Lemma 15.80.3. Let $R$ be a ring. Let $A \to B$ be a finite map of finite type $R$-algebras. Let $M$ be a $B$-module. Then $M$ is an $A$-module finitely presented relative to $R$ if and only if $M$ is a $B$-module finitely presented relative to $R$.

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose $y_1, \ldots , y_ m \in B$ which generate $B$ over $A$. As $A \to B$ is finite each $y_ i$ satisfies a monic equation with coefficients in $A$. Hence we can find monic polynomials $P_ j(T) \in R[x_1, \ldots , x_ n][T]$ such that $P_ j(y_ j) = 0$ in $B$. Then we get a commutative diagram

$\xymatrix{ R[x_1, \ldots , x_ n] \ar[d] \ar[r] & R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/(P_ j(y_ j)) \ar[d] \\ A \ar[r] & B }$

Since the top arrow is a finite and finitely presented ring map we conclude by Algebra, Lemma 10.36.23 and the definition. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).