Lemma 10.76.6. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$I$ is nilpotent,

$M/IM$ is a projective $R/I$-module,

$M$ is a flat $R$-module.

Then $M$ is a projective $R$-module.

Lemma 10.76.6. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$I$ is nilpotent,

$M/IM$ is a projective $R/I$-module,

$M$ is a flat $R$-module.

Then $M$ is a projective $R$-module.

**Proof.**
By Lemma 10.76.4 we can find a projective $R$-module $P$ and an isomorphism $P/IP \to M/IM$. We are going to show that $M$ is isomorphic to $P$ which will finish the proof. Because $P$ is projective we can lift the map $P \to P/IP \to M/IM$ to an $R$-module map $P \to M$ which is an isomorphism modulo $I$. By Nakayama's Lemma 10.19.1 the map $P \to M$ is surjective. It remains to show that $P \to M$ is injective. Since $I^ n = 0$ for some $n$, we can use the filtrations

\begin{align*} 0 = I^ nM \subset I^{n - 1}M \subset \ldots \subset IM \subset M \\ 0 = I^ nP \subset I^{n - 1}P \subset \ldots \subset IP \subset P \end{align*}

to see that it suffices to show that the induced maps $I^ aP/I^{a + 1}P \to I^ aM/I^{a + 1}M$ are injective. Since both $P$ and $M$ are flat $R$-modules we can identify this with the map

\[ I^ a/I^{a + 1} \otimes _{R/I} P/IP \longrightarrow I^ a/I^{a + 1} \otimes _{R/I} M/IM \]

induced by $P \to M$. Since we chose $P \to M$ such that the induced map $P/IP \to M/IM$ is an isomorphism, we win. $\square$

Your email address will not be published.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: