The Stacks project

[Theorem 2, Serre-projective]

Proposition 15.127.4. In Situation 15.127.1 assume $\Omega $ is a Noetherian topological space. Let $s_1, \ldots , s_ h \in M$. Let $Z(s_1, \ldots , s_ h) \subset F \subset \Omega $ be closed. Let $x_1, \ldots , x_ n \in F$ be pairwise distinct. Let $v_ i \in V(x_ i)$. Let $k \geq 0$ be an integer such that

\[ (*)\quad h + k \leq \dim _{\kappa (x)} V(x)\text{ for all }x \in \Omega \]

Then there exist $s \in M$ and $F' \subset \Omega $ closed such that

  1. $s(x_ i)$ maps to $v_ i$,

  2. $Z(s_1, \ldots , s_ h, s) \subset F \cup F'$, and

  3. every irreducible component of $F'$ has codimension $\geq k$ in $\Omega $.

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. $\square$

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