The Stacks project

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

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

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

is the desired splitting. $\square$