The Stacks project

Proof. To see (1) use that the double dual of $M$ is torsion free by Lemma 15.22.12. Assume that $M$ is finite. Choose generators $x_1, \ldots , x_ n$ of $M$ over $R$. Let $K$ be the fraction field of $R$. After renumbering we may assume that for some $0 \leq r \leq n$ the elements $x_1, \ldots , x_ r$ map to a basis of $M \otimes _ R K$ over $K$. For $r < i \leq n$ write $x_ i \otimes 1 = \sum _{1 \leq j \leq r} x_ j \otimes k_{ij}$ with $k_{ij} \in K$. Choose $c \in R$ such that $a_{ij} = ck_{ij} \in R$ and such that $cx_ i = \sum _{1 \leq j \leq r} a_{ij} x_ j$ in $M$ (small detail omitted). Then we get maps

with both $\alpha \circ \beta $ and $\beta \circ \alpha $ given by multiplication with $c$. Here $\alpha $ uses $x_1, \ldots , x_ r$ and $\beta $ sends $x_ i$ to $c$ times the $i$th basis vector for $1 \leq i \leq r$ and to the vector $(a_{i1}, \ldots , a_{ir})$ for $r < i \leq n$. Consider the commutative diagram

Of course the compositions of consecutive arrows on the bottom row are equal to multiplication by $c$. The diagram shows that the kernel and cokernel of $j$ are annihilated by $j$ which proves (2). For (3) note that if $M$ is torsion free, then $\beta $ is injective which implies that $j$ is injective. The converse is clear because we've seen already that the double dual of $M$ is torsion free. $\square$