= s j Ui U j. i, j, then s F(U) with s Ui F(U) G(U) F(V ) G(V )

Transcription

1 1 Lecture 2 Recap Last tme we talked about presheaves and sheaves. Preshea: F on a topologcal space X, wth groups (resp. rngs, sets, etc.) F(U) or each open set U X, wth restrcton homs ρ UV : F(U) F(V ) or all open V U, satsyng certan condtons (ρ UU = d U, ρ UW = ρ V W ρ UV, F( ) = 0). Shea: In addton, we also requre dentty and glung. Identty: Gven U = U, s F(U) s.t. s U = 0, then s = 0. (locally zero everywhere zero) Glung: Gven s F(U ) s.t. s U U j = s j U U j, j, then s F(U) wth s U = s. (suces to gve a uncton locally, agreeng on overlaps) Remark: F( ) = 0 s not mpled by ether the preshea condtons, nor by consderng presheaves as a contravarant unctor Op(X) Ab. It s appled axomatcally by conventon. For sheaves, t s mpled by glung (e.g. consder a cover o the empty set). Moral: don t worry about t. Morphsms o (Pre)sheaves: φ : F G. A seres o morphsms φ(u) : F(U) G(U) or every open set U, that commute wth restrcton maps: F(U) ρ φ(u) G(U) ρ F(V ) G(V ) A morphsm s an somorphsm t has an nverse. φ(v ) Stalks: F p = lmf(u) = F(U)/ = {(s, U) s F(U)}/, where (s, U) (s, U ) V U U p U such that s V = s V. More sheaves Stalks package most o the normaton we want. A morphsm o (pre)sheaves φ : F G nduces a homomorphsm on stalks, φ p : F p G p,.e. φ p (s, U) = (φ(u)s, U). Ths s well-dened, or (s, U) = (s, U ), then because φ commutes wth restrctons, φ p (s, U) = φ p (s, U ). Ths gves another characterzaton o somorphsms. Proposton. (H, Prop.II.1.1): Let F G be a morphsm o sheaves. Then φ s an somorphsm and only φ p s an somorphsm p X. Sheacaton: Every preshea has an assocated shea. Gven a preshea F, we dene a shea F + called the sheacaton o F, as ollows: F + (U) = {unctons s : U F p s(p) F p, and p U, V wth p V U and t F(V ) such that s(q) = t q q V, where t q = (t, V ) F q } These are unctons whch map nto the dsjont unon o the stalks, where we requre that p s mapped nto the stalk at p, and locally the choces o stalks are related. Usually presheaves al to be sheaves

2 2 when they don t satsy the glung axom. The + constructon xes ths by addng n enough elements to allow glung. Exercse: 1. F + s a shea. 2. There s a natural morphsm o presheaves θ : F F + such that s F(U) s + : U F p, wth s + (p) = s p = (s, U). 3. Unversal property. For any shea F and a morphsm o presheaves φ : F G there exsts a unque ψ : F + G such that: F F + F + s the smplest possble shea we could assocate to F. The + constructon s useul because sometmes when we do an operaton on a shea, we only get a preshea, so then we need to sheay. Denton: Let φ : F G be a morphsm o (pre)sheaves. The preshea kernel o φ s U ker φ(u) The preshea cokernel o φ s U coker φ(u) = G(U)/φ(U) The preshea mage o φ s U φ(u) Each o the φ(u) are group homomorphsms, so t makes sense to talk about kernels, etc. These are subpresheaves o F (or ker φ) or G (or the other two). I F and G are sheaves, then preshea kernel s n act a shea (check ths!), and the shea kernel (ker φ) s dened as ths. But the shea cokernel (coker φ) and shea mage (m φ) are the sheacatons o the preshea cokernel and mages. Surjectvty doesn t mean what you thnk t does. We say φ : F G s njectve ker φ = 0. We say φ : F G s surjectve m φ = G. G!ψ Ane Schemes From now on, R wll always be a commutatve rng wth dentty (not necessarly Noetheran). To talk about geometry, we need a topologcal space and a noton o unctons. Sheaves gve us a way o talkng about unctons. Denton: An ane scheme, Spec R, s a topologcal space, wth a shea o rngs O SpecR, ts structure shea. Examples to keep n mnd when thnkng about these dentons: ) thnk o R as a coordnate rng o a varety - or even k[x 1,..., x n ]; ) thnk o R as the most pathologcal example magnable. A shea o rngs means a shea where O SpecR (U) s a rng open U and the homs are rng homs. The topologcal space Spec R s the set o all prme deals n R. We place the Zarsk topology on Spec R : the closed sets are V (I) = {P : P I} or an deal I R. (Check!) Ths s a topology. V (I J) = V (I) V (J) and V (I + J) = V (I) V (J). Proo s the same as or varetes. Examples:

