Algebraic Geometry I - Wintersemester 2016/17 Taught by Prof. Dr. Peter Scholze. Typed by Jack Davies. 1 Contents 1 Affine Algebraic Varieties 18/10/2016 4 2 Affine Schemes 20/10/2016 10 3 Topological Properties of SpecA 25/10/2016 14 4 SpecA has a Natural Sheaf 27/10/2016 18 5 General Sheaf Theory 03/11/2016 22 6 Locally Ringed Spaces and Schemes 08/11/2016 26 7 Affine Schemes with Strucutre Sheaf are Rings 10/11/2016 30 8 Scheme Valued Points and Fibre Products 15/11/2016 34 9 Examples, History, and Motivation 17/11/2016 37 10 Projective Space - 22/11/2016 39 11 Quasi-Coherant Sheaves and Closed Immersion 24/11/2016 42 12 Vector Bundles and the Picard Group 29/11/2016 47 13 Finiteness Conditions and Dimension 01/12/2016 51 14 Krull Dimension and (Pre)Varieties 06/12/2016 55 15 Separated Schemes and Locally Closed Immersions 08/12/2016 58 16 Proper Maps of Schemes 13/12/2016 62 17 Internal Hom Sheaves and Affine Morphisms 15/12/2016 65 18 Valuations and Valuation Rings 20/12/2016 69 19 Separated and Properness Criterions 22/12/2016 73 20 Normal Schemes and Curves 10/01/2017 77 21 Normalisations 12/01/2017 80 22 Function Fields 17/01/2017 84 23 Topology on the Adic Spectrum 19/01/2017 87 24 Proper Normal Curves are Field Extensions 24/01/2017 91 25 An Example and Ample Line Bundles 26/01/2017 94 26 Ample Line Bundles and Quasi-Projectiveness 31/01/2017 97 27 Separated Curves are Quasi-Projective 02/02/2017 100 28 Divisors and the Picard Group of a Curve 07/02/2017 103 29 Towards Sheaf Cohomology and Next Semester 09/02/2017 107 2 Introduction ThiscoursewastaughtinBonn,GermanyovertheWintersemester2016/17,byProf. Dr. PeterScholze. Our plan was to learn the basics of algebraic geometry, so about sheaves, schemes, O -modules, X affine/separated/proper morphisms, and eventually to show that proper normal curves over k can be naturally associated to a type of field extension of k, and separated curves are quasi-projective. The author of these notes would like to thank Dr. Johannes Anschu¨tz and Alice Campigotto for reading through and editing these notes. The author also takes full responsibility for any and all inac- curacies, mistakes and typos in this write up. Peter had a lot more to say in lectures than what could be captured here. 3 1 Affine Algebraic Varieties 18/10/2016 Algebraic geometry is the study about solution sets to systems of polynomial equations. The algebra and the geometry play a sort of dual role to each other. To explore this, we’ll first revisit the (now outdated) mathematical objects that are varieties. For this lecture we fix an algebraically closed field k. Definition 1.1. A subset V ⊆ kn is an affine algebraic set if it can be written as the set of common zeros of a set of polynomials. In other words, if there is a set M ⊆ k[X ,...,X ] of polynomials in 1 n n-variables such that V =V(M):={(x ,...,x )∈kn | ∀f ∈M :f(x ,...,x )=0}. 1 n 1 n There are two simple (given Hilbert’s basis theorem) consequences of this definition. Proposition 1.2. 1. For any subset M ⊆ k[X ,...,X ], let a = a(M) = Mk[X ,...,X ] be the 1 n 1 n ideal generated by M, then V(M)=V(a). 2. ForanyM ⊆k[X ,...,X ],thereexistsafinitesubset{f ,...,f }⊆M suchthatV(f ,...,f )= 1 n 1 n 1 n V(M). Proof. 1. The containment V(M) ⊇ V(a) is obvious. The conditions of the set V(a) are stronger than the conditions of the set V(M), because M ⊆a. For the converse, write f ∈a as (cid:80)m f g i=1 i i with f ∈ M and g ∈ k[X ,...,X ], then for all x = (x ,...,x ) ∈ V(M) we have f(x) = i i 1 n 1 n (cid:80) f (x)g (x)=0, i.e. f =0. i i 2. OnetranslationofHilbert’sbasistheoremsaysthatk[X ,...,X ]isnoetherian,sincekisnoethe- 1 n rian(itisafield). LetM besomearbitrarysubsetofk[X ,...,X ],andleta=a(M)betheideal 1 n itgenerates. RecallthataringR isnoetherianifeveryidealisfinitelygenerated, orequivalently, ifeverynon-emptysetofidealshasamaximalelement. Letf ,...,f ∈M beasetofgenerating 1 m elements of a, then V(M)=V(a)=V(f ,...,f ) 1 n follows from part 1 above. Let us now consider a handful of examples of affine algebraic sets. Example 1.3. Any finite subset of kn.1) Conversely, any affine algebraic set of k is either finite or all of k. In fact, k[x] is a principal ideal domain and every f ∈k[x] factors as f =(cid:81)n (x−α ), so then i=1 i V(f)={α ,...,α }. 