Introduction to Algebraic Geometry

By Serge Lang

Author Serge Lang defines algebraic geometry as the study of systems of algebraic equations in several variables and of the structure that one can give to the solutions of such equations. The study can be carried out in four ways: analytical, topological, algebraico-geometric, and arithmetic. This volume offers a rapid, concise, and self-contained introductory approach to the algebraic aspects of the third method, the algebraico-geometric. The treatment assumes only familiarity with elementary algebra up to the level of Galois theory.
Starting with an opening chapter on the general theory of places, the author advances to examinations of algebraic varieties, the absolute theory of varieties, and products, projections, and correspondences. Subsequent chapters explore normal varieties, divisors and linear systems, differential forms, the theory of simple points, and algebraic groups, concluding with a focus on the Riemann-Roch theorem. All the theorems of a general nature related to the foundations of the theory of algebraic groups are featured.

• Language: English
• Publisher: Dover Publications
• Release date: Mar 20, 2019
• ISBN: 9780486839806

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Index

CHAPTER I

General Theory of Places

We are concerned here with extending the ordinary theory of homomorphisms. We know that a homomorphism of a field must be trivial, i.e. must be an isomorphism. However, by generalizing the notion and allowing mappings which send some of the elements of the field onto a suitable infinity, we get places which prove very useful in number theory and algebraic geometry. The notion of a place was already known to Dedekind and Weber, but it is the extension theorem which gives us a powerful tool, and this has become available only comparatively recently.

If we consider an algebraic curve, then it is well known that through each point we can pass an analytic branch on the Riemann surface. The functions of the curve then have a well determined value (which may be a complex number, or ∞) at the place on the Riemann surface over the point. Furthermore the zeros and infinities of the functions can be measured by their orders.

We shall obtain a generalization of these notions, which can be formulated completely abstractly for arbitrary rings and fields. A homomorphism of a ring generalizes the notion of point on a curve, and a place of a field will generalize the place on a Riemann surface. The notion of valuation of a field will generalize the notion of order of a zero or pole.

1. Definition of places

Let F be a field. We shall work with an element ∞ satisfying the following algebraic rules. For a F we define

and ∞/∞ are undefined.

A place (ϕ of a field K into a field F is a mapping ϕ : K → {F, ∞} of K into the set consisting of F and ∞, satisfying the usual rules for a homomorphism namely ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ{a)ϕ(b) whenever the expressions on the right sides of these formulas are defined, and such that ϕ(1) = 1. We shall also say that the place is F-valued. The elements of K which are not mapped into 00 will be called finite under the place, and the others will be called infinite. If ϕ(a) = 0 we say that a has a zero.

be the set of finite elements of K under the place ϕ. is a ring. If ϕ(a) ≠ 0, ∞ then aunder ϕ

satisfies the following important property: If a K If K is a valuation ring of K.

We shall now show that valuation rings give rise to places and that there is essentially a 1 — 1 correspondence between them.

Let K a valuation ring of K. are non-units. Say a/b . Then 1 + a/b = (a + b)/b . If a + b , contradicting the fact that b is not a unit. One sees trivially that for c , ca is not a unit. This proves our contention.

. We define a mapping ϕ of K by letting ϕ and by putting ϕ(a) = ∞ if a K, It is easily verified that ϕ is an F-valued place.

It is convenient to introduce an equivalence relation between places. Recall first that if we have two homomorphisms ϕ1 : R → S1 and ϕ12 : R → S2 of a ring R onto rings S1 and S2 respectively, then we say that these homomorphisms are equivalent if there exists an isomorphism λ : S1 → S2 such that ϕ2 = λϕ1.

be two places. Taking their restrictions to their images, we may assume that they are surjective. We shall say that they are equivalent if there exists an isomorphism λ : F1 → F2 such that ϕ2 = λϕ1 (We put λ(∞) = ∞.) One sees that two places are equivalent if and only if they have the same valuation ring.

