INTRODUCTION TO ALGEBRAIC GEOMETRY CONTENTS

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2 INTRODUCTION TO ALGEBRAIC GEOMETRY JAMES D. LEWIS Abstract. Algebraic geometry is a mixture of the ideas of two Mediterranean cultures. It is the superposition of the Arab science of the lightening calculation of the solutions of equations over the Greek art of position and shape. This tapestry was originally woven on European soil and is still being refined under the influence of international fashion. Algebraic geometry studies the delicate balance between the geometrically plausible and the algebraic possible. Whenever one side of this mathematical teeter-totter outweighs the other, one immediately loses interest and runs off in search of a more exciting amusement. George R. Kempf CONTENTS 1. Algebra prerequisites 2. Motivation 3. Classical story I 4. Classical story II 5. Classical story III 6. Local study of varieties 7. Affine schemes 8. Examples of affine schemes 9. Appendix: The two Hilbert theorems 1. Algebra prerequisites Students should be familiar with certain basic ideas in commutative algebra, such as the first few pages of Atiyah-Macdonald, Introduction to Commutative Algebra. For example, commutative rings A, ideals, A-modules and A-algebras, tensor products, integral domains, and unique factorization domains UFD). From field theory, one should know the meaning of algebraic extensions, algebraic closure, algebraically independent sets, transcendence base and degree some knowledge of the relationship between transcendence degree of finitely generated separable field extensions and derivations would be helpful, but not required). All rings A are assumed commutative with unity 1 A. We sometimes write A = 1). Any homomorphism of rings f : A B must satisfy f1) = 1. By definition, such a homomorphism f turns B into an A-algebra.) We recall the following. Definition 1.1. i) An ideal A is said to be prime if 1) and the following equivalent conditions hold: a, b A, ab a or b. Date: May 25,

3 2 JAMES D. LEWIS A/ is an integral domain. ii) An ideal M A is said to be maximal if M 1) and the following equivalent conditions hold: If U A is an ideal such that M U, then either U = M or U = 1). A/M is a field. Remark 1.2. A Zorn s lemma argument shows that any ring A contains a maximal ideal. [Exercise. Fill in the details of this, as well as the proofs of the equivalent statements in the definition above.] Consider the polynomial and the solution set 2. Motivation fz, w) = 2ew 2 πz 3 + πz C[z, w], X := V f) := {p C 2 fp) = 0}. Calculating the gradient leads to a vector perpendicular to X. We have f = 3πz 2 + π, 4ew ) = 0, 0) z, w) = ± 1 ), 0 3 π In particular fp) = 0, 0) p R 2, hence fp) 0 otherwise 1 R). Since fp) 0, 0) p X, one can argue from the point of view of analysis that X is locally a disk in the strong topology on C 2 implicit function theorem), i.e. a complex open) manifold of dimension 1. X is an example of an open Riemann surface, and a very special case of a Stein manifold. Its tangent space at any p X is given by T p X) = {p + v C 2 v fp) = 0} C. From the point of view of a complex geometer, X has a complex structure with an underlying differentiable structure X R, and it would be of interest to study the family of complex structures with underlying X R a real orientable surface). Put differently, we would like to think of X as part of a family of Riemann surfaces, and we will comment on this shortly. As an algebraist, one can study the ring R X := C[z, w] fz, w)). Then X can be identified with the C-algebra homomorphisms X = Alg C RX, C ). Every such homomorphism ψ Alg C RX, C ) has ker ψ = a maximal ideal of R X. The maximal ideals are of the form M p := z p 1, w p 2 ) R X, where p = p 1, p 2 ) X. So one can interpret X = MaxR X ), the space of maximal ideals maximal spectrum). An algebraic geometer make take this one step further. Instead of R X, create a new algebra over a number field that encodes all the information about X, as part of

4 ALGEBRAIC GEOMETRY 3 a family of Riemann surfaces, and which incorporates some arithmetic information about X. Consider the number field K := Q[ 1], and replace R X by Now put with quotient field R Y := K[z, w, u, v, t] 2tw2 uz 3 + vz + 2 1, u v 2). R S := K[u, v, t] u v 2 ) K[v, t], KS) := QuotR S ). An embedding KS) C corresponds to a general complex point in S := V u v 2 ) C 3, via R S KS) C. For example, the assignment of u, v, t) = π, π, e) determines an embedding. In particular the points in S parameterize a family of open Riemann surfaces, where X corresponds to π, π, e) S. One has an inclusion R S R Y inducing a map of complex spaces h : Y C := MaxR Y K C) S C := MaxR S K C), where R Y K C amounts to changing K in R Y by C same for R S K C). Here we have h 1 π, π, e) = X. Intuitively, dim S = 2 and dim Y = 3. The fiber dimension of h is 3 2 = 1. An arithmetic geometer is inclined to dig further. Note that a choice of 1 being a solution of x = 0 enables us to encode everything into what is called a proper model over Z. Namely, put R Y,Z := Z[z, w, u, v, t, x] 2tw2 uz 3 + vz + 2x, u v 2, x ). One has Z R Y,Z and instead of maximal ideals, one considers prime ideal spaces prime spectrum): Y Z := SpecR Y,Z ) SpecZ). As dim SpecZ) = 1, dim Y Z = 4. For example, over the prime ideal 0) Z, we are really studying Y Z as a complex space, where over p) Z we can work modulo p. This leads to the concept of reduction mod p. For example, over Z 2, we are studying Z 2 [z, w, u, v, t, x] uz3 + vz, u v 2, x ). So in summary, the basic object we considered, after compactifying, is an elliptic curve. To a topologist, an elliptic curve is a torus. To a complex geometer, it is a torus with a family of complex structures, represented by a family over a surface S. To an algebraic geometer, there is arithmetical information that comes equipped with all of this, leading to complexity of points heights), Frobenius actions mod p), and so forth. 3. Classical story I Let k be a field. and K k an overfield. Denote by k[x 1,..., X n ] the polynomial algebra in n letters. As mentioned earlier, all rings that we will consider will be commutative with the identity element 1. And to re-iterate, that all ring resp. algebra) homomorphisms map 1 to 1.

