# nLab flat module

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Idea

A module over a ring $R$ is called flat if its satisfies one of many equivalent conditions, the simplest to state of which is maybe: forming the tensor product of modules with $N$ preserves submodules.

Under the dual geometric interpretation of modules as generalized vector bundles over the space on which $R$ is the ring of functions, flatness of a module is essentially the local triviality of these bundles, hence in particular the fact that the fibers of these bundles do not change, up to isomorphism. See prop. below for the precise statement. On the other hand there is no relation to “flat” as in flat connection on such a bundle.

## Definition

We first state the definition and its equivalent reformulations over a commutative ring abstractly in In terms of exact functors and Tor-functors. Then we give an explicit element-wise characterization in Explicitly in terms of identities.

Much of this discussion also works in the more general case where the ring is not-necessarily taken to be commutative and not necessarily required to be equipped with a unit, this we indicate in For more general rings.

### In terms of exact functors and Tor-functors

Let $R$ be a commutative ring.

###### Definition

An $R$-module $N$ is flat if tensoring with $N$ over $R$ as a functor from $R$Mod to itself

$(-)\otimes_R N : R Mod \to R Mod$

is an exact functor (sends short exact sequences to short exact sequences).

###### Definition

A module as above is faithfully flat if it is flat and tensoring in addition reflects exactness, hence if the tensored sequence is exact if and only if the original sequence was.

###### Remark

The condition in def. has the following immediate equivalent reformulations:

1. $N$ is flat precisely if $(-)\otimes_R N$ is a left exact functor,

because tensoring with any module is generally already a right exact functor;

2. $N$ is flat precisely if $(-)\otimes_R N$ sends monomorphisms (injections) to monomorphisms,

because for a right exact functor to also be left exact the only remaining condition is that it preserves the monomorphisms on the left of a short exact sequence;

3. $N$ is flat precisely if $(-)\otimes_R N$ is a flat functor,

because Mod is finitely complete;

4. $N$ is flat precisely if the degree-1 Tor-functor $Tor_1^{R Mod}(-,N)$ is zero,

because by the general properties of derived functors in homological algebra, $L_1 F$ is the obstruction to a right exact functor $F$ being left exact;

5. $N$ is flat precisely if all higher Tor functors $Tor_{\geq 1}(R Mod)(-,N)$ are zero,

because the higher derived functors of an exact functor vanish;

6. $N$ is flat precisely if $N$ is an acyclic object with respect to the tensor product functor;

because the Tor functor is symmetric in both arguments and an object is called tensor-acyclic object if all its positive-degree $Tor$-groups vanish.

The condition in def. also has a number of not so immediate equivalent reformulations. These we discuss in detail below in Equivalent characterizations. One of them gives an explicit characterization of flat modules in terms of relations beween their elements. An exposition of this we give now in In terms of identities.

### Explicitly in terms of identities

There is a characterisation of flatness that says that a left $A$-module $M$ is flat if and only if “everything (that happens in $M$) happens for a reason (in $A$)”. We indicate now what this means. Below in prop. it is shown how this is equivalent to def. above.

The meaning of this is akin to the existence of bases in vector spaces. In a vector space, say $V$, if we have an identity of the form $\sum_i \alpha_i v_i = 0$ then we cannot necessarily assume that the $\alpha_i$ are all zero. However, if we choose a basis then we can write each $v_i$ in terms of the basis elements, say $v_i = \sum_j \beta_{i j} u_j$, and substitute in to get $\sum_{i j} \alpha_i \beta_{i j} u_j = 0$. Now as $\{u_j\}$ forms a basis, we can deduce from this that for each $j$, $\sum_i \alpha_i \beta_{i j} = 0$. These last identities happen in the coefficient field, which is standing in place of $A$ in the analogy.

When translating this into the language of modules we cannot use bases so we have to be a little more relaxed. The following statement is the right one.

Suppose there is some identity in $M$ of the form $\sum_i a_i m_i = 0$ with $m_i \in M$ and $a_i \in A$. Then there is a family $\{n_j\}$ in $M$ such that every $m_i$ can be written in the form $m_i = \sum_j b_{i j} n_j$ and the coefficients $b_{i j}$ have the property that $\sum_i a_i b_{i j} = 0$.

The module $M$ being flat is equivalent to being able always to do this.

There is an alternative way to phrase this which is less element-centric. The elements $m_i$ correspond to a morphism into $M$ from a free module, say $m \colon F \to M$. The $a_i$ correspond to a morphism $a \colon F \to F$, multiplying the $i$th term by $a_i$. That we have the identity $\sum_i a_i m_i = 0$ says that the composition $m a$ is zero, or that $m \colon F \to M$ factors through the coequaliser of $a$ and $0$.

Now we consider the elements $n_j$. These define another morphism from a free module, say $n \colon E \to M$. That the $m_i$ can be expressed in terms of the $n_j$ says that the morphism $m$ factors through $n$. That is, there is a morphism $b \colon F \to M$ such that $m = n b$. We therefore have two factorisations of $m$: one through $n$ and one through the cokernel $\coker a$. The question is as to whether these have any relation to each other. In particular, does $\coker a \to M$ factor through $n$? We can represent all of this in the following diagram.

Saying that $M$ is flat says that this lift always occurs.

