The Stacks project

Lemma 15.15.4. Let $R$ be a ring. The following are equivalent

  1. $R$ has property (P) of Lemma 15.15.2,

  2. any injective map of projective $R$-modules is universally injective,

  3. if $u : N \to M$ is injective and $N$, $M$ are finite projective $R$-modules then $\mathop{\mathrm{Coker}}(u)$ is a finite projective $R$-module,

  4. if $N \subset M$ and $N$, $M$ are finite projective as $R$-modules, then $N$ is a direct summand of $M$, and

  5. any injective map $R \to R^{\oplus n}$ is a split injection.

Proof. The implication (1) $\Rightarrow $ (2) is Lemma 15.15.3. It is clear that (3) and (4) are equivalent. We have (2) $\Rightarrow $ (3), (4) by Algebra, Lemma 10.82.4. Part (5) is a special case of (4). Assume (5). Let $I = (a_1, \ldots , a_ n)$ be a proper finitely generated ideal of $R$. As $I \not= R$ we see that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots , xa_ n)$ is not a split injection. Hence it has a nonzero kernel and we conclude that $\text{Ann}_ R(I) \not= 0$. Thus (1) holds. $\square$


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Your email address will not be published.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