Commutative case of [Theorem 3.1, Bass]

Theorem 15.128.6. Let $R$ be a ring with Jacobson radical $J$ such that $R/J$ is Noetherian. Let $P$ be a countably generated projective $R$-module such that $P_\mathfrak m$ has infinite rank for all maximal ideals $\mathfrak m$ of $R$. Then $P$ is free.

Proof. We first prove that $P$ is a countable direct sum of finite stably free modules. Let $x_1, x_2, \ldots$ be a countable set of generators for $P$. We inductively construct finite stably free direct summands $F_1, F_2, \ldots$ of $P$ such that for all $n$ we have that $F_1 \oplus \ldots \oplus F_ n$ is a direct summand of $P$ which contains $x_1, \ldots , x_ n$. Namely, given $F_1, \ldots , F_ n$ with the desired properties, write

$P = F_1 \oplus \ldots \oplus F_ n \oplus P'$

and let $s \in P'$ be the image of $x_{n + 1}$. By Lemma 15.128.5 we can find a finite stably free direct summand $F_{n + 1} \subset P'$ containing $s$. Then $P = \bigoplus _{i = 1}^{\infty } F_ i$.

Assume that $P$ is an infinite direct sum $P = \bigoplus _{i = 1}^{\infty } F_ i$ of nonzero finite stably free modules. The stable freeness of the modules $F_ i$ will be used in the following manner: the rank of each $F_ i$ is constant (and positive). Hence we see that $P_\mathfrak m$ is free of countably infinite rank for each maximal ideal $\mathfrak m$ of $R$. By Lemma 15.128.4 applied with $s = 0$ and $M = P$, we can find a $t_1 \in P$ such that $Rt_1$ is a free direct summand of $P$. Then $t_1$ is contained in $F_1 \oplus \ldots \oplus F_{n_1}$ for some $n_1 > n_0 = 0$. The same reasoning applied to $\bigoplus _{n > n_1} F_ n$ produces an $n_1 < n_2$ and $t_2 \in F_{n_1 + 1} \oplus \ldots \oplus F_{n_2}$ which generates a free direct summand. Continuing in this fashion we obtain a free direct summand

$\bigoplus \nolimits _{i \geq 1} t_ i : \bigoplus \nolimits _{i \geq 1} R \longrightarrow \bigoplus \nolimits _{i \geq 1} \bigoplus \nolimits _{n_ i \geq n > n_{i - 1}} F_ n = P$

of infinite rank. Thus we see that $P \cong Q \oplus F$ for some free $R$-module $F$ of countable rank. Since $Q$ is countably generated it follows that $Q \oplus Q' \cong F$ for some module $Q'$. Then the Eilenberg swindle (Lemma 15.128.1) implies that $Q \oplus F \cong F$ and $P$ is free. $\square$

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