Lemma 10.77.6. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $\overline{P}$ be a finite projective $R/I$-module. Then there exists a finite projective $R$-module $P$ such that $P/IP \cong \overline{P}$.

**Proof.**
Recall that $\overline{P}$ is a direct summand of a free $R/I$-module $\bigoplus _{\alpha \in A} R/I$ by Lemma 10.77.2. As $\overline{P}$ is finite, it follows that $\overline{P}$ is contained in $\bigoplus _{\alpha \in A'} R/I$ for some $A' \subset A$ finite. Hence we may assume we have a direct sum decomposition $(R/I)^{\oplus n} = \overline{P} \oplus \overline{K}$ for some $n$ and some $R/I$-module $\overline{K}$. Choose a lift $p \in \text{Mat}(n \times n, R)$ of the projector $\overline{p}$ associated to the direct summand $\overline{P}$ of $(R/I)^{\oplus n}$. Note that $p^2 - p \in \text{Mat}(n \times n, R)$ is nilpotent: as $I$ is locally nilpotent and the matrix entries $c_{ij}$ of $p^2 - p$ are in $I$ we have $c_{ij}^ t = 0$ for some $t > 0$ and then $(p^2 - p)^{tn^2} = 0$ (by looking at the matrix coefficients). Hence by Lemma 10.32.7 we can modify our choice of $p$ and assume that $p$ is a projector. Set $P = \mathop{\mathrm{Im}}(p)$.
$\square$

## 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Your email address will not be published.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)

There are also: