Recall. (Flat)
Let f : X → Y be a morphism of schemes, and let F be an O x - module. We say that F is flat over Y at a point x ∈ X, if the stalk at x is a flat O y -module, where y = f (x).
We say simply F is flat over Y if it is flat at every point of X.
We say X is flat over Y if O X is.
Definition. (Smooth)
Let f : X → Y be a morphism of schemes of finite type over a field k, we say it’s smooth of relative dimension n if:
(1) f is flat.
(2) If X 0 ⊆ X, Y 0 ⊆ Y are irreducible components such that f (X 0 ) ⊆ Y 0 , then dim X 0 = dim Y 0 + n.
(3) For each point x ∈ X (closed or not),
dim k(x) (Ω X/Y ⊗ k(x)) = n
Proposition. 0.1 If X is integral, (3) is equivalent to Ω X/Y is locally free of rank n.
Recall (II 8.9): Let A be a noetherian local integral domain, with residue field k and quotient field K. If M is a finitely generated A-module and if dim k M ⊗ A k = dim K M ⊗ A K = r, then M is free of rank r.
Proposition. 0.2 If Y = Spec k, k = k, then X is smooth over k if and only if X is regular of dimension n.
Recall (II 8.8): Let B be a local ring containing a field k isomorphic to its residue field. Assume furthermore that k is perfect, and that B is a localization of a finitely generated k-algebra. Then
Ω B/k is a free B-module of rank dim B ⇐⇒ B is a regular local ring.
In particular, if X is irreducible separated over k, then it’s smooth if and only if it’s a nonsingular variety.
Recall (II 8.15): Let X be an irreducible separated scheme of finite type over an algebraically closed field k. Then
Ω X/k is locally free of rank dim X ⇐⇒ X is a nonsingular variety over k.
Remark. Over a nonperfect field, the last proposition is false: let k 0 be a field of characteristic p > 0, let k = k 0 (t), and let X ⊆ A 2 k be the curve defined by y 2 = x p − t, then every local ring of X is a regular local ring, but X is not smooth over k.
Proposition. 1
(a) An open immersion is smooth of relative dimension 0.
1
(b) If f : X → Y is smooth of relative dimension n, and g : Y 0 → Y is any morphism, then the morphism f 0 : X 0 → Y 0 obtained by base extension is also smooth of relative dimension n.
(c) If f : X → Y is smooth of relative dimension n, and g : Y → Z is smooth of relative dimension m, then g ◦ f : X → Z is smooth of relative dimension n + m.
(d) If f : X → Z is smooth of relative dimension n, and g : Y → Z is smooth of relative dimension m, then X × Z Y → Z is smooth of relative dimension n + m.
Proof. (a) Just check the condition.
(b) (1) By (III 9.2), f 0 is flat.
(2) Recall (III 9.6): Let f : X → Y be a flat morphism of schemes of finite type over a field k, and assume that Y is irreducible. Then the following conditions are equivalent:
(i) every irreducible component of X has dimension equal to dim Y + n.
(ii) for any point y ∈ Y (closed or not), every irreducible component of the fibre X y has dimension n.
In this case, (2) is equivalent to say every irreducible component of every fibre X y of f 0 has dimension n, which, by (II Ex. 3.20), is preserved under base change.
(3) By (II 8.10), Ω X/Y is stable under base extension, thus (3) holds for f 0 . (c) (1) By (III 9.2).
(2) If X 0 ⊆ X, Y 0 ⊆ Y, Z 0 ⊆ Z are irreducible components such that f (X 0 ) ⊆ Y 0 , g(Y 0 ) ⊆ Z 0 , then dim X 0 = dim Y 0 + n = dim Z 0 + n + m.
(3) Use the exact sequence in (II 8.11)
f ∗ Ω Y /Z −→ Ω X/Z −→ Ω X/Y −→ 0 Tensoring k(x) to get:
f ∗ Ω Y /Z ⊗ k(x) −→ Ω X/Z ⊗ k(x) −→ Ω X/Y ⊗ k(x) −→ 0
Then dim Ω X/Z ⊗ k(x) ≤ dim f ∗ Ω Y /Z ⊗ k(x) + dim Ω X/Y ⊗ k(x) = n + m.
