1 Flat, Smooth, Unramified, and Étale Morphisms


 Lorraine Reed
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1 1 Flat, Smooth, Unramified, and Étale Morphisms 1.1 Flat morphisms Definition 1.1. An Amodule M is flat if the (rightexact) functor A M is exact. It is faithfully flat if a complex of Amodules P N Q is exact if and only if P M N M Q M is exact. (Or equivalently, N M = 0 if and only if N = 0). Furthermore, a morphism A B is (faithfully) flat if B is (faithfully) flat as an Amodule. Example 1.1. For A = Z, we can completely describe flat and faithfully flat modules. M is flat if and only if M is torsionfree (the forward direction always holds), and M is faithfully flat if and only if M is free (the reverse direction always holds). Furthermore, we can also give an example of a flat module which is not faithfully flat by considering M = Q. While Q is torsionfree and flat by the above, it is not faithfully flat since it s not free. Before proceeding to globalize the above for morphisms of schemes, we ll make note of a few properties: Proposition If f : A B is faithfully flat, then f is injective. 2. If S A is a multiplicatively closed subset, then A S 1 A is flat (but not faithfully flat in general as the previous example demonstrates). 3. A ring morphism f : A B is faithfully flat if and only if it is flat and ϕ := f # : Spec(B) Spec(A) is surjective. Proof. The first is a routine argument, and the second follows from that localization is exact, so we ll focus on the third statement. For the forward direction, suppose p Spec(A), then the fiber of ϕ over p is Spec(B A k(p)) where k(p) = A p /pa p. Since k(p) is necessarily nonzero and B is faithfully flat, B A k(p) 0 implies Spec(B A k(p)) 0. In other words, ϕ is surjective. For the reverse direction, let M 0 be arbitrary, then since B is flat, for any N M, B N B M. Hence, to show B M 0, it suffices to show B N 0 where N is some submodule. In particular, let N be a cyclic submodule generated by x M, then N = A/I. In this case, B A/I = B/IB and we claim that IB B. For if I m where m is maximal, then since ϕ is surjective, ϕ 1 (m) 0. Let q ϕ 1 (m) or equivalently q A = m, then I q A implies IB q which is properly contained in B. Definition 1.2. Let ϕ : X Y be a morphism of schemes, then 1. It is flat at x X if ϕ # x : O Y,ϕ(x) O X,x is flat. It is flat if it is flat at every point. 2. It is faithfully flat if it is flat and surjective. 1
2 Remark 1.3. Notice that the definition of faithful flatness isn t just that O Y,y O X,x is faithfully flat. The reason for this is that any flat morphism of local rings is necessarily faithfully flat. By closely considering our proof above, it was really only sufficient to know every closed point was in the image of ϕ # to prove faithful flatness. But for a morphism of local rings (R, m) (S, n), n R = m; hence, the above condition is immediately satisfied. Hence, to globalize the notion of faithful flatness, we ve utilized the previous proposition. Example The projection Spec(k[x, y]) Spec(k[y]) given by (x, y) y is surjective and k[y] k[x, y] is flat (A[x 1,..., x n ] is always flat over A), this implies the above map is faithfully flat. 2. The normalization of the cusp Spec(k[t]) Spec(k[t 2, t 3 ]) is surjective but not flat. More generally, for any integral domain A such that the integral closure A is not A, then the normalization Spec(A) Spec(A) is not flat. 3. Branched covers of curves are faithfully flat Properties of Flatness Finally, we ll list a collection of useful properties of flat and faithfully flat morphisms. 1. Any open immersion is flat. 2. Any flat morphism which is locally of finite presentation is open. (Without being locally of finite type, this is not necessarily the case as Spec(Q) Spec(Z) demonstrates.) 3. Flatness and faithful flatness are preserved by base change. 4. If f : X Y is finitely presented, then {x : f is flat at x} is open (but possibly empty). 5. Generic Flatness:) If f : X Y is of finite type, Y is integral, then there exists nonempty open subset U Y such that f 1 (U) U is flat. We can see that the integrality of Y is necessary since the map C[x]/x 2 C is not flat, and the only nonempty open subset of Y is Y itself. 6. Dimension of Fibers:) If f : X Y is flat at x and f(x) = y, then dim O X,x = dim O Y,y + dim ϕ 1 (y). 7. If f : X Y is a map of irreducible schemes over a field k, then flatness implies dim ϕ 1 (y) is constant by the above. Example 1.5. Consider the blowup of A 2 at the origin: π : Bl 0 (A 2 ) A 2. This is not a flat map since the fiber over the origin is isomorphic to P 1, while the the fiber over any other point is a single point; hence, it cannot satisfy the above condition on the dimension of fibers. 8. Finite maps of nonsingular curves are flat. 9. If ϕ : X Y is a map of nonsingular varities whose fibres have constant dimension, then ϕ is flat. 2
