The Stacks project

Proof. We will deduce this lemma from Algebra, Lemma 10.128.4 (please note that we exchanged the roles of $M$ and $N$). To do the reduction we will use More on Algebra, Lemma 15.25.7 to reduce to the finitely presented case.

By assumption we can write $B$ as a quotient of the localization of a polynomial algebra $P = A[x_1, \ldots , x_ n]$ at a prime ideal $\mathfrak q$. Then we can think of $u : N \to M$ as a map of finite $P_\mathfrak q$-modules. Hence we may and do assume that $B$ is essentially of finite presentation over $A$.

Next, the $B$-module $N$ is finite but perhaps not of finite presentation. Write $N = \mathop{\mathrm{colim}}\nolimits N_\lambda$ as a filtered colimit of finitely presented $B$-modules with surjective transition maps. For example choose a presentation $0 \to K \to B^{\oplus r} \to N \to 0$, write $K$ as the union of its finite submodules $K_\lambda$, and set $N_\lambda = \mathop{\mathrm{Coker}}(K_\lambda \to B^{\oplus r})$. The module $N/\mathfrak m_ A N$ is of finite presentation over the Noetherian ring $B/\mathfrak m_ A B$. Hence for $\lambda$ large enough we have $N_\lambda /\mathfrak m_ A N_\lambda = N/\mathfrak m_ A N$. Now, if we can show the lemma for the composition $u_\lambda : N_\lambda \to M$, then we conclude that $N_\lambda = N$ and the result holds for $u$. Hence we may and do assume $N$ is of finite presentation over $B$.

By More on Algebra, Lemma 15.25.7 the module $M$ is of finite presentation over $B$. Thus all the assumptions of Algebra, Lemma 10.128.4 hold and we conclude. $\square$