On the other hand, let z = g(f (x)), we have Ω X/Z ⊗ k(x) = Ω Xz/k(z) ⊗ k(x), let X 0 be an irreducible component of X z containing x, with its reduced induced structure. Then by (II 8.12),
Ω Xz/k(z) ⊗ k(x) −→ Ω X
0/k(z) ⊗ k(x) is surjective.
But again by (III 9.6), X 0 is an integral scheme of finite type over k(z) of dimension n + m, thus by (II 8.6A), we see Ω X0/k(z) is locally generated by at least n + m elements. Combining above, (3) is verified.
(d) Use (b), (c).
Theorem. 2 Let f : X → Y be a morphism of schemes of finite type over a field k, then f is smooth of relative dimension n if and only if:
(i) f is flat.
(ii) The fibres of f are geometrically regular of equidimension n, i.e., for point y ∈ Y ,
let X y = X y ⊗ k(y) k(y), then X y is equidimensional of dimension n and regular.
Proof. (⇒) By Proposition (0.2),(1(b)).
(⇐) (1) By assumption.
(2) By (III 9.6).
(3) By Proposition (0.2), regularity implies Ω X
y
/k(y) is locally free of rank n.
By the fact states below, it is equivalent to Ω Xy/k(y) is locally free of rank n, thereby (3) holds.
Fact. If M is an A-module with A being a local ring, B is a faithfully flat A-algebra, then
M is free ⇐⇒ M ⊗ A B is free.
Recall. The Zaraski tangent space T x for a point in a scheme X be the dual of k-vector space m x /m 2 x . If f : X → Y is a morphism, y = f (x), then there is a natural induced mapping
T f : T x → T y ⊗ k(y) k(x)
Lemma. 3-1 Suppose that R → S is a local homomorphism of Noetherian local rings.
Denote m the maximal ideal of R. Let M be a flat R-module and N a finite S-module.
Let u : N → M be a map of R-modules. If u 1 : N/mN → M/mM is injective then u is injective. In this case M/u(N ) is flat over R.
Proof. Claim: u k : N/m k N → M/m k M is injective ∀ k ∈ N.
The case k = 1 is just the assumption. To prove by induction, assuming the case k = n holds. Consider the diagram:
M/mM ⊗ R/m m n /m n+1
||
0 → M ⊗ R m n /m n+1 → M/m n+1 M → M/m n M → 0
↑ ↑ ↑
N ⊗ R m n /m n+1 → N/m n+1 N → N/m n N → 0
||
N/mN ⊗ R/m m n /m n+1
The first and the last map are injective, which imply the middle one is also injective.
Therefore the claim is proved by induction.
By Krull’s intersection theorem , T m n N = 0, thus the injectivity of u n ∀ n ∈ N implies u is injective.
To show that M/u(N ) is flat over R, it suffices to show that I ⊗ R M/u(N ) → M/u(N ) is injective for every ideal I in R. Consider the diagram
0 0 0
↑ ↑ ↑
N/IN → M/IM → M/(IM + u(N )) → 0
↑ ↑ ↑
0 → N → M → M/u(N ) → 0
↑ ↑ ↑
N ⊗ R I → M ⊗ R I → M/u(N ) ⊗ R I → 0
M ⊗ R I → M is injective, then by the snake lemma, it suffices to prove that N/IN injects into M/IM . Note that R/I → S/IS is a local homomorphism of Noetherian local rings, N/IN → M/IM is a map of R/I-modules, N/IN is finite over S/IS, and M/IM is flat over R/I and u : N/IN → M/IM is injective modulo m. Thus we may apply the first part of the proof to u to conclude.
Definition. Let A be a commutative ring, I be an ideal of A. An A-module M is ideally Hausdorff w.r.t. J if ∀ a is a finitely generated ideal of A, a ⊗ A M is Hausdorff with J -adic topology.
Lemma. 3-2 Let A be a commutative ring, I, J be ideals of A, M be an A-module, gr(A) be the graded ring with I-adic filtration, gr(M ) be the graded gr(A)-module associated with M with I-adic topology. Consider the following:
(i) M is a flat A-module.
(ii) Tor A 1 (N, M ) = 0 ∀N is an A-module annihilated by J .
(iii) M/IM is a flat A/I-module and the canonical map I ⊗ A M → IM is bijective.
