Lemma 15.21.6. Let $R \to S$ be an injective integral ring map. Let $M$ be a finitely presented module over $R[x_1, \ldots , x_ n]$. If $M \otimes _ R S$ is flat over $S$, then $M$ is flat over $R$.

Proof. Choose a presentation

$R[x_1, \ldots , x_ n]^{\oplus t} \to R[x_1, \ldots , x_ n]^{\oplus r} \to M \to 0.$

Let's say that the first map is given by the $r \times t$-matrix $T = (f_{ij})$ with $f_{ij} \in R[x_1, \ldots , x_ n]$. Write $f_{ij} = \sum f_{ij, I} x^ I$ with $f_{ij, I} \in R$ (multi-index notation). Consider diagrams

$\xymatrix{ R \ar[r] & S \\ R_\lambda \ar[u] \ar[r] & S_\lambda \ar[u] }$

where $R_\lambda$ is a finitely generated $\mathbf{Z}$-subalgebra of $R$ containing all $f_{ij, I}$ and $S_\lambda$ is a finite $R_\lambda$-subalgebra of $S$. Let $M_\lambda$ be the finite $R_\lambda [x_1, \ldots , x_ n]$-module defined by a presentation as above, using the same matrix $T$ but now viewed as a matrix over $R_\lambda [x_1, \ldots , x_ n]$. Note that $S$ is the directed colimit of the $S_\lambda$ (details omitted). By Algebra, Lemma 10.168.1 we see that for some $\lambda$ the module $M_\lambda \otimes _{R_\lambda } S_\lambda$ is flat over $S_\lambda$. By Lemma 15.21.5 we conclude that $M_\lambda$ is flat over $R_\lambda$. Since $M = M_\lambda \otimes _{R_\lambda } R$ we win by Algebra, Lemma 10.39.7. $\square$

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