, f = (f_{0}, …, f_{n}) is regular on an open set which has a non-trivial intersection with Z. This contradicts the fact that U is the largest open set on which f is regular. QED

Corollary 4.8 Any rational map from a smooth and projective curve to a projective variety is a morphism.

Corollary 4.9 A birational map between smooth and projective curves is an isomorphism.

Hence birationally equivalent smooth projective curves are isomorphic. The same is very far from being true in higher dimensions (there are many examples, e.g. the projection P_{k}^{2} − − > P_{k}^{1} from a point (0 : 0 : 1) given by (x_{0} : x_{1} : x_{2}) ↦ (x_{0} : x_{1}), or the stereographic projection of a quadric). The smoothness assumption is also very important. Indeed, let C be the image of the morphism f : P_{k}^{1} → P_{k}^{2} defined by (x_{0} : x_{1}) ↦ (x_{0}^{3} : x_{0}x_{1}^{2} : x_{1}^{3}). Then f : P_{k}^{1} → C is a birational map, but C is not isomorphic to P_{k}^{1} (C contains a singular point (1 : 0 : 0), whereas P_{k}^{1} is smooth).

5 Divisors

5.1 The Picard group

Let X be a smooth variety. Let Div(X) = {∑_{Y} n_{Y} Y } be the free abelian group generated by all subvarieties of codimension 1 Y ⊂ X. The elements of Div(X) are called divisors. Divisors of the form Y, where Y ⊂ X is a subvariety of codimension 1, are called irreducible. A divisor is called effective if all the coefficients n_{Y} ≥ 0, and at least one coefficient is positive. Then one writes ∑_{Y} n_{Y} Y > 0. The union of the subvarieties Y such that n_{Y} ≠ 0 is called the support of ∑_{Y} n_{Y}.

Corollary 4.6 allows us to define the divisor of a rational function f ∈ k(X)^{∗} as

div(f) = ∑_{codim(Y)=1} val_{Y} (f).Y,

where val_{Y} : k(X)^{∗} → Z is the valuation attached to the irreducible divisor Y ⊂ X. The following lemma shows that this sum is finite.

Lemma 5.1 Let f ∈ k(X)^{∗}. For almost all subvarieties Y ⊂ X of codimension 1 we have val_{Y} (f) = 0.

Proof. Let U be a dense open set where f is regular, and U′ be a dense open set where f^{−1} is regular. Then f ∈ O_{Y}^{∗} for any subvariety Y ⊂ X of codimension 1 that has a non-empty intersection with U ∩ U′. Hence if val_{Y} (f) ≠ 0, then Y is one of the finitely many irreducible components of X \ (U ∩ U′) of codimension 1. QED

The divisors of rational functions are called principal.

Proposition 5.2 Let f ∈ k(X)^{∗} be such that div(f) = 0. Then f is regular on X.

Proof. Let P ∈ X be a point where f is not regular. Since O_{P} is a UFD we can write f = uf_{1}^{i_{1}} … f_{n}^{i_{n}}, where u ∈ O_{P}^{∗}, and the f_{i} are irreducible elements. Since f is not regular at P we must have i_{s} < 0 for some s. But f_{s} comes from a regular function on some afffine open neighbourhood U of P. Let Y ⊂ U be given by f_{s} = 0. Its Zariski closure is a subvariety of X of codimension 1 (cf. Theorem 2.13). But then val_{Y} (f) = n_{s} < 0 contrary to our assumption. QED

Corollary 5.3 Let X be a smooth and projective variety. Then a rational function is determined by its divisor up to a constant.

Proof. The ratio of two functions with the same divisor is regular by Proposition 5.2. But Proposition 3.5 says that any regular function on a projective variety is a constant. QED

Definition. The group of classes of divisors modulo principal divisors is called the Picard group of the variety X, and is denoted by Pic(X).

Remark. For a smooth and projective variety X there is an exact sequence of abelian groups

1 → k^{∗} → k(X)^{∗} → Div(X) → Pic(X) → 0.

The words ‘exact sequence’ mean that the kernel of each homomorphism is the image of the previous one.

Remark. Note the analogy with the construction of the class group Cl(K) of a number field K. In that case we have an exact sequence

1 → O_{K}^{∗} → K^{∗} → Frac(K) → Cl(K) → 0,

where Frac(K) is the group of fractional ideals of K with respect to multiplication.

The Picard group contains a lot of information about X.

Examples. Pic(A_{k}^{n}) = 0.

Pic(P_{k}^{n}) = Z is generated by the class of a hyperplane.

Pic(P_{k}^{n_{1}} × … × P_{k}^{n_{m}}) = Z^{m}.

Pic(P_{k}^{2} \ C) = Z/d where C is a curve of degree d.

The Picard group of a smooth cubic surface is isomorphic to Z^{7} (and is generated by 27 lines on it, cf. [Reid, UAG], Ch. 3).

Functoriality of the Picard group. Associating the abelian group Pic(X) to a smooth variety X is a natural construction in the sense that to any morphism f : X → Y there corresponds a homomorphism f^{∗} : Pic(Y) → Pic(X). This f^{∗} is functorial: if g : Y → Z is another morphism, then (gf)^{∗} = f^{∗}g^{∗}. We only outline the construction of f^{∗} for smooth and projective curves. See Appendix B.2 for a sketch of a more complicated definition of f^{∗} in the case of varieties of arbitrary dimension.

Consider a surjective morphism f : X → Y of smooth and projective curves. We know by Corollary 4.6 that the local ring at any point of X or Y is a DVR. This makes it possible to define the inverse image homomorphism f^{∗} : Div(Y) → Div(X) as follows. Let P ∈ Y, Q ∈ X such that f(Q) = P. Then we have an injective homomorphism of local rings f^{∗} : O_{P} → O_{Q} which allows us to think about O_{P} as a subring of O_{Q}. Let u_{P} be the local parameter at P, and let val_{Q} be the valuation defined by the local ring O_{Q} ⊂ k(Y). We define

f^{∗}(P) = ∑_{Q∈X,f(Q)=P} val_{Q}(u_{P}).Q,

and then extend this to Div(Y) by linearity. It can be checked directly that if g ∈ k(Y)^{∗} is a rational function, then div(g ∘ f) = f^{∗}(div(g)). Thus f^{∗} gives rise to a homomorphism of Picard groups:

f^{∗} : Pic(Y) → Pic(X).