3 1.2 Smooth morphisms Geometrically, smooth maps are the analogues of submersions between manifolds (the induced derivative map on tangent vectors is surjective at every point). However, we ll also impose the condition that these maps are flat which can be motiviated by various reasons. For one reason, if we have flatness and ϕ : X Y is a smooth map of nonsingular varieties, then dϕ will be surjective on tangent spaces. Example 1.6. Consider the blowup at the origin of A 2, π : Bl 0 (A 2 ) = V (xv yu) A 2 where V (xv yu) A 2 P 1 with coordinates (x, y) (u : v). On the affine open set u 0, this map corresponds to the map k[a, b] k[s, t]: a s, b st. In particular, on this open set, [ ] 1 0 dπ = t s Thus, when s = 0, the differential is not surjective, and in particular, this morphism should not be smooth. (As established previously, this morphism is not flat which gives another reason for this morphism to not be smooth.) Next, we ll list several equivalent conditions for a morphism and define smoothness by requiring that any of the following hold. Proposition 1.7. Suppose ϕ : X Y is locally of finite presentation, then the following are equivalent: 1. ϕ is flat, and for every geometric point y = Spec(k) Y, y Y X is a nonsingular k(y)scheme (it satisfies the Jacobian criterion). 2. y X Y is regular where y Y. 3. For every x X, there exists an open affine V containing x and an open affine U containing ϕ(x) such that if O V = O U[t 1,..., t n ] (p 1,..., p m ), m n, then 1 is in the ideal generated by all m m minors of the Jacobian matrix. 4. ϕ is flat, the sheaf of relative differentials Ω X/Y is locally free of rank equal to dim(x/y ) Definition 1.3. If any of the above hold for ϕ : X Y, then ϕ is smooth Formal Smoothness Since tangent vectors at a point p correspond to morphisms Spec(k[x]/(x 2 )) X where k = k(p), for a morphism to be surjective on tangent vectors (i.e., a smooth morphism), there ought to exist a morphism g such that the following diagram commutes: 3
4 Spec k[x] (x 2 ) Spec(k) g Y X In fact, a more generalized statement holds true. Definition 1.4. A morphism ϕ : X Y is formally smooth if for every morphism Y Y and Y 0 Y defined by a nilpotent ideal, there exists a function g such that the following commutes: Y g Y Y 0 X The idea is that Y corresponds to Spec(k[x]/(x 2 )) and Y 0 corresponds to Spec(k) (defined by the nilpotent ideal (x)). While this seems much more general than smoothness, the following relates both and gives another equivalent condition for a morphism to be smooth: Theorem 1.8. A morphism ϕ : X Y is smooth if and only if it is formally smooth and locally of finite presentation. 1.3 Unramified Morphisms Geometrically, unramified morphisms are the analogues of immersions in differential geometry (the induced derivative map is injective on tangent vectors). Definition 1.5. A morphism locally of finite type f : X Y is unramified at x if the fiber X y := y X Y over y satisfies the following: O Xy,x = O X,x /m y O X,x is a finite separable algebra over k(y) (in other words, it is a finite dimensional, separable k(y) vector space). Example 1.9. Consider the morphism A 1 k A1 k, where k is algebraically closed, z z2, then the fiber over any a A 1 is k[t]/(t 2 a). For a 0, this splits as k[t] (t 2 a) = k[t] (t + a) k[t] (t a). In particular, since each piece is separable over k(a) = k, this morphism is unramified at a. However, if a = 0, then the fiber is k[t]/(t 2 ) which implies this morphism is ramified at a = 0. From another perspective, the derivative map dϕ = [2z]; this map is injective on tangent vectors only when z 0 giving another reason for it being ramified at 0 and unramified elsewhere. 4
