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Lectures on classical algebraic geometry (C3.4 Algebraic Geometry).∗ † Damian RÖSSLER‡ Oxford, Michaelmas Term PLEASE LET ME KNOW OF ANY MISTAKES OR TYPOS T...

Lectures on classical algebraic geometry (C3.4 Algebraic Geometry).∗ † Damian RÖSSLER‡ Oxford, Michaelmas Term PLEASE LET ME KNOW OF ANY MISTAKES OR TYPOS THAT YOU FIND IN THESE NOTES Contents 1 Introduction 3 2 Hilbert’s Nullstellensatz and algebraic sets 4 3 Regular maps between algebraic sets 6 4 Varieties 10 5 Open and closed subvarieties 15 6 Projective space 16 7 Projective varieties 18 8 Dimension 21 9 Rational maps 24 10 Products 28 11 Intersections in affine and projective space 35 ∗I am so very grateful to Yutong Dai, who carefully read the text and provided me with a long list of typos. † Lastmodified on 2024/11/24 at 17:36:21 ‡ Mathematical Institute, University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, United Kingdom 1 12 Separatedness and completeness 37 13 Smoothness 42 14 Blowing up 47 15 Appendix. Proof of Theorem 8.7. 50 Exercise sheet 1. Chapters 1-4. 54 Exercise sheet 2. Chapters 1-8. 55 Exercise sheet 3. Chapters 1-12. 57 Exercise sheet 4. All lectures. 59 2 1 Introduction Classical algebraic geometry is the study of the sets of of simultaneous solutions of collections of polynomial equations in several variables with coefficients in an algebraically closed field. Such sets are called algebraic varieties. So eg the set of simultaneous solutions of the equations x2 + y 2 − 1 = 0, xy = 0 in C2 is an algebraic variety. Because they are so easy to define, algebraic varieties appear in almost every area of mathematics. They play a crucial role in number theory, in topology, in differential geometry and complex geometry (ie the theory of complex manifolds). When the base field is C, an algebraic variety defines a complex manifold provided it has ”no kinks” (we shall give a precise definition later). A basic reference for classical algebraic geometry is chap. I of D. Mumford’s book The Red Book of Varieties and Schemes (Springer Lecture Notes in Mathematics 1358). Another reference is chap. I of R. Hartshorne’s book Algebraic Geometry (Springer). One might also consult the book by M. Reid Undergraduate algebraic geometry (London Mathematical Society Student Texts 12, Cambridge University Press 1988). An updated free online version of M. Reid’s lectures can be found under https://homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf The natural generalisation of classical algebraic geometry is the theory of schemes, which will be taught in Hilary Term. In Grothendieck’s theory of schemes, the base field can be replaced by any commutative ring but the absence of Hilbert’s Nullstellensatz, which is at the root of the material presented here, means that different techniques have to be used. There are three important tools, which will not be presented in this course: The theory of sheaves Cohomological techniques The technique of base change These tools are very powerful but there will not be enough time to present them in these lectures. Also, the best framework for them is the theory of schemes (although they could also be used in the restricted setting of this text). There is also a tool from Commutative Algebra, which will not be used here but which is very useful in Algebraic Geometry: the tensor product of modules over a ring. Tensor products are ubiquitous in the theory of schemes. The prerequisites for this course are the part A course Rings and Modules and the part B course Commutative Algebra. It is assumed that the reader is familiar with the terminology used in the notes of the commutative algebra course. We shall often quote results proven in that course, referring to it as ”CA”. I have put the CA notes on the web page of the present course for easy reference. Throughout the course, we shall work over a fixed algebraically closed field k. As in the CA course, a ring will be a commutative ring with unit, unless stated otherwise. The reader may assume that for any n > 1, the ring of polynomials k[x1 ,... , xn ] is a UFD (Unique Factorisation Domain). It can also be assumed that the localisation k[x1 ,... , xn ]S is a UFD for any multiplicative set S ⊆ k[x1 ,... , xn ]. 3 2 Hilbert’s Nullstellensatz and algebraic sets Let n > 0 and let Rn := k[x1 ,... , xn ]. Let Σ ⊆ Rn. The algebraic set associated with Σ is Z(Σ) = zero set of Σ := {(t1 ,... , tn ) ∈ k n | ∀P ∈ Σ : P (t1 ,... , tn ) = 0} Note the following simple fact. Let ΣRn be the ideal generated by Σ in Rn. Then we have Z(Σ) = Z(ΣRn ). We now recall two basic results in commutative algebra. Theorem 2.1 (Hilbert’s basis theorem; see Th. 7.6 in CA). The ring k[x1 ,... , xn ] is noetherian. Recall that a noetherian ring is a ring all of whose ideals are finitely generated. In particular, by the remark above any algebraic set in k n is the zero set of a finite number of polynomials. Theorem 2.2 (Hilbert’s strong Nullstellensatz; see Cor. 9.5 in CA). For any ideal I ⊆ Rn we have r(I) = {P ∈ Rn | ∀(t1 ,... , tn ) ∈ Z(I) : P (t1 ,... , tn ) = 0} Recall that r(I) is the radical (or nilradical) of I, ie the intersection of all the prime ideals of Rn containing I. One can show that r(I) consists of all the elements Q of Rn , such that Ql ∈ I for some l = l(Q) (see Prop. 3.2 in CA). Recall also that a radical ideal is an ideal which coincides with its own radical. If A ⊆ k n is subset, we shall write I(A) := {P ∈ Rn | ∀(t1 ,... , tn ) ∈ A : P (t1 ,... , tn ) = 0}. The set I(A) is clearly an ideal in Rn. Note that in terms of the operator I(·), the strong HNS implies that I(Z(I)) = r(I) for any ideal I of Rn. We may now prove the basic Proposition 2.3. Let V ⊆ k n be an algebraic set and let I ⊆ Rn be an ideal. Then the identities Z(I) = Z(r(I)), I(Z(I)) = r(I) and Z(I(V )) = V hold. In particular, the two maps I {algebraic sets in k n }  {radical ideals in Rn } Z are inverse to each other. Note that in this correspondence, we have V1 ⊆ V2 iff Z(V1 ) ⊇ Z(V2 ) for any two algebraic sets V1 and V2 (why?). Proof. (of Proposition 2.3) The identity Z(I) = Z(r(I)) follows from the definitions and the identity I(Z(I)) = r(I) was noted just before the statement of the proposition. We thus only have to prove that Z(I(V )) = V. To see this, note that by definition we have V ⊆ Z(I(V )). On the other hand, by definition V = Z(J) for some ideal J in k[x1 ,... , xn ]. By construction, we have J ⊆ I(V ), so Z(J) = V ⊇ Z(I(V )). Hence V = Z(I(V )). We also note the following identities, whose proof is left as an exercise for the reader: (1) I(V1 ∪ V2 ) = I(V1 ) ∩ I(V2 ) 4  P (2) I(∩i Vi ) = r( i I(Vi )) (3) Z(I1 ∩ I2 ) = Z(I1 ) ∪ Z(I2 ) P (4) Z( i Ii ) = ∩i Z(Ii ) P (where the Vi are algebraic sets, the Ii are ideals and the symbol refers to the sum of ideals). n In view of the properties (4) and (3) above, the algebraic sets in k can be viewed as the closed sets of a topology on k n , called the Zariski topology. If V ⊆ k n is an algebraic set, we endow V with the topology induced by the Zariski topology of k n. This topology is called the Zariski topology of V. We can refine the correspondence above as follows. Say that an algebraic set V ⊆ k n is reducible if V = V1 ∪ V2 , where V1 , V2 ⊆ k n are non empty algebraic sets, V1 6⊆ V2 and V2 6⊆ V1. An algebraic set V ⊆ k n is said to be irreducible if it is not reducible. One verifies from the definition that an algebraic set is irreducible iff all its non empty open subsets are dense. For the following two lemmata, we shall need the following result from CA: Theorem 2.4. Let R be a noetherian commutative ring and let I ⊆ R be a radical ideal. Then there is a unique finite set of prime ideals {pl } such that I = ∩l pl and such that for all indices l we have pl 6⊇ ∩j6=l pj. Furthermore, the pl are the prime ideals of R, which are minimal for the inclusion relation among the prime ideals containing I. Proof. This follows from the Lasker-Noether theorem (see Prop. 7.8 in CA) and the remark after Th. 6.7 in CA. Lemma 2.5. Let V ⊆ k n be an algebraic subset. Then V is irreducible iff I(V ) is a prime ideal. Proof. ”⇐”: Suppose that V is reducible. Then V = V1 ∪V2 , where V1 and V2 are two algebraic subsets not contained in each other (and in particular not empty). By property (1) above, we have I(V ) = I(V1 )∩I(V2 ), where I(V1 ) and I(V2 ) are two ideals not contained in each other. In particular, there is a1 ∈ I(V1 ) such that a1 6∈ I(V2 ) and a2 ∈ I(V2 ) such that a2 6∈ I(V1 ). In particular a1 , a2 6∈ I(V ). On the other hand a1 a2 ∈ I(V ) so that I(V ) is not prime. ”⇒”: Suppose that I(V ) is not prime. Let {pl }l∈Λ be the set of prime ideals in R, which are minimal among the prime ideals containing I(V ). By Theorem 2.4 we know that Λ is finite and that I(V ) = ∩l pl. Hence #Λ > 1 since I(V ) is not prime. Let l1 be any element of Λ. By Theorem 2.4 again (or Prop. 6.1 (ii) in CA and the minimality of the pl ) we have pl1 6⊇ ∩l6=l1 pl. On the other hand, we also have pl1 6⊆ ∩l6=l1 pl by minimality. Hence Z(pl1 ) 6⊆ Z(∩l6=l1 pl ) and Z(pl1 ) 6⊇ Z(∩l6=l1 pl ). Finally, we have Z(I(V )) = V = Z(pl1 ) ∪ Z(∩l6=l1 pl ) by (3) above and Proposition 2.3 so that V is reducible. Lemma 2.6. Let V ⊆ k n be an algebraic set. Then there is a unique finite collection {Vl }l∈Λ of irreducible algebraic subsets of k n such that (1) V = ∪l Vl ; (2) ∀l : Vl 6⊆ ∪j6=l Vj. Furthermore, the Vl are the irreducible algebraic sets in k n , which are maximal among the irreducible alge- braic sets contained in V. 5 Proof. In view of the remark after Prop. 2.3, the properties (1)...