Example. Let X be the conic in P_{k}^{2} given by x_{0}x_{1} = x_{2}^{2}, Y = P_{k}^{1}, and f : X → Y be given by (x_{0} : x_{1} : x_{2}) ↦ (x_{0} : x_{1}).

Let P = (1 : 0) ∈ Y. We compute f^{∗}(P). It is clear that f^{−1}(P) = Q = (1 : 0 : 0). We first check that x_{2}/x_{0} is a local parameter at Q. Indeed, in the affine plane given by x_{0} ≠ 0 where the coordinates are t_{1} = x_{1}/x_{0} and t_{2} = x_{2}/x_{0}, the equation of X is t_{1} = t_{2}^{2}. Hence the maximal ideal of Q is (t_{1}, t_{2}) = (t_{2}). Next, t_{1} is a local parameter at P. In k(X) we have t_{1} = t_{2}^{2}, thus val_{Q}(u_{P}) = 2, and finally f^{∗}(P) = 2Q.

Now we want to compute f^{∗}(R), where R = (1 : 1) ∈ Y. It is clear that f^{−1}(R) = {Q_{+}, Q_{−}}, where Q_{±} = (1 : 1 : ±1). Obviously t_{1} − 1 is a local parameter at R. Let us find a local parameter at Q_{+}. The maximal ideal at this point is (t_{1} − 1, t_{2} − 1). But in k(X) we have t_{1} − 1 = (t_{2} − 1)(t_{2} + 1) so that (t_{1} − 1, t_{2} − 1) = (t_{2} − 1) = (t_{1} − 1) since t_{2} − 1 is regular and invertible at Q_{+}. A similar computation shows that t_{1} − 1 is also a local parameter at Q_{−}. Thus val_{Q_{±}}(u_{P}) = 1, hence f^{∗}(P) = Q_{+} + Q_{−}.

5.2 Automorphisms of P_{k}^{n} and of A_{k}^{n}

As an application of the fact that Pic(P_{k}^{n}) is isomorphic to Z we prove the following useful statement.

Proposition 5.4 Any automorphism of P_{k}^{n} is given by a linear transformation of the corresponding n + 1-dimensional vector space, so that Aut P_{k}^{1} = PGL(n + 1).

Proof. The group of non-degenerate matrices GL(n + 1) acts on k^{n+1}. Lines through the origin are mapped to lines through the origin, hence GL(n + 1) also acts on P_{k}^{n}. The only matrices that act trivially on the lines through the origin are scalar matrices. Hence PGL(n + 1) = GL(n + 1)/k^{∗} acts faithfully on P_{k}^{n} (the only element acting trivially is the identity). We must show that any automorphism of P_{k}^{n} is of such a form.

We first consider P_{k}^{1}. Any rational map P_{k}^{1} − − > P_{k}^{1} is given by a bijective rational function t ↦ f(t). Then f(t) = (at + b)/(ct + d) for some a, b, c, d ∈ k. (It is easy to check that other rational functions are never bijective.) This morphism clearly comes from the linear transformation with matrix

( \begin{matrix}a & b \\ c & d \end{matrix} )

Since Pic(P_{k}^{n}) = Z the induced morphism f^{∗} sends a generator to a generator. Let H ⊂ P_{k}^{n} be the projective subspace of codimension 1, also called a hyperplane. We denote by [H] its class in Pic(P_{k}^{1}). The classes [H] and −[H] are the only classes that generate Pic(P_{k}^{n}). But the hyperplane class [H] ∈ Pic(P_{k}^{n}) = Z contains effective divisors, whereas −[H] does not. Hence f^{∗}([H]) = [H]. Therefore the inverse image of a hyperplane is a hyperplane. Now any rational map P_{k}^{n} − − > P_{k}^{n} is given by a collection of n rational functions f_{1}, …, f_{n} in the affine coordinates t_{1}, …, t_{n}. The condition f^{∗}(H) = H says that ∑ a_{i}f_{i} = a defines a hyperplane for any a, a_{1}, …, a_{n} ∈ k, where not all of the a_{i} are equal to 0. Write each f_{i} as a fraction in lowest terms F_{i}/G_{i}. Then, for a, a_{1}, …, a_{n} general enough, we get a hyperplane only if all the G_{i} are of degree 1 and proportional, and the F_{i} are of degree 1 (the details of this argument are left to the reader). This is precisely what we needed to prove. QED

Exercise. Compute Aut(A_{k}^{1}) by proving that every automorphism of A_{k}^{1} extends uniquely to an automorphism of P_{k}^{1} which fixes the point at infinity.

Remark. In contrast, Aut(A_{k}^{n}) for n ≥ 2 is very big. Note that (x, y) ↦ (x, y+g(x)), where g(x) is any polynomial, is an automorphism of A_{k}^{2}. There are also automorphisms of A_{k}^{2} defined by the elements of PGL(2) that preserve the line at infinity. It can be proved that these automorphisms generate the whole group. See [Shafarevich] for more details.

The Jacobian Conjecture. Let F (x, y), G(x, y) be polynomials in x and y. When is the map f : A_{k}^{2} → A_{k}^{2} given by (x, y) ↦ (F (x, y), G(x, y)) an automorphism of the affine plane?

Consider the Jacobian matrix

J(f) = ( \begin{matrix}\frac{∂F}{∂x} & \frac{∂F}{∂y} \\ \frac{∂G}{∂x} & \frac{∂G}{∂y} \end{matrix} )

If g : A_{k}^{2} → A_{k}^{2} is another polynomial map, then one checks by using the chain rule for partial differentiation that J(fg) = J(f)J(g). If f is an automorphism, then we can find g such that fg is the identity map, so that J(f)J(g) is the identity matrix. This implies that det(J(f)) · det(J(g)) = 1. This says that the polynomial det(J(f)) is invertible in the polynomial algebra k[x, y], and so must be a constant. The famous Jacobian Conjecture asserts that if det(J(f)) ∈ k, then f ∈ Aut(A_{k}^{2}). Many cases of polynomials of small degree are known, but the general case of this hard conjecture remains open.

