Lemma 15.80.1. Let $R \to A$ be a ring map of finite type. Let $M$ be an $A$-module. The following are equivalent

1. for some presentation $\alpha : R[x_1, \ldots , x_ n] \to A$ the module $M$ is a finitely presented $R[x_1, \ldots , x_ n]$-module,

2. for all presentations $\alpha : R[x_1, \ldots , x_ n] \to A$ the module $M$ is a finitely presented $R[x_1, \ldots , x_ n]$-module, and

3. for any surjection $A' \to A$ where $A'$ is a finitely presented $R$-algebra, the module $M$ is finitely presented as $A'$-module.

In this case $M$ is a finitely presented $A$-module.

Proof. If $\alpha : R[x_1, \ldots , x_ n] \to A$ and $\beta : R[y_1, \ldots , y_ m] \to A$ are presentations. Choose $f_ j \in R[x_1, \ldots , x_ n]$ with $\alpha (f_ j) = \beta (y_ j)$ and $g_ i \in R[y_1, \ldots , y_ m]$ with $\beta (g_ i) = \alpha (x_ i)$. Then we get a commutative diagram

$\xymatrix{ R[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \ar[d]^{x_ i \mapsto g_ i} \ar[rr]_-{y_ j \mapsto f_ j} & & R[x_1, \ldots , x_ n] \ar[d] \\ R[y_1, \ldots , y_ m] \ar[rr] & & A }$

Hence the equivalence of (1) and (2) follows by applying Algebra, Lemmas 10.6.4 and 10.36.23. The equivalence of (2) and (3) follows by choosing a presentation $A' = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and using Algebra, Lemma 10.36.23 to show that $M$ is finitely presented as $A'$-module if and only if $M$ is finitely presented as a $R[x_1, \ldots , x_ n]$-module. $\square$

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