3 3 1. R = Z. Spec Z = {0} pz, p prme. V((2)) V((3)) V((5)) V((7)) [(0)] generc pont Notce we have closed ponts (e.g. 2, 3, 5,...) and not closed ponts (0, whose closure s Spec Z because 0 (p), (p)). 2. R = k[x]. The prme deals are (), rreducble. I k = k, then Spec R = (0), (x a), a k. (x a) are the closed ponts; navely they look lke A 1. We also have the generc pont (0). I k = R, then Spec R = (0), (x a), and (x 2 + ax + b) such that a 2 4b < 0. Agan the (x a) and (x 2 + ax + b) are closed ponts, and the closure o (0) s Spec R. I k = Z/2Z, then Spec R s nnte = (0), (x), (x + 1), (x 2 + x + 1),... Agan (0) s a generc pont, the others are closed. Observe that A 1 here s not a two pont set, as mght be expected. The space remembers the orbts o ponts over the algebrac closure. 3. R = k[x 1,..., x n ], k = k. I (0) s prme (ntegral doman) t wll always be dense. (x 1 a 1,..., x n a n ) are closed ponts (0satz). Any rreducble subvarety o A n corresponds to a prme deal. The closure o a such a prme contans all the ponts on t. The shea o rngs. To gve the shea o rngs on our topologcal space, we make use o a bass or the topology. For any R, we have the basc open set D() = {P SpecR : / P }. The set o all D() orms a bass or the Zarsk topology. (c. HW1 Q4) Gven a shea we can recover t by just knowng t on a bass. We ll dene the structure shea O SpecR by denng t on the bass,.e. we ll gve a rng or every D() such that t gves a shea on the base (check restrctons, d, glung). Dene: O SpecR (D()) = R. R s R localzed at the multplcatvely closed set (a set closed under products, ncludng the empty product 1) {1,, 2,...}. That s, R = { r r r R, m 0}/, where m m r j such that n j ( n r m r ) = 0. It s lke ractons, but a more general equvalence relaton to take care o zero dvsors. In partcular, = 1, O SpecR (D(1)) = R, called the global sectons. In the case R s the coordnate rng o a varety, the polynomal unctons on a varety, then n a nave varety sense, D() s the set o ponts where s nonzero, so ratonal unctons wth n the denomnator are stll well-dened unctons on the varety. Now we want to check ths denton actually gves us a shea on the base. For ths we start wth a couple o commutatve algebra results. Lemma. I I s an deal o R dsjont rom a multplcatvely closed set U, then an deal P, maxmal wth respect to contanng I and dsjont rom U, s prme. Proo. I, g / P, then P + (), P + (g) are larger deals contanng I, so u 1 P + (), u 2 P + (g), wth u 1, u 2 U. So, u 1 = p 1 + r 1, u 2 = p 2 + r 2 g. Snce u 1 u 2 / P, then g / P. Corollary. I D() D(g), then m > 0 such that m (g). Proo. I not, (g) s dsjont rom U = {1,, 2,...} so by the lemma P D() \ D(g).

4 4 Thus, D() D(g), we have m = gr or some r, so we have a homomorphsm R g R, a arn gn g n r n = arn mn so restrcton s straghtorward. When D() D(g) D(h), ths satses R h R R g Ths s a preshea on the base. We now check ths preshea s a shea on the base, that t sates dentty and glung. Suppose now that we have an open cover D() = I D( ). Frst, we clam that m > 0 such that m (, I). Otherwse, the lemma gves P ( ) avodng U = {1,, 2,...}, whch s a contradcton. Also, as beore, snce D( ) D(), we can wrte m = r or some m, r I. Identty: (secton vanshes all ts restrctons do). Suppose a/ n R satses ar n/ m n = 0 1 R. Then N such that N ar n = 0. So 0 = n N ar n = N +m n a. Snce m (, I), then or some N 0, N ( N +m n, I). So N = b N +m n or some b R. Then N a = b N +m n a = 0 and a/ n = 0/1 n R. Glung: (Gven sectons o D( ) agreeng on overlaps, can construct a secton o D()). Notce that our open cover D() = I D( ) may be assumed nte, although we haven t made any assumptons about Noetheranness o rngs. Elements o deals are only nte sums, m = J b, J <. Ths mples D() = J D( ), because any prme not contanng m, and thereore, must omt one o the, J. Every cover does have a nte subcover ( quas-compactness ); but we wll see later that properness s a more useul generalzaton o compactness. Clam: It suces to check glung only on nte covers. Suppose we are gven s = a a n j ( j ) = a j nj n n R, I wth ( j ) n j R j. It then suces to show there s s R wth s D( ) = s, J. Identty wll mply s D( ) = s, I (thnk about ths). (clam) The nte cover enables us to take a maxmum over the exponents that show up. Snce s D( ) D( j ) = s j D( ) D( j ),, j J, Ñ such that ( j )Ñ( n j j a n a j ) = 0 n R. We use nteness here - or pars, we can make ractons have the same denomnator, so assume all pars have the same denomnator. Then snce J s nte, renamng a s necessary, we may assume that n = n j = N,, j and Ñ = 0,.e. j Na N a j = 0 n R. We can wrte m = J c N or some m 0, and set g = J c a. We clam g m D( ) = s. Indeed N g = j J N c j a j = j J c ja j N = a m, and so (recall m = r ) g m D( ) = grm m m D( ) = a = s N. So we have glung. We have now shown that we do have a shea on the base. We could also have done ths another way. The approach n H s to gve the ollowng denton, and ths s then proved to be the same as the way we dened t usng a base. We wll dscuss ths

5 5 urther next tme. Also, recall that the localzaton at a prme deal R p s dened wth R \ p as the multplcatvely closed set. Proposton. a) The stalk (O SpecR ) p o (O SpecR ) at p s the localzaton R p. b) (O SpecR )(U) s the unctons {s : U R p s(p) R p ; and p U, V Uand a, R, such that Q V and / Q, s(q) = a/ R q }