# 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.

# 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 $K=0$. ♦

# 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.

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

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 $\mathfrak a M \oplus \mathfrak b M \to (\mathfrak a + \mathfrak b) M$?

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