Taking this a step further, we consider the filtered family of all finite subsets of $M$. This generates a filtered family of finitely generated free modules with compatible morphisms to $M$. So there is a morphism from the colimit of this family to $M$. This morphism is surjective by construction. To show that it is injective, we need to show that any element in one of the terms in the family that dies by the time it reaches $M$ has actually died on the way. This is precisely what the above characterisation of flatness is saying: the element corresponding to $\sum_i a_i m_i$ that dies in $M$ is already dead by the time it reaches $E$.

We have thus arrived at the following result:

###### Theorem

A module is flat if and only if it is a filtered colimit of free modules.

This observation (Wraith, Blass) can be put into the more general context of modelling geometric theories by geometric morphisms from their classifying toposes, or equivalently, certain flat functors from sites for such topoi.

### For more general rings

Even if the ring $R$ is not necessarily commutative and not necessarily unital, we can say:

A left $R$-module is flat precisely if the tensoring functor

$(-)\otimes_R \colon Mod_R \to Ab$

from right $R$ modules to abelian groups is an exact functor.

## Properties

### Equivalent characterizations

By def. , or its immediate consequence, remark . $N \in R Mod$ is flat if for every injection $i \colon A \hookrightarrow$ also $i \otimes_R N \colon A \otimes_R N \to B \otimes_R N$ is an injection. The following proposition says that this may already be checked on just a very small subclass of injections.

###### Theorem

An $R$-module $N$ is flat already if for all inclusions $I \hookrightarrow R$ of a finitely generated ideal into $R$, regarded as a module over itself, the induced morphism

$I \otimes_R N \to R \otimes_R N \simeq N$

is an injection.

(…)

###### Proposition

A module $N$ is flat precisely if for every finite linear combination of zero, $\sum_i r_i n_i = 0\in N$ with $\{r_i \in R\}_i$, $\{n_i \in N\}$ there are elements $\{\tilde n_j \in N\}_j$ and linear combinations

$n_i = \sum_j b_{i j} \tilde n_j \;\;\in N$

with $\{b_{i j} \in R\}_{i,j}$ such that for all $j$ we have

$\sum_i r_i b_{i j} = 0\;\;\; \in R \,.$
###### Proof

A finite set $\{r_i \in R\}_i$ corresponds to the inclusion of a finitely generated ideal $I \hookrightarrow R$.

By theorem $N$ is flat precisely if $I \otimes_R N \to N$ is an injection. This in turn is the case precisely if the only element of the tensor product $I \otimes_R R$ that is 0 in $R \otimes_R N = N$ is already 0 on $I \otimes_R N$.

Now by definition of tensor product of modules an element of $I \otimes_R N$ is of the form $\sum_i (r_i ,n_i)$ for some $\{n_i \in N\}$. Under the inclusion $I \otimes_R N \to N$ this maps to the actual linear combination $\sum_i r_i n_i$. This map is injective if whenever this linear combination is 0, already $\sum_i (r_i, n_i)$ is 0.

But the latter is the case precisely if this is equal to a combination $\sum_j (\tilde r_j , \tilde n_j)$ where all the $\tilde r_j$ are 0. This implies the claim.

### Relation to projective modules

###### Proposition

Every projective module is flat.

###### Proof

Clearly every ring $R$ is a flat module over itself, and direct sums as well as direct summands of flat modules are flat. Hence direct summands of free modules are flat, and these are precisely the projective modules (prop.)

###### Proposition

(Lazard's criterion)

A module is flat if and only if it is a filtered colimit of free modules.

This is due to (Lazard (1964)).

###### Proof

(…) For the moment see the above discussion. (…)

### Relation to (locally) free modules

###### Definition

An $R$-module $N$ over a Noetherian ring $R$ is called a locally free module if there is a cover by prime ideals $I \hookrightarrow R$ such that the localization $N_I$ is a free module over the localization $R_I$.

###### Proposition

For $R$ a Noetherian ring and $N$ a finitely generated module over $R$, $N$ is flat precisely if it is locally free module, def. .

###### Proposition

If

1. $R$ is a local ring,

2. $N$ is a finitely generated module,

3. $N$ is a flat module

then $N$ is a free module.

This is Matsumara, Theorem 7.10

## Examples

###### Example

An abelian group is flat (regarded as a $\mathbb{Z}$-module) precisely if it is torsion-free.

###### Proof

By the general discussion at derived functor in homological algebra, the obstruction to $A \in Ab$ being flat are the first Tor-groups $Tor_1^{\mathbb{Z}}(-,A)$. By the discussion at Tor – relation to torsion subgroups these a filtered colimits and direct sums of the torsion subgroups of $A$. In particular for $Tor_1^\mathbb{Z}(\mathbb{Z}_n,A)$ is the $n$-torsion subgroup of $A$. Hence $Tor_1^\mathbb{Z}(-,A)$ vanishes and hence $A$ is flat precisely if all torsion subgroups of $A$ are trivial.

Original articles include

• Shizuo Endo, On flat modules over commutative rings, J. Math. Soc. Japan Volume 14, Number 3 (1962), 284-291. (EUCLID)

• Michel Raynaud, Laurent Gruson, Critères de platitude et de projectivité, Techniques de “platification” d’un module. Invent. Math. 13 (1971), 1–89.

• S. Jondrup, Flat and projective modules, Math, Scand. 43 (1978) (pdf)

The characterization of flat modules as filtered colimits of projective modules is due to

• Daniel Lazard, Sur les modules plats C. R. Acad. Sci. Paris 258, 6313–6316 (1964)

For a general account see for instance section 3.2 of

For more details see

• Matsumura’s CRT book

Lecture notes include

• Arthur Ogus, Flatness – a brief overview (pdf)

Further resources include