5.3 The degree of the divisor of a rational function on a projective curve

From this point on all our varieties are smooth and projective curves. For curves one defines the degree of the divisor ∑ n_{P} P as the integer ∑ n_{P}. It is clear that this produces a surjective homomorphism deg : Div(X) → Z which sends any point to 1.

Lemma 5.5 Let f : X → Y be a surjective morphism of curves. Then k(X) is a finite extension of k(Y).

The proof is omitted. The number [k(X) : k(Y)] = dim_{k(Y)}k(X) is called the degree of f, and is denoted by deg(f). A sketch of proof of the following theorem can be found in Appendix B.1.

Theorem 5.6 Let f : X → Y be a surjective morphism of smooth and projective curves. Then the degree of the divisor f^{∗}(P) equals deg(f), for any P ∈ Y.

In the notation of the example in the end of Subsection 5.1 we have deg(f^{∗}(P)) = deg(2Q) = 2, deg(f^{∗}(R)) = deg(Q_{+} + Q_{−}) = 2. This agrees with the fact that the degree of the map f in that example is 2: indeed, k(X) = k(\sqrt{t_{1}}) is a quadratic extension of k(Y) = k(t_{1}).

Corollary 5.7 The degree of the divisor of a non-zero rational function on a smooth projective curve is zero.

Proof. If f is constant there is nothing to prove. By Corollary 4.8 any rational non-constant function f ∈ k(X)^{∗} gives rise to a dominant morphism f : X → P_{k}^{1}. Since X is projective, f(X) is closed, hence f(X) = P_{k}^{1}. A comparison of definitions shows that the divisor div(f) equals f^{∗}(0)−f^{∗}(∞). By Theorem 5.6 deg(f^{∗}(0)) and deg(f^{∗}(∞)) both equal to deg(f), hence deg(div(f)) = 0. QED

Due to Corollary 5.7 deg descends to a surjective homomorphism deg : Pic(X) → Z. Define Pic^{0}(X) as the kernel of deg.

Proposition 5.8 Let X be a smooth and projective curve.

(i) The divisor P −Q is principal for some P ≠ Q if and only if X = P_{k}^{1}. (ii) Pic^{0}(X) = 0 if and only if X = P_{k}^{1}.

Proof. We already know that Pic(P_{k}^{1}) = Z is generated by the class of a point. Hence deg is an isomorphism in this case, so that Pic^{0}(P_{k}^{1}) = 0.

To prove (i) let P ≠ Q be points on X such that P − Q = div(f). As in the proof of Corollary 5.7 f defines a morphism f : X → P_{k}^{1}. If we write div(f) = f^{∗}(0) − f^{∗}(∞), then the divisors f^{∗}(∞) and f^{∗}(0) are effective and disjoint. This implies that f^{∗}(0) = P. By Theorem 5.6 the degree of the morphism f is 1, hence f is birational, hence is an isomorphism (Corollary 4.9).

The degree of P − Q is 0. Thus (ii) follows from (i). QED

Among the plane curves this proposition applies to lines and conics.

5.4 Bezout theorem for curves

Let X ⊂ P_{k}^{n} be a smooth curve, and let F be a homogeneous form in n + 1 variables not vanishing identically on X. Cover X by open subsets U_{i} = X ∩ {G_{i} ≠ 0}, where deg(G_{i}) = deg(F). Then F/G_{i} is a regular function on U_{i}. We define the divisor of F by the formula

div(F) = ∑_{P ∈X} val_{P} (F/G_{i}).P,

where in the term corresponding to P we take any G_{i} such that P ∈ U_{i}, or equivalently, G_{i}(P) ≠ 0. The definition does not depend on what U_{i} we choose for a given point P, because G_{i}/G_{j} is an invertible rational function on U_{i} ∩ U_{j}, so that for P ∈ U_{i} ∩ U_{j} we have val_{P} (F/G_{i}) = val_{P} (F/G_{j}). Similarly, one shows that another family of G_{i}’s produces the same divisor div(F).

Definition. The intersection index (X.F) is the degree of div(F).

Since the function F/G_{i} is regular on U_{i} we see from the definition that div(F) ≥ 0, therefore (X.F) ≥ 0. The intersection index counts the number of intersection points of X with the hypersurface F = 0 with ‘correct multiplicities’.

Definition. The degree of X in P_{k}^{n}, denoted by deg(X), is the intersection index of X with the hyperplane.

Of course, we need to show that this definition makes sense, that is, it does not matter which hyperplane we take.

Theorem 5.9 (Bezout) The intersection index (X.F) depends only on X and the degree of F. We have (X.F) = deg(X)deg(F).

Proof. If F′ is another form of the same degree, then div(F) − div(F′) is clearly the divisor of the rational function F/F′. Thus (X.F) = (X.F′) since by Corollary 5.7 deg(div(F/F′)) = 0. This shows that the intersection index depends only on deg(F) and not F itself. To compute (X.F) we can now replace F by any other form of the same degree, for example by H^{deg(F)}, where H is linear. This proves the second statement. QED

Exercise. Let X ⊂ P_{k}^{2} be a plane curve, P ∈ X be a smooth point, and L ⊂ P_{k}^{2} be a line passing through P. Let div(L) be the divisor on X given by a linear form defining L.

1. Show that the multiplicity of P in div(L) is 1 if and only if L is not the tangent line to X at P. Solution: This question is local. Assume that P = (0, 0) ∈ X ⊂ A_{k}^{2}, then X is given by 0 = ax + by+ terms of higher degree in x and y. Since P is smooth, a and b are not both 0. To fix ideas assume that b ≠ 0. Then x is a local parameter at P. Indeed, the maximal ideal of O_{P} is (x, y), but modulo the equation of X we can write y as the product of x and a rational function that is regular at P. Hence m_{P} = (x). If L = μx + νy, then L = (μ − \frac{a}{b}ν)x+terms of degree at least 2 in x. Therefore, if L′ is a linear form such that L′(P) ≠ 0, then val_{P} (L/L′) = 1 if and only if bμ − aν ≠ 0, that is, when the vectors (a, b) and (μ, ν) are not proportional. On the other hand, T_{X,P} is given by ax + by = 0.

