A nonzero finite projective module over a Dedekind domain splits off a rank-one summand

The $K$-vector space $M_K:=M{\otimes }_{R}K$ is nonzero because $M$ is a nonzero projective module over a domain. Choose a nonzero $K$-linear functional $\lambda :M_K\to K$. Its image on the finite module $M$ is a finitely generated nonzero $R$-submodule $I:=\lambda (M)\subseteq K$, so $I$ is a fractional ideal. As a submodule of the field $K$, the module $I$ is torsion-free; since it is finitely generated, [L1] makes it projective. Because ${\lambda }_K:M_K\to K$ is surjective, one has $I{\otimes }_{R}K=K$, so $I$ has rank one. Therefore [L2] makes $I$ an invertible fractional ideal.

The restricted map $M\to I$ is surjective by construction. Since $I$ is projective by step 1.1, this surjection splits. Therefore $M\cong \ker (\lambda {\mathrm{\mid }}_M)\oplus I$. Put $N:=\ker (\lambda {\mathrm{\mid }}_M)$. Then $N$ is finite projective as a direct summand of $M$.