(page 1) Chapter 6

flat

Flatness

A long time ago, I promised that we would try to understand the notion of flatness. The only useful definition is algebraic; however, I want to try to convince you that flatness is a fundamentally geometric concept. Now, some concepts pass very easily between the realms of commutative algebra and algebraic geometry. An integral domain corresponds to an irreducible, reduced scheme. A projective module corresponds to a locally free sheaf. A regular ring corresponds to a nonsingular scheme. In the same way, a flat morphism corresponds to a continuous family of schemes.

Algebraic Preliminaries

Definition 6.1. Let R be any commutative ring with 1, and let M be an R-module. We say that M is a flat R-module if M ⊗_{R} − is an exact functor on the category of R-modules.

Example 6.2. Consider the projection from the coordinate axes in the plane to the x-axis. Algebraically, this corresponds to the homomorphism k[x] → R = k[x, y]/(xy). We claim that R is not a flat module over k[x]. To see this, tensor the injection

0 → k[x]⟶^{x}k[x]

with R. The resulting map

R⟶^{x}R

is not injective, since it takes the nonzero element y to xy = 0. Geometrically, this failure of flatness can be seen by looking at the dimensions of the fibres of the map. For nonzero values of x, the fibre is a point. When x goes to 0, however, the fibre becomes an entire line.

Example 6.3. Let’s modify the previous example so that at least the dimension of fibres stays constant. Consider the map of rings k[x] → R = k[x, y]/(xy, y(y − 1)). So, Spec(R) has two components, a line and a point. When x is nonzero, the fibre consists of one point. When x = 0, the fibre consists of two points. The same

◯^{c}Coombes 1996  1

(page 2)

2 Chapter 6: Flatness

free

localization

algebraic analysis shows that R is not flat over k[x], because x becomes a zero divisor in R.

Example 6.4. We now consider the projection of a parabola onto a line, given algebraically by the homomorphism k[x] → R = k[x, y]/(x − y^{2}). In this case, R is a free k[x]-module, and hence flat.

Example 6.5. The previous example can be reinterpreted as the inclusion of the elements of even degree in the polynomial ring k[y]. The analogue of this example in one dimension higher is the homomorphism k[x^{2}, xy, y^{2}] → k[x, y]. Geometrically, this corresponds to a map from the plane onto a cone. This example is not flat, as you can see by examining a neighborhood of the singular point on the cone.

Proposition 6.6. For arbitrary modules over arbitrary rings,

free ⇒ projective ⇒ flat.

Proof: Since a module is projective if and only if it is a direct summand of a free module, it is easy to see that free implies projective. Now suppose that M is projective. Find another module N so that M ⊕ N = F is free. Given an exact sequence of modules

0 → X → Y → Z → 0,

we can tensor it with both M and N separately. Since tensor product commutes with direct sums, the sum of the tensored sequences is the same thing you would get by tensoring with F. Since free clearly implies flat, we know that the map (M ⊗X)⊕(N ⊗X) = F ⊗X → F ⊗Y is injective. Hence, the restriction to M ⊗X must also be injective, as desired.

Remark 6.7. If R is any ring and S is any multiplicative set in R, then the localization S^{−1}R is a flat R-module. This follows directly from the definition of localization. It can be used to give examples of flat modules that are not projective (such as the rational numbers as a Z-module).

Proposition 6.8. Flatness is preserved under base extension.

Proof: More precisely, this means that if M is a flat R-module and if S is any Ralgebra, then M ⊗_{R} S is a flat S-module. The result now follows from the equality of the functors (M ⊗_{R} S) ⊗_{S} − and M ⊗_{R} −.

Proposition 6.9. For arbitrary modules over arbitrary rings, flatness is a local property (which can be checked by localizing at all maximal ideals).

Proof: Since localization of modules is the same as base extension to the local ring, we see immediately that flat implies locally flat. Now suppose M is a locally flat R-module and let

0 → X → Y → Z → 0

◯^{c}Coombes 1996 annihilator residue field Nakayama’s Lemma

be any exact sequence of R-modules. Tensoring with M, we get a new exact sequence

0 → K → M ⊗ X → M ⊗ Y → M ⊗ Z → 0.

This sequence defines the kernel K. The problem is to show that K = 0.

Since localization is an exact functor, we can transport the four-term exact sequence to any local ring of R. Since M is locally flat, we know that all the localizations K_{p} are trivial. In particular, if x ∈ K is any element and if p is any prime ideal of R, then there exists an element s ∈ R − p such that sx = 0. Now let I be the annihilator ideal in the ring R of the element x. We have just seen that no (maximal) prime ideal of R can contain I. Therefore, I = R. Since the element 1 lies in the annihilator of x, we must have x = 0, and therefore K = 0. The result follows.

