The Stacks project

Lemma 10.82.12. Let $R \to S$ be a ring map. Let $M \to M'$ be a map of $S$-modules. The following are equivalent

  1. $M \to M'$ is universally injective as a map of $R$-modules,

  2. for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R$-modules,

  3. for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R$-modules,

  4. for each prime $\mathfrak q$ of $S$ the map $M_{\mathfrak q} \to M'_{\mathfrak q}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak q$ in $R$, and

  5. for each maximal ideal $\mathfrak m$ of $S$ the map $M_{\mathfrak m} \to M'_{\mathfrak m}$ is universally injective as a map of $R_{\mathfrak p}$-modules, where $\mathfrak p$ is the inverse image of $\mathfrak m$ in $R$.

Proof. Let $N$ be an $R$-module. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak p$ of $R$. Then we have

\[ (M \otimes _ R N)_{\mathfrak q} = M_{\mathfrak q} \otimes _ R N = M_{\mathfrak q} \otimes _{R_{\mathfrak p}} N_{\mathfrak p}. \]

Moreover, the same thing holds for $M'$ and localization is exact. Also, if $N$ is an $R_{\mathfrak p}$-module, then $N_{\mathfrak p} = N$. Using this the equivalences can be proved in a straightforward manner.

For example, suppose that (5) holds. Let $K = \mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N)$. By the remarks above we see that $K_{\mathfrak m} = 0$ for each maximal ideal $\mathfrak m$ of $S$. Hence $K = 0$ by Lemma 10.23.1. Thus (1) holds. Conversely, suppose that (1) holds. Take any $\mathfrak q \subset S$ lying over $\mathfrak p \subset R$. Take any module $N$ over $R_{\mathfrak p}$. Then by assumption $\mathop{\mathrm{Ker}}(M \otimes _ R N \to M' \otimes _ R N) = 0$. Hence by the formulae above and the fact that $N = N_{\mathfrak p}$ we see that $\mathop{\mathrm{Ker}}(M_{\mathfrak q} \otimes _{R_{\mathfrak p}} N \to M'_{\mathfrak q} \otimes _{R_{\mathfrak p}} N) = 0$. In other words (4) holds. Of course (4) $\Rightarrow $ (5) is immediate. Hence (1), (4) and (5) are all equivalent. We omit the proof of the other equivalences. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 10.82: Universally injective module maps

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Your email address will not be published.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 05CL. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.82.12 as pdf