5 4 JAMES D. LEWIS Definition 3.1. An affine closed algebraic set V K n is the zeros of a finite set of polynomials f 1,..., f m k[x 1,..., X n ]. It is often written as V = V f 1,..., f m ); i.e V is cut out by the zeros of f j = 0 for j = 1,..., m, i.e {ξ K n f 1 ξ) =... = f m ξ) = 0}. Exercise. Consider the vector space k n2 of n n matrices. Show that X := {A k n2 deta) 0}, is an affine closed algebraic set, where K = k. Remark 3.2. V f 1,..., f m ) = V U) := {ξ K n fξ) = 0 f U} where U = f 1,..., f m ) k[x 1,..., X n ] is the ideal generated by f 1,..., f m. Recall that the ring A is called Noetherian if it does not admit an infinitely ascending chain of ideals, U 1 U 2 U 3 U m U m+1 ; equivalently every ideal of A is finitely generated see the Appendix for the proof). If A is Noetherian then Hilbert Basis theorem says that A[t] is also Noetherian, for an indeterminate t. Hence we arrive at: Corollary 3.3. k[x 1,..., X n ] is Noetherian. Proof. k has only two ideals: 0 and itself k = 1), hence is obviously Noetherian. Now apply Hilbert Basis Theorem and induction on n using k[x 1,..., X n ] = k[x 1,..., X n 1 ])[X n ]. Let U k[x 1,..., X n ] be an ideal. Then U = f 1,..., f m ) for some f 1,..., f m by Hilbert Basis Theorem. Definition 3.4. k[x 1,..., x n ] := k[x 1,..., X n ], U is called a k-algebra of finite type. Remark 3.5. Since ascending chains of ideals pullback to ascending chains of ideals under k[x 1,..., X n ] k[x 1,..., x n ], it is obvious that k[x 1,..., x n ] is Noetherian an exercise!). Furthermore, it is clear that V U) = Alg k k[x1,..., x n ], K ). Proposition 3.6. i) U 1 U 2 = V U 2 ) V U 1 ). ii) V U 1 ) V U 2 ) = V U 1 U 2 ) = V U 1 U 2 ), where { U 1 U 2 = f 1α f 2α, f iα U i }. α I finite) iii) V α I U α) = α I V U α) where I is any index set and { } U α = f α J finite in I. α I iv) V 0) = K n, V 1) =. α J

6 ALGEBRAIC GEOMETRY 5 v) Let p = p 1,..., p n ) k n and M p = X 1 p 1,..., X n p n ) a maximal ideal). Then V M p ) = {p}. vi) Let U = {f k[x 1,..., X n ] f m U for some m = mf) N} = radical of U. Then V U) = V U). Proof. Let us prove e.g. the property ii). Note that so by property i) we get Also we have U 1 U 2 U 1 U 2, V U 1 U 2 ) V U 1 U 2 ). V U 1 ) V U 2 ) V U 1 U 2 ). Now let us show the other inclusion. Let ξ V U 1 U 2 ) and suppose ξ / V U 1 ) then f 1 U 1 such that f 1 ξ) 0. By our choice of ξ we have f 2 U 2 f 1 ξ)f 2 ξ) = 0, hence f 2 ξ) = 0 f 2 U 2. So ξ V U 2 ) which proves the property ii). Definition 3.7. Properties i) - iv) implies that Zariski) closed subsets of K n form a topology of closed sets in K n called the Zariski topology. Equivalently, a Zariski open subset is given by U = K n \V, where V is Zariski closed. Remark ) For any Zariski closed subset W K n, one has the induced Zariski topology on W. 2) A basis of open sets for the Zariski topology can be arrived at as follows: Let W = V U 0 ) K n, U 0 k[x 1,..., X n ] an ideal, be a Zariski closed set and let U k[x 1,..., X n ] be another ideal where U 0 U. Put E = V U) W. Note that U is finitely generated, so U = f 1,..., f r ), hence E = V f 1,..., f r ) = V f 1 )... V f r ). We have W \E is open and W \E = W \{V f 1 )... V f r )} = {W \V f 1 )}... {W \V f r )}. For f R W := k[x 0,..., X n ]/U 0 set W f = W \V f). Hence {W f f R W } form a basis of open sets for the Zariski topology of W. 4) In the case K = C, it is known that any non-empty Zariski open set U C n is a dense subset of C n, in the strong topology. Now recall that an ideal A is prime if 1) := A and A/ is an integral domain, and that for any ideal U A, the radical is given by U := {a A a m U, for some m = ma) N}. Proposition 3.9. i) For a prime ideal A, =. ii) For any ideal U A, U =. U Proof. Part i) is obvious. The proof of part ii) is in the Appendix. Exercise. Give a direct proof that U A is an ideal. Definition Let V = V U) K n. The ideal of polynomials vanishing on V is given by IV ) = { f k[x 1,..., X n ] f 0 on V }.