(4) above and Lemma 2.5, this is equivalent to Theorem 2.4 for R = Rn. Example. The algebraic set defined by the equation x1 x2 = 0 in k 2 has the irreducible components x1 = 0 and x2 = 0. Indeed, the two components are clearly not contained in each other and xi = 0 defines an irreducible algebraic set because k[x1 , x2 ]/(xi ) ' k[x] and thus the ideal (xi ) is prime in k[x1 , x2 ] (use Lemma 2.5). Proposition 2.7. Let V ⊆ k n be an algebraic set defined by a radical ideal I. Let v̄ = hv1 ,... , vn i ∈ V and let m be a maximal ideal of Rn. Suppose that m ⊇ I. Then (1) I({v̄}) ⊇ I and I({v̄}) is a maximal ideal of Rn ; (2) Z(m) consists of one point ū = hu1 ,... , un i ∈ V ; (3) m = (x1 − u1 ,... , xn − un ) where ū is as in (2). Proof. Unravel the definitions and apply Proposition 2.3. The last proposition in particular provides a correspondence between the points of V and the maximal ideals of Rn containing I(V ), or equivalently with the maximal ideals of Rn /I(V ). In other words, if we write for any ring R Spm(R) := {maximal ideals of R} then there is a natural bijection Spm(Rn /I(V )) ' V. Lemma 2.8. Let V ⊆ k n be an algebraic set. Under the bijection Spm(Rn /I(V )) ' V , the closed subsets of V correspond to the subsets of Spm(Rn /I(V )) of the form Z(S) := {m ∈ Spm(Rn /I(V )) | m ⊇ S} where S ⊆ Rn /I(V ). The closed subsets of V are in one to one correspondence with the radical ideals of Rn /I(V ) via Z(·). Proof. Left to the reader. Unroll the definitions. Note that the set {m ∈ Spm(Rn /I(V )) | m ⊇ S} corresponds in V to the set Z(S 0 ) ∩ V for any lifting S 0 of S to Rn. So the notation Z(S) will not lead to any confusion. Also, if C ⊆ V is a closed subset, then we have C = Z(I(C) (mod I(V ))) = Z(I(C)) ∩ V. So we will sometimes use the shorthand I(C) for I(C) (mod I(V )) ⊆ Rn /I(V ) if C is a closed subset of V. With this notation, the properties (1),... , (4) listed above are also valid for the correspondence described in Lemma 2.8. 3 Regular maps between algebraic sets Let n, t > 0. A map φ : k n → k t is said to be polynomial if there are elements P1 (x1 ,... , xn ),... , Pt (x1 ,... , xn ) ∈ Rn = k[x1 ,... , xn ], such that φ(a1 ,... , an ) = hP1 (a1 ,... , an ),... , Pt (a1 ,... , an )i for all ha1 ,... , an i ∈ k n. 6 Note that the polynomials Pi define a map of k-algebras φ∗ : Rt → Rn by the formula φ∗ (Q(y1 ,... , yt )) := Q(P1 (x1 ,... , xn ),... , Pt (x1 ,... , xn )) and on the other hand, if we are given a map of k-algebras Φ : k[y1 ,... , yt ] = Rt → Rn = k[x1 ,... , xn ], then we can define polynomials T1 (x1 ,... , xn ),... , Tt (x1 ,... , xn ) ∈ Rn by the formula Ti (x1 ,... , xn ) := Φ(yi ) and these two processes are obviously inverse to each other. So to give polynomials Pi as above is equivalent to giving a map of k-algebras Rt → Rn. Note that from definitions we see that the composition of two polynomial maps is a polynomial map. If Φ : Rt → Rn is a map of k-algebras, we shall write Spm(Φ) : k n → k t for the corresponding polynomial map (defined as above from the polynomials arising from Φ). Lemma 3.1. The map Spm : {maps of k-algebras Rt → Rn } → {polynomial maps k n → k t } is bijective. Proof. The surjectivity of Spm is a tautology so we only have to prove injectivity. Let Φ1 , Φ2 : Rt → Rn be two maps of k-algebras. Suppose that Spm(Φ1 ) = Spm(Φ2 ). We have to prove that Φ1 = Φ2. Suppose that Φ1 (resp. Φ2 ) is defined by polynomials P11 (x1 ,... , xn ),... , P1t (x1 ,... , xn ) (resp. P21 (x1 ,... , xn ),... , P2t (x1 ,... , xn )). Let i ∈ {1,... , t}. If Spm(Φ1 ) = Spm(Φ2 ) then the polynomial P1i − P2i vanishes for all the values of its variables. This implies that P1i = P2i (why?). Since i was arbitrary, we conclude that Φ1 = Φ2. In view of the lemma, for any polynomial map φ : k n → k t , there is a unique map of k-algebras φ∗ : Rt → Rn such that Spm(φ∗ ) = φ. Note that the operation (·)∗ (resp. Spm(·)) is compatible with composition of polynomial maps (resp. composition of maps of k-algebras). This follows from the definitions. Let now V ⊆ k n and W ⊆ k t be algebraic sets in k n and k t , respectively. A map ψ : V → W is said to be regular if there is a polynomial map φ : k n → k t such that φ(V ) ⊆ W and such that ψ(v) = φ(v) for all v ∈ V. Note that if ψ is given, there might be several different φ inducing ψ (what is an obvious example of this phenomenon?). Note also that a regular map is continuous for the Zariski topology (why?). Also, a composition of regular maps is regular (unroll the definitions). We shall attempt to generalise Lemma 3.1 to algebraic sets. For this, we make the following definition. Definition 3.2. Let V ⊆ k n be an algebraic set. The coordinate ring C(V ) of V is the ring C(V ) := Rn /I(V ). Note that since I(V ) is a radical ideal (see above - this also follows directly from the definitions), the ring C(V ) is a reduced ring, ie the only nilpotent element of C(V ) is the zero element. We also recall that any finitely generated algebra over a field is a Jacobson ring (see Cor. 9.4 in CA). In particular, C(V ) is a Jacobson ring. Recall that a Jacobson ring R is a ring such that for any ideal I ⊆ R, we have ∩m∈Spm(R),m⊇I = ∩p∈Spec(R),p⊇I =: r(I) 7 where Spec(R) is the set of prime ideals of R (see section 4 in CA). Let again V ⊆ k n and W ⊆ k t be algebraic sets in k n and k t , respectively. Let ψ : V → W be a regular map and let φ : k n → k t be a polynomial map inducing ψ, as above. Suppose that φ = Spm(Φ) for the map of k-algebras Φ : Rt → Rn. Lemma 3.3. We have Φ(I(W )) ⊆ I(V ). Proof. I am grateful to one of the members of the audience for pointing out a simplified proof of this lemma. Suppose Φ is given by elements P1 (x1 ,... , xn ),... , Pt (x1 ,... , xn ) ∈ Rn = k[x1 ,... , xn ], as above. By assumption, for all v̄ ∈ V , we have hP1 (v̄),... , Pt (v̄)i ∈ W and so for any Q(y1 ,... , tt ) ∈ I(W ) and any v̄ ∈ V , we have Q(P1 (v̄),... , Pt (v̄)) = 0. In other words, Φ(Q) = Q(P1 (x1 ,... , xn ),... , Pt (x1 ,... , xn )) ∈ I(V ) as required. From the lemma, we see that Φ induces a map of k-algebras ΦV,W : C(W ) → C(V ). The next lemma is needed in the next proposition. Lemma 3.4. If v̄ := hv1 ,... , vn i ∈ V then the maximal ideal of C(W ) corresponding to ψ(v̄) is the ideal Φ−1 V,W ((x1 − v1 ,... , xn − vn ) (mod I(V ))) = Φ −1 ((x1 − v1 ,... , xn − vn )) (mod I(W )). In particular, Φ−1 V,W sends maximal ideals to maximal ideals and ΦV,W entirely determines ψ : V → W. Proof. Note first that Φ−1 ((x1 − v1 ,... , xn − vn )) is maximal in Rt because there is by construction an injection of k-algebras Rt /Φ−1 ((x1 − v1 ,... , xn − vn )) ,→ Rn /(x1 − v1 ,... , xn − vn ) ' k so that Rt /Φ−1 ((x1 − v1 ,... , xn − vn )) ' k (isomorphism of k-algebras). On the other hand, any maximal ideal in Rt = k[y1 ,... , yt ] is likewise of the form (y1 − u1 ,... , yt − ut ) by Proposition 2.7. So in order to determine the ideal Φ−1 ((x1 − v1 ,... , xn − vn )) we only need to find u1 ,... , ut ∈ k such that Φ(yi − ui ) ∈ (x1 − v1 ,... , xn − vn ). (1) By the correspondence between algebraic sets and radical ideals, condition (1) is equivalent to the condition that the polynomial Φ(yi − ui ) vanishes on hv1 ,... , vn i. We compute Φ(yi − ui )(hv1 ,... , vn i) = Φ(yi )(hv1 ,... , vn i) − ui = φi (hv1 ,... , vn i) − ui where φi is the projection of the map φ : k n → k t to the i-th coordinate. We thus see that Φ(yi − ui ) vanishes on hv1 ,... , vn i for all i ∈ {1,... , t} iff φ(hv1 ,... , vn i) = hu1 ,... , ut i. Hence Φ−1 ((x1 − v1 ,... , xn − vn )) = (y1 − φ1 (v̄),... , yt − φt (v̄)). In particular, the maximal ideal of C(W ) corresponding to ψ(v̄) is the ideal Φ−1 V,W ((x1 − v1 ,... , xn − vn ) (mod I(V ))). We now have the 8 Proposition 3.5. The map ΦV,W : C(W ) → C(V ) depends only on ψ. Proof. Suppose that ψ is also induced by another polynomial map φ0 : k n → k t , associated with a map of k-algebras Φ0 : Rt → Rn. Let Φ0V,W : C(W ) → C(V ) be the map of k-algebras induced by φ0 via Lemma 3.3. Let r ∈ C(W ) and let m ∈ Spm(V )). By the above remark and the assumptions, we have (Φ0 )−1 −1 0 −1 −1 V,W (m) = ΦV,W (m) ∈ Spm(C(W )). Let n := (Φ )V,W (m) = ΦV,W (m). We have commutative diagrams ΦV,W C(W ) / C(V )   C(W )/n / C(V )/m O O ' ' k = /k and also Φ0V,W C(W ) / C(V )   C(W )/n / C(V )/m O O ' ' k = /k In particular, we see that ΦV,W (r) (mod m) = Φ0V,W (r) (mod m). Since m was an arbitrary maximal ideal of C(V ), we conclude that ΦV,W (r) − Φ0V,W (r) lies in the Jacobson radical of C(V ). Since C(V ) is a Jacobson ring and is reduced, we thus see that ΦV,W (r) = Φ0V,W (r). Since r ∈ C(W ) was arbitrary, we conclude that ΦV,W = Φ0V,W. From the last lemma, we see that we may write ΦV,W =: ψ ∗. Lemma 3.6. Let Λ : C(W ) → C(V ) be a map of k-algebras. Then there is a regular map λ : V → W such that λ∗ = Λ. Proof. Let Λ0 : Rt → Rn be a map of k-algebras such that the diagram Rt = k[y1 ,... , yt ] Λ0 / Rn = k[x1 ,... , xn ]   C(W ) Λ / C(V ) commutes. We may obtain such a map by choosing representatives in Rn of Λ(yi (mod I(W ))) for each i ∈ {1,... , xt }. By construction, we then have Λ0 (I(W )) ⊆ I(V ). Applying Lemma 3.4, Lemma 3.1 and Proposition 2.7, we conclude that Λ0 arises from a polynomial map Spm(Λ0 ) : k n → k t such that Spm(Λ0 )(V ) ⊆ W. By construction, we have (Spm(Λ0 )|V )∗ = Λ so we may choose λ = Spm(Λ0 )|V. From the last lemma, Lemma 3.4 and Proposition 3.5, we see that given a map of k-algebras Λ : C(W ) → C(V ), there is a unique regular map Spm(Λ) : V → W such that Spm(Λ)∗ = Λ (note that this generalises the operator Spm(·) defined before Lemma 3.1). On the other hand, by Proposition 3.5, Lemma 3.4 and the previous lemma, given a regular map λ : V → W , the map of k-algebras λ∗ : C(W ) → C(V ) is the unique one such that Spm(λ∗ ) = λ. 9 We conclude that there is a bijection from the set of regular maps V → W to the set of maps of k-algebras C(W ) → C(V ), which sends λ : V → W to λ∗ and who inverse is given by Spm(·). Finally note that any finitely generated reduced k-algebra is isomorphic as a k-algebra to the coordinate ring of some algebraic set (why?). All this leads to an intrinsic characterisation of algebraic sets and regular maps between them. We may view algebraic sets as a category whose objects are pairs (V, n) (n > 0), where V is the zero set in k n of a set of k-polynomials in n variables, and where the arrows from (V, n) to (W, t) are the maps from V to W which are restrictions of polynomial maps from k n to k t. The following theorem summarises the previous discussion. Theorem 3.7. The category of algebraic sets is antiequivalent to the category of finitely generated reduced k-algebras. Note that in this antiequivalence, a finitely reduced k-algebra R is not naturally associated with an algebraic set. However if V1 ⊆ k n1 and V2 ⊆ k n2 are two algebraic sets such that C(V1 ) ' C(V2 ), then the algebraic sets V1 ⊆ k n1 and V2 ⊆ k n2 are isomorphic, and any such isomorphism corresponds to precisely one