Injective module but not flat module?

Question

We know a projective module is a flat module.

I want to know whether an injective module can also imply that it is a flat module.

Is there any counter example?

Thank you!

Answer 1

Here's an example from abelian groups $=$ $\mathbb{Z}$-modules: $\mathbb{Q}/\mathbb{Z}$ is an injective $\mathbb{Z}$-module because it's a divisible abelian group, but it's not a flat $\mathbb{Z}$-module because $\require{cancel}\xcancel{\mathbb{Q}/\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}=0}$ (... sorry for the mistake ...) $\mathbb{Q}/\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Q}=0$.

Answer 2

Field of fraction of an integral domain always is a flat module. Take $\mathbb Q$ as a $\mathbb Z$-module. Since $\mathbb Z$ is a PID, the injective $\mathbb Z$-modules are exactly the divisible modules, then $\mathbb Q$ is a flat injective $\mathbb Z$-module.

Answer 3

$\newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}}$

$\Q/\Z$ is the first example of an injective $\Z$ module. You should think of this group as the rational numbers on the unit circle=$\R/\Z$. Alternatively $\Q/\Z$ can be thought of as the image of $e^{2\pi i (-)}|_{\Q}$. Every cyclic group sits inside here and this actually characterizes the group.

$\Q/\Z$ is not an injective $\Z$-module:

$0 \to \Z \to \Z \to \Z/2 \to 0$ is an exact sequence but
$\Z \otimes_{\Z} \Q/\Z \to \Z \otimes_{\Z} \Q/\Z \to \Z/2 \otimes_{\Z} \Q/\Z$ is not.

(b/c tensoring by $\Q/\Z$ picks out the torsion in an abelian group and discards the free part. Thus you end up with the sequence $0 \to 0 \to \Z/2$ which isn't exact)

This contradicts $\Q/\Z$ being a flat $\Z$ module.