7 6 JAMES D. LEWIS [Note. Observe that I ) = k[x 1,..., X n ].] Corollary 3.11 to the definition). If V = V U), then i) U IV ), ii) V = V IV )), iii) IV ) = IV ). iv) I V IV ) ) = IV ) Immediate from ii).) 4. Classical story II The inclusion in statement i) in the above corollary can be replaced by an equality if K = k = algebraic closure of k). This is a consequence of the Hilbert Nullstellensatz theorem. It is also the case when K = Ω, where Ω is a universal overfield. Definition 4.1. Let Ω k be an overfield. Then Ω is universal over k if it is algebraically closed and has infinite transcendence degree over k. The utility of universality can be seen as follows. Let us assume that K = Ω is universal over k. Let U k[x 1,..., X n ] be an ideal, X = V U) Ω n, and Note the identification: R X = k[x 1,..., X n ]. U X = Alg k RX, Ω ). Further, any ψ Alg k RX, Ω ) determines a prime ideal := ker ψ R X. Conversely, let R X be a prime ideal, and put L = QuotR X / ). Then L/k is a field extension of finite transcendence degree over k. The universality of Ω guarantees the existence of a embedding L Ω of fields over k. Now the composite R X R X / L Ω, defines an element of Alg k RX, Ω ). Thus one can think of the prime spectrum SpecR X ) as defining an equivalence class of solutions in V U), namely for ψ 1, ψ 2 Alg k RX, Ω ), ψ 1 ψ 2 ker ψ 1 = ker ψ 2. Note that we can regard k Ω. A solution ψ Alg k RX, Ω ) is said to be algebraic if ψr X ) k. A version of the Hilbert Nullstellensatz says that if B is a k-algebra of finite type, then B is a field iff B is algebraic, viz., B /k k, equivalently, every element of B is a root of a polynomial in k[t]. In particular, if M R X is a maximal ideal, then B := R X /M is a field, hence algebraic. Thus M = ker ψ for some ψ Alg k R X, k). Thus the algebraic solutions correspond in the same way to the maximal ideals in R X. Proposition 4.2. For K = Ω universal, IV U)) = U. Proof. We need only show the inclusion IV U)) U. Let U be a prime ideal, with image R X, which is likewise prime in R X why?). By the above discussion, ψ Alg k RX, Ω ) such that = ker ψ. If π : k[x 1,..., X n ] R X is the quotient homomorphism, then π 1 ) =, and hence = kerψ π). Now let

8 ALGEBRAIC GEOMETRY 7 p j = ψ πx j ). Then p := p 1,..., p n ) V U). Now suppose that f IV U)). Then f 0 on V U) fp) = 0, hence f. We have in fact shown that f = U), a fortiori IV U)) U. U Corollary 4.3. Suppose k = K = C. Then IV U)) = U. Proof. Let X = V U) C n, and f IX) be given. By Hilbert s basis theorem, U = f 1,..., f r ) C[X 1,..., X n ], is finitely generated. For the moment, let k be the subfield of C made up of Q adjoined with the coefficients {c 1,..., c M } of {f, f 1,..., f r }, i.e. k = Qc 1,..., c M ). Then C/k is a universal overfield. Thinking of now {f 1,..., f r } k[x 1,..., X n ], we set U 0 = f 1,..., f r ) k[x 1,..., X n ], and accordingly V U 0 ) = X. By the previous proposition, f IV U 0 )) = U 0, and hence via the inclusion U 0 U, f U. We have thus shown that IV U)) U, and hence equality holds. Definition 4.4. V is irreducible if V V 1 V 2, where V 1, V 2 closed algebraic and V i V j for i j. Proposition 4.5. V is irreducible IV ) = prime. Proof. = ): Set U = IV ) V = V U)). Assume to contrary U is not prime. Thus f, g k[x 1,..., X n ], f, g / U such that f g U. Then V = V U) V f g) = V f) V g). Now set V 1 = V f) V, V 2 = V g) V. Note that V 1 = V f IV 1 ) = IV ) = U, a contradiction. Hence V 1 V. Similarly V 2 V. =): Suppose IV ) = prime but that V = V 1 V 2. Then V 1 = V U 1 ), V 2 = V U 2 ) and V ) = V U 1 U 2 ) U 1 U 2 IV ) =. If U 1 then f U 1 \ with f U 2. But prime implies U 2, so V 2 V V = V 2. In summary, we have the following picture for A a k-algebra of finite type, with Ω/k universal and k Ω: SpecA) = {prime ideals in A}) Alg k A, k) Alg k A, Ω) ker ker MaxA) SpecA) By universality of Ω, the second vertical arrow is surjective. By Hilbert, the first vertical arrow is likewise surjective and in fact a bijection if k = k). [The proof injectivity in the bijection goes as follows: Let ψ 1, ψ 2 Alg k A, k) with ψ 1 ψ 2. Then ξ A such that ψ 1 ξ) ψ 2 ξ). But ψ 2 ξ) k = k ψ j ψ 2 ξ)) = ψ 2 ξ). Thus ξ ψ 2 ξ) ker ψ 2, whereas ξ ψ 2 ξ) ker ψ 1. Thus ker ψ 1 ker ψ 2 define two different maximal deals in A.]