isomorphism of k-algebras C(V1 ) ' C(V2 ). Also, if V ⊆ k n is an algebraic set, there is a canonical identification between the set V and the set Spm(C(V )). Finally, we see from Lemma 2.8 that the topology induced on Spm(C(V )) by this identification is determined by C(V ) only. 4 Varieties Let V ⊆ k n be an algebraic set. Note that from Theorem 3.7, there is a natural identification between the regular maps from V to k (where k is viewed as an algebraic set) and the elements of C(V ). Indeed the elements of C(V ) are in one-to-one correspondence with the morphisms of k-algebras k[x] → C(V ) and in turn these morphisms correspond to regular maps V → k. More concretely, let f ∈ C(V ) = Rn /I(V ) and let fe be an arbitrary lifting of f to Rn = k[x1 ,... , xn ]. The regular function V → k corresponding to f is then the restriction of the map k n → k given by the polynomial fe. We would also like to make sense of regular maps from open subsets of V to k. We first note the Lemma 4.1. Any open set in V is a union of open subsets of the form V \Z(f ), for f ∈ C(V ). Proof. Left to the reader. Unroll the definitions. Definition 4.2. Let U ⊆ V be an open subset. A function u : U → k is said to be regular if for any regular map of algebraic sets τ : T → V such that τ (T ) ⊆ U , the function τ ◦ u is regular on T (ie corresponds to an element of C(T )). To show that this definition is useable, we shall need the following Lemma 4.3. Suppose that the regular map h : V 0 → V makes C(V 0 ) isomorphic to C(V )[f −1 ] as a C(V )- algebra for some f ∈ C(V ). Then (1) h is injective and h is a homeomorphism onto V \Z(f ); 10 (2) if g : V 00 → V is a regular map such that g(V 00 ) ⊆ V \Z(f ), then there is a unique regular map g 0 : V 00 → V 0 such that g = h · g 0. Proof. (1) The injectivity follows from the fact that for any maximal ideal m of C(V )[f −1 ], m is generated by the image of m ∩ C(V ) = (h∗ )−1 (m) in C(V )[f −1 ] (recall that (h∗ )−1 (m) is maximal by Lemma 3.4 - one could also appeal to Cor. 10.4 in CA). See Lemma 5.6 in CA for this. Next, we show that h(V 0 ) ⊆ V \Z(f ). In terms of maximal ideals, this translates to the statement that f 6∈ (h∗ )−1 (m) = m ∩ C(V ) for all the maximal ideals of C(V )[f −1 ]. By the general properties of localisations (see Lemma 5.6 in CA), m ∩ C(V ) does not meet the multiplicative set generated by f , so in particular f 6∈ m ∩ C(V ). This shows that h(V 0 ) ⊆ V \Z(f ). We now show that h|V 0 : V 0 → V \Z(f ) is surjective. For this, note that if n is a maximal ideal of C(V ) such that f 6∈ n then there is a unique prime ideal n0 of C(V )[f −1 ] such that n0 ∩ C(V ) = (h∗ )−1 (n0 ) = n (again use Lemma 5.6 in CA). The ideal n0 is also maximal. To see this, let n00 ⊇ n0 be a maximal ideal. Since we have (h∗ )−1 (n00 ) ⊇ n, we thus have (h∗ )−1 (n00 ) = n. Since as before n0 (resp. n00 ) is generated by the image of n (resp. (h∗ )−1 (n00 )) in C(V )[f −1 ], we see that n0 = n00 , ie n0 is maximal. To show that h is a homeomorphism onto V \Z(f ), it is sufficient to show that image of any closed subset of V 0 maps to a closed subset of V \Z(f ). In terms of ideals, this translates to the statement that for any ideal J of C(V )[f −1 ], there exists an ideal I = I(J) of C(V ) such that m ⊇ J iff m ∩ C(V ) ⊇ I for all m ∈ Spm(C(V )[f −1 ]). Letting J be an ideal of C(V )[f −1 ], define I := J ∩ C(V ). If m ⊇ J we clearly have m ∩ C(V ) ⊇ I. On the other hand, if m ∩ C(V ) ⊇ I, we have m = (m ∩ C(V )) · C(V )[f −1 ] ⊇ I · C(V )[f −1 ] = J (again use Lemma 5.6 in CA, in particular (ii) in the proof of Lemma 5.6). So for any ideal J of C(V )[f −1 ], we may choose I = I(J) = J ∩ C(V ). (2) We first translate this into a statement of commutative algebra. We are given a map of k-algebras g ∗ : C(V ) → C(V 00 ) such that for any maximal ideal m of C(V 00 ), we have f 6∈ (g ∗ )−1 (m). We would like to show that there is a map of C(V )-algebras from C(V )[f −1 ] to C(V 00 ). In view of the universal property of localisation (see Lemma-Definition 5.1 in CA), it is sufficient for this to show that g ∗ (f ) is a unit in C(V 00 ). Now suppose for contradiction that g ∗ (f ) is not a unit in C(V 00 ). Then g ∗ (f ) is contained in a maximal ideal m of C(V 00 ). Hence f ∈ (g ∗ )−1 (m), which is a contradiction. We give a description of regular functions in terms of the ambient space in the next corollary. Corollary 4.4. Let f ∈ C(V ). The regular functions on V \Z(f ) are the restrictions of the functions k n → k which are of the form (FP(x (x1 ,...,xn ) 1 ,...,xn )) l ( l > 0 ), where P (x1 ,... , xn ) ∈ Rn and F (x1 ,... , xn ) ∈ Rn is any lifting of f to Rn. Proof. Note first that C(V )[f −1 ] ' C(V )[t]/(tf − 1) as a C(V )-algebra (see Lemma 5.3 in CA). Hence C(V )[f −1 ] corresponds to the algebraic set Z in k n+1 given by the ideal generated by the sets I(V ) and tF (x1 ,... , xn ) − 1 in k[x1 ,... , xn , t]. The polynomial map φ : k n+1 → k n inducing the map of k-algebras C(V ) → C(V )[t]/(tf − 1) is simply given by the formula φ(hv1 ,... , vn , zi) = hv1 ,... , vn i. The inverse of φ|Z the map Z → V \Z(f ) is given by the formula hv1 ,... , vn i 7→ hv1 ,... , vn , F (v1 ,... , vn )−1 i (it must be this φ|Z map since the map Z → V \Z(f ) is bijective and (v1 ,... , vn , F (v1 ,... , vn )−1 ) ∈ Z by construction). Hence a regular map on V \Z(f ) is given by the evaluation of a polynomial in the variables x1 ,... , xn , t on the vector hv1 ,... , vn , F (v1 ,... , vn )−1 i (for hv1 ,... , vn i ∈ V \Z(f )). This is the conclusion of the corollary. Note that the last lemma implies that the regular functions on V \Z(f ) are all quotients of restrictions of 11 regular functions on V by powers of f. Also, Lemma 4.3 implies that if h ∈ C(V ) is a regular function on V and h|V \Z(f ) = 0, then f · h = 0 (by the definition of localisation and the fact that C(V ) is reduced). Proposition 4.5. Let U be an open subset of the algebraic set V ⊆ k n. A function a : U → k is regular iff for any point ū ∈ U , there is a polynomial F ∈ Rn , such that F (ū) 6= 0 and a polynomial P ∈ Rn such that a coincides with P/F in a neighbourhood of ū. This implies in particular that if a function a : U → k is regular and nowhere vanishing, then 1/a : U → k is also a regular function. In other words, the units in the ring of regular functions U → k are the nowhere vanishing regular functions. Proof. (of Proposition 4.5). We first show the following. Let W ⊆ k t be an algebraic set. Let f1 ,... , fl ∈ C(W ) and suppose that (f1 ,... , fl ) = C(W ). Let h : W → k be a function (not assumed regular) and suppose that for each i ∈ {1,... , l} there is an integer ni > 0 and an element ci ∈ C(W ) such that h|W \Z(fi ) = ci /fini. We claim that the function h is then regular on W (ie arises from an element of C(W ), or in other words by restriction to W of a polynomial map k t → k). To prove this, note first that we may assume that all the ni are equal to some m > 1. Indeed, if we let m := 1+supi ni then we may write h|W \Z(fi ) = ci fim−ni /fim for all i. Now notice that for all i, j ∈ {1,... , l} we have h|W \Z(fi fj ) = ci /fim = cj /fjm so that (fi fj )m (ci /fim − cj /fjm ) = fjm ci − cj fim = 0 on W \Z(fi fj ). We deduce from the remark preceding the proposition that (fi fj )fjm ci = (fi fj )cj fim on V. Now let bi ∈ C(W ) be functions such that X bi fi2m = 1 i (note that we also have (f12m ,... , fl2m ) = C(W ) - prove this or see Lemma 12.2 in CA). Let X eh := bi fim ci. i We compute X X X X hfj2m = e bi fim fj2m ci = bi (fi fj )m fjm ci = bi (fi fj )m fim cj = ( bi fi2m )fjm cj = fjm cj i i i i h|W \Z(fj ) = cj /fjm. Hence e so that e h = h. This completes the proof of the claim. Coming back to the proposition, note that the ”⇒” direction of the equivalence stated in the proposition is clear from Lemma 4.1 and Corollary 4.4. Thus we only have to prove the ”⇐” direction of the equivalence. Since the topology of U is quasi-compact (this will be proven in exercise sheet 2, question 2.5 (4) - you can also prove this directly), we may reword this implication as follows. Let g1 ,... , gl ∈ C(V ) and suppose 0 that U = ∪i (V \Z(gi )). Let V 0 ⊆ k n be an algebraic set and let H : V 0 → V be a regular map such that H(V 0 ) ⊆ U. Suppose that for all i ∈ {1,... , l} we have a|V \Z(gi ) = di /gi for some ni > 0 and some di ∈ C(V ). The ”⇐” direction of the equivalence of the proposition is then the statement that a◦H = H ∗ (a) is a regular function on V 0. So we only have to prove this last statement under the just stated assumptions. 12 Note first that by construction, for all i ∈ {1,... , l} we have H ∗ (a)|V 0 \Z(H ∗ (gi )) = H ∗ (di )/H ∗ (gi ). Also, since H(V 0 ) ⊆ U , we have (H ∗ (g1 ),... , H ∗ (gl )) = C(V 0 ). Hence we may apply the preceding claim to W = V 0 , fi = H ∗ (gi ) and h = H ∗ (a) to conclude that H ∗ (a) is regular on V 0. Note that in view of the previous proposition, the following property holds trivially: if U 0 ⊆ U is an inclusion of open subsets of V , then the restriction to U 0 of a regular function on U is also regular. We encapsulate this property in the following Definition 4.6. Let T be a topological space. A sheaf of functions OT on T with values in k is an as- signement, which associates with each open subset O of T a sub k-algebra OT (O) of Maps(O, k), with the following property: for any open covering {Oi } of an open subset O, a function f : O → k lies in OT (O) iff f |Oi ∈ OT (Oi ) for all i. Here Maps(O, k) is the set of functions from O to k, with its natural k-algebra structure (given by pointwise multiplication and addition). Note that if O is an open subset of topological space endowed with a sheaf of k-valued functions, O inherits a sheaf of k-valued functions from T. Proposition 4.5 implies that for any algebraic set V ⊆ k n , the regular functions on Zariski open subsets of V define a sheaf of functions OV with values in k on V. There is a natural notion of mapping between topological spaces endowed with sheaves of k-valued functions: Definition 4.7. Let (T, OT ) and (T 0 , OT 0 ) be two topological spaces endowed with sheaves of functions with values in k. A morphism (sometimes loosely called a map) from (T, OT ) to (T 0 , OT 0 ) is a continuous map a : T → T 0 such that for any open subset U 0 ⊆ T 0 and any