2. The set of lines in P_{k}^{2} has a natural structure of the projective variety P_{k}^{2}. Show that the map P ↦ T_{X,P} defines a rational map f : X − − > P_{k}^{2} (called the Gauss map). Conclude that dim(f(X)) = 1.

3. By comparing the dimensions we see that P_{k}^{2} \ f(X) ≠ ∅. Thus there are infinitely many lines L such that div(L) is the sum of distinct points taken with multiplicity 1. (Recall that k is algebraically closed hence infinite.)

Therefore, deg(X) equals the maximal number of points in the intersection X ∩ L. This is also true for the curves in P_{k}^{n}, where we need to replace lines by hyperplanes. See [Shafarevich] for details. For plane curves, however, we can also say that deg(X) equals the degree of the form by which X is defined. Since a curve in P_{k}^{n} is defined by two or more forms, this property has no higher dimensional analogue. (Warning: two forms may not be enough to define a curve in P_{k}^{3}! An example is provided by the rational normal cubic curve, that is, the image of P_{k}^{1} under the map (x : y) ↦ (x^{3} : x^{2}y : xy^{2} : y^{3}).)

Exercise. Let X ⊂ P_{k}^{2} be a plane curve, P ∈ X be a smooth point. Let L ⊂ P_{k}^{2} be the tangent line to X at P. Show that the multiplicity of P in div(L) is 2 if and only if the Hessian matrix of X at P is non-degenerate. The Hessian matrix of the curve given by f(x_{0}, x_{1}, x_{2}) = 0 is defined as follows:

H(f) = ( \begin{matrix}\frac{∂^{2}f}{∂x_{0}^{2}} & \frac{∂^{2}f}{∂x_{0}∂x_{1}} & \frac{∂^{2}f}{∂x_{0}∂x_{2}} \\ \frac{∂^{2}f}{∂x_{1}∂x_{0}} & \frac{∂^{2}f}{∂x_{1}^{2}} & \frac{∂^{2}f}{∂x_{1}∂x_{2}} \\ \frac{∂^{2}f}{∂x_{2}∂x_{0}} & \frac{∂^{2}f}{∂x_{2}∂x_{1}} & \frac{∂^{2}f}{∂x_{2}^{2}} \end{matrix} )

Solution (char(k) ≠ 2): The question is local. Assume that P = (0, 0) ∈ X ⊂ A_{k}^{2} and that L is given by y = 0. Then X is given by 0 = y+ax^{2}+bxy+cy^{2}+ terms of higher degree in x and y. As in the previous example one shows that x is a local parameter at P. A direct computation shows that the Hessian of f at P is non-degenerate if and only if a ≠ 0. On the other hand, y can be written as the product of the polynomial in one variable ax^{2}+terms of higher degree in x, and a rational function that is regular at P.)

We define deg(X), where X is a reducible curve, as the sum of the degrees of components. We then extend the Bezout theorem to reducible curves X, and get the same formula (X.F) = deg(X)deg(F). Note that (X.F) is only defined when F does not vanish identically on some component of X.

In the case of plane curves the Bezout theorem says that the number of common points of two (possibly reducible) curves without common components is the product of their degrees, provided we count points with correct multiplicities. As an illustration of this we now prove Pascal’s mysterious theorem.

Theorem 5.10 (Pascal) Let C ⊂ P_{k}^{2} be a conic. For any six distinct points of C consider the hexagon with vertices in these points. Then the common points of the three pairs of opposite lines of the hexagon are collinear.

Proof. (Plücker) Let the sides of the hexagon be l_{1}, m_{2}, l_{3}, m_{1}, l_{2}, m_{3} in this order. We denote their equations by the same letters. Then Q_{x} = l_{1}l_{2}l_{3} + xm_{1}m_{2}m_{3} is a cubic form that depends on the parameter x ∈ k. We observe that for any x this cubic passes through the six vertices of the hexagon. Let P ∈ C be a point different from these six points. Then l_{i}(P) ≠ 0, m_{i}(P) ≠ 0 for i = 1, 2, 3, as no line contains more than two points of a conic. Thus we can find x ∈ k such that Q_{x}(P) = 0. Write Q = Q_{x}. Either C is a component of Q, or it isn’t. In the second case by the Bezout theorem C ∩ Q consists of at most six points, but there are visibly seven points in this intersection. Hence the cubic form defining Q is a product of the quadratic form defining C and some linear form L. Let A_{i} = l_{i} ∩ m_{i}, i = 1, 2, 3. Since a line intersects a conic in at most two points, we have A_{i} ∉ C. On the other hand, A_{i} ∈ Q, hence these points must be in L. QED

Let X ⊂ P_{k}^{2} be a smooth curve of degree d given by f = 0. A point P ∈ X is a called a flex if the tangent line T_{X,P} intersects X in P with multiplicity at least 3. We have seen that P is a flex of X if and only P belongs to the closed set given by det(H(f)) = 0, called the Hessian H_{X} of X. It can be shown that if d ≥ 3, then the Hessian H_{X} is a curve of degree 3(n − 2). By applying the Bezout theorem we see that every curve of degree at least 3 has a flex, and at most 3n(n − 2) of them. (Check that X is not a component of H_{X}.)

We now study smooth plane cubics X ⊂ P_{k}^{2} in some more detail.

Exercise. (char(k) ≠ 2, 3) Choose a flex P on smooth plane cubic X. Suppose that P = (0 : 1 : 0), and T_{X,P} has the equation z = 0. Show that X is then given by the affine equation y^{2} + axy = f(x), where f(x) is a polynomial in x of degree 3. Now show how to reduce to the equation y^{2}z = x^{3} + axz^{2} + bz^{3}.

Let us consider Pic^{0}(X) where X is a smooth plane cubic. In the following proposition we do not assume that k is algebraically closed.

