Determinant of a projective module

Remarks

The module $\det P$ is invertible, since localizing at a prime gives $\Lambda^n(R_\mathfrak{p}^n) \cong R_\mathfrak{p}$. For finitely generated projective modules $P$ and $Q$ of constant ranks, $\det(P \oplus Q) \cong \det P \otimes_R \det Q$. When $\mathrm{Spec}\, R$ is connected, every finitely generated projective module has constant rank, and $[P] \mapsto [\det P]$ extends to a group homomorphism $K_0(R) \to \mathrm{Pic}(R)$. An $R$-linear map $f \colon P \to P$ induces $\Lambda^n f$ on $\det P$, which is multiplication by an element $\det f$ of $R$; for $P = R^n$ this is the usual determinant of a matrix.

Example: $\det R^n \cong R$, generated by $e_1 \wedge \cdots \wedge e_n$ for the standard basis. For a Dedekind domain $R$ and a nonzero ideal $I$, the module $R^{n-1} \oplus I$ has determinant $R^{\otimes (n-1)} \otimes_R I \cong I$.

Non-example: the wrong degree does not give an invertible module. For $P = R^n$ over a nonzero ring, $\Lambda^{n+1} P = 0$, and $\Lambda^{n-1} P \cong R^n$ is not of rank $1$ when $n \ge 2$.