element f ∈ OT 0 (U 0 ), the function f ◦ a|a−1 (U 0 ) on a−1 (U 0 ) lies in OT (a−1 (U 0 )). We will also need the following definition. Let T be a topological space endowed with a sheaf of functions OT with values in k. Let t ∈ T. Let ObT,t := ∪O open, t∈O OT (O) (where all the OT (O) are considered to be disjoint from each other). Define an equivalence relation on O bT,t by declaring two functions in O bT,t equivalent if they coincide in some open neighbourhood of t. The set of equivalence classes in O bT,t has a natural k-algebra structure and we denote it by OT,t. The k-algebra OT,t is called the local ring at t. Note that by definition, for any open neighbourhood O of t, there is a natural map of k-algebras OT (O) → OT,t. Also, there is a natural map of k-algebras OT,t → k, which is given by evaluation at t. If we are given a morphism from (T, OT ) to (T 0 , OT 0 ) as in the last definition, the pull-back of functions gives a map of k-algebras OT,a(t) → OT,t for any t ∈ T. From the very definition of regularity, we see that any regular map from an algebraic set to another induces a morphism between the associated topological spaces with sheaves of k-valued functions. We are now ready to define a general variety. Definition 4.8. Let T be a topological space endowed with a sheaf of functions with values in k. We say that T is a variety if there is a finite open covering {Ui } of T , such that Ui with its induced sheaf of k-valued 13 functions is isomorphic to an algebraic set endowed with its sheaf of regular functions. A morphism of varieties is a morphism of the corresponding topological spaces with sheaves of k-valued functions. Lemma 4.9. Let V ⊆ k n be an algebraic set and let (V, OV ) be the associated topological space with sheaf of k-valued functions. Let v̄ ∈ V. Then the natural map of k-algebras C(V ) = OV (V ) → OV,v̄ extends (necessarily uniquely) to an isomorphism of k-algebras C(V )v̄ ' OV,v̄. Here we identified v̄ with the corresponding maximal ideal I({v̄}) when writing C(V )v̄ (so that C(V )v̄ is the localisation of C(V ) at the multiplicative set C(V )\I({v̄})). Proof. We first show that the map C(V ) → OV,v̄ extends to a map of k-algebras C(V )v̄ → OV,v̄. To show this, we have to show that a regular function f ∈ C(V ), which does not vanish at v̄, maps to a unit in OV,v̄. Now by definition a unit in OV,v̄ is represented by a regular function in a neighbourhood of v̄, which vanishes nowhere in that neighbourhood (see the remark before Definition 4.6). Now since f does not vanish at v̄, it is nowhere vanishing in the set V \Z(f ), which is a neighbourhood of v̄. So the image of f in OV,v̄ is a unit. So we have a unique extension of the map C(V ) → OV,v̄ to a map of k-algebras C(V )v̄ → OV,v̄. We still have to show that this last map is injective and surjective. We first show injectivity. Let f /s ∈ C(V )v̄ (where s ∈ C(V )\I({v̄})). Suppose that the image of f /s in OV,v̄ vanishes. By definition, this means that the function f vanishes in a neighbourhood of v̄. In particular, there exists an h ∈ C(V ) such that f vanishes in V \Z(h), where h does not vanish at v̄ (use Lemma 4.1). In other words, the image of f in C(V )[h−1 ] vanishes (use Lemma 4.4 and the commentary thereafter). Since h 6∈ I({v̄}), the natural map C(V ) → C(V )v̄ factors through C(V )[h−1 ] and hence the image of f in C(V )v̄ also vanishes. This settles injectivity. Now for surjectivity. By Lemma 4.1, an element ee ∈ OV,v̄ is represented by a regular function on V \Z(h), for some h which does not vanish at v̄. Such a function corresponds to an element of C(V )[h−1 ] and again since the natural map C(V ) → C(V )v̄ factors through C(V )[h−1 ], we see that ee lies in the image of C(V )v̄. Since ee ∈ OV,v̄ was arbitrary, the natural map C(V )v̄ → OV,v̄ is surjective. Note the following consequences of the last lemma. With the terminology of the lemma, the ring OV,v̄ is local. Also, note that the natural evaluation map OV,v̄ → k is surjective, because all constant functions are regular on V. Hence the kernel of the map OV,v̄ → k is maximal. Hence this kernel coincides with the unique maximal ideal of OV,v̄. For Definition 4.8 to be coherent, we need to check that we can recover an algebraic set from its associated topological space with sheaf of k-valued functions: Lemma 4.10. Let V ⊆ k n and W ⊆ k t be two algebraic sets. Let (V, OV ) and (W, OW ) be the associated topological spaces with sheaves of k-valued functions. Let g be a morphism from (V, OV ) to (W, OW ). Then g is induced by a regular map ψ : V → W. Proof. By definition, the morphism g provides a map of k-algebras C(W ) → C(V ). Furthermore, for any v̄ ∈ V , we have a commutative diagram of k-algebras 14  g∗ C(W ) / C(V )  g∗  OW,g(v̄) / OV,v̄ From the remark after Lemma 4.9, the ring OV,v̄ is a local ring and its maximal ideal consists of the elements represented by the regular functions h defined in a neighbourhood of v̄ such that h(v̄) = 0. A similar statement is true for OW,g(v̄) and g(v̄) in place of v̄. In particular, the map g ∗ : OW,g(v̄) → OV,v̄ sends the maximal ideal of OW,g(v̄) into the maximal ideal of OV,v̄. Since the involved rings are local, this implies that the inverse image by g ∗ of the maximal ideal of OV,v̄ is the maximal ideal of OW,g(v̄). Using standard properties of localisations and Lemma 4.9, we conclude that the inverse image of I({v̄}) ⊆ C(V ) by g ∗ : C(V ) → C(W ) is I({ḡ(v̄)}). In particular, g(v̄) = Spm(g ∗ )(v̄) (use Lemma 3.4). Hence g is induced by the map of k-algebras g ∗ : C(W ) → C(V ) and hence by a regular map V → W (by Theorem 3.7). In categorical terms, this implies that the category of algebraic sets embeds in the category of topological spaces with sheaves of k-valued functions by a fully faithful functor. We shall from now on call affine variety a variety isomorphic to a variety associated with an algebraic set. We shall often abbreviate ”topological space with sheaf of k-valued functions” as ”Topskf” from now on. 5 Open and closed subvarieties Proposition 5.1. Let (V, OV ) be a variety. Let U ⊆ V be an open subset and let OU be the sheaf of k-valued functions induced by OV. Then (U, OU ) is a variety and the inclusion map is a morphism of varieties. Proof. Let {Vi } be an open covering of V such that each Vi is isomorphic as a Topskf to an affine variety (where Vi is endowed with the sheaf of k-valued functions induced by V ). Then {Vi ∩ U } is an open covering of U. Since Vi ∩ U is open in Vi , there is for each i a subset Ei ⊆ C(Vi ) such that ∪e∈Ei (Vi \Z(e)) = Vi ∩ U (use Lemma 4.1). Hence we only have to show that the open subset Vi \Z(e) of Vi is isomorphic as a Topskf to an affine variety. But this follows from Lemma 4.3. An open subset of a variety is called an open subvariety if it is endowed with the structure of Topskf described in the last Proposition. Let (V, OV ) be a variety. Let Z ⊆ V be a closed subset. Endow Z with the topology induced by V. For any open subset O of Z, define a function f : O → k to be regular if there is collection of open subsets {Ui } of V and regular functions gi : Ui → k such that - (∪i Ui ) ∩ Z = O; - gi |O∩Ui = f |O∩Ui. In words, f : O → k is regular iff in the neighbourhood of every point of O the function f is the restriction of a regular function on some open subset of V. This endows Z with a structure of topological space with k-valued functions. We shall write OZ for the corresponding sheaf of k-valued functions. The sheaf of k-valued functions OZ on Z is said to be induced by OV. Proposition 5.2. The topological space Z with sheaf of k-valued functions OZ is a variety. The inclusion map Z → V is a morphism of varieties. 15 Proof. The inclusion map Z → V provides us with a morphism (Z, OZ ) → (V, OV ) of Topskf by construc- tion. Hence we only have to show that (Z, OZ ) is a variety (see Definition 4.8). Let {Vi } be a covering of V by open subsets such that (Vi , OVi ) is isomorphic as a Topskf to an affine variety. By definition, it is sufficient to show that for each i, the Topskf Z ∩ Vi is isomorphic to an affine variety. Hence we may assume that V is affine to begin with. Hence we are reduced to the situation where V ⊆ k n is an algebraic set and Z ⊆ k n is another algebraic set such that Z ⊆ V. Endow Z with the sheaf of functions OZ induced by OV. We would like to show that (Z, OZ ) is isomorphic to an affine variety as a Topskf. Now note that by Proposition 4.5 the sheaf OZ is precisely the sheaf of regular functions on Z viewed as an algebraic subset of k n. So (Z, OZ ) is isomorphic to an affine variety as a Topskf. An closed subset of a variety V is called a closed subvariety if it is endowed with the structure of Topskf induced by V. Lemma 5.3. Let (W, OW ) and (V, OV ) be two varieties. Let Z (resp. O) be a closed subset (resp. open subset) of V. Endow Z (resp. O) with its structure of closed (resp. open) subvariety. Let λ : W → V be a morphism of Topskf such that λ(W ) ⊆ Z (resp. λ(W ) ⊆ O). Then the induced map W → Z (resp. W → O) is a morphism of Topskf. Proof. Left to the reader. Unroll the definitions. We also record a consequence of the proof of Proposition 5.2: Lemma 5.4. Let V ⊆ W ⊆ k n , where V and W are algebraic sets in k n. Let (V, OV ) → (W, OW ) be the corresponding morphism of topological spaces with sheaves of k-valued functions. Then OV is induced by OW. 6 Projective space Projective varieties arise when one tries to find an algebraic counterpart of the topological notion of com- pactness. We will revisit this later when we consider complete varieties. Let n > 0. A line through the origin of k n+1 is by definition the vector subspace [v̄] of k n+1 generated by a vector v̄ ∈ k n+1 \{0}. We define projective space of dimension n to be the set Pn (k) of lines through the origin of k n+1. If v̄ = hv0 ,... , vn i ∈ k n+1 \{0}, we shall write [v0 ,... , vn ] for [hv0 ,... , vn i]. We shall endow Pn (k) with a variety structure. For i ∈ {0,... n}, define Ui = {[v0 ,... , vn ] ∈ Pn (k) | vi 6= 0}. In the following, we shall write the symbol q over a term that is to be omitted. The map ui : k n → Ui such that ui (hv0 ,... , vqi ,... , vn i) := [v0 ,... , vi−1 , 1, vi+1 ,... vn ] is clearly a bijection and we have v0 vi | vn u−1 i ([v0 ,... , vn ]) = h ,... , ,... , i. vi vi