Proposition 5.11 Let X be a smooth cubic curve in P_{k}^{2} with a k-point P_{0}. Then the map P ↦ P − P_{0} is a bijection from the set of k-points X(k) to Pic^{0}(X).

Proof. By virtue of example (e) in Subsection 3.1 X is not isomorphic to P_{k}^{1}. (Note that the previous exercise shows how to reduce the equation of X to y^{2}z = x^{3} + axz^{2} + bz^{3}, at least when k is algebraically closed.) By Proposition 5.8 (i) the divisor P − P_{0} is never equivalent to 0 for P ≠ P_{0}. This proves injectivity. To prove surjectivity use secants and tangents to decrease the number of points in the support of the divisor. QED

For any smooth and projective curve X it can be proved that Pic^{0}(X) is isomorphic to the group of k-points on the so called Jacobian variety of X.

Further exercises: Directly prove the associativity of the group law on a cubic curve.

5.5 Riemann–Roch theorem

Definition. Let X be a smooth projective curve, D ∈ Div(X). The space of functions associated with D is the subset L(D) ⊂ k(X) consisting of 0 and rational functions f such that div(f) + D ≥ 0 (that is, is 0 or an effective divisor).

Note that any function f ∈ L(D) is regular away from the support of D.

Proposition 5.12 Let D be a divisor on a smooth and projective curve X.

(i) Let g ∈ k(X)^{∗}. Then the map f ↦ f/g identifies L(D) with L(D + div(g)).

(ii) L(D) = 0 if deg(D) < 0.

(iii) L(0) = k.

(iv) L(D) is a vector space of dimension at most deg(D) + 1.

Proof. (i) is obvious.

(ii) follows from Corollary 5.7 and the trivial observation that an effective divisor has positive degree.

(iii) reflects the fact that div(f) = 0 implies that f is a constant function. (iv) Let D = ∑_{P} n_{P} P. Then f ∈ L(D) if and only if val_{P} (f) ≥ −n_{P}.

The ultrametric inequality (Proposition 4.2) shows that L(D) is closed under addition; and L(D) is obviously closed under multiplication by constants.

Now we prove dim(L(D)) ≤ deg(D) + 1 for all D of non-negative degree by induction on deg(D). Because of (i) we can assume that D is effective or zero. In the last case we conclude by (ii). Now D is effective, hence of positive degree. Suppose that the inequality is proved for all divisors of degree at most deg(D) − 1. We can write D = n_{Q}Q + ∑_{P ≠Q} n_{P} P where n_{Q} > 0 and n_{P} ≥ 0. Then D′ = D − Q ≥ 0, and L(D′) ⊂ L(D). It suffices to show that the codimension of L(D′) in L(D) is 0 or 1. Choose a local parameter u_{Q} at Q. Then it is immediate from the definition of L(D) that fu_{Q}^{n_{Q}} is regular at Q. Consider the linear form on L(D) given by the value of fu_{Q}^{n_{Q}} at Q. The set of zeros of this form is precisely L(D′), hence the codimension of this subspace is at most 1. Now we use the inductive assumption to finish the proof. QED

Let ℓ(D) = dim(L(D)).

Exercise. If the curve is P_{k}^{1}, then ℓ(D) = deg(D) + 1 if deg(D) ≥ 0, and ℓ(D) = 0 otherwise. To show this it is enough to assume that D = n(∞). In this case L(D) is the space of polynomials of degree at most n in x, where x is the coordinate on A_{k}^{1} = P_{k}^{1} \ ∞.

Exercise. If the curve X is not isomorphic to P_{k}^{1}, then ℓ(D) ≤ deg(D). Hint: combine the proof of (iv) above with Proposition 5.8 (i).

The Riemann–Roch theorem is the following formula:

ℓ(D) − ℓ(K − D) = deg(D) − g + 1.

Here K is the canonical class, defined as the class of the divisor of any differential form on X, and g is the genus of X, defined as g = ℓ(K). By letting D = K we see that deg(K) = 2g − 2. If deg(D) > 2g − 2, then the degree of K − D is negative implying that ℓ(K − D) = 0, so that in this case we have ℓ(D) = deg(D) − g + 1. In general, we only have Riemann’s inequality ℓ(D) ≥ deg(D) − g + 1.

The main reason to introduce L(D) is that this is a nice way to build morphisms f : X → P_{k}^{n}, where n = ℓ(D) − 1. Indeed, let D be an effective divisor on X, and let 1, f_{1}, …, f_{n} be a basis of L(D). Consider the rational map f which sends P ∈ X to the point (1 : f_{1}(P) : …: f_{n}(P)). It is actually a morphism, as is any rational map from a smooth and projective curve to a projective space.

Exercise. Let H be a hyperplane in P_{k}^{n}. Show that deg(f^{∗}(H)) = deg(D). In particular, if f is injective, then f(X) is a curve of degree deg(D).

It can be shown that if deg(D) > 2g, then f is an embedding, that is, f defines an isomorphism of X with its image f(X).

5.6 From algebraic curves to error correcting codes

In this subsection k is a finite field. Recall that if p is the characteristic of k, then k has p^{s} elements for some s > 0. (Indeed, k contains F_{p}, and is a vector space over it.)

The vector space k^{n} is turned into a metric space with the Hamming distance between two vectors v and v′ defined as the number of coordinates where v and v′ differ:

|v, v′| = #{i = 1, …, n|v_{i} − v′_{i} ≠ 0}.

Let D(v, r) be the disc of radius r with centre in v ∈ k^{n}. A subset C ⊂ k^{n} is a code correcting at least r errors if the discs D(v, r) with centres in v ∈ C are disjoint. Then one can correct up to r errors occuring in the elements of C, called the code words. The number

dist(C) = min_{v,v′∈C,v≠v′}|v, v′|

is called the minimum distance of C. It is easy to see that C corrects r errors if 2r + 1 ≤ dist(C).

A linear code is a vector subspace C ⊂ k^{n}. For a linear code dist(C) is simply the minimal number of non-zero coordinates of a code word.

