Lemma 15.128.5. Let $R$ be a ring with Jacobson radical $J$ such that $R/J$ is Noetherian. Let $P$ be a projective $R$-module such that $P_\mathfrak m$ has infinite rank for all maximal ideals $\mathfrak m$ of $R$. Let $s \in P$. Then we can find a finite stably free direct summand $M \subset P$ such that $s \in M$.

Proof. By Lemma 15.128.3 we can find a finite free module $F$ and a finite free direct summand $K \subset F \oplus P$ such that $(0, s) \in K$. By induction on the rank of $F$ we reduce to the case discussed in the next paragraph.

Assume there exists a finite stably free direct summand $K \subset R \oplus P$ such that $(0, s) \in K$. Choose a complement $K'$ of $K$, i.e., such that $R \oplus P = K \oplus K'$. The projection $\pi : R \oplus P \to K'$ is surjectve, hence by Lemma 15.128.4 we find a $p \in P$ such that $\pi (1, p) \in K'$ generates a free direct summand. Accordingly we write $K' = R\pi (1, p) \oplus K''$. We see that

$R \oplus P = K \oplus K' = K \oplus R\pi (1, p) \oplus K''$

The projection $\pi ' : P \to K''$ is surjective1 and hence split (as $K''$ is projective). Thus $\mathop{\mathrm{Ker}}(\pi ') \subset P$ is a direct summand containing $s$. Finally, by construction we have an isomorphism

$R \oplus \mathop{\mathrm{Ker}}(\pi ') \cong K \oplus R\pi (1, p)$

and hence since $K$ is finite and stably free, so is $\mathop{\mathrm{Ker}}(\pi ')$. $\square$

 Namely, if $k'' \in K''$ then $k''$ viewed as an element of $K'$ can be written as $k'' = \lambda \pi (1, 0) + \pi (0, q)$ for some $\lambda \in R$ and $q \in P$. This means $k'' = \lambda \pi (1, p) + \pi (0, q - \lambda p)$. This in turn means that $q - \lambda p$ maps to $k''$ by the composition $P \to R \oplus P \xrightarrow {\pi } K' \to K''$ since $K' \to K''$ annihilates $\pi (1, p)$.

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