The Stacks project

Proof. This is similar to the proof of part (4) of Lemma 10.6.2. We may assume $S = R[x_1, \ldots , x_ n]/J$. Choose $y_1, \ldots , y_ m \in M$ which generate $M$ as an $R$-module and choose relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ which generate the kernel of $R^{\oplus m} \to M$. For any $i = 1, \ldots , n$ and $j = 1, \ldots , m$ write

for some $a_{ijk} \in R$. Consider the $S$-module $N$ generated by $y_1, \ldots , y_ m$ subject to the relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ and $x_ i y_ j = \sum a_{ijk} y_ k$, $i = 1, \ldots , n$ and $j = 1, \ldots , m$. Then $N$ has a presentation

By construction there is a surjective map $\varphi : N \to M$. To finish the proof we show $\varphi$ is injective. Suppose $z = \sum b_ j y_ j \in N$ for some $b_ j \in S$. We may think of $b_ j$ as a polynomial in $x_1, \ldots , x_ n$ with coefficients in $R$. By applying the relations of the form $x_ i y_ j = \sum a_{ijk} y_ k$ we can inductively lower the degree of the polynomials. Hence we see that $z = \sum c_ j y_ j$ for some $c_ j \in R$. Hence if $\varphi (z) = 0$ then the vector $(c_1, \ldots , c_ m)$ is an $R$-linear combination of the vectors $(a_{i1}, \ldots , a_{im})$ and we conclude that $z = 0$ as desired. $\square$