nLab
flat module

Definition

In terms of exact functors

Definition

Given a ring (or $k$-algebra) $A$, a left $A$-module, say $M$, is flat if tensoring with $M$ over $A$ as a functor from left $A$-modules to left $k$-modules is an exact functor (sends short exact sequences to short exact sequences).

Remarks

• Tensoring with $M$ is automatically right exact, so it is equivalent to require that tensoring with $M$ be left exact. Since Mod is finitely complete, this is also equivalent to requiring that tensoring with $M$ be a flat functor.

• On an abelian category, the notions of exact functor and flat functor coincide.

• The ring may be taken non-unital without any change.

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 $A$)”.

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 $v_i={\sum }_j{\beta }_{ij}{u}_j$, and substitute in to get ${\sum }_{ij}{\alpha }_i{\beta }_{ij}{u}_j=0$. Now as $\{u_j\}$ forms a basis, we can deduce from this that for each $j$, ${\sum }_i{\alpha }_i{\beta }_{ij}=0$. 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 $m_i={\sum }_j{b}_{ij}{n}_j$ and the coefficients $b_{ij}$ have the property that ${\sum }_i{a}_i{b}_{ij}=0$.

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 $m:F\to M$. The $a_i$ correspond to a morphism $a:F\to F$, multiplying the $i$th term by $a_i$. That we have the identity ${\sum }_i{a}_i{m}_i=0$ says that the composition $ma$ is zero, or that $m:F\to M$ factors through the coequaliser of $a$ and $0$. Now we consider the elements $n_j$. These define another morphism from a free module, say $n:E\to M$. 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 $b:F\to M$ such that $m=nb$. We therefore have to factorisations of $m$: one through $n$ and one through $cokera$. The question is as to whether these have any relation to each other. In particular, does $cokera\to M$ 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 $E$.

Hence 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.