The main problem of the theory of linear error-correcting codes is to construct codes with both dist(C) and dim(C) as large as possible. (Another, more practical task is to construct efficient coding and decoding algorithms, but we don’t discuss this here.) There are combinatorial bounds that say that dist(C) and dim(C) can’t both be too big. The contribution of algebraic geometry is an explicit construction of linear codes from algebraic curves. These codes turn out to be very good (better than those provided by the randomly chosen vector subspaces) when n tends to infinity, and p^{s} is not too small.

Let X be a smooth and projective curve over k, and let {P_{1}, …, P_{n}} be k-rational points of X. Choose a divisor D on X of degree at most n. Assume for simplicity that the points P_{i} are not in the support of D. Then we can evaluate any function f ∈ L(D) at the points P_{1}, …, P_{n} since f is regular at these points. The collection of values {f(P_{1}), …, f(P_{n})} is an element of k^{n}. This gives a linear map

ev : L(D) → k^{n}.

Let C = ev(L(D)).

Theorem 5.13 We have dim(C) ≥ deg(D) − g + 1 and dist(C) ≥ n − deg(D).

Proof. We have

dim(C) = dim(L(D)) − dim(L(D − ∑_{i=1}^{n} P_{i})) = ℓ(D) − ℓ(D − ∑_{i=1}^{n} P_{i}).

Since deg(D) < n the second term is 0, and so the Riemann-Roch theorem gives the first inequality. By Corollary 5.7 the degree of div(f) is 0. Hence the degree of the divisor of zeros of f outside the support of D is at most deg(D). In particular, a non-zero code word has at least n−deg(D) non-zero coordinates. Thus we obtain the second inequality. QED

Exercise. Work out C in the case when X = P_{k}^{1}, P_{1}, …, P_{n}, n = p^{s}, are the k-points in A_{k}^{1} ⊂ P_{k}^{1}, and D = d∞, 0 < d < n. In this way we get Reed–Solomon codes (much used in practice).

A More algebra

A.1 Krull’s intersection theorem

The proof of the following general property of Noetherian rings uses primary decomposition of ideals ([Lang], VI.5, [Reid, UCA], 7).

Lemma A.1 (Krull’s intersection theorem) Let R be a Noetherian integral domain with an ideal m. Then ∩_{i=1}^{∞}m^{i} = 0.

Proof. Let M = ∩_{i=1}^{∞}m^{i}. This is an ideal in R, and hence has a finite basis, say, μ_{1}, …, μ_{n}. It is clear that m.M ⊂ M. We claim that we also have M ⊂ m.M. Let us assume this and show how to conclude the proof. We can write μ_{i} = ∑_{j=1}^{n} α_{ij}μ_{j} for all i = 1, …, n, where α_{ij} ∈ m. Let f(t) be the characteristic polynomial of the matrix (α_{ij}). Then f(1) ∈ R annihilates M (Hamilton–Cayley). But f(t) is monic with coefficients in m, hence f(1) is invertible modulo m, hence f(1) ≠ 0. Since R is an integral domain we conclude that M = 0.

The proof of the claim. Here is a proof borrowed from [Van der Waerden]. We first prove that every ideal in a Noetherian ring can be represented as the intersection of finitely many primary ideals. (An ideal I ⊂ R is called primary if every zero divisor in R/I is nilpotent. For example, prime ideals are primary.)

Let us call an ideal I irreducible if whenever I = J_{1} ∩J_{2} we have I = J_{1} or I = J_{2}. By Noetherian induction one proves that every ideal is an intersection of finitely many irreducible ideals (the set of ideals that do not have this property contains a maximal element). Now it suffices to show that every irreducible ideal I is primary. [Otherwise we can find a ∉ I, and b^{i} ∉ I for all i such that ab ∈ I. Let J_{i} := {x ∈ R|xb^{i} ∈ I}. The ascending chain of ideals J_{1} ⊂ J_{2} ⊂ … stabilizes, say, at J_{l} = J_{l+1}. We trivially have I ⊂ (a, I) ∩ (b^{l}, I). Let us show that the inverse inclusion is also true. Let x+b^{l}r be an element of this ideal, where x ∈ I. Then bx+b^{l+1}r ∈ (ab, bI) ⊂ I, hence b^{l+1}r ∈ I. This means that r ∈ J_{l+1} = J_{l}, thus b^{l}r ∈ I. Hence x + b^{l}r ∈ I. Now I = (a, I) ∩ (b^{l}, I), hence I is reducible. Contradiction.]

We write m.M = I_{1} ∩ … ∩ I_{s}, where all the I_{j} are primary. Then for every j we have either m^{s} ⊂ I_{j} for some s, or M ⊂ I_{j}. In both cases we conclude that M ⊂ I_{j}. Therefore M ⊂ m.M. QED

A.2 Completion

Let R be a local ring with maximal ideal m. We have a topology on R such that the ideals m^{i} form a base of neighbourhoods of 0 (i.e., x ∈ R is close to 0 is x belongs to a high power of m). Let the ring R̂ be the completion of R with respect to this topology (defined via Cauchy sequences, as usual). Then R̂ is also a local ring, with maximal ideal mR̂, and has the same dimension as R.

A standard example of a complete local ring is the ring of formal power series k[[x_{1}, …, x_{n}]]. It is regular of dimension n. The units of this ring are the power series with non-zero constant term.

Theorem A.2 (Weierstrass preparation theorem) Let F ∈ k[[x_{1}, …, x_{n}]] be such that the lowest non-zero homogeneous form of F contains the monomial cx_{1}^{m}, c ≠ 0. Then F = UG, where U ∈ k[[x_{1}, …, x_{n}]]^{∗} and G is a monic polynomial in x_{1} of degree m over k[[x_{2}, …, x_{n}]].

It can be deduced form this result (by induction in n) that k[[x_{1}, …, x_{n}]] is a UFD.

Note by the way that there is not much variety of complete local rings of geometric origin. This is shown by the following result, which shall not be used.

Theorem A.3 (Cohen structure theorem) A complete regular local ring R of dimension n containing some field is isomorphic to k[[x_{1}, …, x_{n}]], where k is the residue field of R.