9 8 JAMES D. LEWIS 5. Classical story III We will now consider the situation where K = k. Theorem 5.1 Hilbert Nullstellensatz). If is a prime ideal, then is precisely the ideal of polynomials f k[x 1,..., X n ] that vanish identically on V ) k n, i.e. = IV )). More generally, if U is any ideal, then U = IV U)). Proof. See the appendix for the proof. Exercise. Consider the ideal U = x 2 + 1) R[x], with k = K = R. Show that U IV U)). Corollary 5.2. Suppose that k = k is algebraically closed. Every maximal ideal in k[x 1,..., X n ] of the form M p = X 1 p 1,..., X n p n ) where p k n. Proof. Let M be a maximal ideal. By Nullstellensatz V M), otherwise 1) = M = M. Choose p = p1,..., p n ) V M). Then {p} = V M p ) V M) M = IV M) IV M p )) = M p. By maximality, M = M p. Corollary 5.3. Suppose k = K = k, and U k[x 1,..., X n ] an ideal. Put X = V U). Then X MaxR X ). Proof. Exercise! Now suppose that K = k, U k[x 1,..., X n ] an ideal, X = V U), with R X = k[x 1,..., X n ]/U. For f R X, put X f = X\V f). Exercise. Show that X f R Xf = R X [t]/t f 1). is an affine closed algebraic set with coordinate ring Recall that {X f f R X } forms a basis of open sets for the Zariski topology of X. Then one can also prove the following quasi-compactness property: Proposition 5.4. Let {f α } α I R X be any collection satisfying X = α I X fα. Then X = X f1 X fn, for some finite sub-collection {f 1,..., f N } {f α } α I. Proof. We have By the Nullstellensatz, { = X\ α I } X fα = V U + α If ) α ). U + α If α ) = 1) = k[x 1,..., X n ], hence g 1,..., g N R X such that N g j f j = 1 R X. j=1 Hence for any p X, f j p) 0 for some j.

10 ALGEBRAIC GEOMETRY 9 From now on, we will assume that K = k = k. Definition 5.5. A closed algebraic set V ) where k[x 1,..., X n ] is a prime ideal, is called an affine algebraic variety. Corollary 5.6. There is an order reversing bijection Ideals U V Algebraic k[x 1,..., X n ] U = I subsets U X k n where the prime ideals correspond to varieties. Lemma 5.7. Let X be a variety and U X be non-empty Zariski open set. Then U Zar = X Proof. Set V = U Zar and W = X\U X. Then X = W V. X is irreducible so X W or X V. If X W then W = X. But this contradicts U. So X V. But also V X, hence X = V. Definition 5.8. Let X = V ) k n be a variety. R X := k[x 1,..., X n ]/, is called the affine coordinate ring associated to X. It is precisely the functions on X which are the restrictions to X of polynomials on k n. Proposition 5.9. Let X = V U) k n be an algebraic set. Then varieties V 1,..., V N k n such that X = V 1 V N. If we assume moreover that V i V j for all i j, then up to relabelling), this decomposition is unique. Proof. Outline.) First of all, if we have two irreducible decompositions: then e.g.: V 1 V N = X = W 1 W M, V 1 = {V 1 W 1 } {V 1 W N }. Hence up to relabelling, V 1 W 1. But W 1 V j by the same reasoning, and hence 1 = j. Thus the real issue is existence. If X is irrducible viz., a variety), them we are done. Otherwise X = X 1 X 2, where X 1 X 2 are algebraic. If X 1 and X 2 are both irreducible then we are done. Otherwise we end up with IX) IX j ) IX j1 ), which must terminate by the ascending chain condition on k[x 1,..., X n ]. Example. Let f k[x 1,..., X n ] be a non-unit. Then since k[x 1,..., X n ] is a UFD, one has a unique decomposition of f into irreduibles, viz., N f = f kj j. j=1

11 10 JAMES D. LEWIS Note that f j ) is a prime ideal for all j. Put V j = V f j ). Then X = V 1 V N. Exercise. Let hx, y) be a homogeneous polynomial of degree d 1. Show that V h) k 2 is a union of d lines. Exercise. Decompose X := V x 2 yz, xz x) k 3 into a union of irreducible varieties. Let X = V U 1 ) k n, Y = V U 2 ) k m be algebraic sets, with corresponding R X and R Y. Let h Alg k R Y, R X ). For ψ Alg k R X, k) = X, we have ψ h Alg k R Y, k) = Y. I.e. h induces a map T h : X Y. Such a map is called a morphism from X to Y. If p X, then h 1 M p ) = M q for some q Y. If we write R X = k[x 1,..., x n ], R Y = k[y 1,..., y m ], and if we write g j x 1,..., x n ) = hy j ) k[x 1,..., x n ], Then T h is induced by Y 1,..., Y m ) = g 1 X 1,..., X n ),..., g m X 1,..., X n )). In particular, q = g 1 p),..., g m p)). Definition The morphisms from X to Y are precisely those of the form T h, where h Alg k R Y, R X ). Example. Let X = V U), where U R k n := k[x 1,..., X n ] is an ideal, and R X = k[x 1,..., x n ] := k[x 1,..., X n ]/U. Let π : k[x 1,..., X n ] R X be the quotient map. Then π Alg k R k n, R X ) induces the inclusion T π : X k n. Exercise!) Exercise. Let X = V y 2 x 3 ) k 2. Show that the inclusion k[x] k[x, y] y 2 x 3 ), defines the projection X k onto the x-axis. Definition Two varieties X and Y are said to be biregular if morphisms f : X Y and g : Y X such that f g is the identity on Y and g f is the identity on X. Exercise. Let X = V y x 2 ) k 2, and Y = k. Show that X and Y are biregular. Let X := V ) k n be a variety, with coordinate ring R X and quotient field kx) := QuotR X ). kx) is called the rational function field of X. Definition The dimension of X is given by dim X := tr deg k kx). Example. tr deg k kx 1,..., X n ) = n, hence dim k n = n. Remark For any algebraic set V, we have the unique decomposition V = V 1 V N into subvarieties. We define dim V = max{dim V 1,..., dim V N }. We say V has pure dimension if dim V = dim V 1 = = dim V N.