5 Proposition If f : X Y is locally of finite presentation, then the following are equivalent: 1. f is unramified 2. Ω X/Y = 0 3. X/Y : X X Y X is an open immersion Formally Unramified Just as before for smoothness, tangent vectors corresponds to morphisms Spec(k[x]/(x 2 )) Y, so for a morphism to be injective on tangent vectors (i.e. unramified), there ought to exist at most one function g such that the following commutes: Spec k[x] (x 2 ) g Y Spec(k) X More generally, a morphism is formally unramified if the analogous statement holds in the general situation described above in the section on formal smoothness. And furthermore, we have essentially the same relation: Theorem A morphism is unramified if and only if it is formally unramified and locally of finite presentation. 1.4 Étale morphisms Finally, étale morphisms correspond to the differential geometric analogue of local diffeomorphisms, which is equivalent (by the inverse function theorem) to the induced morphism on tangent spaces being an isomorphism. Hence, the following definition is motiviated by the above: Definition 1.6. A morphism ϕ : X Y is étale at x if it s smooth and unramified at x. 5
6 Proposition The following are equivalent: 1. ϕ is flat and unramified 2. ϕ is smooth and unramified 3. ϕ is smooth of relative dimension 0 4. ϕ is formally unramified, flat, and locally of finite presentation 5. (Jacobian Criterion for that Étaleness) For every x X, there exists U = Spec(C) such C = A[t 1,..., t n ]/(p 1,.., p n ) with det( P i t j ) a unit. Just as before, the concept of formally étale can be defined, and in this case, we require that there exists a unique function g such that the above diagrams commute. 2 Smoothness of Group Schemes Definition 2.1. A group scheme G/S is flat/smooth/unramified/étale if the morphism G S is flat/smooth/unramified/étale. Remark 2.1. If G/S is flat, then it is necessarily faithfully flat since the identity e : S G produces a section of G S giving surjectivity. Definition 2.2. Let k be a field. A scheme X/k is geometrically reduced if any of the following occur: 1. X k k is reduced for every k k. 2. X k k is reduced where k is the algebraic closure of k. 3. X k k s is reduced, where k s is the separable closure of k. Remark 2.2. If char(k) = 0 (more generally when k is perfect), then X is reduced if and only if it is geometrically reduced. This follows from the algebraic statement that if A is a reduced kalgebra, then for any field extension k k, A k is reduced. Example 2.3. While geometrically reduced always implies reduced, the converse fails in positive characteristic. Let X = Spec(F p (t)[x]/(x p t)) = Spec(k), then k Fp(t) k = F p(t)[x, y] (x p t, y p t). However, x p y p = 0 implies (x y) p = 0; thus, this ring is not reduced and the above is not geometrically reduced. 6
7 Proposition 2.4. Suppose G/S is a flat group scheme which has geometrically reduced fibers, then G/S is smooth. Proof. Recall, smoothness is equivalent to G/S being flat and every geometric fiber being nonsingular; hence, it suffices to show G s Spec(k(s)) is smooth where k(s) is an algebraically closed field. By hypothesis G s is reduced and k(s) is perfect generic smoothness on the source, implies that there exists an open set U G s such that U is smooth over Spec(k(s)). In particular, O U,x = O G,x is a regular local ring for every x U. For a general point y G, let g G be such that gy = x U (such a point exists since the action of leftmultiplication on G is transitive), then since multiplication by g is an automorphism, O G,y = OG,g 1 x is regular. Example 2.5. Consider the family of elliptic curves parametrized by k[t], y 2 = x(x 1)(x t): Spec k[x, y, t] (y 2 x(x 1)(x t)) Spec(k[t]). This is a flat family with geometrically reduced fibers. If t 0 then the fiber is a smooth elliptic curve which is a group schemes. However, the fiber over t = 0 is a singular elliptic curve and the previous result implies that the total family cannot have a group scheme structure. However, the complement of the node in the central fiber is isomorphic to G m. If we delete the node then the global family has a global group scheme structure which degenerates to a G m at the origin. Example 2.6. From the above, we have that GL(n, Z) is smooth over Spec(Z); since it s a localization of a polynomial ring, it is flat over Spec(Z), and for every prime p Z, the fiber is GL(n, F p ). Since GL(n, F p ) is reduced, this implies it is geometrically reduced, and hence, by the previous proposition, it is smooth. However, if we consider the nth roots of unity, µ n = Spec(Z[t]/(t n 1)), this is not smooth over Spec(Z). The fiber over p is Spec(F p [t]/(t n 1)). However, if n = mp, then t mp 1 = (t m 1) p which implies the above is not geometrically reduced. Remark 2.7. For a group scheme over a field G/k, if char(k) = 0 and G is reduced, then G is necessarily smooth. However, by the following example, if char(k) = p and k is not perfect, then there exists reduced groups which are not geometrically reduced, and hence, they are not smooth. Example 2.8. (B. Conrad/Milne) Let k = F p (t), c k which doesn t have a pth root (e.g. let c = t), and let I = (x p + cy p ) k[x, y]. Since I is a prime ideal, k[x, y]/i is reduced. Furthermore, Spec(k[x, y]/i) is a subgroup of Spec(k[x, y]) = G 2 a (essentially because the above equation preserves addition). But if we adjoin a pth root of c and consider k(c 1/p ), then over this field, x p + cy p = (x + c 1/p y) p is not reduced implying it is not smooth. 