The tensor product functor and flat modules

#module_theory

Suppose $D$ is an $(R, S)$ -bimodule. We can then consider the two possible functors corresponding to tensoring with $D$ , namely the left tensor product functor $D \otimes_{S} -$ and the right tensor product functor $- \otimes_{R} D$ . The first can be used as a functor from the category of $(S, T)$ -bimodules to the category of $(R, T)$ -bimodules (for any ring $T$ ); the latter can be used as a functor from the category of $(T, R)$ -bimodules to the category of $(T, S)$ -bimodules (for any ring $T$ ). Both functors will have similar properties, so we'll focus on the former.

The tensor product and direct sums

The tensor product commutes with direct sums

Suppose $M$ is an $(R, S)$ -bimodule and ${N_{i} ∣ i \in I}$ is a family of $(S, T)$ -bimodules. Then there is a unique isomorphism of $(R, T)$ -bimodules
$M \otimes_{S} (⨁_{i \in I} N_{i}) ≃ ⨁_{i \in I} (M \otimes_{S} N_{i})$

The tensor product and exact sequences

The tensor product functor is right exact

Let $D$ be an $(R, S)$ -bimodule, and suppose we have a short exact sequence of $(R, T)$ -bimodules

0 \to L \overset{f}{\to} M \overset{g}{\to} N \to 0.

Then the corresponding sequence of $(R, T)$ -bimodules

D \otimes_{S} L \overset{1_{D} \otimes f}{\to} D \otimes_{S} M \overset{1_{D} \otimes g}{\to} D \otimes_{S} N \to 0

is exact.

Notice that the $0$ on the far left of the sequence is gone! We have lost the "left end" of our exact sequence. Because of the above property, we say that the functor $D \otimes_{S} -$ is right exact.

Is it ever the case that the functor $D \otimes_{S} -$ is exact, in other words sends short exact sequences to short exact sequences?

Definition of a flat module

An $S$ -module $D$ is flat if for every short exact sequence of $S$ -modules

0 \to L \overset{f}{\to} M \overset{g}{\to} N \to 0

the corresponding sequence of abelian groups is also exact:

0 \to D \otimes_{S} L \overset{1_{D} \otimes f}{\to} D \otimes_{S} M \overset{1_{D} \otimes g}{\to} D \otimes_{S} N \to 0

Projective modules are flat

Every projective $R$ -module is also flat.

In particular, free modules are flat.

Examples of flat modules

The abelian group $Z$ is projective and hence also flat.
The abelian group $Q$ is flat.
Any direct sum of flat modules is flat; e.g., the abelian group $Z \oplus Q$ is flat (but neither injective nor projective).

Examples of non-flat modules

The abelian group $Z_{2}$ is not flat.
The quotient group $Q / Z$ is not flat (although it is injective).