Lemma 10.127.5. Let $A$ be a ring and let $M, N$ be $A$-modules. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras.

1. If $M$ is a finite $A$-module, and $u, u' : M \to N$ are $A$-module maps such that $u \otimes 1 = u' \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ then for some $i$ we have $u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$.

2. If $N$ is a finite $A$-module and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is surjective, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective.

3. If $N$ is a finitely presented $A$-module, and $v : N \otimes _ A R \to M \otimes _ A R$ is an $R$-module map, then there exists an $i$ and an $R_ i$-module map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$ such that $v = v_ i \otimes 1$.

4. If $M$ is a finite $A$-module, $N$ is a finitely presented $A$-module, and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is an isomorphism, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is an isomorphism.

Proof. To prove (1) assume $u$ is as in (1) and let $x_1, \ldots , x_ m \in M$ be generators. Since $N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i$ we may pick an $i \in I$ such that $u(x_ j) \otimes 1 = u'(x_ j) \otimes 1$ in $M \otimes _ A R_ i$, $j = 1, \ldots , m$. For such an $i$ we have $u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$.

To prove (2) assume $u \otimes 1$ surjective and let $y_1, \ldots , y_ m \in N$ be generators. Since $N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in M \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $N \otimes _ A R$ equal $y_ j \otimes 1$. For such an $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective.

To prove (3) let $y_1, \ldots , y_ m \in N$ be generators. Let $K = \mathop{\mathrm{Ker}}(A^{\oplus m} \to N)$ where the map is given by the rule $(a_1, \ldots , a_ m) \mapsto \sum a_ j x_ j$. Let $k_1, \ldots , k_ t$ be generators for $K$. Say $k_ s = (k_{s1}, \ldots , k_{sm})$. Since $M \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i M \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in M \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $M \otimes _ A R$ equal $v(y_ j \otimes 1)$. We want to use the $z_ j$ to define the map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$. Since $K \otimes _ A R_ i \to R_ i^{\oplus m} \to N \otimes _ A R_ i \to 0$ is a presentation, it suffices to check that $\xi _ s = \sum _ j k_{sj}z_ j$ is zero in $M \otimes _ A R_ i$ for each $s = 1, \ldots , t$. This may not be the case, but since the image of $\xi _ s$ in $M \otimes _ A R$ is zero we see that it will be the case after increasing $i$ a bit.

To prove (4) assume $u \otimes 1$ is an isomorphism, that $M$ is finite, and that $N$ is finitely presented. Let $v : N \otimes _ A R \to M \otimes _ A R$ be an inverse to $u \otimes 1$. Apply part (3) to get a map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$ for some $i$. Apply part (1) to see that, after increasing $i$ we have $v_ i \circ (u \otimes 1) = \text{id}_{M \otimes _ R R_ i}$ and $(u \otimes 1) \circ v_ i = \text{id}_{N \otimes _ R R_ i}$. $\square$

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