(iv) M/IM is a flat A/I-module and canonical homomorphism r : gr(A) ⊗ gr0(A) gr 0 (M ) is bijective.
(v) ∀ n ∈ N, M/I n is a flat A/I n -module.
Then (i)⇒(ii)⇔(iii)⇒(iv)⇔(v). Furthermore, if A is noetherian and M is ideally Haus- dorff, all of them are equivalent.
Proof.
(i)⇒(ii)⇔(iii) Omitted.
(ii)⇔(ii)’ (ii)’:Tor A 1 (N, M ) = 0 ∀ N is an A-module annihilated by some power of J . (⇐) is trivial, so let’s focus on (⇒).
In particular Tor A 1 (I n N/I n+1 N, M ) = 0 ∀ n ∈ N. Then the exact sequence 0 → I n+1 N → I n N → I n N/I n+1 N → 0
which induces
Tor A 1 (I n+1 N, M ) → Tor A 1 (I n N, M ) → Tor A 1 (I n N/I n+1 N, M )
Notice that ∃ m ∈ N s.t. I m N = 0, using descending induction, we have Tor A 1 (I n N, M ) = 0 ∀ n ≤ m, especially m = 0.
(ii)⇒(iv) First we have a lemma:
(a) Tor A 1 (A/I n , M ) = 0 ∀ n ∈ N.
(b) The canonical homomorphism θ n : I n ⊗ A M → I n M is bijective.
(c) r : gr(A) ⊗ gr0(A) gr 0 (M ) is bijective.
Then (a)⇔(b)⇒(c), (b)⇐(c) when I is nilpotent.
Proof. (a)⇔(b) by the exact sequence
0 = Tor A 1 (A/I, M ) → Tor A 1 (A/I n , M ) → I n ⊗ M → M Consider the diagram:
I n+1 ⊗ A M → I n ⊗ A M → I n /I n+1 ⊗ A/I M/IM → 0
↓ θ n+1 ↓ θ n ↓ r n
0 → I n+1 M → I n M → gr n (M )
Where r n is the canonical surjective map. It’s commutative by the definition of r n . In this case, θ n is bijective, thus r n is bijective. For the second statement, notice that I n ⊗ A M = I n M = 0 for some n, thus using descending induction we see it’s true.
Thus by lemma, (ii)⇒(ii)’⇒ r is bijective, also (ii)⇒(iii), a fortiori M/IM is a flat A/I-module, thereby (iv) stands.
Also, (iii) ⇐(iv) when I is nilpotent is also shown by the above process.
(iv)⇔(v) Observe gr m (M/I n M ) =
( gr m (M ), m < n
0, m ≥ n For k ∈ N, let (iv) k and (v) k be the statement replacing A, I, M by A/I k A, I/I k , M/I k M . Obviously
(iv) ⇒ (iv) k ∀ k ∈ N (v) ⇒ (v) k ∀ k ∈ N
Thus it’s sufficient to show that (iv) k ⇒ (v) k ∀ k ∈ N or also (iv)⇔(v) for the case I is nilpotent, which will be assumed below. Since M/I n M ∼ = M ⊗ A A/I n , we know (v) ⇒ (i), more apparently, (i) ⇒ (v), so the statement is proved.
(v)⇒(i) (For the case A is noetherian and M is ideally Hausdorff) It’s proved in Lemma 3-1.
With above we now conclude:
Lemma. 3 Let A → B be a local homomorphism of local noetherian rings. Let M be a finitely generated B-module, and let t ∈ A be a nonunit that is not a zero divisor.
Then M is flat over A if and only if t is not a zero divisor in M and M/tM is flat over A/tA.
Proof. Take I = J = (t) in Lemma 3-1, 3-2.
Proposition. 4 Let f : X → Y be a morphism of nonsingular varieties over an algebraically closed field k. Let n = dim X − dim Y . Then the following are equivalent:
(i) f is smooth of relative dimension n.
(ii) Ω X/Y is locally free of rank n on X.
(iii) For every closed point x ∈ X, T f : T x → T y ⊗ k(y) k(x) is surjective.
Proof.
(i)⇒(ii) By Proposition (0.2).
