The Stacks project

Lemma 33.38.8. Let $X$ be a scheme. Let $Z_1, \ldots , Z_ n \subset X$ be closed subschemes. Let $\mathcal{L}_ i$ be an invertible sheaf on $Z_ i$. Assume that

  1. $X$ is reduced,

  2. $X = \bigcup Z_ i$ set theoretically, and

  3. $Z_ i \cap Z_ j$ is a discrete topological space for $i \not= j$.

Then there exists an invertible sheaf $\mathcal{L}$ on $X$ whose restriction to $Z_ i$ is $\mathcal{L}_ i$. Moreover, if we are given sections $s_ i \in \Gamma (Z_ i, \mathcal{L}_ i)$ which are nonvanishing at the points of $Z_ i \cap Z_ j$, then we can choose $\mathcal{L}$ such that there exists a $s \in \Gamma (X, \mathcal{L})$ with $s|_{Z_ i} = s_ i$ for all $i$.

Proof. The existence of $\mathcal{L}$ can be deduced from Lemma 33.38.7 but we will also give a direct proof and we will use the direct proof to see the statement about sections is true. Set $T = \bigcup _{i \not= j} Z_ i \cap Z_ j$. As $X$ is reduced we have

\[ X \setminus T = \bigcup (Z_ i \setminus T) \]

as schemes. Assumption (3) implies $T$ is a discrete subset of $X$. Thus for each $t \in T$ we can find an open $U_ t \subset X$ with $t \in U_ t$ but $t' \not\in U_ t$ for $t' \in T$, $t' \not= t$. By shrinking $U_ t$ if necessary, we may assume that there exist isomorphisms $\varphi _{t, i} : \mathcal{L}_ i|_{U_ t \cap Z_ i} \to \mathcal{O}_{U_ t \cap Z_ i}$. Furthermore, for each $i$ choose an open covering

\[ Z_ i \setminus T = \bigcup \nolimits _ j U_{ij} \]

such that there exist isomorphisms $\varphi _{i, j} : \mathcal{L}_ i|_{U_{ij}} \cong \mathcal{O}_{U_{ij}}$. Observe that

\[ \mathcal{U} : X = \bigcup U_ t \cup \bigcup U_{ij} \]

is an open covering of $X$. We claim that we can use the isomorphisms $\varphi _{t, i}$ and $\varphi _{i, j}$ to define a $2$-cocycle with values in $\mathcal{O}_ X^*$ for this covering that defines $\mathcal{L}$ as in the statement of the lemma.

Namely, if $i \not= i'$, then $U_{i, j} \cap U_{i', j'} = \emptyset $ and there is nothing to do. For $U_{i, j} \cap U_{i, j'}$ we have $\mathcal{O}_ X(U_{i, j} \cap U_{i, j'}) = \mathcal{O}_{Z_ i}(U_{i, j} \cap U_{i, j'})$ by the first remark of the proof. Thus the transition function for $\mathcal{L}_ i$ (more precisely $\varphi _{i, j} \circ \varphi _{i, j'}^{-1}$) defines the value of our cocycle on this intersection. For $U_ t \cap U_{i, j}$ we can do the same thing. Finally, for $t \not= t'$ we have

\[ U_ t \cap U_{t'} = \coprod (U_ t \cap U_{t'}) \cap Z_ i \]

and moreover the intersection $U_ t \cap U_{t'} \cap Z_ i$ is contained in $Z_ i \setminus T$. Hence by the same reasoning as before we see that

\[ \mathcal{O}_ X(U_ t \cap U_{t'}) = \prod \mathcal{O}_{Z_ i}(U_ t \cap U_{t'} \cap Z_ i) \]

and we can use the transition functions for $\mathcal{L}_ i$ (more precisely $\varphi _{t, i} \circ \varphi _{t', i}^{-1}$) to define the value of our cocycle on $U_ t \cap U_{t'}$. This finishes the proof of existence of $\mathcal{L}$.

Given sections $s_ i$ as in the last assertion of the lemma, in the argument above, we choose $U_ t$ such that $s_ i|_{U_ t \cap Z_ i}$ is nonvanishing and we choose $\varphi _{t, i}$ such that $\varphi _{t, i}(s_ i|_{U_ t \cap Z_ i}) = 1$. Then using $1$ over $U_ t$ and $\varphi _{i, j}(s_ i|_{U_{i, j}})$ over $U_{i, j}$ will define a section of $\mathcal{L}$ which restricts to $s_ i$ over $Z_ i$. $\square$

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