Proposition 6.10. Let R be a noetherian local ring, and let M be a finitely generated flat R-module. Then M is free.

Proof: Let I be the maximal ideal of R and let k = R/I be the residue field. Write M̄ = M ⊗ k = M/IM for the reduction of M to a module over the residue field. This is a finitely generated vector space; choose a basis and lift it arbitrarily to x_{1}, x_{2}, …, x_{n} ∈ M. By Nakayama’s Lemma, the x_{i} generate M. Now let F be free on n-generators, mapping to M in the obvious way, and let K = ker(F → M). Tensoring with the exact sequence

0 → I → R → k → 0,

we get an exact commutative diagram:

0  0
↓  ↓
K ⊗ I ⟶ F ⊗ I ⟶ M ⊗ I ⟶ 0
↓  ↓  ↓
0 ⟶ K ⟶ F  ⟶ M ⟶ 0
↓  ↓  ↓
K̄ ⟶ F̄ ⟶ M̄ ⟶ 0
↓  ↓  ↓
0  0  0

By chasing the diagram, we see that the left map on the bottom row is injective. Because we chose the generators by lifting a basis, we know that the induced map

◯^{c}Coombes 1996 Nakayama’s Lemma flat free projective locally free Snake Lemma

F̄ → M̄ is an isomorphism. So, we now know that K̄ = 0. Because R is noetherian and F is finitely generated, K is also finitely generated. So, we can apply Nakayama’s Lemma again to conclude that K = 0, and thus F ≈ M.

Remark 6.11. Putting together the results we have obtained so far, we see that for finitely generated modules over local rings, the concepts of flat, free, and projective are equivalent. We certainly know examples of finitely generated projective modules over noetherian rings that are not free. The earlier examples (using localization) give infinitely generated flat modules over noetherian rings that are not projective. That’s the best we can do, however, because of the following result.

Proposition 6.12. Let M be a finitely generated module over a noetherian ring. Then M is flat ⟺ M is projective ⟺ M is locally free.

Proof: We already know, under these hypotheses, that flat ⟺ locally flat ⟺ locally free, so we only need to check their equivalence with projective. First, choose a finite set of generators for M, so that we have an exact sequence

0 → K → F → M → 0,

where F is free. The goal is to try to split this exact sequence. If we use the noetherian hypothesis, we can conclude that K is finitely generated. Now the problem reduces to the following lemma.

Lemma 6.13. Let 0 → K⟶^{ι}F → M → 0 be an exact sequence of modules over an arbitrary ring, with F free and M flat. Let x_{1}, …, x_{n} be any finite set of elements of K. Then there exists a partial splitting h : F → K such that hι(x_{j}) = x_{j} for all j.

Proof: By induction on the number of specified elements in K. When only one element x ∈ K is given, we begin by choosing a basis {e_{i}} for F. Write x = ∑_{1}^{m} a_{i}e_{i} as a finite linear combination of basis elements, with a_{i} ∈ R. Let I be the ideal of R generated by the elements {a_{i}}. We get an exact commutative diagram. (See diagram (6.14) on the next page.)

The zeros at the top of diagram (6.14) use flatness. Using the Snake Lemma, we see that K/IK injects into F/IF. Since x ∈ K maps to zero in F/IF by definition, the injectivity tells us that it also maps to zero in K/IK. So, there exist elements y_{i} ∈ K such that x = ∑ a_{i}y_{i}. We define the desired partial splitting by sending the basis elements of F to K via e_{i} ↦ y_{i}.

Now suppose n > 1. Begin by choosing a map f : F → K such that f(x_{n}) = x_{n}. Then define new elements of K by setting y_{i} = x_{i} − f(x_{i}). By induction, we can find another homomorphism that satisfies g(y_{i}) = y_{i} for y = 1, …, n − 1. Now the desired map is defined by

h(x) = f(x) + g(x − f(x)).

◯^{c}Coombes 1996

One verifies this by computing flat morphism

h(x_{n}) = f(x_{n}) + g(x_{n} − f(x_{n})) = x_{n} + g(0) = x_{n}

and

h(x_{i}) = f(x_{i}) + g(x_{i} − f(x_{i})) = f(x_{i}) + g(y_{i}) = f(x_{i}) + x_{i} − f(x_{i}) = x_{i}.

