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 M, N 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.

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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 $K=0$ . ♦

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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 $\mathfrak c M = \mathrm{ker } f_M$ .

warning The 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

Commutative Algebra 31

Flat Modules

Flatness is Local

Local Property for A-Modules

Local Property for A-Algebras

Ideals and Submodules

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Commutative Algebra 31

Flat Modules

Flatness is Local

Local Property for A-Modules

Local Property for A-Algebras

Ideals and Submodules

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