The Stacks project

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 $(u \otimes 1) \circ v_ i = \text{id}_{N \otimes _ R R_ i}$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05LI. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.127.5 as pdf