The Stacks project

Remark 12.29.2. Let $\mathcal{A}$, $\mathcal{B}$, $u : \mathcal{A} \to \mathcal{B}$ and $v : \mathcal{B} \to \mathcal{A}$ be as in Lemma 12.29.1. In the presence of assumption (1) assumption (2) is equivalent to requiring that $v$ is exact. Moreover, condition (2) is necessary. Here is an example. Let $A \to B$ be a ring map. Let $u : \text{Mod}_ B \to \text{Mod}_ A$ be $u(N) = N_ A$ and let $v : \text{Mod}_ A \to \text{Mod}_ B$ be $v(M) = M \otimes _ A B$. Then $u$ is right adjoint to $v$, and $u$ is exact and $v$ is right exact, but $v$ does not transform injective maps into injective maps in general (i.e., $v$ is not left exact). Moreover, it is not the case that $u$ transforms injective $B$-modules into injective $A$-modules. For example, if $A = \mathbf{Z}$ and $B = \mathbf{Z}/p\mathbf{Z}$, then the injective $B$-module $\mathbf{Z}/p\mathbf{Z}$ is not an injective $\mathbf{Z}$-module. In fact, the lemma applies to this example if and only if the ring map $A \to B$ is flat.


Comments (0)

There are also:

  • 1 comment(s) on Section 12.29: Injectives and adjoint functors

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Your email address will not be published.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 03B8. Beware of the difference between the letter 'O' and the digit '0'.