1 n Example 1.4. For n = 2 and k = C, with x,y coordinates, we have a range of classical examples. Keep in mind that we are really just looking at the purely real solutions. The reader is asked to graph these solution sets by hand or using wolfram alpha etc. 1. The equation x+y =0 gives us a straight line through the origin of gradient −1. 2. The unit circle in R2 can be represented by x2+y2 =1. 3. For the equation x2+y2 = −1, we of course have a non-empty solution set in C2, but there are no real solutions. 1Seeexercisesheet1problem1(i): LetZ⊆An(k))beafiniteset. ProvethatZ isZariskiclosedinAn(k). 4 4. When we consider curves of degree three (the degree of a curve now will be the degree of the equation that defines it), we have non-singular elliptic curves such as y2 =x3−x. 5. Theellipticcurvey2 =x3−x2 isquitedifferent,andhasasingularitycalledanodeattheorigin. 6. Another elliptic curve with a singularity is y2 = x3, which has a cusp at the origin (somehow even worse than a node). 7. Theequationx2 =y2 canbefactorisedas(x+y)(x−y)=0,sothesolutionsetofthispolynomial are two lines through the origin of gradient −1 and 1. The intersection of the two lines is still considered to be a singularity of this curve. Let’s now jump into something that we should have been expecting all the lecture so far: the Zariski topology on kn. Proposition 1.5. There is a unique topology on kn for which the closed subsets are exactly the affine algebraic sets. Proof. Of course we can just claim we have a topology on kn by letting the closed sets be exactly the affine algebraic sets, but now we have to check these sets contain ∅ and all of kn, that affine algebraic sets are closed under arbitrary intersection and finite unions. Firstly we note that V(∅)=kn and V(1)=∅, so we’re done with that part. We also have the equality (cid:32) (cid:33) (cid:92) (cid:91) V(M )=V M i i i∈I i∈I once we unjumble the set-theoretic definitions of objects above. Finally we have to show that affine algebraic sets are closed under finite unions, but by induction we need only worry about the union of two affine algebraic sets. It is clear that V(M )∪V(M )⊆V(M ·M ), where M ·M ={f ·g | f ∈M , g ∈M }. 1 2 1 2 1 2 1 2 To check the converse, take x ∈ V(M ·M )\V(M ). Take f ∈ M such that f(x) (cid:54)= 0, then for all 1 2 1 1 g ∈M we have fg ∈M ·M , so 2 1 2 0=(fg)(x)=f(x)g(x). Since f(x)(cid:54)=0 and we’re in a field (which is necessarily an integral domain), we have g(x)=0 for each g ∈M , so x∈V(M ). 2 2 Definition 1.6. When we write An(k), we mean the space kn with the Zariski topology. We call An(k) an n-dimensional affine space. Given a closed subset V of An(k) (so an affine algebraic subset), we have V = V(a) for some ideal a∈k[X ,...,X ]. There is a fundamental question we now want to ask ourselves about varieties. 1 n How tight is the relationship between V and a? The answer to this question is: quite tight. To be precise about this, let us remember a definition from commutative algebra. Definition 1.7. The radical of an ideal a contained in a (commutative unital) ring R is √ rad(a)= a:={x∈R | ∃m>0:xm ∈a}. 