 Solution

If (i) holds, the submodule theorem for free modules over a principal ideal domain says that any submodule $N\subseteq R^n$ is free. Its rank is at most $n$, since tensoring the inclusion with $F$ embeds $N{\otimes }_{R}F$ into $F^n$. Hence $N$ has at most $n$ generators, proving (iv).

Conversely, apply (iv) with $n=1$. Every ideal is an $R$-submodule of $R$ and therefore has one generator, which is (i). Thus all four conditions are equivalent.