It is clear that there is a 1 — 1 correspondence between equivalent places and valuation rings of a field, this correspondence being given by the above constructions which are inverse to each other. A place is said to be trivial if it is an isomorphism. Its valuation ring is simply the whole field.

2. Valuations

Let Γ be a multiplicative commutative group. We shall say that an ordering is defined in Γ if we are given a subset S of Γ (disjoint), i.e. Γ is the disjoint union of S, the unit element 1, and the set S-1 consisting of all inverses of elements of S.

If α β Γ we define α < β We have a < 1 if and only if α S. One easily verifies the following properties of the relation <:

1. For any two elements α, β of Γ we have α < β or α = β or β < α, and these possibilities are mutually exclusive.

2. α < β for any λ Γ.

3. α < β and β < λ implies α < λ.

Conversely, a relation satisfying the three properties gives rise to a semigroup S consisting of all elements < 1. We leave the verification of this step to the reader.

It is convenient to attach to an ordered group formally an extra element 0, such that 0 · α = 0, and such that 0 < α for all α Γ. This is useful in valuation theory. The ordered group is then analogous to the multiplicative group of the positive reals, except that it may have non-archimedean ordering.

The following proposition is trivial but useful.

PROPOSITION 1. If αn = 1, then α = 1, or n = 0.

This means that the map α → αn is an isomorphism of Γ into itself. Indeed, if α ≠ 1, we may assume α S, and hence αn cannot be 1 for any n, because S is a semigroup which does not contain 1.

We shall now consider valuations. A valuation v of a field K is a map a → |a| = v(a) of K into an ordered group Γ, with the extra element 0, such that

1. |a| = 0 if and only if a = 0.

2. |ab| = |a| · |b|.

3.|a + b| ≦ max (|a|, |b|).

Let K* be the multiplicative group of non-zero elements of K. Then the valuation gives a homomorphism of K* into Γ. The valuation is called trivial if it maps K* into 1. If the map is not surjective, we may restrict our attention to the precise image of K, and this gives a valuation onto an ordered subgroup of Γ. If we have two valuations v1 and v2 of a field K, with ordered groups Γ1 and Γ2 both assumed to be surjective, we shall say that they are equivalent if there is an order-preserving isomorphism À : → Γ11 → Γ2 such that v2 = λv1.

Valuations have the following additional properties:

4. = |1| = 1, because |1| = |1|².

5. |±a| = |a|. (The proof is obvious.)

If |a| < |b| then |a| + b| = |b|.

Proof: Under our hypothesis, we have

7. In a sum a1 + a2 + . . . + an = 0, at least two elements of the sum have the same value.

This is an immediate consequence of the preceding property.

We shall now connect the valuations with valuation rings and places.

Let ϕ be a place of a field K, . Let U and the union is disjoint. Consider the factor group K*/U and denote a coset a U by |a|. We also put |0| = 0. We define an ordering in the group K*/U by putting |a| The set S of values |a| < 1 is clearly a semigroup and if we let Γ = K*/U (disjoint).

We obviously have

Hence the place gives rise to a valuation of K if we put |0| = 0 < |a| for all a ≠ 0.

Conversely, let Γ be an ordered group and | | a valuation of K in Γ. be the subset of K consisting of all elements a such that |a| is a valuation ring. If |a| = 1 then |a-1| = 1 also. In that case a is a unit of , and is also called a unit for the valuation. From this we see that the maximal ideal of consists of all a such that |a| < 1.

We see that there is a 1 — 1 correspondence between equivalent valuations, equivalent places, and valuation rings.

3. Examples

We shall determine all places of the rationals.

Let Q be the rational numbers and Z the integers. Let ϕ be a non-trivial place of Q. Then one sees immediately that ϕ cannot be trivial on the integers. Since ϕ(1) = 1, it follows that ϕ is finite on Z. with Z is a prime ideal of Z which is not 0, because we assumed ϕ is non-trivial. All prime ideals of Z are of type (p) where p is a prime number. We have ϕ(p) = 0. If m then ϕ(m) ≠ 0 and is finite. In fact ϕ induces a homomorphism on Z, namely Z → Z/pZ, and ϕ(m) is the residue class mod p. If we write a rational number in the form prm/n, then

The value group associated with this place is infinite cyclic. The valuation ring consists of all fractions whose donominator is not divisible by p, and its maximal ideal consists of those fractions whose numerator is divisible by p.