5 If we consider An(k) we have V(a) = V(rad(a)). The containment V(a) ⊇ V(rad(a)) is obvious, as a⊆rad(a). For the converse note that fn(x)=0 implies f(x)=0 for each x∈An(k). The following version of Hilbert’s Nullstellensatz (zero-position-theorem) will be proved by the end of the lecture. Theorem 1.8 (Hilbert’s Nullstellensatz). The map Φ:{ideals a⊆k[X ,...,X ] with a=rad(a)}−→{closed sets V ⊆An(k)} 1 n defined by a(cid:55)→V(a) is a bijection, with inverse Φ−1 defined by V (cid:55)−→{f ∈k[X ,...,X ] | ∀x∈V :f(x)=0}. 1 n Before we prove this theorem, let’s see an easy corollary, which shows us that we can always find non-empty affine algebraic sets for all proper ideals a⊆k[X ,...,X ]. 1 n Corollary 1.9. Let a(cid:36)k[X ,...,X ] be a proper ideal, then there exists x∈kn such that f(x)=0 1 n for all f ∈a. Proof. Since a is a proper ideal, 1 (cid:54)∈ a, so clearly 1 (cid:54)∈ rad(a), which also implies that rad(a) is a proper ideal of k[X ,...,X ]. By Hilbert’s Nullstellensatz, this implies that V(a) = V(rad(a)) (cid:54)= 1 n V(k[X ,...,X ])=∅, hence we have x∈kn with x∈V(a). 1 n A consequence of this corollary tells us how a set of polynomials has to behave. Given a finite collection of polynomials f ,...,f ∈ k[X ,...,X ], then exactly one of the follow- 1 m 1 n ing two things happens: 1. There exists x∈kn such that f (x)=...=f (x)=0. 1 m 2. There exists g ,...,g ∈k[X ,...,X ] such that 1 m 1 n f g +···+f g =1. 1 1 m m Part 2 is a clear obstruction to this collection of polynomials having a common zero. Soon we will see a proof Hilbert’s Nullstellensatz. In the proof we will first make a claim, and fin- ish the proof assuming this claim to be true. We will then prove this claim, which will require a small Lemma that we will prove now. Lemma 1.10. Let R be a non-zero, finitely generated k-algebra, then there is always a map R→k of k-algebras. Proof. We know from commutative algebra that R has a non-zero maximal ideal m, so by replacing R with R/m (which is also a non-zero finitely generated k-algebra) we can assume that R is a field. By Noether normalisation2 we have a finite injective map k[X ,...,X ](cid:44)→R for some m≥0. If m>0, 1 m then the image of the element X becomes invertible in R (it’s a field don’t forget), so X−1 ∈ R. 1 1 SinceR isafinitelygeneratedk[X ,...,X ]-module, R isintegraloverk[X ,...,X ]andamountthe 1 m 1 m X ,...,X we have some a ,...,a ∈k[X ,...,X ] such that 1 m 1 n 1 m (cid:0)X−1(cid:1)n+a (cid:0)X−1(cid:1)n−1+···+a =0. 1 1 1 n If we multiply the above equation by Xn we obtain an algebraic relation amoung X and 1, which 1 1 contradicts the fact these elements are algebraically independent. 2HereisareminderofNoethernormalisation: Givenafieldkandafinitelygeneratedk-algebraA,thenthereexistsa non-negativeintegermandasetofalgebraicallyindependentelementsX1,...,Xm∈AsuchthatAisafinitelygenerated moduleoverthepolynomialringk[X1,...,Xm]. Thiscanberephrasedasdoneinourfirstlecturebysayingthereexists afinite(sothatRisafinitelygeneratedk[X1,...,Xm]-module)injectivemapk[X1,...,Xm](cid:44)→R. 6 This means m = 0 and, by the finiteness of the map k → R, that R is a finite field extension of k. Recall though from the very beginning of the lecture that k is algebraically closed, so there does not exist a non-trivial finite field extension of k. Hence k ∼= R and we have our map R → k of k-algebras. Note that if k is not algebraically closed, then we fall flat right at the end of that proof, and we can only conclude that R is some finite field extension of k. Now we are ready to prove the Nullstellensatz. Proof of Hilbert’s Nullstellensatz. Weclaimthefollowingformulaholdsforallidealsa∈k[X ,...,X ], 1 n and where V =V(a), rad(a)={f ∈k[X ,...,X ] | ∀x∈V(a):f(x)=0}=Φ−1(V(a)). (1.11) 1 n If we know this, then it is clear that Φ−1◦Φ is the identity, Φ−1◦Φ(V(a))=Φ−1(V(a))=rad(a)=a, since a = rad(a) from our hypotheses, and Claim 1.11 says that Φ−1(a) = rad(a). The map Φ is also clearly surjective by