Idea of proof of Theorem 4.4. Let P be a smooth point of a variety X, O_{P} the local ring of P in X, and Ô_{P} be the completion of O_{P}. The canonical homomorphism O_{P} → Ô_{P} is injective by Theorem A.1. It is clear that this map associates to a function its Taylor series with respect to a system of local parameters at P, that is, n functions u_{1}, …, u_{n} such that n = dim(X) and (u_{1}, …, u_{n}) is the maximal ideal of O_{P}. This also proves that Ô_{P} = k[[x_{1}, …, x_{n}]]. This is a UFD. Finally, it can be deduced from here that O_{P} is also a UFD.

A.3 The topological space Spec(R)

Let R be a commutative ring with 1. We define the set Spec(R) as the set of prime ideals of R. One endows Spec(R) with the topological space structure, where a subset Z ⊂ Spec(R) is closed if there is an ideal I ⊂ R such that Z is the set of prime ideals that contain I. This identifies Z with Spec(R/I). If f : R_{1} → R_{2} is a ring homomorphism, then f^{−1}(I), where I ⊂ R_{2} is a prime ideal, is a prime ideal of R_{1}.

Exercise. Prove that this gives a continuous map of topological spaces Spec(R_{2}) → Spec(R_{1}).

Thus Spec is a functor from the category of rings and their homomorphisms to the category of topological spaces and their continuous maps.

Let us explore in more detail the topological space Spec(k[X_{1}, …, X_{n}]) where k is any field.

Definition. The topological space Spec(k[X_{1}, …, X_{n}]) is called the affine space over k, and is denoted by A_{k}^{n}.

If n = 0 we obtain a one-point space Spec(k), called a k-point. Let n = 1 and k is algebraically closed. The prime ideals of k[X] (which is a principal ideal ring) are the zero ideal (0) and the principal ideals (X − a), for a ∈ k. The ideals (X −a) naturally correspond to the points of the usual affine line, with finite subsets as closed sets. The ideal (0) is called the generic point of Spec(k[X]): its closure is the whole space Spec(k[X]).

For n > 1 the situation becomes dramatically different. As for n = 1 we can consider k^{n} as the subset of A_{k}^{n} corresponding to maximal ideals (X_{1} − a_{1}, …, X_{n} − a_{n}) ⊂ k[X_{1}, …, X_{n}]. Obviously the Zariski topology of k^{n} is induced by the topology of A_{k}^{n}. When k is algebraically closed, then by Nullstellensatz the points of Spec(k[X_{1}, …, X_{n}]) bijectively correspond to irreducible closed subsets of k^{n}. Let us check when a point of A_{k}^{n} is closed. Let I ⊂ k[X_{1}, …, X_{n}] be a prime ideal. Let Z(J) be a closed set containing the point corresponding to I, where J is an ideal. Then I contains J. Hence the smallest closed subset corresponds to I = J, that is, Z(I) is the closure of I. This means that only the maximal ideals correspond to closed points of A_{k}^{n}, and all the other points, for example, those corresponding to principal ideals (f(X)), are not closed! If J is not a closed point, then the closed points of its closure correspond to maximal ideals containg J. If k is algebraically closed than this is precisely the closed subset Z(J) ⊂ k^{n}. The point J is called the generic point of Z(J). For example, (0) is the generic point of A_{k}^{n}.

A fundamental example of a scheme is Spec(Z) (= (0) and (p), where p is a prime number, the closure of (0) is the whole space, other closed sets are finite unions of closed points). Note that Spec(Z) is connected. Let us also consider Spec(Z[T]). We have a natural continuous surjective map Spec(Z[T]) → Spec(Z) (sending a prime ideal I ⊂ Z[T] to the prime ideal I ∩ Z). There are several kinds of prime ideals in Spec(Z[T]): 0, then (p) and (f(T)), where p is a prime, and f(T) is a polynomial (all these are not closed points!), and, finally, the closed points. Every closed point is located “over” a closed point of Spec(Z), that is, over some prime p, and the quotient by the corresponding maximal ideal is a finite field F of characteristic p. Such an ideal of Z[T] can be given by (p, f(T)) for some polynomial f(T) ∈ Z[T], where F = F_{p}[T]/(f(T)). The closure of (p) is A_{F_{p}}^{1} (the “fibre” of Spec(Z[T]) → Spec(Z) at p). The closure of (f(T)) consists of (p, g(T)) where g(T) ∈ Z[T], considered modulo p, is an irreducible divisor of f(T).

Other instructive examples are Spec(O_{k}) where O_{k} is the ring of integers in a number field k. The natural map Z → O_{k} gives rise to a surjection Spec(O_{k}) → Spec(Z). The inverse image of (p) ∈ Spec(Z) consists of the prime ideals of O_{k} that lie over p.

Exercise: make this explicit when k = Q(\sqrt{−1}).

Let X be a topological space. The dimension of X can be defined as the supremum of all integers n such that there exists a chain Z_{0} ⊂ Z_{1} ⊂ … ⊂ Z_{n} of distinct closed irreducible subsets of X. Then trivially dim(Spec(R)) is the Krull dimension of R.

B More geometry

B.1 Finite morphisms

We begin with defining finite morphisms of affine varieties.

Definition. A dominant morphism f : X → Y of affine varieties is called finite if k[X] is integral over f^{∗}(k[Y]) ≃ k[Y].

Proposition B.1 Let f : X → Y be a finite morphism of affine varieties, then

(i) for any P ∈ Y the set f^{−1}(P) is finite,

(ii) f is surjective.

Proofs. (i) Coordinates of points in f^{−1}(P), as elements of k[X], satisfy monic polynomial equations, and hence for any values of (non-leading) coefficients have only finitely many roots.