2.1 Algebraic Groups in Characteristic 0 Theorem 2.9. (Cartier) In characteristic 0, any affine algebraic group over k is smooth. Corollary If G/S is a flat affine group over S, where S is finite type over k and char(k) = 0, then G is smooth. 7
8 In order to prove the above, we ll need to know the following series of results. Proposition If k is perfect and G/k is an affine kgroup, then G red is a closed k subgroup of G. Proof. Let G = Spec(A), µ # : A A k A the comultiplication map, and let N = (0) A (in particular, Spec(A/N) = G red ). To show G red is a ksubgroup, we ll simply show that the comultiplication, coidentity, and coinverse factor through A/N. For e # : A k and i # : A A, they both immediately factor through A/N. Since k is perfect, (A A)/N(A A) = A/N A/N which allows µ # to factor as A/N A/N A/N. Example If k is not perfect, then the above is false. For let k be a nonperfect field of characteristic p and c k k p (c does not have a pth root), then consider the closed subscheme of G a = Spec(k[x]) defined by x p2 cx p = x p (x p2 p c) = 0. Since the above equation preserves addition, this implies the above defines a closed subgroup of G a. However, G red is just defined by x(x p2 p c) = 0 since the latter product is irreducible. While G red,0 is smooth, the other local rings are not smooth implying this cannot be a subgroup. Alternatively, since this reduced equation doesn t preserve addition, this gives an alternative reason for it not forming a subgroup. Corollary Over an algebraically closed field, G is smooth if and only if every nilpotent element is contained in m 2 e where m e = ker(e # : O(G) k) Proof. Let G = Spec(A), then if G is smooth, A is a regular ring which is necessarily reduced. Conversely, since G red is a reduced algebraic group over an algebraically closed field, it is smooth and hence, (where N is the nilradical of A) dim T e (Spec(A/N)) = dim(a/n) me = dim A me. However, by our assumption on G, the natural map m/m 2 (m/n)/(m/n) 2 is injective (it s always surjective). For if a = st where s, t m/n, then a = st + n where n N. But since n m 2 and st m 2, this implies a m 2 giving injectivity (and hence an isomorphism). Thus, this imples G is smooth: dim T e (Spec(A) = dim T e (Spec(A/N)) = dim A me. Lemma Let (A, µ #, e # ) be a Hopf algebra over k, let m e be as above, then 1. A = k m e as kvector space. 2. µ # (a) = a a modulo m e m e. Proof. Since the composition k A k is the identity, this implies A splits as k m e. For part (b), by one of the axioms of the Hopf algebra, we have that the following holds: 8
9 (id A e # )(µ # (a)) = a (id A e # )(a 1) = a. (id A e # )(1 a) = 0 (id A e # )(µ # (a) a 1 1 a) = 0. Thus, µ # (a) a 1 1 a ker(id A e # ) = A m e and by symmetry, it is in m e A. However, A m e m e A = (k m e ) m e m e (k m e ) = m e m e. Finally, we have the results necessary to prove Cartier s theorem. Proof. Our strategy will be as in the above corollary, we ll show that every nilpotent element must be contained in m 2 e. Let a be a nilpotent element, then note that if a 0 in A me, this immediately implies a m 2 e. Hence, let n be the smallest exponent such that a n = 0 A and a n 1 0 in A me. By the previous lemma, there exists y m e m e such that µ # (a) = a a + y. Furthermore, µ # (a n ) = (µ # (a)) n = 0. By expanding this latter product, it consists of terms of the following form: a n 1, n(a n 1 1)y + n(a n 1 1)(1 a), (a 1) i (1 a) j y k. The first term is 0 since a n = 0, and the last terms are contained in A m 2 e since (1 a) A m e and y m e m e. Thus, n(a n 1 a) + n(a n 1 1)y A m 2 e. Since char(k) = 0, n is a unit and can be dropped; thus, by combining the above, (a n 1 a) a n 1 m e m e + A m 2 e. Now, a n 1 / a n 1 m e simply because this would imply (1 l)a n 1 = 0 A me for some l m e, but 1 l is a unit and this equation would contradict our assumption on n that a n 1 0 in A me. Hence, by reducing to A A/m 2 e, in order for this element a n 1 a a n 1 m e k A/m 2, we must have that a = 0 and a m 2. 9
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