# $\mathbb{Z}_m$ as a $\mathbb{Z}_n$-module

It is routine to check that $\mathbb{Z}_m=\mathbb{Z}/m\mathbb{Z}$ is a $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$-module if and only if $m \mid n$. This leads to lots of questions such as, "Is $\mathbb{Z}_2$ a projective or free $\mathbb{Z}_6$-module?". This uses the fact that $\mathbb{Z}_6=\mathbb{Z}_2 \oplus \mathbb{Z}_3$. This made me do a bit of thinking about these as 'test modules' and brought up the following questions:

$1$) Is this type of decomposition always valid? That is, if $\mathbb{Z}_n=\bigoplus_{i=1}^k \mathbb{Z}_{m_i}$ as $\mathbb{Z}$-modules, is it always valid as a $\mathbb{Z}_n$-module?

Here, I imagine this should be true since the operation as a $\mathbb{Z}$-module is really "the same" as a $\mathbb{Z}_n$-module.

$2$) It is easy to check that by cardinality that $\mathbb{Z}_2$ is not a free $\mathbb{Z}_6$-module. But then what are the free $\mathbb{Z}_6$-modules? Generally, what are the free $\mathbb{Z}_n$-modules? Is it really the case that the examples are the trivial sums of copies of $\mathbb{Z}_n$? (This is of course generally true since this is what it means to be free, I mean this in sense that supposing $\mathbb{Z}_2$ were free, it would not be immediately obvious whereas $\oplus\mathbb{Z}_6$ is clearly free).

$3$) These ideas give easy checks for whether a $\mathbb{Z}_n$-module $\mathbb{Z}_m$ is free or projective. But is it also simple to check whether any of these summands are injective?

• You need to be careful because the universal properties that characterize free modules etc. may hold when you restrict attention to $\Bbb{Z}_n$-homomorphisms but fail to hold for all $\Bbb{Z}$-homomorphisms. Hence $\Bbb{Z}_2$ is free, projective and injective as a $\Bbb{Z}_2$-module but is not free, projective or injective as $\Bbb{Z}$-module. For direct sums, it so happens that this difficulty does not arise: a direct sum decomposition $A = B \oplus C$ for $\Bbb{Z}_n$-modules will also be a direct sum decomposition of $\Bbb{Z}$-modules and vice versa (given that $A$ is a $\Bbb{Z}_n$-module). – Rob Arthan Jan 7 '17 at 21:16