The Stacks project

Proof. Define $B(x) \subset M(x)$ as the perpendicular of the image of the map

and set $V(x) = M(x)/B(x)$. Then any $R$-linear map $\varphi : M \to R$ induces a map $\overline{\varphi } : V(x) \to \kappa (x)$ and conversely any $\kappa (x)$-linear map $\lambda : V(x) \to \kappa (x)$ is equal to $\overline{\varphi }$ for some $\varphi $. Let $s_1, \ldots , s_ r \in M$.

Suppose $s_1, \ldots , s_ r$ map to linearly independent elements of $V(x)$. Then we can find $\varphi _1, \ldots , \varphi _ r \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ such that $\varphi _ i(s_ j)$ maps to $\delta _{ij}$¹ in $\kappa (x)$. Hence the matrix of the composition

has a determinant $f \in R$ which maps to $1$ in $\kappa (x)$ Clearly, this implies that $s_1, \ldots , s_ r : R^{\oplus r} \to M$ is the inclusion of a direct summand after inverting $f$.

Conversely, suppose that we have an $f \in R$, $f \not\in x$ such that $s_1, \ldots , s_ r : R^{\oplus r} \to M$ is the inclusion of a direct summand after inverting $f$. Hence we can find $R_ f$-linear maps $\varphi _ i : M_ f \to R_ f$ such that $\varphi _ i(s_ j) = \delta _{ij} \in R_ f$. Since $\mathop{\mathrm{Hom}}\nolimits _ R(M, R)_ f = \mathop{\mathrm{Hom}}\nolimits _{R_ f}(M_ f, R_ f)$ by Algebra, Lemma 10.10.2 we conclude that we can find $n \geq 0$ and $\varphi '_ i \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ such that $\varphi '_ i(s_ j) = f^ n\delta _{ij} \in R$. It follows that $s_1, \ldots , s_ r$ map to linearly independent elements of $V(x)$ as $\overline{\varphi }'_ i(s_ j) = f^ n\delta _{ij}$. $\square$