vi if [v0 ,... , vn ] ∈ Ui. 16 If j < i and vj 6= 0, we compute v0 vj | 1 vi+1 vn (u−1 qi ,... , vn i) = u−1 j ◦ ui )(hv0 ,... , v j ([v0 ,... , vi−1 , 1, vi+1 ,... vn ]) = h ,..., ,..., , ,..., i vj vj vj vj vj and if j > i and vj 6= 0, we have similarly v0 vi−1 1 vj | vn (u−1 j ◦ ui )(hv0 ,... , v qi ,... , vn i) = h ,..., , ,..., ,..., i vj vj vj vj vj Hence, if i 6= j, the map u−1 j ◦ ui gives a map from the open subset of k n Uij := {hv0 ,... , vqi ,... , vn i ∈ k n | vj 6= 0} into the open subset of k n Uji := {hv0 ,... , vqj ,... , vn i ∈ k n | vi 6= 0} and ui (Uij ) = Ui ∩ Uj = uj (Uji ). Let uij := u−1 j ◦ ui : Uij → Uji. Note that if one sees Uij as an open subvariety of k n , then Uij is an affine variety associated with the coordinate ring k[x0 ,... , xqi ,... , xn ][x−1 j ] ' k[x0 ,... , xi ,... , xn ][t]/(txj − 1) q and similarly, Uji is an affine variety associated with the coordinate ring k[y0 ,... , yqj ,... , yn ][yi−1 ] ' k[y0 ,... , yqj ,... , yn ][t]/(zyi − 1) One checks from the definitions that uij arises from the polynomial map which sends z to xj and yl to xl · t if l 6= i and to t if l = i. Hence uij defines a morphism of varieties from Uij to Uji. One checks from the just given formula that uij and uji are inverse to each other, so uij is an isomorphism of varieties. Now we define a topology on Pn (k) by declaring a subset O ⊆ Pn (k) to be open iff u−1 i (O) is open in k n for all i ∈ {0,... , n} (why does this define a topology?). Furthermore, if O ⊆ Pn (k) is open, we define a k-valued function f : O → k to be regular iff f ◦ ui |u−1 (O) is a regular function on u−1 i (O) for all i. i n n Since (k , Ok ) is a Topskf, we see that with this definition, P (k) becomes a Topskf (why? - unroll the n definitions). We shall write OPn (k) for the just defined sheaf of k-valued functions on Pn (k). Proposition 6.1. The sets Ui are open in Pn (k) for all i ∈ {0,... , n} and the maps ui : k n → Pn (k) restrict to isomorphisms of Topskf between k n and (Ui , OUi ), where OUi is the sheaf of k-valued functions induced on Ui by OPn (k). In particular, the Topskf (Pn (k), OPn (k) ) is a variety. The Ui are called the standard coordinate charts of Pn (k). We shall sometimes write Uin for Ui to emphasise the dependence on n. Proof. To show that Ui is open, we have to show that u−1 n j (Ui ) is open in k for all j. We have shown above that u−1 j (Ui ) = Uji is open, so Ui is open. Next, we have to show that the map ui is a homeomorphism onto its image. The map ui is continuous and injective by definition so we only have to show that ui is an open map. So let O ⊆ k n be an open set. We have to show that ui (O) is open, or in other words that u−1 j (ui (O)) is open for all j. Now we have u−1 −1 −1 j (ui (O)) = uj (ui (O) ∩ (Ui ∩ Uj )) = uj (ui (O ∩ Uij )) = uij (O ∩ Uij ) 17 and uij (O ∩ Uij ) is open in Uji since uij : Uij → Uji is a homeomorphism by the above. On the other hand Uji is open in Uj , so uij (O ∩ Uij ) is also open in Uj. So ui is a homeomorphism onto its image. Finally, we have to show that if O ⊆ Ui is an open set then f : O → k is a regular function iff f ◦ ui |u−1 (O) i is a regular function on the open subset u−1 n i (O) of k. By definition, if f : O → k is a regular function, then f ◦ ui |u−1 (O) is regular. So suppose that f ◦ ui |u−1 (O) is regular. We have to show that for all j the i i function f ◦ uj |u−1 (O) is regular on u−1 n j (O) viewed as an open subset of k. Now we have by definition j uj |Uji = ui |Uij ◦ uji and since u−1 j (O) ⊆ Uji we thus have uj |u−1 (O) = ui |Uij ∩u−1 (O) ◦ uji |u−1 (O) j i j where uji |u−1 (O) is viewed as a map from u−1 −1 j (O) to Uij ∩ ui (O). The map uji |u−1 is a morphism of j j (O) Topskf since uji is a morphism of Topskf by the above. Also, the function f ◦ ui |Uij ∩u−1 (O) is a regular i function by the definition of a sheaf of k-valued functions. Hence f ◦ uj |u−1 (O) = (f ◦ ui |Uij ∩u−1 (O) ) ◦ uji |u−1 (O) j i j is a regular function. This completes the proof. Example. The space P1 (k) only has two coordinate charts, the charts U0 and U1. By inspection, we see that P1 (k)\Ui consists of only one point. So one can see P1 (k) as the ”compactification” of k obtained by adding a ”point at ∞” to k. If k = C, the space P1 (k) can be naturally identified (as a set) with the Riemann sphere of complex analysis. 7 Projective varieties What are the closed subsets of projective space? To answer this question, we shall need the following definitions. A polynomial P (x0 ,... , xn ) ∈ k[x0 ,... , xn ] is said to be homogenous if it is a sum of monomials of the same degree. Any polynomial P (x0 ,... , xn ) has a canonical decomposition deg(P ) X P = P[i] i=0 where P[i] is the sum of the monomials of degree i appearing in P (so that in particular P[i] is homogenous). Example. The polynomials x0 , x20 + x0 x1 are homogenous but x20 + x1 is not. We have a decomposition of k[x0 ,... , xn ] as an internal direct sum M k[x0 ,... , xn ] = k[x0 ,... , xn ][l] l>0 where k[x0 ,... , xn ][l] is the k-vector space of homogenous polynomials of degree l. In particular, we have k[x0 ,... , xn ] = k. This decomposition into a direct sum makes k[x0 ,... , xn ] into a graded ring in the sense of section 11.2 of CA. 18 Example. We have (x20 + x1 ) = x20 , (x20 + x1 ) = x1 , (x20 + x1 ) = 0. Note the following elementary fact. If P (x0 ,... , xn ) ∈ k[x0 ,... , xn ] is homogenous then P (s·x0 ,... , s·xn ) = sdeg(P ) P (x0 ,... , xn ) for all s ∈ k. We thus see that if P (x0 ,... , xn ) ∈ k[x0 ,... , xn ] is a homogenous polynomial and v̄ ∈ k n+1 is non zero, we have P (v̄) = 0 iff P (s · v̄) = 0 for all s ∈ k ∗. This gives rise to the following definition. Let S ⊆ k[x0 ,... , xn ] be a set of homogenous polynomials. We define Z(S) := {[v̄] ∈ Pn (k) | v̄ ∈ k n+1 \{0}, ∀P ∈ S : P (v̄) = 0}. A projective algebraic set in Pn (k) is a subset of the form Z(S), where S ⊆ k[x0 ,... , xn ] is a set of homogenous polynomials. For convenience, we shall extend the operator Z(·) to non homogenous polynomials. For any set S ⊆ k[x0 ,... , xn ] (not necessarily consisting of homogenous polynomials), we set Z(S) := {[v̄] | v̄ ∈ k n+1 \{0}, P[i] (v̄) = 0 ∀i > 0}. Just as in the affine case, we have Z(S) = Z(S · k[x0 ,... , xn ]) (why?). Hence the projective algebraic sets in Pn (k) are the sets of the type Z(I), where I ⊆ k[x0 ,... , xn ] is an ideal generated by homogenous elements. We shall say that an ideal of k[x0 ,... xn ] is homogenous if it is generated by homogenous elements. Lemma 7.1. Let I ⊆ k[x0 ,... xn ] be an ideal. Then I is homogenous iff for all P ∈ I and all i > 0, we have P[i] ∈ I. If I is homogenous then its radical r(I) is also homogenous. In other words, a homogenous ideal is a graded ideal in k[x0 ,... , xn ] (ie a graded k[x0 ,... , xn ]-submodule of k[x0 ,... , xn ]). Proof. See exercises. Proposition 7.2. Projective algebraic sets are closed in Pn (k). Furthermore, if C ⊆ Pn (k) is a closed subset and J is the ideal generated by the homogenous polynomials which vanish on C, then Z(J) = C. In particular, the closed subsets of Pn (k) are precisely the projective algebraic sets. Proof. Let S := {Pl } be a set of homogenous polynomials in k[x0 ,... , xn ]. By construction, we have u−1 i (Z(S)) = Z({Pl (x0 ,... , xi−1 , 1, xi+1 ,... , xn )}) so that u−1 n i (Z(S)) is closed in k. By Proposition 6.1, the set Z(S) ∩ Ui is thus closed in Ui (for the induced topology). Since the Ui cover Pn (k), we thus see that Z(S) is closed in Pn (k). As to the second assertion, we clearly have Z(J) ⊇ C. So we need to prove that Z(J) ⊆ C. In other words, we have to prove that if [v̄] 6∈ C, then there is a homogenous polynomial H ∈ J, such that H([v̄]) 6= 0. Now let j ∈ {0,... , n} and suppose that [v̄] ∈ Uj. We then have [v̄] 6∈ C ∩ Uj. Since u−1 j (C) is the zero set of an ideal in k[x0 ,... xqj ,... , xn ], there is a polynomial P (x0 ,... , xqj ,... xn ) ∈ k[x0 ,... xqj ,... , xn ] such that P (u−1 j ([v̄])) 6= 0 and such that P ∈ I(uj (C)). Let −1 deg(Pj ) x0 xj−1 xj+1 xn βj (P ) := xj P( ,..., ,..., ). xj xj xj xj 19 This is a homogenous polynomial (the ”homogenisation” of P with respect of the variable xj ) such that (βj (P ))(x0 ,... , xj−1 , 1, xj ,... , xn ) = Pj. In particular we have Z(βj (P )) ⊇ C ∩ Uj and (βj (P ))([v̄]) = P (u−1 j ([v̄])) 6= 0. Now let Qj = xj βj (P ). Then Qj is still homogenous and we have Qj ([v̄]) 6= 0 and Z(Qj ) ⊇ C (because xj vanishes on Pn (k)\Uj ). Hence we may set H = Qj. This completes the proof. If A ⊆ Pn (k) is a subset, we shall write I(A) ⊆ k[x0 ,... , xn ] for the ideal generated by the homogenous polynomials vanishing on A. This notation clashes with the notation in the affine case but the context should make it clear which definition of I(·) we use. Now we have the analogue of Proposition 2.3: Proposition 7.3. Let C ⊆ Pn (k) be a closed subset and let J ⊆ k[x0 ,... , xn ] be a homogenous radical ideal. Suppose that Z(J) 6= ∅. Then I(C) is a (by definition homogenous) radical ideal and we have Z(I(C)) = C and I(Z(J)) = J. Proof. We first show that I(C) is a radical ideal. To see this, let H ⊆ r(I(C)) be the subset of r(I(C)) consisting of the homogenous elements of r(I(C)). By the definition of the nilradical of an ideal, all the elements of H vanish on C. On the other hand, r(I(C)) is a homogenous ideal by Lemma 7.1 and so H generates r(I(C)). Hence r(I(C)) ⊆ I(C). Hence r(I(C)) = I(C). The equality Z(I(C)) = C is contained in Proposition 7.2. For the second equality, note first that the inclusion J ⊆ I(Z(J)) follows from the definitions. We thus only have to prove that J ⊇ I(Z(J)). So let Q be a non zero homogenous polynomial vanishing on Z(J). We need to show that Q ∈ J. Note that deg(Q) > 0. Indeed, if deg(Q) = 0 then Q is a non zero constant polynomial and then Z(Q) = ∅, which implies that Z(J) = ∅. This is not possible by assumption. Also, note that J does not contain any constant polynomial, for otherwise Z(J) = ∅. Now consider the map q : k n+1 \{0} → Pn (k) given by the formula q(v̄) := [v̄]. Note that q −1 (Z(J)) is by construction the set of zeroes of J in k n+1 \{0}. Hence the set of zeroes of J in k n is the set q −1 (Z(J))∪{0} (since every non constant homogenous polynomial vanishes at the 0 vector). Now Q also vanishes on q −1 (Z(J)) ∪ {0} and so by the strong Nullstellensatz we have Q ∈ r(J) = J. Lemma 7.4. Let J ⊆ k[x0 ,... , xn ] be a homogenous radical ideal. Then the subset Z(J) of Pn (k) is empty iff J = k[x0 ,... , xn ] or J = k[x0 ,... , xn ]+. Here k[x0 ,... , xn ]+ is the homogenous ideal of k[x0 ,... , xn ] generated by all the non constant homogenous polynomials. Proof. We first prove the ⇐ direction of the equivalence. So let v̄ = hv1 ,... , vn i ∈ k n+1 \{0}. Suppose that vi0 6= 0 for some i0 ∈ {0,... , n}. The homogenous polynomial xi0 ∈ k[x0 ,... , xn ]+ does not vanish at [v̄]. Since v̄ ∈ k n+1 \{0} was arbitrary, we see that Z(J) is empty if J = k[x0 ,... , xn ]+ or J = k[x0 ,... , xn ]. We now prove the ⇒ direction. So suppose that Z(J) = ∅. To avoid notational confusion, write Zaff (I) for the set of common zeroes in k n+1 of the elements of a (not necessarily homogenous) ideal I ⊆ k[x0 ,... , xn ]. By using the map q : k n+1 \{0} → Pn (k) described in the proof of Proposition 7.3, we see that Zaff (J) ∩ (k n+1 \{0}) = ∅. 