Commutative Algebra 31

Flat Modules

Recall from proposition 3 here: for an A-module M, M \otimes_A - is a right-exact functor.


We say M is flat over A (or A-flat) if M\otimes_A - is an exact functor, equivalently, if

N_1\stackrel f \longrightarrow N_2 \text{ injective } \implies M\otimes_A N_1\stackrel {1_M \otimes f} \longrightarrow M\otimes_A N_2 \text{ injective. }

A flat A-algebra is an A-algebra which is flat as an A-module.

We say a ring homomorphism f:A\to B is flat if the resulting A-algebra structure on B is flat.

Flat modules and algebras are a little hard to get a handle on, so we will first go through some properties before offering a few examples.

Proposition 1.

If (M_i) is a collection of A-flat modules, then \oplus_i M_i is A-flat.

Conversely, if M\oplus N is flat, so is M.


This follows from the natural isomorphism in N:

\left(\oplus_i M_i \right) \otimes_A N \cong \oplus_i (M_i \otimes_A N).

Proposition 2.

The localized ring S^{-1}A is A-flat. In particular, A is A-flat.


We have the natural isomorphism in M given by S^{-1}A \otimes_A M \cong S^{-1}M. Then use the fact that localization functors are exact (theorem 1 here). ♦

Corollary 1.

A free module is flat. More generally, a projective module is flat.


Since A is flat, by proposition 1, any free module A^{\oplus I} is also flat.

If P is projective, P \oplus Q is free for some A-module Q. Thus P \oplus Q is flat. By proposition 1 again, P is flat. ♦

Proposition 3.

If B is a flat A-algebra and M is a flat B-module, then M is a flat A-module.

In particular, if C is a flat B-algebra and B is a flat A-algebra, then C is a flat A-algebra.


Indeed, we have a natural isomorphism for any A-module N:

M\otimes_A N \cong M\otimes_B (B \otimes_A N).

Exercise A

Prove the following.

  • If MN are A-flat, so is M\otimes_A N.
  • If M is a flat A-module, so is S^{-1}M.
  • If M is A-flat, then M^B = B\otimes_A M is B-flat for any B-algebra.

As a special case of the last property, we have:

M \text{ is } A\text{-flat} \implies \begin{cases}S^{-1}M \text{ is } S^{-1}A\text{-flat} \\ M/\mathfrak a M \text{ is } A/\mathfrak a\text{-flat}.\end{cases}

for any multiplicative subset S\subseteq A and ideal \mathfrak a \subseteq A.


The above properties can be summarized as follows.

  • Flatness is preserved by direct sum and decomposition.
  • Localization is flat.
  • Projective modules are flat.
  • Flatness is transitive.


The property of flatness may seem mysterious and obscure for the first-time reader. Roughly speaking, flat algebras can be interpreted geometrically as follows: if A\to B is a flat ring homomorphism, the fibres B\otimes_A k(\mathfrak p) maintain some consistency over various \mathfrak p. We will see some concrete examples in the next article.


Flatness is Local

Before we state and prove the local properties, first recall that tensor product commutes with localization:

(S^{-1}M) \otimes_{S^{-1}A} (S^{-1}N) \cong S^{-1}(M\otimes_A N).

Also we need the following.

Lemma 1.

If N is an S^{-1}A-module and we treat it as an A-module, then S^{-1}N \cong N as S^{-1}A-modules.


Easy exercise. ♦

Local Property for A-Modules

Proposition 4.

Flatness is a local property, i.e. M is A-flat if and only if for each maximal ideal \mathfrak m\subset A, M_{\mathfrak m} is A_{\mathfrak m}-flat.


(⇒) Let M be A-flat; to show M_{\mathfrak m} is A_{\mathfrak m}-flat let f : N_1 \to N_2 be an injective map of A_{\mathfrak m}-modules. Now

M_{\mathfrak m} \otimes_{A_{\mathfrak m}} N_i \cong M_{\mathfrak m} \otimes_{A_{\mathfrak m}} (N_i)_{\mathfrak m} \cong (M \otimes_A N_i)_{\mathfrak m}.

Since M is A-flat, 1 \otimes f : M\otimes_A N_1 \to M\otimes_A N_2 is injective, and hence so is (M\otimes_A N_1)_{\mathfrak m}\to (M\otimes_A N_2)_{\mathfrak m}.

For (⇐) let f : N \to N' be an injective homomorphism of A-modules. From corollary 2 here, 1_M \otimes f : M\otimes_A N \to M\otimes_A N' is injective if and only if (1_M \otimes f)_{\mathfrak m} is injective for each \mathfrak m. But this map is just:

M_{\mathfrak m} \otimes_{A_{\mathfrak m}}N_{\mathfrak m} \cong (M\otimes_A N)_{\mathfrak m} \longrightarrow (M\otimes_A N')_{\mathfrak m} \cong M_{\mathfrak m} \otimes_{A_{\mathfrak m}}N'_{\mathfrak m},

obtained by 1_{M_{\mathfrak m}} \otimes f_{\mathfrak m}. Since M_{\mathfrak m} is A_{\mathfrak m}-flat and f_{\mathfrak m} is injective, we are done. ♦

Local Property for A-Algebras

Next, we have an even better result for A-algebras.