12 ALGEBRAIC GEOMETRY 11 Example. Geometric form of the Normalization Lemma of Noether. Let X := V ) k n be a variety of dimension d, with coordinate ring R X = k[x 1,..., x n ] = k[x 1,..., X n ]/. The normalization lemma asserts that one can find linear combinations n y i = a ij x j, a ij k, i = 1,..., d, j=1 such {y 1,..., y d } are algebraically independent over k hence ranka ij ) = d), and that R X is integral over the polynomial ring k[y 1,..., y d ] see the Appendix for the definition of integrality). Let {Y 1,..., Y d } be variables. Then the inclusion k[y 1,..., Y d ] k[x 1,..., X n ], defined by Y i n j=1 a ijx j defines a surjective linear map π := a ij ) : k n k d. Consider the following diagram: k[y 1,..., Y d ] k[x 1,..., X n ] k[y 1,..., y d ] R X Now let ψ Alg k k[y 1,..., y d ], k). The integrality property guarantees that ψ extends to ψ Alg k R X, k). It then follows that the restriction of the linear projection π to X defines a surjective morphism π : X k d, whose fibers are all zero dimensional. Example. dim k n = n. Let f k[x 1,..., X n ] be a non-unit. Then X := V f) k n is called a hypersurface. If f is irreducible, then X is a variety of dimension n 1. In general, X is a finite union of varieties of dimension n 1. Proposition Let X k n be a variety and R X a non-zero prime ideal. Let Y = V ) X. Then dim Y < dim X. Proof. Let d = dim X, R X be the coordinate ring of X. Assume to the contrary. Then an algebraically independent set {x 1,..., x d } R X over k, such that the images form an algebraically independent set {x 1,..., x d } R X / over k. Let b be a nonzero element. Then {x 1,..., x d, p} cannot be algebraically independent over k. Hence a polynomial pt, x 1,..., x n ) k[t, x 1,..., x d ] such that P b, x 1,..., x d ) = 0 R X. Since R X is an integral domain, we may assume that P is irreducible. But P cannot be of the form P = a t m for any m N, with a k, as P b) = a b m 0. Hence P 0, x 1,..., x d ) = 0 R X /, defines a nontrivial relation among the {x 1,..., x d }, a contradiction. Corollary If X k n is a variety of dimension n 1, then X = V f) is a hypersurface. Proof. Write X = V ), for some prime ideal k[x 1,..., X n ]. Since 0) is prime, and k[x 1,..., X n ] is a UFD, 0 f such that f is irreducible. Thus X V f) k n. By the above proposition dim X = dim V f), and hence X = V f).

13 12 JAMES D. LEWIS 6. Local study of varieties Definition 6.1. i) For a ring A, we define the Krull dimension of A by: dim Krull A := Sup N { 0 1 N A }, where the supremum is over all ascending chains of prime ideals in A. ii) A ring A is said to be a local ring if it has only one maximal ideal. Let M A be the unique maximal ideal. k := A/M is called the residue field. The local ring A is said to be regular if dim k M/M 2 = dim Krull A. Remark 6.2. In the case where X/k is a classical variety, using the normalization lemma together with some standard results on integral rings Cohen-Seidenberg,...), this definition agrees with the previous definition of dimension, viz., dim X = dim Krull R X. Example. Let X be a variety with coordinate ring R X. Let p X with maximal ideal M p R X. The set S := R X \{M p } is multiplicatively closed. In particular g R X \{M p } gp) 0. The localization R X is a well-known concept in )S commutative algebra. Define O X,p = { ) f R X = S g }. gp) 0 Then O X,p is a local ring with maximal ideal M p, where M p is identified with its image O X,p M p. Note that O X,p kx). Proposition 6.3. Let X be a variety. Then R X = O X,p, p X where the intersection on the RHS is in kx). Proof. Clearly R X p X O X,p. Let h p X O X,p, and let U = {g R X g h R X }. Note that in O X,p, h = f p /g p where f p, g p R X and g p p) 0. Thus g p U. In particular V U) =, and hence by the Nullstellensatz, 1 p X O X,p. This means that f = 1 f R X. Definition 6.4. Let X = V ) k n be a variety and p X. The tangent space of X at p is given by T p X) = { p + v k n fp) v = 0, f }. Exercise. Write = f 1,..., f m ). Show that T p X) = { p + v k n fj p) v = 0, j = 1,..., m }. To arrive at a coordinate invariant version of T p X) we introduce the following. Definition 6.5. X a variety and p X. A p-centered derivation is given by a k-linear map D : R X k satisfying Leibniz rule, viz., Df g) = gp)df) + fp)dg).

14 ALGEBRAIC GEOMETRY 13 Remark 6.6. i) A p-centered derivation naturally extends to a derivation on O X,p by the quotient rule ) f gp)df) fp)dg) D = g gp) 2. The p-centered derivations are denoted by Der p R X, k) = Der k O X,p, k). ii) A derivation D : kx) kx) is a k-linear map satisfying Leibniz rule, viz., Df g) = g Df) + f Dg) kx). We denote this space of derivations by Der k kx), kx)). It is well-known that dim k Der k kx), kx)) = tr deg k kx) =: dim X. This uses the fact that k = k kx)/k is separably generated.) Exercise. Let D Der k kx), kx)). Show that Dk) = 0. Proposition 6.7. There is a canonical identification T p X) Der p O X,p, k). Proof. Let p + v T p X), and write v = ξ 1,..., ξ n ). The corresponding derivation is given by n D v = ξ j x j. p Going the other way, let D Der p R X, k). Via the quotient map j=1 k[x 1,..., X n ] R X = k[x 1,..., X n ], it is clear that D extends to a p-centered derivation on k[x 1,..., X n ] that kills both and k. For f, we can write n f f = fp) + p)x j p j ) + g α h α, x j α j=12 where g α p) = h α p) = 0. Leibniz rule implies that Dg α h α ) = 0. Then if we set ξ j = Dx j ) k, we have n f n 0 = Df) = p) ξ j = ξ j x j x j f). p j=12 j=1 Hence D = D v, where v = ξ 1,..., ξ n ) and p + v T p X). Now assume given varieties X k n, Y k m and a morphism T h : X Y induced by some h Alg k R Y, R X ) with q = T h p). Let D Der p O X,p, k) and g O Y,q be given. Note that g T h O X,p. [In fact g T h = hg).] Define dt h D)g) := Dg T h ) k. This defines a linear map called the differential of T h ), dt h p) : T p X) T q Y ). Exercise. Express dt h p) in terms of the jacobian matrix of a map from k n k m, sending T p X) k n to T q Y ) k m. Exercise. Let X = V y 2 x 3 ) k 2. Show that X and k are not biregular. Exercise. Let M p O X,p be the unique) maximal ideal. Show that any D Der p O X,p, k) kills M 2 p. Deduce that ) Mp T p X) hom k M 2, k. p