20 Now suppose first that J does not contain any non zero constant polynomials. Then 0 ∈ Zaff (J) (because J is generated by non constant homogenous polynomials) so that Zaff (J) = {0}. Using the correspondence described after Proposition 2.3, we conclude that J is the radical ideal of k[x0 ,... , xn ] associated with the point 0, which is k[x0 ,... , xn ]+. If J contains a non zero constant polynomial then J = k[x0 ,... , xn ] (because J contains a unit). So we conclude that if Z(J) = ∅ then either J = k[x0 ,... , xn ]+ or J = k[x0 ,... , xn ]. We shall call the ideal k[x0 ,... , xn ]+ the irrelevant ideal of k[x0 ,... , xn ]. We conclude from Lemma 7.4 and Proposition 7.3 that there is a correspondence I {closed sets in Pn (k)}  {non irrelevant homogenous radical ideals in Rn } Z where the maps Z(·) and I(·) are inverse to each other. A projective variety is a variety isomorphic (as a variety) to a closed subvariety of Pn (k) (for some n > 0). A quasi-projective variety is a variety isomorphic to an open subvariety of a projective variety. 8 Dimension Let T be a topological space. Then T is said to be noetherian if for any descending sequence C1 ⊇ C2 ⊇ C3 ⊇... of closed subsets of T , there is an i0 > 0 such that Ci0 = Ci0 +1 =.... In this situation, we say that the sequence stabilises at i0. Note that any subset of a noetherian topological space is also noetherian (in the induced topology) (why?). Finally, note that a noetherian topological space is quasi-compact (ie any covering of the space has a finite subcovering). See exercises. The topological space T is said to be irreducible if T is not empty and any open subset of T is dense in T. Example. The Zariski topology on k n is noetherian. Indeed any descending sequence C1 ⊇ C2 ⊇ C3 ⊇... of closed subsets of k n corresponds uniquely to a sequence I(C1 ) ⊆ I(C2 ) ⊆ I(C3 ) ⊆... (see the first section) and such a sequence stabilises for some index because k[x1 ,... , xn ] is a noetherian ring (by Hilbert’s basis theorem). Consequently, the topology of any algebraic set is noetherian. A closed subspace Z of k n is irreducible iff Z is irreducible as an algebraic set (why?). Lemma 8.1. Let T be a non empty noetherian topological space. Then there is a unique finite collection {Ti } of irreducible closed subsets of T such that (1) T = ∪i Ti (2) Ti 6⊆ ∪j6=i Tj for all i. 21 Note that a consequence of the lemma is that the Ti are the irreducible closed subsets of T which are maximal for the relation of inclusion among all the irreducible closed subsets contained in T (why?). Proof. See exercises. The closed subsets Ti described in Lemma 8.1 are called the irreducible components of T. If T is an algebraic set, the decomposition of T into irreducible components coincides with the decomposition given by Lemma 2.6 (why?). Lemma 8.2. A variety is noetherian. Proof. Let V be a variety. Let C1 ⊇ C2 ⊇ C3 ⊇... be a descending sequence of closed subsets of V. Let {Ui } be a finite covering of V by open affine subvarieties. Since the Ui are noetherian (as topological spaces) by the remark above and since there are only finitely many Ui , there is an integer l > 1 such that Cl ∩ Ui = Cl+1 ∩ Ui =... for all i. Since the Ui cover V , this implies that Cl = Cl+1 =... Now consider again a non empty topological space T. The dimension dim(T ) of T is dim(T ) := sup{t | there are irreducible closed subsets C0 ,... , Ct ⊆ T such that C0 ( C1 ( · · · ( Ct }. Note that dim(T ) might be infinite. Dimension is not defined for the empty topological space (note that some authors define the dimension of the empty topological space to be −1). Lemma 8.3. Let V ⊆ k n be an algebraic set. Then dim(V ) = dim(C(V )). Here dim(C(V )) is the dimension of C(V ) as a ring (see Def. 11.1 in CA). Recall that by definition we have dim(R) := sup{n | ∃ p0 ,... , pn ∈ Spec(R) : p0 ) p1 ) · · · ) pn } for any ring R. Proof. We have already seen that irreducible closed subsets of V correspond to prime ideals of C(V ) (see Lemma 2.5). Hence the definition of dim(C(V )) corresponds with the definition of dim(V ) under the correspondence between radical ideals of C(V ) and closed subsets of V described at the beginning of section one. Theorem 8.4. (1) The dimension of k n is n. (2) The dimension of Pn (k) is n. Proof. (1) We saw in CA that dim(k[x1 ,... , xn ]) = n (see Cor. 11.27 in CA). Hence dim(k n ) = n by Lemma 8.3. (2) Apply question 2.7 to the open covering of Pn (k) by its standard coordinate charts and use (1). Definition 8.5. Let T be a topological space. Let C ⊆ T be a closed irreducible subspace. The codimension, or height of C is cod(C, T ) = ht(C, T ) := sup{t | there are irreducible closed subsets C1 ,... , Ct ⊆ T such that C ( C1 ( · · · ( Ct } 22 We shall sometimes write cod(C) and ht(C) instead of cod(C, T ) and ht(C, T ), respectively, when the ambient topological space T is clear from the context. Note that from the definitions, we have dim(T ) = sup ht(C, T ). C closed irreducible subset of T Suppose that C, V ⊆ k n are algebraic sets in k n and that C ⊆ V. Suppose that C is irreducible. Then the height of C in V is the height of the prime ideal I(C) (mod I(V )) of C(V ) (in the sense of section 11 of CA). The proof is similar to the proof of Lemma 8.3 (we leave the details to the reader). Proposition 8.6. Let V be a variety. Let C ⊆ V be an irreducible closed subset. Then dim(V ) and cod(C, V ) are finite. Proof. See question 2.7 (4). Finally, we also have the following difficult result of commutative algebra, which justifies the use of the word ”codimension”. Theorem 8.7. Let R be a finitely generated k-algebra. Suppose that R is an integral domain. Let p ⊆ R be a prime ideal. Then we have ht(p) + dim(R/p) = dim(R) The proof of this theorem is given in the Appendix. The proof is in several steps and is structured as an exercise with model solution. We suggest the reader go through the steps by themselves first without looking at the model solution. Note that the proof of Theorem 8.7 is not examinable. Corollary 8.8. Let V be an irreducible variety. Let C ⊆ V be an irreducible closed subset. Then cod(C, V ) + dim(C) = dim(V ) Note first that from the definitions, we have cod(C, V ) + dim(C) 6 dim(V ) (why?). So we only have to to prove that cod(C, V ) + dim(C) > dim(V ). Proof. Let {Vi } be a finite open covering of V. We suppose that each Vi is isomorphic to an affine variety when viewed as an open subvariety of V. Note that since V is irreducible, each Vi is irreducible as well (why?). We use question 2.7 and we obtain sup cod(C ∩ Vi , Vi ) = cod(C, V ) i,C∩Vi 6=∅ and sup dim(C ∩ Vi ) = dim(C) i,C∩Vi 6=∅ 23 Let Ri be the coordinate ring associated with Vi (so that Vi ' Spm(Ri ) as a set). Then Ri is integral by question 2.5 (3). Hence we may apply Theorem 8.7 and we compute cod(C ∩ Vi , Vi ) + dim(C ∩ Vi ) = dim(Vi ) if C ∩ Vi 6= ∅. Applying supi,C∩Vi 6=∅ (·) to both sides of this equality and using question 2.7 again, we see that there is an index i0 such that C ∩ Vi0 6= ∅ and such that cod(C ∩ Vi0 , Vi0 ) + dim(C ∩ Vi0 ) = dim(V ) and hence cod(C, V ) + dim(C) > cod(C ∩ Vi0 , Vi0 ) + dim(C ∩ Vi0 ) = dim(V ). which is what we wanted to prove. Here is another fundamental result from the CA course, which is relevant to the theory of dimension. Theorem 8.9. Let n > 0 and let V, W ⊆ k n be algebraic sets. Suppose that V ⊆ W. Suppose that I ⊆ k[x1 ,... , xn ] is such that Z(I) = V. Let l > 1 and suppose that the ideal I (mod I(W )) ⊆ C(W ) is generated by l elements. Then every irreducible component of V has codimension 6 l in W. Furthermore, if C is an irreducible component of V then there is an ideal J ⊆ I(C) ⊆ C(W ) which is generated by cod(C, W ) elements and such that C is an irreducible component of Z(J) ⊆ W (in other words I(C) is a prime ideal, which is minimal among the prime ideals which contain J). See Cor. 11.15 and Cor. 11.17 in CA for the proof. This is a consequence of Krull’s principal ideal theorem. 9 Rational maps Let V, W be varieties. Consider the set H = HV,W whose elements are morphisms f : U → W , where U is a non empty open subvariety of V. Let ∼ = ∼V,W be the relation on H, such that f : U → W and g : O → W are related by ∼ iff there is a open subvariety U O of U ∩O, which is dense in V and which is such that f |U O = g|U O. The relation ∼ is easily seen to be an equivalence relation. We shall write Rat(V, W ) for the set of equivalences classes of H under the relation ∼. We call elements of Rat(V, W ) rational maps from V to W. Beware that rational maps are not actual maps but equivalence classes of maps. We will use the notation f : V 99K W to denote a rational map f from V to W. Suppose now until further notice that V is irreducible. Note the following. Let f : U → W be a representative of a rational map from V to W. If f is dominant, then any other representative of the same rational map is dominant as well. Indeed, let g : O → W be another representative of the rational map defined by f. Then f |U O = g|U O for some open subset U O ⊆ U ∩ O (which is automatically dense in V ). Suppose for contradiction that g is not dominant. Then W \g(O) contains a non empty open subset W1. Since f : U → W is dominant, we know that f −1 (W1 ) 6= ∅. Thus, since V is irreducible, we have f −1 (W1 ) ∩ U O = g −1 (W1 ) ∩ U O 6= ∅. In particular g −1 (W \g(O)) 6= ∅, which is a contradiction. So g is also dominant. 