(ii)⇒(iii) Use the exact sequence in (II 8.11) tensoring k(x) ∼ = k(since x is a closed point), we have:
f ∗ Ω Y /k ⊗ k(x) − → Ω φ X/k ⊗ k(x) −→ Ω X/Y ⊗ k(x) −→ 0
since the dimension of each terms are dim Y, dim X, n, φ is actually an injection.
Recall (II 8.7): Let (B, m) be a local ring with a field k ∼ = B/m containing in B. Then
m → m/m 2 d − → Ω B/k ⊗ B k b 7−→ b 7−→ db ⊗ 1 is an isomorphism.
So φ is actually m y /m 2 y ,→ m x /m 2 x , take dual to get our result.
(iii)⇒(i) Recall (III 9.1A(d)): M is an A-module, M is flat over A if and only if M p is flat over A p ∀ p ∈ Spec A. So to prove f is flat, we only have to verify O X,x is flat over O Y,y ∀ x ∈ X is a closed point, y = f (x).
Since X, Y both nonsingular, O x , O y are regular local rings. Also by taking dual on T f , m y /m 2 y ,→ m x /m 2 x is injective.
Take a regular system of parameters of O y : (t) = (t 1 , . . . , t r ), then (t) forms a part of regular system of parameters of O x . Since O x /(t) is flat over O y /(t) ∼ = k, by Lemma 3 and induction, O x /(t 1 , . . . , t i ) is flat over O y /(t 1 , . . . , t i ) ∀ i = 0 r, in particular it’s already proved f is flat. Proceed similar to (ii)⇒(iii) but backward, we see for x is a closed point,
dim k(x) (Ω X/Y ⊗ k(x)) = n
On the other hand, f is flat, by (III Ex 9.1), is also dominant, by (II 8.6A), the generic point ξ has the property:
dim k(x) (Ω X/Y ⊗ k(x)) ≥ n
In either case, we can conclude Ω X/Y is a coherent sheaf of rank ≥ n, so by (II 8.9), it’s locally free of rank n, thus f is smooth of relative dimension n.
Lemma. 5 Let f : X → Y be a dominant morphism of integral schemes of finite type over an algebraically closed field k of characteristic 0. Then there is a nonempty open set U ⊆ X such that f : U → Y is smooth.
Proof. By (II 8.16), every variety over k has an open dense nonsingular subset, thus we may assume X, Y are nonsingular. Since char k = 0, thus perfect, by (I,4.8A), K(X) is a separably generated field extension of K(Y ). Therefore by (II 8.6A), Ω X/Y is free of rank n at the generic point, thus locally free of rank n on some nonempty open set U . By Proposition 4, f : U → Y is smooth.
Remark. The lemma may fail when characteristic of field is not zero. Let char k = p > 0, k = k, f : P k 1 → P k 1 be the Frobenius morphism , then f is not smooth on any open set.
Proposition. 6 Let f : X → Y be a morphism of schemes of finite type over an algebraically closed field k of characteristic 0. For any r, let
X r = {closed points x ∈ X | rank T f,x ≤ r}
Then dim f (X r ) ≤ r.
Proof. Let Y 0 be any irreducible component of f (X r ) , and let X 0 be an irreducible component of X r which dominates Y 0 . Give X 0 and Y 0 their reduced induced structures, and consider the induced dominant morphism f 0 : X 0 → Y 0 . By (Lemma 5), there is a nonempty open subset U 0 ⊆ X 0 such that f : U 0 → Y 0 is smooth. Let x ∈ U 0 ∩ X r , then consider the diagram
T x,U0 → T x,X
↓ T f0,x ↓ T f,x
T y,Y0 → T y,Y
→ T y,Y
The horizontal arrows are injective, because U 0 and Y 0 are locally closed subschemes of X and Y , respectively. Also T f0,x is surjective by Proposition 4. Since dim T f,x ≤ r, dim T y,Y
0 ≤ r, we get dim Y 0 ≤ r.
Proposition. 7 Let f : X → Y be a morphism of varieties over an algebraically closed field k of characteristic 0, suppose X is nonsingular. Then there exists a nonempty open subset V ⊆ Y such that f : f −1 V → V is smooth.
Proof. Again may assume Y is nonsingular by (II 8.16). Let dim Y = r, X r−1 ⊆ X as defined in Proposition 6, then dim f (X r−1 ) ≤ r − 1, moving it from X, therefore we can assume for every closed point in X, rank T f ≥ r = dim Y , which means they’re all surjective, thus f is smooth by Proposition 4.