Here is the commutative diagram that was used in the first part of the proof.

\begin{aligned}& & & & 0 & & 0 \\ & & & & ↓ & & ↓ \\ & & K ⊗ I & ⟶ & F ⊗ I & ⟶ & M ⊗ I & ⟶ & 0 \\ & & ↓ & & ↓ & & ↓ \\ 0 & ⟶ & K & ⟶ & F & ⟶ & M & ⟶ & 0 \\ & & ↓ & & ↓ & & ↓ \\ & & K/IK & ⟶ & F/IF & ⟶ & M/IM & ⟶ & 0 \\ & & ↓ & & ↓ & & ↓ \\ & & 0 & & 0 & & 0 \end{aligned}  (6.14)

Geometric Applications

Let f : X → Y be a morphism of schemes, and let F be a sheaf of O_{X}-modules. We say that F is flat over Y at a point x ∈ X if F_{x} is a flat module over O_{Y,y} where y = f(x). We say that F is flat over Y if it is flat at every point x ∈ X. If O_{X} is flat over Y then we either say that X is flat over Y, or we say that f is a flat morphism.

Remark 6.15. Since flatness is a local property in the case of modules over rings, we see that an A-module M is a flat module if and only if the quasicoherent sheaf F = M̃ is flat over X = Spec(A). Moreover, if f : X → Y is any morphism and if F is any sheaf of O_{X} modules, then F is flat over Y if and only if for all affine open subsets U ⊂ X and V ⊂ Y with f(U) ⊂ V, we have Γ(U, F) a flat Γ(V, O_{Y})-module.

Proposition 6.16. (a) Open immersions are flat.

(b) Flatness is stable under base extension.

(c) The composite of flat morphisms is flat.

(d) A coherent module on a noetherian scheme is flat if and only if it is locally free.

◯^{c}Coombes 1996

Proof: (a) is immediate from the local characterization. (b) follows immediately _{Čech cohomology} from 6.8, and (c) follows from a similar affine argument. (d) is the sheaf theoretic version of 6.10.

Proposition 6.17. Let f : X → Y be a separated morphism of finite type between noetherian schemes, and let F be a quasicoherent sheaf on X. Let u : Y′ → Y be any morphism from a noetherian scheme. Form the fibre product diagram

\begin{matrix}X′ & ⟶^{v} & X \\ ↓^{g} & & ↓^{f} \\ Y′ & ⟶^{u} & Y. \end{matrix}

For all i ≥ 0, there exist natural maps of sheaves

u^{∗}R^{i}f_{∗}(F) → R^{i}g_{∗}(v^{∗}(F)).

If u is flat, then these morphisms are all isomorphisms.

Proof: The result is local on both Y and Y′, so we can assume that Y = Spec(A) and Y′ = Spec(A′) are spectra of noetherian rings. Now both sides of the claimed map are quasicoherent sheaves. Writing F′ = v^{∗}(F), the result reduces to proving the existence of maps

H^{i}(X, F) ⊗_{A} A′ → H^{i}(X′, F′)

that are isomorphisms when A′ is flat over A.

The technical hypotheses are included so we can use Čech cohomology for our computations. Let U = Spec(B) ⊂ X be an open affine. Then its inverse image f^{−1}(U) = Spec(B ⊗_{A} A′) ⊂ X′ is also an open affine, and Γ(f^{−1}(U), F′) = Γ(U, F) ⊗_{A} A′. So, if U is a cover of X by open affines, then its inverse image U′ is a cover of X′ by open affines, and the Čech complexes satisfy

C(U′, F′) = C(U, F) ⊗_{A} A′.

It is easy to check that this relation on the Čech complexes induces maps on cohomology, and that these maps are isomorphisms when the ring extension is flat.

Corollary 6.18. Let f : X → Y and F be as above. Let Y be affine. For each y ∈ Y, let X_{y} denote the fibre, F_{y} the induced sheaf, and k(y) the constant sheaf on {y}^{−}. Then for all i ≥ 0, there are isomorphisms

H^{i}(X_{y}, F_{y}) ≈ H^{i}(X, F ⊗ k(y)).

Proof: We cannot immediately apply the proposition, since the inclusion of the point y in Y need not be flat. Instead, we let Y′ denote the reduced induced subscheme structure on the closed set {y}^{−}, and let X′ = X ×_{Y} Y′. Both sides of the claimed isomorphism only depend on Y′ and X′. So, we can assume that

◯^{c}Coombes 1996 generic point generic point Hilbert polynomial Čech cohomology Serre’s Finiteness Theorem

Y is an integral scheme and y is its generic point. In this case, the inclusion Spec(k(y)) → Y is flat, and the result follows from the proposition.

