The Stacks project

Lemma 15.23.12. Let $R$ be a Noetherian ring. Let $\varphi : M \to N$ be a map of $R$-modules. Assume that for every prime $\mathfrak p$ of $R$ at least one of the following happens

  1. $M_\mathfrak p \to N_\mathfrak p$ is injective, or

  2. $\mathfrak p \not\in \text{Ass}(M)$.

Then $\varphi $ is injective.

Proof. Let $\mathfrak p$ be an associated prime of $\mathop{\mathrm{Ker}}(\varphi )$. Then there exists an element $x \in M_\mathfrak p$ which is in the kernel of $M_\mathfrak p \to N_\mathfrak p$ and is annihilated by $\mathfrak pR_\mathfrak p$ (Algebra, Lemma 10.63.15). This is impossible in both cases. Hence $\text{Ass}(\mathop{\mathrm{Ker}}(\varphi )) = \emptyset $ and we conclude $\mathop{\mathrm{Ker}}(\varphi ) = 0$ by Algebra, Lemma 10.63.7. $\square$


Comments (4)

Comment #2471 by Raymond Cheng on

Third sentence: "This is impossible in all three cases." There are only two cases!

Comment #8274 by Laurent Moret-Bailly on

In the proof, "is annihilated by " should be "whose annihilator is ".


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Your email address will not be published.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 0AV7. Beware of the difference between the letter 'O' and the digit '0'.