Proposition. 8 A morphism f : X → Y of schemes of finite type over k is ´ etale if it is smooth of relative dimension 0. It is unramified if for every x ∈ X, letting y = f (x), we have m y O X = m x , and k(x) is a separable algebraic extension of k(y). Show that the following conditions are equivalent:
(i) f is ´ etale.
(ii) f is flat, Ω X/Y = 0.
(iii) f is flat and unramified.
Proof.
(i)⇔(ii) By the definition of smooth.
(iii)⇒(ii) We have
m y /m 2 y ⊗ k(y) k(x) = (m y ⊗ Oyk(y)) ⊗ k(y) k(x) = m y ⊗ Oy(O x /m y O x ) Also, m y ⊗ A O x ∼ = m y O x = m x , thus
(O x /m y O x ) Also, m y ⊗ A O x ∼ = m y O x = m x , thus
m y ⊗ (O x /m y O x ) = (m y ⊗ O x )/m 2 y = m x /m 2 x
Thus the map T f , x is an isomorphism, i.e. it’s smooth of relative dimension 0.
(ii)⇒(iii) Fact: Let O x = B, O y = A, if ˆ A → ˆ B is an isomorphism then f is unramified at x.
Proof. We know m n y O x = m n x ∀ n ∈ N, and the composition A/m n y → B/m n y B → ˆ A/m n y A ˆ
is an isomorphism, it’s left to show ˆ A → ˆ B is injective, in other words, m n y A ∩ B = m ˆ n y B ∀ n ∈ N
Notice that B = A + m n y B and m n y B ⊆ m n y A, so ∀ b ∈ B, it can be represented ˆ as a + , a ∈ A, b ∈ m n y B m n y A ∩ B ⊆ m ˆ n y B.
Conversely, if b ∈ m n y A, a ∈ m ˆ y A ∩ A = m ˆ n y , thus b ∈ m n y B. Thus we only need to prove ˆ A → ˆ B is an isomorphism, but O x = O y +m n x , m n x = m n+1 x +m n y ∀ n ∈ N, thus ˆ A → ˆ B is an isomorphism. Also, by (II 8.6A), dim k(x) Ω X/Y ⊗ k(x) ≥tr.deg k(x)/k(y), equality holds if and only if k(x) is separately extended over k(y).
Thus in this case, k(x) is separately extended over k(y) of transcendental degree
0, i.e. a separable algebraic extension.
Recall. A group variety G is a variety G over an algebraically closed field k, together with morphisms µ : G × G → G, ρ : G → G s.t. G(k), k-rational points of G, becomes a group under the operation induced by µ, with ρ giving the inverses.
We say that a group variety G acts on a variety X if we have a morphism θ : G×X → X which induces a homomorphism G(k) → AutX of groups.
A homogeneous space is a variety X, together with a group variety G acting on it, such that the group G(k) acts transitively on X(k) .
Remark. A homogeneous space is necessarily a nonsingular variety.
Theorem. 9 Let X be a homogeneous space with group variety G over an algebraically closed field k of characteristic 0. Let f : Y → X, g : Z → X are two morphisms between nonsingular varieties. For σ ∈ G(k), let Y σ be the same variety with Y but with morphism σ ◦ f : Y σ → X. Then ∃ V ⊆ G s.t. ∀ σ ∈ V (k), Y σ × X Z is nonsingular and either empty or has the dimension dim Y + dim Z − dim X.
Proof. Define h : G × Y → X as the composition of f and θ : G × X → X. We first prove that it’s smooth. Now nonsingular is equivalent to smooth over k by Proposition 0.2, and G is nonsingular by the Remark above, therefore G × Y is nonsingular by Proposition 1(d).
Apply Proposition 7, we see that ∃ U ⊆ X s.t. h −1 (U ) → U is smooth.
Now G acts on G×Y by left multiplication on G; G acts on X by θ, and these two actions are compatible with the morphism h, by construction. Thus ∀ σ ∈ G(k), h −1 (U σ ) → U σ is smooth. Since U σ cover X, we conclude h is smooth.
Next consider the diagram:
W := (G × Y ) × X Z h
0