Theorem 6.19. Let T be an integral noetherian scheme, and let X ⊂ P_{T}^{n} be a closed subscheme. For all t ∈ T, let p_{t}(z) ∈ Q[z] denote the Hilbert polynomial of the fibre X_{t}, viewed as a closed subscheme of P_{k(t)}^{n}. Then the morphism f : X → T is flat if and only if the Hilbert polynomials p_{t} are independent of t.

Proof: Replace O_{X} by any coherent sheaf F on P_{T}^{n}, and work with the Hilbert polynomials of F_{t}. To show that the Hilbert polynomial is constant, we can compare an arbitrary point t with the generic point of T. Since flatness is also a local property, we can assume that T = Spec(A) is the spectrum of a local noetherian domain. Now the proof follows from the following lemma.

Lemma 6.20. Let T = Spec(A), where A is a local noetherian domain. Let X = P_{T}^{n}, and let F be a coherent sheaf on X. Then the following are equivalent:

(i) F is flat over T.

(ii) H^{0}(X, F(m)) is a free A-module of finite rank for m >> 0.

(iii) The Hilbert polynomials p_{t} of F_{t} are independent of t.

Proof: (i ⇒ ii) We compute the cohomology of F(m) using Čech cohomology with respect to the standard affine open cover of X. By flatness, each C^{i}(U, F) is a flat finitely generated A-module. By Serre’s Finiteness Theorem, the higher cohomology vanishes for m >> 0. In this range, then, we have a resolution

0 → H^{0}(X, F(m)) → C^{0}(U, F) → ⋯ C^{n}(U, F) → 0.

By downward induction, H^{0}(X, F(m)) is flat and finitely generated over the local noetherian ring A, and hence must be free.

(ii ⇒ i) Write S = A[x_{0}, …, x_{n}]. Choose m_{0} large enough that H^{0}(X, F(m)) is free for all m ≥ m_{0}. Define

M = ⨁_{m≥m_{0}} H^{0}(X, F(m)).

Then M̃ = Γ_{∗}(F)˜ = F. Since M is free, it is flat, and therefore F is flat. (ii ⇒ iii) It is enough to show that

P_{t}(m) = rank_{A}H^{0}(X, F(m))

for all m >> 0. In turn, it is enough to show that

H^{0}(X_{t}, F_{t}(m)) = H^{0}(X, F(m)) ⊗ k(t)

for all m >> 0,

So, write T′ = Spec(A_{P}) where t ∈ T corresponds to the prime ideal P ⊂ A. Make the flat base extension from T to T′. So, we can assume that t is the closed

◯^{c}Coombes 1996 Serre’s Finiteness Theorem Nakayama’s Lemma generic point dimension degree arithmetic genus Hilbert polynomial

point. Write X_{0} = X_{t}, F_{0} = F_{t}, and k = k(t). Since A is noetherian, we have a finite presentation

A^{q} → A → k → 0

of k as an A-module. Tensoring with F, we get an exact sequence of sheaves

F^{q} → F → F_{0} → 0.

By Serre’s Finiteness Theorem, we can take m >> 0 large enough that all the higher cohomology of F(m) and of F_{0}(m) vanishes. Then we have an exact commutative diagram

H^{0}(X, F(m)) ⊗_{A} A^{q} → H^{0}(X, F(m)) → H^{0}(X, F(m)) ⊗_{A} k → 0
↓  ↓  ↓
H^{0}(X, F(m)^{q})  → H^{0}(X, F(m)) → H^{0}(X_{0}, F_{0}(m)) → 0

Now the first two vertical arrows are clearly isomorphisms, and hence the final vertical arrow is also an isomorphism.

(iii ⇒ ii) By Nakayama’s Lemma, you can check that a finitely generated module over a noetherian local domain is free by comparing its rank at the generic point to its rank at the closed point. Thus, the previous argument is reversible.

Corollary 6.21. If X ⊂ P_{T}^{n} is a flat family of subschemes of projective space parametrized by a connected, noetherian scheme T, then the dimension, degree, and arithmetic genus of the fibres X_{t} are independent of t.

Proof: Make a base extension to an irreducible, reduced component of T. Then the result follows from the theorem and the fact that each of these invariants can be computed in terms of the Hilbert polynomial.

◯^{c}Coombes 1996