Why $\mathbb{Z}_p$ over $\mathbb{Z}_{p^2}$ cannot be free?

Question

Makoto Kato in the answer to this question wrote:

Let $p$ be a prime number. Let $R = \mathbb{Z}/p^2\mathbb{Z}$. Let $M = R/pR$. Since the number of elements of $M$ is $p$, $M$ cannot be free. Hence $M$ cannot be projective.

I cannot conclude that $M$ cannot be free.

Answer 1

Because $px=0$ for all $x\in M$, and $p\neq0_R$, we see that every non-empty subset of $M$ is linearly dependent over $R$. But $M$ is not generated by the empty set, so it cannot be free.