24 We thus see from the discussion in the last paragraph that it makes sense to speak of a dominant rational map from V to W : it is a rational map all of whose representative are dominant (or equivalently, it is a rational map with one dominant representative). We shall write Ratdom (V, W ) for the set of dominant rational maps from V to W. We shall write κ(V ) as a shorthand for Rat(V, k). If f : U → k and g : O → k are two elements of HV,k , one may define a new element f + g : U ∩ O → k of HV,k by declaring that (f + g)(u) = f (u) + g(u) for all u ∈ U ∩ O (note that U ∩ O is not empty because V is irreducible). Note that f + g : U ∩ O → k is a morphism of varieties (use Proposition 4.5). Similarly, one may define an element f g = f · g : U ∩ O → k by declaring that (f · g)(u) = f (u) · g(u) for all u ∈ U ∩ O. Again, f · g : U ∩ O → k is a morphism (same reasoning as before). Finally, if f : U → k does not vanish on all of U , then we may define f −1 : U \Z(f ) → k by the formula f −1 (u) = 1/f (u). Here again, f −1 : U \Z(f ) → k is a morphism (reason as before and use the remark after Proposition 4.5). It is again easily verified that these operations are compatible with ∼V,k and we thus obtain a structure of field on κ(V ). This field is called the function field of V. There is an obvious injection k ,→ κ(V ) which makes κ(V ) into a k-algebra. Note finally that for any v ∈ V , there is a natural injection OV,v ,→ κ(V ), which sends any representative of an equivalence class in OV,v to its equivalence class in κ(V ). So κ(V ) naturally contains the local rings at all the points of V. Now suppose that we are given a dominant morphism of irreducible varieties a : V → W. Then we may define a map HW,k → HV,k by the recipe (f : O → k) 7→ (f ◦ a|f −1 (O) : f −1 (O) → k) where O is a non empty open subvariety of W and f : O → k is an element of HW,k. This definition makes sense because f −1 (O) 6= ∅ as f is dominant. One checks (we skip te details) that this map is compatible with the relations ∼W,k and ∼V,k and also with the operations +, (·)−1 and ·. One thus obtains a map of rings a∗,rat : κ(W ) → κ(V ). Note that since κ(W ) is a field, the map a∗,rat is injective. Also, if a : V → W is the inclusion of an open subvariety of V into W , it follows from the definitions that the map a∗,rat is a bijection (check!). Finally, the construction of a∗,rat is clearly compatible with compositions of dominant morphisms (ie if b : W → W1 is another dominant morphism of irreducible varieties, then (b ◦ a)∗,rat = a∗,rat ◦ b∗,rat ). We conclude from all this that the homomorphism a∗,rat only depends on the element of Rat(V, W ) defined by a. In turn, any dominant representative g : O → W of an element of Rat(V, W ) defines a map of k-algebras g ∗,rat : κ(W ) → κ(V ) ' κ(O) and again it follows from the definitions that this map only depends on the class of g in Rat(V, W ). So all in all, any dominant rational map ρ ∈ Ratdom (V, W ) gives rise to an injection of fields ρ∗,rat : κ(W ) → κ(V ). Lemma 9.1. Let X be an irreducible affine variety. Let V ⊆ k n be an algebraic set giving rise to X. Then there is a canonical isomorphism of k-algebras κ(X) → Frac(C(V )). This isomorphism is compatible with dominant regular maps between irreducible algebraic sets and the corresponding morphisms of varieties. Note that by question 2.5, the fact that V irreducible implies that the ring C(V ) is an integral domain. So it makes sense to talk about the fraction field Frac(C(V )) of C(V ). Proof. Define a map τ : C(V ) → κ(X) 25 by sending an element of C(V ) to the equivalence class of the corresponding morphism X → k. By construc- tion, this is a map of k-algebras. Now suppose that f : V → k is a regular map and suppose that τ (f ) = 0. Then by definition f vanishes on an open subset of V. However the vanishing set of f is equal to Z(f ) and so is closed. Hence Z(f ) contains the closure in V of an open subset of V and thus Z(f ) = V (because V is irreducible). Hence f is the map with constant value 0 and thus τ (f ) = 0. We thus see that h is injective. Hence τ extends to a (necessarily injective) map of fields Frac(C(V )) → κ(X) by the universal property of localisation (see Lemma-Definition 5.1 in CA). We have to show that this last map is surjective. To see this, let O be an open subset of V and let g : O → k be a representative of an element of κ(X). By Lemma 4.1, we may assume without restriction of generality that O = V \Z(f ), where f ∈ C(V ). By Corollary 4.4, we know that g = gg21 |O , where g1 , g2 ∈ C(V ). Hence τ (g1 /g2 ) = g. Since g was arbitrary, we have shown that the map Frac(C(V )) → κ(X) is surjective, and thus an isomorphism. The fact that this isomorphism is compatible with dominant regular maps between irreducible algebraic sets follows directly from the definition of τ. Proposition 9.2. Let V be an irreducible variety. Then κ(V ) is finitely generated over k as a field and the dimension of V is equal to the transcendence degree of κ(V ) over k. Recall that the transcendence degree of κ(V ) over k is the largest integer n > 0 such that there exists an injection of k-algebras k[x1 ,... , xn ] ,→ κ(V ) See section 11.1 of CA for details. Proof. (of Proposition 9.2) Let {Vi } be a finite open covering of V and suppose that each Vi is an affine variety. By a remark at the beginning of this section, the function field of Vi is isomorphic to the function field of V as a k-algebra. On the other hand, we have dim(V ) = supi dim(Vi )) by question 2.7. Hence it is sufficient to show that the transcendence degree of κ(Vi ) over k is equal to dim(Vi ) for all i. So we may suppose without restriction of generality that V is affine. In that case, the statement is a consequence of Lemma 8.3, Lemma 9.1 and Cor. 11.28 in CA (which follows from the computation of the dimension of polynomial rings and the Noether normalisation lemma). Proposition 9.3. Let a : V → W be a dominant morphism of irreducible subvarieties. Then a∗,rat : κ(W ) → κ(V ) is an isomorphism iff there exist open subvarieties V0 ⊆ V and W0 ⊆ W such that a(V0 ) ⊆ W0 and such that the induced morphism a|V0 : V0 → W0 is an isomorphism. Proof. The ⇐ direction of the equivalence is clear (why?) so we only have to establish the ⇒ direction. Let W00 ⊆ W be an open affine subvariety and let V00 be an open affine subvariety of a−1 (W0 ) (this exists by Proposition 5.1). We claim that the map V00 → W00 induced by a is also dominant. To prove this claim, suppose for contradiction that the map V00 → W00 is not dominant. Then there is a non empty subset O of W00 such that O ⊆ W00 \a(V00 ). Hence a−1 (O) ∩ V00 = ∅. Now a−1 (O) 6= ∅ since a is dominant, so this contradicts the irreducibility of V. We have thus established the claim. Since the inclusions V00 → V and W00 → W induce isomorphisms of function fields, we may thus assume without restriction of generality that V and W are affine to begin with. In view of Lemma 9.1 and question 2.5 (3), it is thus sufficient to prove the following statement of commutative algebra. Let φ : A → B be a homomorphism of finitely generated integral k-algebras. Suppose that Spm(φ)(Spm(B)) is dense in Spm(A) and suppose that the induced map Frac(φ) : Frac(A) → Frac(B) is an isomorphism. Then there is an element f ∈ A such that the induced map A[f −1 ] → B[φ(f )−1 ] is an isomorphism. 26 To prove this assertion, note that by question 1.5 we already know that under the given assumptions, φ must be injective. Note also that since we have a commutative diagram Frac(φ) Frac(A) / Frac(B) O O A φ /B all whose maps are injective, the induced map A[f −1 ] → B[φ(f )−1 ] is injective for any choice of f ∈ A\{0} (remember that A and B are integral domains). Thus we only have to show that there is f ∈ A\{0} such that the induced map A[f −1 ] → B[φ(f )−1 ] is surjective. Now let b1 ,... , bl be generators of B as a k-algebra. Let a1 /c1 ,... , al /cl ∈ Frac(A) such that bi /1 = φ(ai )/φ(ci ) =: Frac(φ)(ai /ci ) Q Q for all i ∈ {1,... , l}. Let f := i ci. Then bi /1 = Frac(φ)(ai ( j6=i cj )/f ). Hence the image of A[f −1 ] → B[φ(f )−1 ] contains bi /1 for all i ∈ {1,... , l} and also contains 1/φ(f ) = Frac(φ)(1/f ). Since B[φ(f )−1 ] is generated as a k-algebra by 1/φ(f ) and by the elements bi /1 (use Lemma 5.3 in CA), we see that A[f −1 ] → B[φ(f )−1 ] is surjective. If V and W are irreducible varieties, and V0 ⊆ V and W0 ⊆ W are open subvarieties such that V0 ' W0 , we shall say that V and W are birational, or birationally isomorphic. A birational map from V to W is a rational map from V to W which has a representative f : O → W , such that f (O) is open and such that the induced map O → f (O) is an isomorphism (where f (O) is endowed with its structure of open subvariety of W ). A birational morphism from V to W is a morphism V → W which induces a birational map. Proposition 9.3 implies that a dominant rational map ρ ∈ Ratdom (V, W ) is birational iff a∗,rat : κ(W ) → κ(V ) is bijective. Proposition 9.4. Let V, W be irreducible varieties. Let κ(W ) ,→ κ(V ) be a field extension compatible with the k-algebra structures. Then there is an open subvariety V0 of V and a dominant morphism a : V0 → W such that the extension a∗,rat : κ(W ) → κ(V0 ) is isomorphic to κ(W ) ,→ κ(V ) as a κ(W )-extension. A different wording of the conclusion of the proposition is that there is an isomorphism of rings between κ(V ) and κ(V0 ) compatible with the given κ(W )-algebra structures. Proof. We may suppose without restriction of generality that V and W are affine varieties (why?). Let B (resp. A) be the coordinate ring of V (resp. W ). Let ι : Frac(A) ' κ(W ) ,→ κ(V ) ' Frac(B) be the given field extension. We claim that there is an g ∈ B\{0} such that ι(A) ⊆ B[g −1 ] ⊆ Frac(B) (where A is identified with its image in Frac(A)). To prove this, let a1 ,... , al be generators of A as a k-algebra. For all i ∈ {1,... , l} let bi , ci ∈ B be such that bi /ci = ι(ai /1). Let g := i ci. We then have ι(ai /1) ∈ B[g −1 ] and thus Q ι(A) ⊆ B[g −1 ], proving the claim. Now let V0 be the open affine subvariety associated with B[g −1 ]. Let ι0 : A → B[g −1 ] be the map induced by ι and the natural map from A to Frac(A). Since the map ι0 is injective, it induces a dominant map V0 → W by question 1.5. Hence V0 and the map V0 → W satisfy the requirements of the proposition. 