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 $M \cong R \oplus M'$.
[Theorem 1, Serre-projective]
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.
$\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)