We shall determine all places ϕ of a rational field k(x) = K (where x is transcendental over k) which are trivial on k.

Let I be the ring of polynomials k[x]. Suppose ϕ is non-trivial. Then ϕ is non-trivial on I. If ϕ(x) = ∞ then we replace x by 1/x, and consider the ring k[1/x]. Hence we may assume without loss of generality that ϕ is finite on x and hence finite on I. of ϕ with I is a prime ideal which is principal, of type p(x), where p(x) is a polynomial irreducible over k. Hence ϕ maps I on a ring isomophic to the residue class ring I/pi, in analogy with the integers. If we write a rational function as prmjn, where m, n are polynomials prime to p = p(x), then we have the same sort of situation that we had for the rationals. In this case ϕ maps rational functions on their function values. The residue class ring I/pI is isomorphic to a finite algebraic extension of k, namel k(θ), where θ is a root of p(x). The function values lie in k(θ). If k is algebraically closed, then of course an irreducible polynomial is linear, and may be chosen of type (x — θ) with θ k, and ϕ is k-valued, sending x onto θ.

More generally, consider the case where R is an arbitrary unique factorization domain. Let p be a prime of R. An element f of the quotient field can be written f = prg/h, where g and h are elements of R not divisible by p. Then a place, valuation ring, and valuation can be defined on the quotient field K in exactly the same manner as was done before in the case of the rational numbers or of a function field in one variable. An important example of a unique factorization domain is given by the polynomial ring in several variables R = k[x1 . . ., xn], where the xi are algebraically independent over a field k. The units of R are then the elements of k, and the primes are simply the irreducible polynomials p = p(x1 . . ., xn) over k.

All the value groups arising from the preceding examples are infinite cyclic. They are very important, but there are other value groups. We leave it to the reader to prove that given an ordered group there exists a field having a valuation whose value group is the given one.

The method of constructing such a field is illustrated by the following special case. Let A Take a rational field k(x, yin a rational function f(x, y)/g(x, y). Then setting x = 0, we see that we can compute the value of a rational function in an obvious manner. Our place gives us a transcendental parametrization in a neighborhood of the origin (0, 0) in the plane.

4. Extension of places

The theorem proved in this section is fundamental. It is used constantly, and makes possible the simplification of many proofs in algebraic geometry.

Let us start with a definition. By a local ring .

Now let R a prime ideal of R. We consider the set R consisting of all elements a, z with a, z R It is immediately verified to be a local ring, which is called the local ring of R at . Its maximal ideal consists of all quotients a/z with a is said to be at the center of .

If ϕ is a homomorphism of R into a field, say an algebraically closed field Ω, then the kernel of ϕ , and the local ring R is said to be the local ring of ϕ.

If K is a field containing R, we are concerned with the possibility of extending ϕ to an Ω-valued place of K. We shall begin by showing how ϕ extends in a natural way to a homomorphism of the local ring R . First note that if ϕ is so extended to a homomorphism ϕso the extension is unique.

To define the extension, given an element a/z of R , This is well defined because a/z = a’/z’

This extension of ϕ to R is then trivially verified to be a homomorphism. Thus we have shown that a homomorphism of a ring R into Ω can always be extended to a homomorphism at the center of the local ring R , and in a unique manner.

If ϕ is trivial on R, then its kernel is zero, and the local ring is the quotient field K of R. We see that the extension of ϕ is trivial on K. More generally, we have:

PROPOSITION 2. Let R be a ring, ϕ0 an isomorphism of R, L a field containing R and algebraic over the quotient field K of R. Then every place ϕ of L extending ϕ0 is trivial on L.

