[Theorem 1, Serre-projective]

Theorem 15.127.5. Let $R$ be a ring whose max spectrum $\Omega \subset \mathop{\mathrm{Spec}}(R)$ is a Noetherian topological space of dimension $d < \infty$. Let $M$ be a finitely presented $R$-module such that for all $\mathfrak m \in \Omega$ the $R_\mathfrak m$-module $M_\mathfrak m$ has a free direct summand of rank $> d$. Then $M \cong R \oplus M'$.

Proof. For $\mathfrak m \in \Omega$ suppose that $R_\mathfrak m^{\oplus r}$ is a direct summand of $M_\mathfrak m$. Then by Algebra, Lemmas 10.9.9 and 10.127.6 we see that $R_ f^{\oplus r}$ is a direct summand of $M_ f$ for some $f \in R$, $f \not\in \mathfrak m$. Hence the assumption means that $\dim V(x) > d$ for all $x \in \Omega$ where $V(x)$ is as in Lemma 15.127.2. By Proposition 15.127.4 applied with $F = \emptyset$, $h = 0$ and no $s_ i$, $n = 0$ and no $x_ i, v_ i$, and $k = d + 1$ we find an $s \in M$ and $F' \subset \Omega$ such that every irreducible component of $F'$ has codimension $\geq d + 1$ and $Z(s) \subset F'$. Since $d = \dim (\Omega )$ this forces $F' = \emptyset$. Hence $s : R \to M$ is the inclusion of a direct summand at all maximal ideals. It follows that $s$ is universally injective, see Algebra, Lemma 10.82.12. Then $s$ is split injective by Algebra, Lemma 10.82.4. $\square$

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