Let $R$ be the matrix ring $M_2(\mathbb{C})$. Let $M = \mathbb{C}^2$ with its natural $R$-module structure (just given by the usual action of $2 \times 2$ matrices on $2$-dimensional vectors).
My questions are: is $M$ a projective $R$-module? An injective $R$-module? Is it free?
On
$M_2(\mathbb{C})$ is a semisimple ring, so every module over this ring is projective and injective. Moreover, $\mathbb C^2$ is a simple $M_2(\mathbb{C})$-module, and if it is free it must be isomorphic to $M_2(\mathbb{C})$. In fact, $M_2(\mathbb{C})\simeq \mathbb C^2\oplus\mathbb C^2$, and therefore by taking the ranks we get a contradiction.
If $R$ is any ring and $n \geq 1$, then the evident map $M_n(R) \to \bigoplus_{i=1}^{n} R^n$ which decomposes a matrix into its columns is an isomorphism of left $M_n(R)$-modules. Hence, $R^n$ is a projective left $M_n(R)$-module. It is usually not free. If $R$ is commutative and non-zero (more generally, when $R$ has IBN), then free $M_n(R)$-modules of rank $k$ have underlying free $R$-modules of rank $n^2 \cdot k$, but $R^n$ is a free $R$-module of rank $n$, and $n^2 \cdot k = n$ is only solvable in the trivial case $n=1$.
If $R$ is a field, then $R^n$ is an injective left $M_n(R)$-module. One reason for this is that the category of left $M_n(R)$-modules is equivalent to the category of left $R$-modules (i.e. $R$ and $M_n(R)$ are Morita equivalent) and since every left $R$-module is injective, which is a purely categorical notion, the same follows for every left $M_n(R)$-module. Of course, this also shows that every left $M_n(R)$-module is projective.