$\mathbb{Z}\mathrm{/}2\mathbb{Z}$ is projective but not free over $\mathbb{Z}\mathrm{/}6\mathbb{Z}$

The element $3$ is idempotent modulo $6$, and $1-3=4$ is also idempotent, with $3\cdot 4=0$ and $3+4=1$. Hence $R=3R\oplus 4R$.

The module $P$ has two elements. A free module on the empty set has one element; a free module on a nonempty finite set of size $k$ has $6^k$ elements; and a free module on an infinite set has infinitely many distinct standard basis vectors. None has two elements.

Thus $P=3R$ is a direct summand of the rank-one free module $R$, which is projective without AC by [L2]. Precomposing a map from $P$ with the projection $R\to P$, lifting from $R$, and restricting the lift to $P$ proves directly that $P$ is projective. The map $\mathbb{Z}\mathrm{/}2\mathbb{Z}\to P$ sending $1$ to $3$ is an $R$-module isomorphism.

Hence $P\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is projective and nonfree over $\mathbb{Z}\mathrm{/}6\mathbb{Z}$.