Proof: If ϕ is not trivial on L, then there exists an element y of L which is mapped onto 0 by ϕ, and y ≠ 0. We have an equation

with ai K, and a0 ≠ 0. Applying ϕ, we get 0 = ϕ (a0) which contradicts the hypothesis that ϕ is trivial on K.

We now come to our fundamental theorem.

THEOREM 1. Let K be a field containing a ring R, Let ϕ be a homomorphism of R into an algebraically closed field Ω, and suppose ϕ(1) = 1. Then ϕ can be extended to an Ω-valued place of K.

Proof: In view of the discussion at the beginning of this section, we may assume that R and that ϕ i.e. that the kernel of ϕ under ϕ is a subfield F of Ω.

It will be shown that given an element x of K, the homomorphism ϕ into Ω.

LEMMA. Let be a local ring contained in a field K, and let be its maximal ideal. Let x be an element of K, and consider the ideal generated by m in the rings [x] and respectively. Then it cannot happen that in both cases the ideal is equal to the whole ring.

Proof: Suppose it did happen. We could then write an equation

with pi . We assume that n is the smallest degree of such equations for x. Similarly, we would have

with qj , and we assume that the degree m is minimal for such equations for x–1.

Say m ≦ n. We multiply (2) by xm. We then obtain an expressio for xm as a polynomial in x of degree < m, and with coefficients in m. Substituting this expression in (1), we obtain an equation for x of the same type, but of lower degree. This is a contradiction and proves the lemma.

[x. Let σ [xinto some field. Then the restriction of σ is equivalent to ϕ. Without loss of generality we may assume that the restriction of σ is equal to ϕthen t may be transcendental over F, We now take a homomorphism τ of F[t] into Ω mapping t on an element of Ω algebraic over F, and such that τ is identity on F. This can be done because Ω gives us the desired extension of ϕ [x].

We can apply Zorn’s lemma to finish the proof of the extension theorem. We consider the set of all pairs (S, ϕ) consisting of a subring S of K containing R, and a homomorphism ϕ of S into Ω whose restriction to R is ϕif S § S’ and the restriction of ϕ’ to S is ϕ. If (S, ϕ) is a maximal element then S must be a local ring and ϕ must be at the center of S, and, moreover, the preceding arguments show that S must in fact be a valuation ring. We can then extend ϕ to a place by giving it the value infinity on all elements of K which are not in S. This concludes the proof of the theorem.

The reader will note that we have not merely proved that a homomorphism can be extended to a place, but that it can be extended to a place taking its values in the given algebraically closed field Ω.

For some applications it is necessary to complete the extension theorem by giving more precise information of a quantitative nature concerning the extension of the place. This information is contained in Theorem 2. If ϕ is a homomorphism of a ring R, is a polynomial with coefficients in R, the polynomial obtained by applying ϕ to all the coefficients of f, so

THEOREM 2. Let R be a ring, K its quotient field, and ϕ a homomorphism of R into an algebraically closed field Ω. Let be a polynomial with coefficients in R, of degree n, and let . . ., (x1 . . ., xn) be its roots in an algebraic closure of K, each counted with its multiplicity. Assume that is not identically 0, and let m be the degree of . Then for any extension ϕ* of ϕ to a place of K(x1 . . ., xn), the set is uniquely determined up to a permutation, and consists of all the roots of , each taken with its multiplicity, and of the element 00 which appears n — m times.

Proof: Given a place ϕ* of K(x1, . . ., xn) extending ϕthe image under ϕ* of an element y of K(x1} . . ., xn). After a suitable renumbering of the xi, We can write

Then ϕ*(y) can be neither 0 nor ∞. Indeed, if ϕ*(y) = ∞, we divide both sides by y and apply ϕ*. Then the left-hand side would give 0, while the right-hand side would not, and this would be a contradiction. If ϕ*(y= 0, contrary to assumption.

is equal to

are precisely the roots of f. This concludes the proof of our theorem.

Frequently the polynomial f(X) is irreducible over K, and its roots (x1, . . xn) are then called a complete set of conjugates over K. Each one appears with a multiplicity which is equal to a power of the characteristic. From the fact that they are all conjugate over K, we have the following corollary.

