## 15.128 Big projective modules are free

In this section we discuss one of the results of [Bass]; we suggest the reader look at the original paper. Our argument will use the slightly simplified proof given in the papers [Akasaki] and [Hinohara].

Proof.

$F \cong F \oplus F \oplus \ldots \cong P \oplus Q \oplus P \oplus Q \oplus \ldots \cong P \oplus F \oplus F \oplus \ldots \cong P \oplus F$
$\square$

Lemma 15.128.2. Let $R$ be a ring. Let $P$ be a projective module. There exists a free module $F$ such that $P \oplus F$ is free.

Proof. Since $P$ is projective we see that $F_0 = P \oplus Q$ is a free module for some module $Q$. Set $F = \bigoplus _{n \geq 1} F_0$. Then $P \oplus F \cong F$ by Lemma 15.128.1. $\square$

Lemma 15.128.3. Let $R$ be a ring. Let $P$ be a projective module. Let $s \in P$. There exists a finite free module $F$ and a finite free direct summand $K \subset F \oplus P$ with $(0, s) \in K$.

Proof. By Lemma 15.128.2 we can find a (possibly infinite) free module $F$ such that $F \oplus P$ is free. Then of course $(0, s)$ is contained in a finite free direct summand $K \subset F \oplus P$. In turn $K$ is contained in $F' \oplus P$ where $F' \subset F$ is a finite free direct summand. $\square$

Lemma 15.128.4. 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 $s \in P$ and $M \subset P$ such that $Rs + M = P$. Then we can find $m \in M$ such that $R(s + m)$ is a free direct summand of $P$.

Proof. The statement makes sense as $P_\mathfrak m$ is free by Algebra, Theorem 10.85.4.

Denote $M' \subset P/JP$ the image of $M$ and $s' \in P/JP$ the image of $s$. Observe that $R/J s' + M' = P/JP$. Suppose we can find $m' \in M'$ such that $R/J(s' + m')$ is a free direct summand of $M'$. Choose $\varphi ' : P/JP \to R/J$ which gives a splitting, i.e., we have $\varphi '(s' + m') = 1$ in $R/J$. Then since $P$ is a projective $R$-module we can find a lift $\varphi : P \to R$ of $\varphi '$. Choose $m \in M$ mapping to $m'$. Then $\varphi (s + m) \in R$ is congruent to $1$ modulo $J$ and hence a unit in $R$ (Algebra, Lemma 10.19.1). Whence $R(s + m)$ is a free direct summand of $P$. This reduces us to the case discussed in the next paragraph.

Assume $R$ is Noetherian. Let $m \in M$ be an element and let $\varphi _1, \ldots , \varphi _ n : P \to R$ be $R$-linear maps. Denote

$Z(s + m, \varphi _1, \ldots , \varphi _ n) \subset \mathop{\mathrm{Spec}}(R)$

the vanishing locus of $\varphi _1(s + m), \ldots , \varphi _ n(s + m) \in R$.

Suppose $\mathfrak m$ is a maximal ideal of $R$ and $\mathfrak m \in Z(s, \varphi _1, \ldots , \varphi _ n)$. Set $K = M \cap \bigcap \mathop{\mathrm{Ker}}(\varphi _ i)$. We claim the image of

$K/\mathfrak mK \to P/\mathfrak m P$

has infinite dimension. Namely, the quotient $P/K$ is a finite $R$-module as it is isomorphic to a submodule of $P/M \oplus R^{\oplus n}$. Thus we see that the kernel of the displayed arrow is a quotient of $\text{Tor}_1^ R(P/K, \kappa (\mathfrak m))$ which is finite by Algebra, Lemma 10.75.7. Combined with the fact that $P/\mathfrak mP$ has infinite dimension we obtain our claim. Thus we can find a $t \in K$ which maps to a nonzero element $\overline{t}$ of the vector space $P/\mathfrak mP$. By linear algebra, we find an $R$-linear map $\overline{\varphi } : P \to \kappa (\mathfrak m)$ such that $\overline{\varphi }(\overline{t}) = 1$. Since $P$ is projective, we can find an $R$-linear map $\varphi : P \to R$ lifting $\overline{\varphi }$. Then we see that the vanishing locus $Z(s + m + t, \varphi _1, \ldots , \varphi _ n, \varphi )$ is contained in $Z(s + m, \varphi _1, \ldots , \varphi _ n)$ but does not contain $\mathfrak m$, i.e., it is strictly smaller than $Z(s + m, \varphi _1, \ldots , \varphi _ n)$.

Since $\mathop{\mathrm{Spec}}(R)$ is a Noetherian topological space, we see from the arguments above that we may find $m \in M$ and $\varphi _1, \ldots , \varphi _ n : P \to R$ such that the closed subset $Z(s + m, \varphi _1, \ldots , \varphi _ n)$ does not contain any closed points of $\mathop{\mathrm{Spec}}(R)$. Hence $Z(s + m, \varphi _1, \ldots , \varphi _ n) = \emptyset$. Hence we can find $r_1, \ldots , r_ n \in R$ such that $\sum r_ i\varphi _ i(s + m) = 1$. Hence

$R \xrightarrow {s + m} P \xrightarrow {\sum r_ i \varphi _ i} R$

is the desired splitting. $\square$

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 $\mathop{\mathrm{Ker}}(\pi ')$. $\square$

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. $\square$

[1] 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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