The Stacks project

Proof. Define $B(x) \subset M(x)$ as the perpendicular of the image of the map

and set $V(x) = M(x)/B(x)$. Then any $R$-linear map $\varphi : M \to R$ induces a map $\overline{\varphi } : V(x) \to \kappa (x)$ and conversely any $\kappa (x)$-linear map $\lambda : V(x) \to \kappa (x)$ is equal to $\overline{\varphi }$ for some $\varphi $. Let $s_1, \ldots , s_ r \in M$.

Suppose $s_1, \ldots , s_ r$ map to linearly independent elements of $V(x)$. Then we can find $\varphi _1, \ldots , \varphi _ r \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ such that $\varphi _ i(s_ j)$ maps to $\delta _{ij}$¹ in $\kappa (x)$. Hence the matrix of the composition

has a determinant $f \in R$ which maps to $1$ in $\kappa (x)$ Clearly, this implies that $s_1, \ldots , s_ r : R^{\oplus r} \to M$ is the inclusion of a direct summand after inverting $f$.

Conversely, suppose that we have an $f \in R$, $f \not\in x$ such that $s_1, \ldots , s_ r : R^{\oplus r} \to M$ is the inclusion of a direct summand after inverting $f$. Hence we can find $R_ f$-linear maps $\varphi _ i : M_ f \to R_ f$ such that $\varphi _ i(s_ j) = \delta _{ij} \in R_ f$. Since $\mathop{\mathrm{Hom}}\nolimits _ R(M, R)_ f = \mathop{\mathrm{Hom}}\nolimits _{R_ f}(M_ f, R_ f)$ by Algebra, Lemma 10.10.2 we conclude that we can find $n \geq 0$ and $\varphi '_ i \in \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ such that $\varphi '_ i(s_ j) = f^ n\delta _{ij} \in R$. It follows that $s_1, \ldots , s_ r$ map to linearly independent elements of $V(x)$ as $\overline{\varphi }'_ i(s_ j) = f^ n\delta _{ij}$. $\square$

Proof. Since $x_ i$ is a maximal ideal of $R$ we may use Algebra, Lemma 10.15.4 to see that $M(x_1) \oplus \ldots \oplus M(x_ n)$ is a quotient of $M$. $\square$

Proof. We note that codimension was defined in Topology, Section 5.11 and that we will use some results on Noetherian topological spaces contained in Topology, Section 5.9.

The proof is by induction on $k$. If $k = 0$, then we choose $s \in M$ as in Lemma 15.127.3 and we choose $F' = \Omega $.

Assume $k > 0$. By our induction hypothesis we may choose $u \in M$ and $G \subset \Omega $ closed satisfying (a), (b), (c) for $s_1, \ldots , s_ h$, $F$, $x_1, \ldots , x_ n$, $v_1, \ldots , v_ n$, and $k - 1$.

Let $G = G_1 \cup \ldots \cup G_ m$ be the decomposition of $G$ into its irreducible components. If $G_ j \subset F$, then we can remove it from the list. Thus we may assume $G_ j$ is not contained in $F$ for $j = 1, \ldots , m$. For $j = 1, \ldots , m$ choose $y_ j \in G_ j$ with $y_ j \not\in F$ and $y_ j \not\in G_{j'}$ for $j' \not= j$. This is possible as there are no inclusions among the irreducible components of $G$. Choose $w_ j \in V(y_ j)$ not contained in the span of the images of $s_1(y_ j), \ldots , s_ h(y_ j)$; this is possible because $h + k \leq \dim V(y_ j)$ and $k > 0$.

Apply the induction hypothesis to the $h + 1$ sections $s_1, \ldots , s_ h, u$, the closed set $F \cup G$, the points $x_1, \ldots , x_ n, y_1, \ldots , y_ m \in F \cup G$, the elements $0 \in V(x_ i)$ and $w_ j \in V(y_ j)$, and the integer $k - 1$. Note that we have increased $h$ by $1$ and decreased $k$ by $1$ hence the assumption $(*)$ of the proposition remains valid. This produces $t \in M$ and $H \subset \Omega $ closed satisfying (a), (b), (c) for $s_1, \ldots , s_ h, u$, $F \cup G$, $x_1, \ldots , x_ n, y_1, \ldots , y_ m$, $0, \ldots , 0, w_1, \ldots , w_ m$, and $k - 1$.

Let $H_1, \ldots , H_ p \subset H$ be the irreducible components of $H$ which are not contained in $F \cup G$. As before pick $z_ l \in H_ l$, $z_ l \not\in F \cup G$ and $z_ l \not\in H_{l'}$ for $l' \not= l$. Using Algebra, Lemma 10.15.4 we may choose $f \in R$ such that $f(y_ j) = 1$, $j = 1, \ldots , m$ and $f(z_ l) = 0$, $l = 1, \ldots , p$. Claim: the element $s = u + f t$ works.

First, the value $s(x_ i)$ agrees with $u(x_ i)$ because $t(x_ i) = 0$ and hence we see that $s(x_ i)$ maps to $v_ i$. This proves (a). To finish the proof it suffices to show that every irreducible component $Z$ of $Z(s_1, \ldots , s_ h, s)$ not contained in $F$ has codimension $\geq k$ in $\Omega $. Namely, then we can set $F'$ equal to the union of these and we get (b) and (c). We can see that irreducible components $Z$ of $Z(s_1, \ldots , s_ h, s)$ of codimension $\leq k - 1$ do not exist as follows:

1. Observe that $Z(s_1, \ldots , s_ h, s) \subset Z(s_1, \ldots , s_ h, u, t) = F \cup H$ as $s = u + ft$. Hence $Z \subset H$.

2. The irreducible components of $H$ have codimension $\geq k - 1$. Hence $Z$ is equal to an irreducible component of $H$ as $Z$ has codimension $\leq k - 1$. Hence $Z = H_ l$ for some $l \in \{ 1, \ldots , p\} $ or $Z = G_ j$ for some $j \in \{ 1, \ldots m\} $.

3. But $Z = G_ j$ is impossible as $s_1(y_ j), \ldots , s_ h(y_ j)$ map to linearly independent elements of $V(y_ j)$ and $s(y_ j) = u(y_ j) + f(y_ j) t(y_ j) = u(y_ j) + t(y_ j)$ maps to an element of the form

   \[ \text{linear combination images of }s_ i(y_ j) + w_ j \]

   which is linearly independent of the images of $s_1(y_ j), \ldots , s_ h(y_ j)$ in $V(y_ j)$ by our choice of $w_ j$.

4. Also $Z = Z_ l$ is impossible. Namely, again $s_1(z_ l), \ldots , s_ h(z_ l)$ map to linearly independent elements of $V(z_ l)$ and $s(z_ l) = u(z_ l) + f(z_ l) t(z_ l) = u(z_ l)$ maps to an element of $V(z_ l)$ linearly independent of those as $z_ l \not\in F \cup G$.

This finishes the proof. $\square$

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. $\square$