Lemma 15.15.7. Let $R$ be a ring. Suppose that $\varphi : R^ n \to R^ n$ be an injective map of finite free modules of the same rank. Then $\mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Coker}}(\varphi ), R) = 0$.

**Proof.**
Let $\varphi ^ t : R^ n \to R^ n$ be the transpose of $\varphi $. The lemma claims that $\varphi ^ t$ is injective. With notation as in Lemma 15.15.6 we see that the rank of $\varphi ^ t$ is $n$ and that $I(\varphi ) = I(\varphi ^ t)$. Thus we conclude by the equivalence of (1) and (2) of the lemma.
$\square$

## 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Your email address will not be published.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.

## Comments (0)