15 14 JAMES D. LEWIS Definition 6.8. The space of differentials at p X is given by the cotangent space T p X) = M p /M 2 p. Now let X = V ) k n be a variety, with coordinate ring R X and rational function field kx). We want to show that dim k T p X) dim X with equality on a non-empty Zariski open subset of X. We recall that trdeg k kx) = dimension of the k-vector space of derivations on kx) killing k. Write = f 1,..., f m ). tr deg k kx) = dim k Der k kx), kx)) = dim k Der k R X, kx)) = dim k {D Der k R k n, kx)) D = 0} Now for f R k n and D Der k R k n, kx)), n f Df = Dx j ), Dx j ) kx). x j Thus j=1 [ fi tr deg k kx) = dim kx) ker x j fi = n rank x j ) ] : kx) n kx) m So let ) fi r = rank. x j Then A GLn, kx)) and B GLm, kx)) such that ) fi A B = x j ). I r Collecting all the R X -denominators of the kx)-coefficients in A, B, and multiplying them together to get a g R X, it follows that on the open set X g, ) ) fi fi p X g rank p) = rank, x j x j i.e. p X g dim k T p X) = dim X. [ Exercise. Show that {p X dim T p l} is a Zariski closed ] subset of X. f Hint: Look at the ideal of n l) n l) minors of i x j ). So for example, X = {p X dim k T p X) dim X}, and if we put SingX) = {p X dim k T p X) > dim X}, then this is a proper Zariski closed subset of X. In particular dim SingX) < dim X. Tangent cone. LetX = V ) k n be a variety with prime ideal. Let p = p 1,..., p n ) X. Any f k[x 1 p 1,..., X n p n ] = k[x 1,..., X n ] has a Taylor expansion with no constant term. The leading homogeneous term will be denoted by f h. The homogeneous) linear term of f is denoted by f L. Let U L be the ideal generated by the linear terms f L f ), and U h the ideal generated by f h f ). Note that T p X) = V U L ) and we now define the tangent cone to X at p X by C p X) = V U h ). Since U L U h, it is clear that C p X) T p X). One can interpret T p X) as the algebraic linear hull of C p X). In the case k = C,

16 ALGEBRAIC GEOMETRY 15 C p X) is described as the limiting secants to X at p. Such a description enables one to prove that dim C p X) = dim X. In fact it is known that C p X) has pure dimension = dim X. Thus another way of interpreting p as nonsingular, is to say the inclusion C p X) T p X) is an equality. Exercise. Let X = V fx, y)) k 2, and p X. Show that C p X) is a union of lines in k 2 passing through p. Definition 6.9. Let X be a variety. The proper subvariety SingX) X is called the singular set of X. The open subset X sm X is called the smooth or nonsingular part of X. X is said to be a smooth or nonsingular variety if SingX) =. Exercise. Show that any variety X can be stratified into a finite disjoint union of locally closed smooth varieties locally closed means Zariski open in its closure). [Hint: First write X = X sm SingX). Then decompose SingX) into a finite union of varieties, and continue decomposing. This process must terminate because of decreasing dimensions of varieties.] 7. Affine schemes Let A be a commutative ring with unity. Recall that X := SpecA) = {prime ideals in A}. The closed points of X are MaxA). Let E A be a subset. We define V E) = { SpecA) E }. Let U be the ideal generated by E. Since E is an ideal, we deduce that V E) = V U) = V U). Observe that V α I U α) = α I V U α), and that V U 1 U 2 ) = V U 1 ) V U 2 ). Note that SpecA) = V 0) = V N A)), where N A) is the nilradical of A see the Appendix), and that V 1) =. In particular, we have the Zariski topology on SpecA). The beauty behind SpecA) is its apparent functoriality with respect to commutative rings. If f : A B is a ring homomorphism with f1) = 1), then f 1 prime) = prime. In particular we have f : SpecB) SpecA). To arrive at something similar on the level of maximal ideals, requires restricting to k-algebras of finite type, where k = k, using the Nullstellensatz. For example the inclusion f : Z Q has the property that f 1 0) = 0). Note that 0) Q is maximal; however 0) Z is prime, but not maximal. If X = SpecA), then dim X := Krull dimension of A. As we mentioned earlier, in the case where X/k is a classical variety, this coincides with the previous definition of dimension in terms of transcendence degree. Exercise. Define IV U)) where U A is an ideal. Explain why IV U)) = U. Exercise. Suppose that A is an integral domain. The prime ideal 0) X := SpecA) is called the generic point of X. Show that the Zariski closure {0)} = X. Exercise. Let X = SpecA), f A, and let X f = X\V f). Show that {X f f A} forms a basis for the Zariski topology of X Exercise. Let f A and put S = {1, f 2, f 3,...}. The localization of A with respect to S, denoted by A f), is defined in any commutative algebra text. Show that Spec A f) ) = Xf.