COROLLARY. Assume in the preceding theorem that f(X) is irreducible over K. Let x be one of its roots. Then there exists an extension ϕ* of ϕ to K(x) such that ϕ(x) = ∞ if and only if deg < deg f.

is smaller than that of f, then one of the roots of f goes to infinity under a place extending ϕ. There is an automorphism of the algebraic closure of K mapping x on this root. If we follow this automorphism by the place, we get what we want.

Suppose that x is an element of some field containing R. If f(X) is a polynomial in R[X] such that f(xis not identically zero, then any Ω-valued place ϕ* of K(x) extending ϕ must map x . In particular, there is only a finite number of possible values ϕ*(x). It, on the other hand, for every polynomial f(X) in such that f(x= 0, then we can pick an arbitrary element t of Ω and ϕ can be extended to a homomorphism ϕ* of R[x] by putting ϕ* (x) = tIn particular, t can be selected transcendental over R. Summarizing, we get

PROPOSITION 3. Let R be a ring, ϕ a homomorphism into an algebraically closed field F, and x an element of a field L containing R. If there are infinitely many extensions of ϕ to F-valued places ϕi of L such that the ϕ(x) are all distinct, then there exists an extension of ϕ to a place ϕ* of L such that ϕ* (x) is finite and transcendental over F. If in particular x is transcendental over the quotient field of R, then such a place exists.

5. Integral closure

Let K be a field and R a subring. Classically, an element x of K is said to be integral over R if it satisfies an equation

with ai in R and leading coefficient 1. Such an equation will be called an integral equation for x over R. We shall give a characterization of integral elements in terms of places.

PROPOSITION 4. Let K be a field containing a ring R. An element x of K is integral over R if and only if every place of K finite on R is finite on x.

Proof: Suppose x satisfies an equation of the above type, and ϕ is a place of K finite on R. If ϕ(x) = ∞ divide the equation by xn and apply ϕ. This gives 1 = 0, a contradiction.

Assume that every place ϕ finite on R is finite on x. Consider the ring S = R[1/x]. (We may clearly assume x ≠ 0.) If 1/x is a unit in this ring then we can write

with ai R. Multiply by xn–1. This gives an integral equation for x over R.

If 1/x is not a unit in S, then 1/x . The homomorphism ϕ : R → R/ can be extended to a place of K, and this place is finite on R but maps l/x on 0. Hence the place is infinity on x, a contradiction. This proves our proposition.

Our characterization of integral elements in terms of places immediately shows that the set of elements x of K which are integral over R form a ring, which is called the integral closure of R in K. If K is the quotient field of R and R is equal to its integral closure in K, then we shall say that R is integrally closed. The integral closure of R in K is obviously the intersection of all valuation rings of K containing R.

As consequences of our characterization of integral elements, we obtain some corollaries.

COROLLARY 1. If we have three rings such that R2 is integral over R1 and R3 is integral over R3 then R3 is integral over R1.

This principle of transitivity follows immediately from the proposition.

COROLLARY 2. Let be a ring finitely generated over a ring R. If each xi is integral over R, and if ϕ is a homomorphism of R into some algebraically closed field Ω, then there is only a finite number of extensions of ϕ to a homomorphism of S into Q.

Proof: An integral equation for xi over R will be mapped under ϕ into a non-trivial equation which must be satisfied by an image of xi under an extension of ϕ to 5, and such an equation has only a finite number of roots in Ω.

Let R be a ring contained in another ring Sbe an ideal of R. of S will be said to lie above a

COROLLARY 3. If is a prime ideal of a ring R, and S is a ring containing R and integral over R, then there exists always a prime ideal of S lying above .

Proof: This is equivalent to saying that a homomorphism of R into an algebraically closed field can always be extended to a homomorphism of S into that field.

COROLLARY 4. The assumptions being as in Corollary 3, the residue class ring is algebraic over R/ . If is maximal, so is .

Proof: Here we have identified R/ in S/ , and one sees that every element of Sis algebraic over R/ by looking at an integral equation for