definition of the closed sets in An(k) and the fact that V(a) = V(rad(a)) for all ideals a (not just with a=rad(a)). Hence, Φ is injective and surjective, i.e., bijective. Now we have the task of showing Claim 1.11 holds. To show, rad(a)⊆{f ∈k[X ,...,X ] | ∀x∈V(a):f(x)=0}, 1 n let f ∈ rad(a) and take x ∈ V := V(a). Since f ∈ rad(a) we have fm ∈ a for some m > 0, hence 0=fm(x)=(f(x))m in a field k, and we’re done. Conversely, let f (cid:54)∈ rad(a), then by contrapositive we need to find x ∈ V with f(x) (cid:54)= 0. Let R be the following quotient ring, R=k[X ,...,X ,Y]/(f(X ,...,X )Y −1,a). 1 n 1 n In other words, we’ve adjoined an inverse to f(X ,...,X ) and killed a. Along these lines we can 1 n rewrite R as (cid:104) −1(cid:105) R=(k[X ,...,X ]/a) f , 1 n where f is the image of f in the quotient ring k[X ,...,X ]/a. If R=0 we would have 1=0 , i.e. in 1 n this localised ring there would exist an m>0 such that m m m f =f ·1=f ·0=0. Thiswouldimplythatfm ∈ainsidek[X ,...,X ], i.e. f ∈rad(a), acontradiction. HenceR isanon- 1 n zeroring. Alsonoticethatifwehadamapofk-algebrasR→k,thenwewouldobtainx ,...,x ,y ∈k 1 n such that f(x ,...,x )y =1 which implies that f(x ,...,x )(cid:54)=0, but x=(x ,...,x )∈V(a). These 1 n 1 n 1 n facts are just formal consequences from the construction of R. However, we do have a map R → k of k-algebras, since R is a non-zero finitely generated k-algebra and we can apply Lemma 1.10. This concludes that Φ and Φ−1 are mutual inverses. We can construct an even nicer correspondence however, and one that we will see again in scheme theoretic language. Definition 1.12. Let V ⊆An(k) be closed. The algebra of functions on V is defined as O(V):=k[X ,...,X ]/{f | ∀x∈V :f(x)=0}. 1 n 7 Recall that a ring R is called reduced if given x∈R with xm =0 for some m>0, then x=0. Notice that O(V) is always reduced, because if fm =0 in O(V), then fm(x)=0 for all x∈V, which implies thatf(x)=0forallx∈V (sincek isafield), sof =0inO(V). NotealsothatO(V)isclearlyfinitely generated as a k-algebra. Definition1.13. AmapbetweenaffinealgebraicsetsV ⊆An(k)andW ⊆Am(k)isamapf :V →W of sets of solutions, such that there exists a collection of polynomials f ,...,f ∈k[X ,...,X ] with 1 m 1 n f(x)=f(x ,...,x )=(f (x ,...,x ),...,f (x ,...,x )) 1 n 1 1 n m 1 n for all x=(x ,...,x ) in V. 1 n This definition of a morphism clearly has the properties we demand a morphism in a category to have (i.e. composition,associativityandunitialproperties),soforeachalgebraicallyclosedfieldk weobtain a category AffVar(k) (the authors notation for now) of affine algebraic subsets in An(k) for all n≥0. Remark 1.14. The f ’s in the definition above are not determined by f, since they need only have i certainpropertiesonV,andtheclosedsubsetsV ⊆An(k)shouldbethoughtofasquitesmallinAn(k). The images of the f ’s are however well defined in O(V), since the quotient in the definition of O(V) i identifies polynomials with the same value on V. Also note that the map f :V →W determines a map f(cid:101):k[Y1,...,Ym]→O(V) which sends Yi (cid:55)→fi, the image of fi in the quotient algebra O(V). Notice that this map f(cid:101)factors through the quotient O(W). In other words, we have the following commutative diagram, k[Y ,...,Y ] f(cid:101) O(V) 1 m . f∗ O(W) The pullback map f∗ is given by g (cid:55)→g◦f, so simply pre-composition. A corollary of Hilbert’s Nullstellensatz is the following equivalence of categories. Corollary 1.15. ThereisanequivalenceofcategoriesbetweenAffVar(k)andthecategoryofreduced finitely generated k-algebras and k-algebra homomorphisms, defined by the (contravariant) functor F which sends V (cid:55)→O(V) and f :V →W to f∗ :O(W)→O(V). Proof. WewillprovethisequivalenceofcategoriesbyshowingthatF isfully3 faithful4,andessentially surjective5. Fully Faithful