(ii) Let m_{P} ∈ k[Y] be the maximal ideal of P ∈ Y. Then f^{−1}(P) = ∅ iff 1 ∈ m_{P} k[X], that is, m_{P} k[X] = k[X]. But by assumption k[X] is a k[Y]- module of finite type, and Nakayama’s Lemma leads to contradiction. (Let f_{1}, …, f_{n} be a basis of the k[Y]-module k[X], then f_{i} = ∑_{j=1}^{n} m_{ij}f_{j} for all i = 1, …, n, where m_{ij} ∈ m_{P}. Let f(t) be the characteristic polynomial of the matrix (m_{ij}). Then f(1) ∈ k[Y] annihilates k[X]. Since k[X] contains 1, we must have f(1) = 0. But f(t) is monic with non-leading coefficients in m_{P} ≠ k[Y], hence f(1) is non-zero. Contradiction.) QED

One checks that finiteness is a local property, that is, a morphism is finite if and only if it has this property in a neighbourhood of any point (see [Shafarevich]). Then one uses this to define finite morphisms of quasiprojective varieties. (Another general property of finite morphisms is that they send closed subsets to closed subsets.)

Definition. A morphism of quasi-projective varieties f : X → Y is called finite if every point y ∈ Y has an affine neighbourhood U such that f^{−1}(U) is affine, and f : f^{−1}(U) → U is finite.

Definition. The degree of a finite morphism f : X → Y is the degree of the field extension k(X)/k(Y).

Theorem B.2 A dominant morphism of smooth projective curves is finite.

Proof. See [Shafarevich], Ch. 2.

Example. The affine variety X with equation y^{n} = f(x_{1}, …, x_{n}), where f is a polynomial in n variables, is equipped with a morphism π : X → A_{k}^{n} which forgets the coordinate y. This morphism is finite of degree n.

Now we sketch the proof of Theorem 5.6. This theorem is proved as a corollary of Theorems A and B below. We note that by Theorem B.2 f is a finite morphism.

Let P ∈ Y, and f^{−1}(P) = {Q_{1}, …, Q_{r}}. This is a finite set by Proposition B.1 (ii). The ring

Õ = ∩_{i=1}^{r}O_{Q_{i}}

is called the semi-local ring of {Q_{1}, …, Q_{r}}. The first of two following theorems describes the algebraic structure of the semi-local ring Õ, which is very similar to that of a local ring. The second theorem describes the structure of Õ as a module over the local ring O_{P}.

Theorem A. (1) Õ is a PID.

(2) The only prime ideals of Õ are the ideals m_{i} := m_{Q_{i}} ∩ Õ, where m_{Q_{i}} is the maximal ideal of O_{Q_{i}}.

(3) There are elements t_{i} ∈ Õ, i = 1, …, r, such that val_{Q_{j}}(t_{i}) = δ_{ij}, in other words, t_{i} is a local parameter at Q_{i}, (t_{i}) = m_{i}, and a unit at Q_{j} when i ≠ j. Any element of Õ can be uniquely written as ut_{1}^{a_{1}} … t_{r}^{a_{r}}, where u ∈ Õ^{∗} is a unit, and a_{i} ≥ 0.

Theorem B. Let n be the degree of the finite morphism f, n = [k(X) : k(Y)]. Then Õ is a free O_{P} -module of rank n.

Proof of Theorem 5.6. By definition, if t ∈ O_{P} is a local parameter, and we write t = ut_{1}^{a_{1}} … t_{r}^{a_{r}} as in Theorem A, then deg(f^{∗}(P)) = a_{1} + … + a_{r}.

By the Chinese remainder theorem the ring Õ/(t) is isomorphic to the product of Õ/(t_{i}^{a_{i}}) ≃ O_{Q_{i}}/(t_{i}^{a_{i}}). This latter ring is the ring of power series in t_{i} with coefficients in k, considered modulo t_{i}^{a_{i}}. As a vector space over k it has dimension a_{i}. Hence dim_{k}(Õ/(t)) = a_{1} + … + a_{r}.

On the other hand, by Theorem B Õ/(t) is isomorphic to (O_{P} /(t))^{n} as a O_{P} /(t)-module. Since t is a local parameter at P, we have O_{P} /(t) ≃ k, and thus Õ/(t) is a vector space k^{n}. By comparing dim_{k}(Õ/(t)) computed in two different ways we conclude that n = a_{1} + … + a_{r}. QED

B.2 Functoriality of Pic

Let f : X → Y be a morphism of smooth varieties. The construction of f^{∗} relies on a so called “moving lemma”.

Moving lemma. Let X be a smooth variety, and x_{1}, …, x_{n} be points on X. In every class of divisors on X there exists a divisor whose support does not contain the points x_{1}, …, x_{n}.

The inverse image f^{∗}(D) of a divisor D on Y such that Supp(D) ⊄ f(X) is defined as follows. We may assume that D is an irreducible subvariety of codimension 1. We know that locally D is given as the zero set of a rational function. Thus the variety Y can be covered by open affine subsets U_{i} such that D ∩U_{i} is either empty or is the set of zeros of a rational function ϕ_{i} ∈ k(Y) in U_{i}. It is clear that the restrictions of ϕ_{i} and ϕ_{j} to the intersection U_{i} ∩ U_{j} differ by an invertible function. Since Supp(D) ⊄ f(X) we observe that ϕ_{i} ∘f is a well defined rational function on X. Moreover, the restrictions of ϕ_{i} ∘ f and ϕ_{j} ∘ f to f^{−1}(U_{i}) ∩ f^{−1}(U_{j}) differ by an invertible function. Together these function define a divisor f^{∗}(D) on X:

f^{∗}(D) = ∑_{Z⊂X, codim_{X}(Z)=1} val_{Z}(ϕ_{i} ∘ f)

Here Z ⊂ X is irreducible and of codimension 1, val_{Z} is the valuation defined by the local ring O_{Z} of X at Z (recall that O_{Z} is a DVR by Corollary 4.6), and i is such that Z ∩ f^{−1}(U_{i}) ≠ ∅. It is easy to see that all such i give the same result. [Note that the verification that this defintion does not depend on the choice of the affine covering and the functions ϕ_{i} is non-trivial.]

Now by the moving lemma, we can modify D in its divisor class in Pic(Y) such that the condition Supp(D′) ⊄ f(X) is satisfied for some some D′ ∼ D. Then we define f^{∗}([D]) as the class [f^{∗}(D′)]. (Note that in general this homomorphism f^{∗} : Pic(Y) → Pic(X) is not induced by a homomorphism Div(Y) → Div(X)!)