17 16 JAMES D. LEWIS Exercise. Let X := SpecA), and S = A\. Put O X, = A S. Show that O X, is a local ring with maxima ideal M := O X,. The field k := O X, /M is called the residue field. X is called a regular point if dim k M/M 2 = dim O X,. Definition 7.1. X := SpecA) is said to be regular if every point in X is a regular point. Exercise. Show that X f X g = X fg. Exercise. Let U A be an ideal. Show that V U) = U = 1). Exercise. Let X = SpecA). Show that X is irreducible A is an integral domain. Exercise. Show that SpecA B) = SpecA) SpecB). Exercise. Consider two homomorphisms of rings C A, C B. The push-out: A A C B C translates to the fiber product SpecA) SpecC) SpecB) := SpecA C B) on the level of affine schemes. Show that if A, B, C are k-algebras of finite type over k = k, then SpecA) SpecC) SpecB) is bijective to the corresponding fiber product in the category of sets. Remark 7.2. Let K/k be a Galois extension of k of degree m > 1. Then the isomorphism given by B K k K K K, }{{} m times a b ab 1,..., ab m ), where {b 1,..., b m } = {σb) σ GalK/k)}. Hence Spec K k K) is not irreducible. In particular, the category of irreducible affine schemes is not closed under products. i) X = SpecZ) = 8. Examples of affine schemes {p)} p N prime } {{ } closed points ii) k a field. X = Speck) = {0)}, dim X = 0. {0)}. 0) p) dim X = 1. }{{} generic point iii) k a field. X := Speck[x]) = {f) f irreducible} {0)}, dim X = 1. }{{}}{{} closed points generic point In the case k = k X = {x a) a k} {0)} k {0)} = affine line)

18 ALGEBRAIC GEOMETRY 17 iv) For simplicity, assume k = k. Let X = Speck[x, y]) affine plane). The inclusions 0) x) x, y), implies that dim X = 2. We have 0) generic point, codim 2). X = f) f irreducible, codim 1). x a, y b) a, b) k 2, codim 2, closed points). v) Let A = R[x]/x 2 + 1). Then SpecA) is irreducible. However if we perform a base change A C := A R C, we have SpecA C ) reducible. In this situation say that SpecA) is not geometrically irreducible. vi) X = SpecZ[x]), dim X = 2. X consists of the following: a) The generic point 0) b) Principal prime ideals f) where either f = p is a prime in Z, or f is a Q[x] irreducible polynomial such that its coefficients have GCD = 1. c) Maximal ideals p, f), where p Z is prime and f is a monic polynomial, irreducible in Z p [x]. 9. Appendix: The two Hilbert theorems I) The basis theorem. Let A be a commutative ring with unity 1 A. Lemma 9.1. Let M be an A-module. TFAE: i) Every increasing sequence of submodules in M is stationary. ii) Every non-empty) subset of submodules of M has a maximal element. iii) Every submodule is finitely generated. Proof. If N M were not finitely generated, then sequence {f n } n N N such that A f 1 A f 1, f 2 ) A f 1, f 2, f 3 ), which is incompatible with i). Therefore i) iii). On the other hand, if we have an increasing sequence of submodules of M, M 1 M 2 M 3, then set N = j N M j. If N is finitely generated, then we have N = A {f 1,..., f r } where f i M ji. Thus M j = M j+1 =, where j max{j 1,..., j r }. This shows that iii) i). The equivalence i) ii) is left to the reader. Definition 9.2. An A-module M is said to be Noetherian if it satisfies i)-ii)-iii) above. The ring A is Noetherian, if it is a Noetherian A-module, i.e. every ideal is finitely generated equivalently every ascending chain of ideals is stationary). Lemma 9.3. Let N M be A-modules. Then M is Noetherian M & M/N are Noetherian. Proof. This is an exercise!

19 18 JAMES D. LEWIS Corollary 9.4. A is Noetherian A m is Noetherian m N. Theorem 9.5 Basis theorem). If A is Noetherian, then so is A[t]. Proof. Let U A be an ideal. We will show that U is finitely generated. Let U 0 A be the ideal of leading coefficients of elements in U. By assumption U 0 = a 1,..., a n ) is finitely generated. For each a i, we pick an f i U of the form f i = a i t r + lower order terms, where we can insist that r is independent of i, [e.g. If f i = a i t ri +, put f i = t r ri f i, where r = max{r i s}.] Let U = f 1,..., f n ) U, and let f U. Then f = at m +, where a U 0. If m > r, consider g := f i c if i t m r, where a = i c ia i. Obviously deg g < deg f. By repeating this process if necessary), we can assume by relabeling) given g U such that f = g + h, where deg h r. But h U B, where B A[t] is the A-module generated by {1, t,..., t r }. But B A r+1 is a Noetherian A-module, and hence U B is finitely generated, being an A-submodule of B. Let us write B U = A h 1,..., h m ). We have shown that U = A f 1,..., f n, h 1,..., h m ), which is finitely generated. II) Notes on integrality. Let A be a ring. We say that an A-module M is faithful if the annihilator ideal AnnM) := {a A a M = 0} = {0} is zero. Proposition 9.6. Let A B be rings, and ξ B. TFAE: i) monic polynomial P t) A[t] such that P ξ) = 0 B. ii) A[ξ] is a finitely generated A-module. iii) a faithful finitely generated A[ξ]-module M which is finitely generated over A. Proof. i) ii): If P t) = t r + a r 1 t r a 0 A[t], then P ξ) = 0 A[ξ] is generated by {1, ξ,..., ξ r 1 }. ii) iii): Set M = A[ξ] Then 1 M, hence AnnM) = 0. iii) i): Assume given A[ξ]-faithful M with generators {u 1,..., u m } over A. Then ξ M M implies that ξ u 1 : : u m = a ij) for some m m matrix a ij ), a ij A. Thus ξ Im a ij ) ) u 1 : : u m u 1 : : u m, = 0. Cramer s rule tells us that a m m matrix Q with coefficients in A[ξ] such that Q I m a ij ) ) = det I m a ij ) ) I m. Thus det ξ I m a ij ) ) u j = 0 j, hence det I m a ij ) ) M = 0. By faithfullness, det ξ I m a ij ) ) = 0. Now put P t) = det t I m a ij ) ). Let us assume the setting of the above proposition.