Proposition 5.

Let B be an A-algebra. B is A-flat if and only if for any maximal ideal \mathfrak n \subset B and \mathfrak p = \mathfrak n \cap A, the localization B_{\mathfrak n} is A_{\mathfrak p}-flat.


B_{\mathfrak n} is indeed an algebra over A_{\mathfrak p} since any s\in A - \mathfrak p is invertible in B_{\mathfrak n}.


(⇒) S^{-1}B is flat over S^{-1}A. Now B_{\mathfrak q} is a further localization of S^{-1}B so it is flat over S^{-1}B. Apply transitivity of flatness.

(⇐) Now suppose B_{\mathfrak n} is flat over A_{\mathfrak p} for all maximal \mathfrak n and \mathfrak p =\mathfrak n \cap A. Let f : N_1\to N_2 be an injective map of A-modules and let K be the kernel of 1_B \otimes f : B\otimes_A N_1 \to B \otimes_A N_2 so we have an exact sequence of B-modules

0 \longrightarrow K \longrightarrow B\otimes_A N_1 \stackrel {1\otimes f} \longrightarrow B\otimes_A N_2.

For any maximal ideal \mathfrak n \subset B we obtain an exact sequence

0 \longrightarrow K_{\mathfrak n} \longrightarrow (B\otimes_A N_1)_{\mathfrak n} \stackrel {(1\otimes f)_{\mathfrak n}} \longrightarrow (B\otimes_A N_2)_{\mathfrak n}.

But (B\otimes_A N_i)_{\mathfrak n} \cong B_{\mathfrak n} \otimes_{A} N_i (exercise). And since B_{\mathfrak n} is flat over A_{\mathfrak p} and the latter is flat over A, by proposition 3 B_{\mathfrak n} is flat over A. Thus K_{\mathfrak n} = 0 for all maximal \mathfrak n\subset B and K=0. ♦


Ideals and Submodules

Here are some results which are useful for determining when a module is not flat.

Lemma 2.

If M is A-flat, we get an isomorphism \mathfrak a \otimes M\longrightarrow \mathfrak a M by multiplication.


Take the exact sequence 0 \to \mathfrak a \to A \to A/\mathfrak a \to 0 of A-modules. Since tensoring is right-exact we get an exact sequence

\mathfrak a \otimes M \longrightarrow M \longrightarrow M/\mathfrak a M \longrightarrow 0,

where the first map is multiplication. If M is A-flat, the first map is injective; its image is clearly \mathfrak a M. ♦


In fact, the converse is true: if M is such that \mathfrak a \otimes M \cong \mathfrak a M for all finitely generated ideals \mathfrak a, then M is A-flat. But we will need more tools than what we have at the moment to prove this.

Now for any A-module M, we obtain a map as follows.


Such a map satisfies a few nice properties for all modules. Examples:

1. The zero ideal (resp. (1)) maps to the zero submodule (resp. M).

2. If \mathfrak a\subseteq \mathfrak b, then \mathfrak a M \subseteq \mathfrak b N.

3. We have \mathfrak a (\mathfrak b M) = (\mathfrak {ab}) M.

4. For collection of ideals (\mathfrak a_i), we have (\sum_i \mathfrak a_i)M = \sum_i (\mathfrak a_i M).

When M is A-flat, we get many more. Lemma 2 tells us multiplication gives an isomorphism \mathfrak a \otimes_A M \to \mathfrak a M. And since \mathfrak -\otimes_A M is exact, we can apply many of the nice properties of exact functors.

5. For any ideals \mathfrak a, \mathfrak b, we have (\mathfrak a \cap \mathfrak b)M = \mathfrak a M \cap \mathfrak b M.

6. If f : \mathfrak a \to \mathfrak b is an A-linear map of ideals, we obtain an induced A-linear f_M : \mathfrak a M \to \mathfrak b M, xm \mapsto f(x)m.

7. Let \mathfrak c = \mathrm{ker } f; then \mathfrak c M = \mathrm{ker } f_M.

warningThe above correspondence is neither injective nor surjective in general, even when M is A-flat. [Exercise: find a counter-example for each case.]

As a preview of (much) later chapters, when M is faithfully flat, the correspondence is injective.

Exercise B

Find an A-linear map f:\mathfrak a \to \mathfrak b of ideals of A and an A-module M, such that the A-linear map \mathfrak a M \to \mathfrak b M, xm\mapsto f(x)m is not well-defined. Thus property 6 is non-trivial.

What can we say about the kernel of addition \mathfrak a M \oplus \mathfrak b M \to (\mathfrak a + \mathfrak b) M?


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