Let V ⊆An(k) and W ⊆An(k), and consider, F :Hom(V,W)→Hom(O(V),O(W)). V,W To check this map is injective, we take f,f(cid:48) : V → W, such that f∗ = (f(cid:48))∗. In this case we have for all g ∈ O(W),x ∈ V, g(f(x)) = f∗g(x) = (f(cid:48))∗g(x) = g(f(cid:48)(x)). Notice now that g = Y implies i f(x) = (Y f(x),...,Y f(x)) = f(cid:48)(x). To show F is surjective, take a k-algebra homomorphism 1 m V,W G : O(W) → O(V), which we can specify by simply saying where that Y → f for all Y ∈ O(W) i i i (which we remember is a quotient of k[Y ,...,Y ]). We lift these f to f ∈k[X ,...,X ], and obtain 1 m i i 1 n a map, (f ,...,f ):kn →km, 1 m which we restrict to a map f : V → W. It’s then a simple check that f∗ = G. Take some Y ∈ O(W) i and any x=(x ,...,x )∈V, then 1 n f∗Y (x)=Y ◦f(x)=Y (f (x),...,f (x))=f (x)=G(Y )(x). i i i 1 m i i 3AfunctorF :C→D isfulliftheinducedmapsonmorphismclassesarealwaysasurjection. 4Sortofdual,wecallafunctorF :C→D faithfuliftheinducedmaponmorphismclassesisalwaysaninjection. 5AfunctorF :C→D isessentiallysurjectiveifforeachY ∈D thereisanX∈C withF(X)∼=Y inD. 8 Essentially Surjective Let R be some reduced finitely generated k-algebra, so then R is simply R∼=k[X ,...,X ]/a 1 n where the ideal a has the property that a = rad(a), or else R would not be reduced. Then set V =V(a)⊆kn, and we have O(V)∼=R as a consequence of Hilbert’s Nullstellensatz. 9 2 Affine Schemes 20/10/2016 Recall from the last lecture that for a closed subset V ⊆An(k) we have the algebra of functions on V denoted as O(V). Let Map(V,k) be the set of all maps of sets V → k, which we can turn into a ring with point-wise addition and multiplication in k. Then O(V) has the following equivalent description, O(V)=im(k[X ,...,X ]−→Map(V,k)), 1 n where we include polynomials on An(k) into the ring of all set-theoretic functions V →k. We can also re-write V(a) in a similar way, as V(a)=Hom (A,k), k−alg. where A = k[X ,...,X ]/a. We would like to extend Corollary 1.15 to involve general rings, which 1 n is compatible with including the full subcategory of affine algebraic sets into our new, more general category. In general rings do not have an underlying field k. We could now start talking about general varieties, but since we want to work with schemes, it seems more efficient to define an affine scheme and then we’ll get to general schemes. Convention: All rings will now be commutative and unital, unless otherwise specified. All maps will be ring homomorphisms. Definition 2.1. Let A be a ring, then SpecA is defined as the collection of all ring homomorphisms A → K, where K is some field, and where we identify two maps A → K and A → K(cid:48) if there exists the following commutative diagram. K(cid:48) . A K This is a rather categorical definition, and in fact we could rephrase it as SpecA=colim Hom (A,K), K∈Fld Rng where Fld is the category of fields, and Rng is the category of rings. There is a problem here though, thatSpecAmightnotnecessarilybeaset,butweshallrectifythatnow,withanalternativedefinition. Proposition 2.2. The map SpecA → {prime ideals in A} defined by (f : A → K) (cid:55)→ ker(f) is a well-defined bijection. Proof. If f : A → K is a map, where K is a field, then ker(f) is a proper ideal in A, and if we have xy in ker(f), then 0 = f(xy) = f(x)f(y) which imples x or y is in ker(f) since K is a field (and an integral domain). Hence ker(f) is prime. If we have a diagram K(cid:48) f(cid:48) f A K then the map K →K(cid:48) is injective, so ker(f)=ker(f(cid:48)), which concludes our map is well-defined. To check bijectivity, we shall construct an inverse map, which in the process will construct a dis- tinguishedrepresentativefromeachequivalenceclass. Ifp⊆Aisaprimeideal, thenA/pisanintegral domain, so we have an inclusion A/p(cid:44)→Frac(A/p)=:k(p), 10