Lemma 15.80.4. Let $R$ be a ring, $f \in R$ an element, $R_ f \to A$ is a finite type ring map, $g \in A$, and $M$ an $A$-module. If $M$ of finite presentation relative to $R_ f$, then $M_ g$ is an $A_ g$-module of finite presentation relative to $R$.

Proof. Choose a presentation $R_ f[x_1, \ldots , x_ n] \to A$. We write $R_ f = R[x_0]/(fx_0 - 1)$. Consider the presentation $R[x_0, x_1, \ldots , x_ n, x_{n + 1}] \to A_ g$ which extends the given map, maps $x_0$ to the image of $1/f$, and maps $x_{n + 1}$ to $1/g$. Choose $g' \in R[x_0, x_1, \ldots , x_ n]$ which maps to $g$ (this is possible). Suppose that

$R_ f[x_1, \ldots , x_ n]^{\oplus s} \to R_ f[x_1, \ldots , x_ n]^{\oplus t} \to M \to 0$

is a presentation of $M$ given by a matrix $(h_{ij})$. Pick $h'_{ij} \in R[x_0, x_1, \ldots , x_ n]$ which map to $h_{ij}$. Then

$R[x_0, x_1, \ldots , x_ n, x_{n + 1}]^{\oplus s + 2t} \to R[x_0, x_1, \ldots , x_ n, x_{n + 1}]^{\oplus t} \to M_ g \to 0$

is a presentation of $M_ f$. Here the $t \times (s + 2t)$ matrix defining the map has a first $t \times s$ block consisting of the matrix $h'_{ij}$, a second $t \times t$ block which is $(x_0f - )I_ t$, and a third block which is $(x_{n + 1}g' - 1)I_ t$. $\square$

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