Commutative Algebra 31

Flat Modules

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

Definition.

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.

Proof

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.

Proof

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.

Proof

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.

Proof

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.

Summary.

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.

Note

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.

blue-lin

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.

Proof

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.

Proof

(⇒) 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.

Note

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

Proof

(⇒) 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. ♦

blue-lin

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.

Proof

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. ♦

Note

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.

ideals_to_submodules

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?

blue-lin

This entry was posted in Advanced Algebra and tagged , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s