27 Finally, note the following. Let V and W be irreducible varieties. Consider the map Ratdom (V, W ) → homomorphisms of k-algebras κ(W ) → κ(V ) (∗) which sends a ∈ Ratdom (V, W ) to a∗,rat : κ(W ) → κ(V ). Proposition 9.4 implies that this map is surjective. On the other hand we have Lemma 9.5. The map (∗) is injective. Proof. Let a1 , a2 ∈ Ratdom (V, W ) and suppose that a∗,rat 1 = a∗,rat 2. We have to show that a1 = a2. Now there is a by construction an open subset O ⊆ V and morphisms α1 , α2 : O → W which represent a1 and a2 , respectively. Replacing W by one of its open affine subvarieties O0 and replacing V by an open affine subvariety of α1−1 (O0 ), we may assume that both V and W are affine and that a1 (resp. a2 ) is represented by a morphism. Recycling notation, call α1 : V → W (resp. α2 : V → W ) a morphism representing a1 (resp. a2 ). Now let B (resp. A) be the coordinate ring of V (resp. W ). Let ι : Frac(A) ' κ(W ) ,→ κ(V ) ' Frac(B) be the field extension given by a∗,rat 1 = a∗,rat 2. We have by construction a commutative diagram Frac(A) ι / Frac(B) O O α∗ A i /B for i ∈ {1, 2}. Since the vertical maps are injective we thus have α1∗ = α2∗. In view of the last lemma and the comment preceding it, we thus see that the map (∗) is bijective. In particular, there is a one-to-one correspondence between dominant rational maps from V to W and κ(W )- algebra structures on the field κ(V ). We shall from now on often write a∗ for a∗,rat when V and W are irreducible varieties and a ∈ Ratdom (V, W ). This is justified by the proof of Lemma 9.5. 10 Products We wish to endow the cartesian product of two varieties with the structure of a variety. We shall do this for quasi-projective varieties. Q Definition 10.1. Let V and W be varieties. A product of V and W is a triple (V W, πV , πW ), where Q Q Q V W is a variety and πV : V W → V and πW : V W → W are morphisms of varieties. This triple is required to have the following property (PROD). (PROD) If X is a variety and a : X → V and b : X → W are morphisms of varieties, then there is a unique Q Q Q Q morphism of varieties a b : X → V W such that πV ◦ (a b) = a and πW ◦ (a b) = b. Q Note that property (PROD) in Definition 10.1 characterises the triple (V W, πV , πW ) uniquely up to unique isomorphism of triples (an isomorphism of triples is an isomorphism of the underlying varieties which is compatible with the morphisms in play). This is an example of categorical product. Note that if 28 V and W are varieties, it is not clear a priori that they have a product. However, if the product of V and Q W exists, it is uniquely defined. Abusing language, we shall often say that V W is the product of V and W without writing the associated morphisms πV and πW. We first show Theorem 10.2. Let m, n > 0. The product Pm (k) Pn (k) exists. Q Before starting with the proof, we make a construction. We shall consider the projective space Pmn+m+n. The space Pmn+m+n is by definition the set of lines through the origin in the vector space k mn+m+n+1=(m+1)(n+1) and we shall index its standard basis using double indices. Let b1 ,... , bmn+m+n+1 be the standard basis of k mn+m+n+1 , indexed in the usual manner. If i ∈ {0,... , m} and j ∈ {0,... , n}, we shall write bij for the element bj(m+1)+i+1. With this convention, each bl corresponds to precisely one bij. Since we shall exclusively work with double indices, this formula for bij is actually not important. One only needs to know that the bij form a basis of k mn+m+n+1. Let σ : Pm (k) × Pn (k) → Pmn+m+n be the map given by the formula σ(([X0 ,... , Xm ], [Y0 ,... Yn ])) = [(Xi Yj )ij ] where (·)ij means that we put (·) in the coordinate ij (corresponding to bij ). We will write Zij for a quantity in the coordinate ij and we shall write zij for the homogenous variables of Pmn+m+n. Lemma 10.3. The map σ is injective and its image is closed in Pmn+m+n. Proof. (of Lemma 10.3). For each [Zij ] ∈ σ(Pm (k) × Pn (k)) let i0 j0 = i0 ([Zij ])j0 ([Zij ]) be a pair of indices such that Zi0 j0 6= 0. The map τ : σ(Pm (k) × Pn (k)) → Pm (k) × Pn (k) sending [Zij ] to ([Z0j0 , Z1j0 ,... , Zmj0 ], [Zi0 0 , Zi0 1 ,... Zi0 n ]) has the property that τ ◦ σ = IdPm (k)×Pn (k) (why?). Hence σ is injective. To show that σ(Pm (k) × Pn (k)) is closed, consider the subvariety of Pmn+m+n described by the homogenous equations Zij Zrs − Zis Zrj (for all i, r ∈ {0,... , m} and j, s ∈ {0,... , n}. We clearly have Xi Xj Xr Xs = Xi Xs Xr Xj and so Z((Zij Zrs − Zis Zrj )) ⊇ σ(Pm (k) × Pn (k)). On the other hand, if we let [Zij ] ∈ Z((Zij Zrs − Zis Zrj )) and Zi0 j0 6= 0 (say) then σ(([Z0j0 , Z1j0 ,... , Zmj0 ], [Zi0 0 , Zi0 1 ,... Zi0 n ])) = [(Zij0 Zi0 j )ij ] = [(Zij Zi0 j0 )ij ] = [Zij ] so we also have Z((Zij Zrs − Zis Zrj )) ⊆ σ(Pm (k) × Pn (k)). We record one output of the proof of the last lemma: the image of the map σ is the zero set of the set of quadratic equations zij zrs = zis zrj. The map σ is called the Segre embedding. Its image is called the Segre variety (which is a closed subvariety of Pmn+m+n ). Proof. (of Theorem 10.2). Endow Pm (k) × Pn (k) with the variety structure inherited from the Segre variety via the Segre embedding. We will show that Pm (k) × Pn (k) has the properties listed in Definition 10.1. We first show that the projections π1 : Pm (k) × Pn (k) → Pm (k) and π2 : Pm (k) × Pn (k) → Pn (k) are morphisms of varieties. For any i0 ∈ {0,... , m} and any j0 ∈ {0,... , n}, let Ui0 j0 ⊆ Pmn+m+n be the 29 open subset of the elements [Zij ] such that Zi0 j0 6= 0 (this is a standard coordinate chart of Pmn+m+n ). Let πi0 j0 ,1 : Ui0 j0 → Pm (k) be given by the formula πi0 j0 ,1 ([Zij ]) := [Z0j0 , Z1j0 ,... , Zmj0 ] By question 2.3, this defines a morphism from Ui0 j0 to Pm (k). Now suppose that σ(([X0 ,... , Xm ], [Y0 ,... Yn ])) = [(Xi Yj )ij ] ∈ Ui0 j0 In other words, Xi0 , Yj0 6= 0. Then πi0 j0 ,1 (σ(([X0 ,... , Xm ], [Y0 ,... Yn ]))) = πi0 j0 ,1 ([(Xi Yj )ij ]) = [X0 Yj0 , X1 Yj0 ,... , Xm Yj0 ] = [X0 , X1 ,... , Xm ] = π1 (([X0 ,... , Xm ], [Y0 ,... Yn ])) Hence π1 is a morphism on the open subset σ −1 (Ui0 j0 ) of Pm (k) × Pn (k). Now if we vary the indices i0 and j0 , the open subsets σ −1 (Ui0 j0 ) cover all of Pm (k) × Pn (k) and hence π1 is a morphism (by Definition 4.7 and the fact that functions on open subsets of varieties are regular iff there are regular locally (see Definition 4.6]). Similarly π2 is a morphism. Choosing πPm (k) := π1 and πPn (k) := π2 , we shall now verify (PROD) in Definition 10.1. So let X be a variety and a : X → Pm (k) and b : X → Pn (k) be morphisms of varieties. We have to show that there is a unique morphism of varieties c : X → Pm (k) × Pn (k) such that π1 ◦ c = a and π2 ◦ c = b. Now note that the set Pm (k) × Pn (k) is the cartesian product of the sets Pm (k) and Pn (k). Hence, if the morphism c exists, it must be given by the formula c(x) = (a(x), b(x)) for all x ∈ X. Hence we only have to verify that c is a morphism of varieties. Since by the definition of a Topskf, a morphism is a morphism iff it is every locally a morphism, we may assume that X is affine and that a(X) ⊆ UPm (k),i0 and b(X) ⊆ UPn (k),j0 for some indices i0 and j0. Here UPm (k),i0 is the i0 -th standard coordinate chart of Pm (k) (resp. UPn (k),j0 is the j0 -th standard coordinate chart of Pn (k)). So let us suppose that X is associated with an algebraic set V ⊆ k t. The map a is then the restriction to V of a map k t → UPm (k),i0 of the form v̄ ∈ k t 7→ [P0 (v̄),... , Pi0 −1 (v̄), 1, Pi0 +1 (v̄),... Pm (v̄)] where the Ph are polynomials in the entries v1 ,... , vt of the vector v̄. Similarly, the map b is the restriction to V of a map k t → UPn (k),j0 of the form v̄ ∈ k t 7→ [Q0 (v̄),... , Qj0 −1 (v̄), 1, Qj0 +1 (v̄),... Qn (v̄)] where the Pl are polynomials in the entries v1 ,... , vt of the vector v̄. We now compute σ(c(v̄)) = [(Pi (v̄)Qj (v̄))ij ] and since Pi0 (v̄)Qj0 (v̄)) = 1, we see that σ ◦ c factors through a morphism V → Ui0 j0 and in particular is a morphism from V to Pmn+m+n. We conclude from Lemma 5.3 that the morphism c is a morphism of varieties. In the proof above, we have shown that Pm (k) Pn (k) can be realised as the Cartesian product Pm (k) × Q Pn (k) endowed with a certain variety structure. Furthermore, the projections πPm (k) and πPm (k) are then simply the ordinary projections on the two factors. We shall thus often write Pm (k) × Pn (k) instead of Pm (k) Pn (k). Q We shall now use Theorem 10.2 to prove that any two quasi-projective varieties have a product. We start with the 30 Lemma 10.4. Let C1 ⊆ Pm (k) and C2 ⊆ Pn (k) be closed subsets. Let V1 ⊆ Pm (k) and V2 ⊆ Pn (k) be open subsets. Then the Cartesian product C1 × C2 is closed in Pm (k) Pn (k) and the Cartesian product V1 × V2 Q is open in Pm (k) Pn (k). Q Proof. Note that the second statement is a consequence of the first, because the complement of V1 × V2 is (Pm (k)\V1 ) × Pn (k) ∪ Pm (k) × (Pn (k)\V2 ), which is closed according to the first statement. For the proof of the first statement, suppose that C1 (resp. C2 ) is defined by homogenous polynomials P1 (x0 ,... , xm ),... , Pa (x0 ,... , xm ) (resp. Q1 (y0 ,... , yn ),... , Qb (y0 ,... , yn )). Then we have \ σ(C1 × C2 ) = σ(Pm (k) × Pn (k)) T   i=0,...,m j=0,...,n Z P1 (z0j ,... , zmj ),... , Pa (z0j ,... , zmj ), Q1 (zi0 ,... , zin ),... , Qb (zi0 ,... , zin )) and thus C1 × C2 is closed in Pm (k) Pn (k). Q Q Corollary 10.5. Let V and W be two quasi-projective varieties. Then the product V W exists. Proof. By assumption, there are integers m, n > 0 and open subvarieties O1 ⊆ Pm (k) and O2 ⊆ Pm (k) such that V is isomorphic to a closed subvariety of O1 and W is isomorphic to a closed subvariety of O2. We may thus assume that V is a closed subvariety of O1 and that W is a closed subvariety of O2 , where O1 and O2 are as above. Let C1 ⊆ Pm (k) and C2 ⊆ Pn (k) be closed subsets such that C1 ∩ O1 = V and C2 ∩ O2 = W. We then have V × W = (C1 × C2 ) ∩ (O1 × O2 ) and hence V × W is closed in the open set O1 × O2 by Lemma 10.4. We endow the set V × W with the structure of variety which comes from its inclusion into O1 × O2 as a closed subset. We now claim that V × W is a p