20 ALGEBRAIC GEOMETRY 19 Definition 9.7. ξ B is said to be integral over A if it satisfies any of the equivalent statements in i), ii), and iii) above. Proposition 9.8. Let CA, B) := {ξ B ξ is integral over A}. Then CA, B) B is a subring, called the integral closure of A in B. Proof. Let b 1, b 2 B be integral over A. Then A[b 1, b 2 ] is a finitely generated A[b]-module, where b = b 1 + b 2 or b = b 1 b 2. Exercise. Let A be a UFD and K = QuotA) the quotient field. Show that A is integrally closed in K i.e. A is the integral closure of A in K). III) The Nullstellensatz. In this part, we are going to assume that k is any field, with algebraic closure k. For any commutative ring A with 1, we recall the following subrings: Definition 9.9. i) The nilradical of A is given by the intersection of all its prime ideals, viz., N A) :=. ii) The Jacobsen radical of A is given by the intersection of all its maximal ideals, viz., RA) := M. A M A Observe that N A) RA). Now let 1) U k[x 1,..., X n ] be any ideal, and put A := k[x 1,..., X n ] = k[x 1,..., x n ], U which we recall is a k-algebra of finite type. Let π : k[x 1,..., X n ] A be the quotient map. Observe that π 1 N A) ) =. Lemma Let S A be a multiplicatively closed set with 0 S. Then any maximal element of Σ := {Ideals U A U S = }, is prime. Proof. Let be maximal by Zorn s lemma). Take a, b A with a, b. Then + a) and + b) both intersect S. So write s 1 = x 1 + at 1, s 2 = x 2 + bt 2 with s 1, s 2 S, x 1, x 2, t 1, t 2 A. Suppose that ab. Then U S s 1 s 2 = x 1 x 1 + x 1 bt 2 + at 1 x 2 + t 1 t 2 ab, i.e. S, which cannot happen. Hence ab and therefore must be prime. Definition a A is nilpotent if a m = 0 for some m = ma) N. Equivalently, a is nilpotent a 0). Corollary N A) = 0), hence U = U.

21 20 JAMES D. LEWIS Proof. It suffices to show that a N A) a is nilpotent. Assume to the contrary, and put S = {a n n N}. Then 0 S, and hence a prime such that S =. Hence a and so a N A). The Nullstellensatz takes the following equivalent forms: Proposition Let A be a k-algebra of finite type. The following statements are equivalent: i) Let ξ A with ξ N A). Then ψ Alg k A, k) such that ψξ) 0. ii) N A) = RA). iii) Let f, f 1,..., f r k[x 1,..., X n ] be given. If fp) = 0 p k n which satisfy f 1 p) = = f r p) = 0, then for some m N, f m = g 1 f g r f r for some g 1,..., g r k[x 1,..., X n ]. [This is the same thing as saying IV U)) = U, with K = k.] iv) If A is a field, then A is algebraic, i.e. an embedding A k over k. [Thus A is a field A is algebraic.] Proof. The equivalence of i), ii) and iii) is an exercise. To connect iv) with i), consider the following. If ξ A with ξ N A), then a prime A such that ξ. By replacing A by A/, we may assume that A is an integral domain and that ξ A\{0}. Now put A := A[t] tξ 1). Note that A is a k-algebra of finite type. Let M A be any maximal ideal which exists by Zorn s lemma). If we assume iv), then A /M k, i.e. ψ Alg k A, k) such that ker ψ = M. But ξ A, is a unit. Hence ψ ξ) 0. Now put ψ = ψ A Alg k A, k). Hence iv) i). It is easy to see how iii) iv) from Corollary 5.2, and is left as an exercise. Theorem 9.14 Nullstellensatz). Let B be of finite type over a field k. If B is a field, then B is a finite algebraic extension of k. Proof. Write B = k[x 1,..., x n ]. We will argue by induction on n. The case n = 1 is obvious. If we put A = k[x 1 ], then B = A[x 2,..., x n ]. Since B is a field, A B must be an integral domain. Let K = QuotA). Since B is a field, and A B, we must have K B as well. Thus B = K[x 2,..., x n ] is a K-algebra of finite type. By induction on n, B is algebraic over K. Thus we are done if we can show that x 1 is algebraic over k. Assume to the contrary that x 1 is transcendental over k. Then A is a ring of polynomials, hence a UFD. Let f A be the common denominator for the monic coefficient polynomials that occur in the equations for {x 2,..., x n } over K. Put S = {f, f 2, f 3,...} which is a multiplicative set. One can form a ring A S K consisting of elements of the form {a/b K a A & b S}. Since A is a UFD, then so is A S why?). Since each x i is the root of a monic polynomial equation with coefficients in A S, it follows that B is integral over A S. In particular, K B is integral over A S. Since A S, being a UFD, is integrally closed in K = QuotA S ) = QuotA), it follows that A S = K, which is nonsense why?). Therefore x 1 cannot be transcendental over k, and we are done.

22 ALGEBRAIC GEOMETRY Central Academic Building, University of Alberta, Edmonton, Alberta T6G 2G1, CANADA address: