The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 25.4.6. Let $X$, $Y$ be locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals locally generated by sections. Let $i : Z \to X$ be the associated closed subspace. A morphism $f : Y \to X$ factors through $Z$ if and only if the map $f^*\mathcal{I} \to f^*\mathcal{O}_ X = \mathcal{O}_ Y$ is zero. If this is the case the morphism $g : Y \to Z$ such that $f = i \circ g$ is unique.

Proof. Clearly if $f$ factors as $Y \to Z \to X$ then the map $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Conversely suppose that $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Pick any $y \in Y$, and consider the ring map $f^\sharp _ y : \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$. Since the composition $\mathcal{I}_{f(y)} \to \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$ is zero by assumption and since $f^\sharp _ y(1) = 1$ we see that $1 \not\in \mathcal{I}_{f(y)}$, i.e., $\mathcal{I}_{f(y)} \not= \mathcal{O}_{X, f(y)}$. We conclude that $f(Y) \subset Z = \text{Supp}(\mathcal{O}_ X/\mathcal{I})$. Hence $f = i \circ g$ where $g : Y \to Z$ is continuous. Consider the map $f^\sharp : \mathcal{O}_ X \to f_*\mathcal{O}_ Y$. The assumption $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero implies that the composition $\mathcal{I} \to \mathcal{O}_ X \to f_*\mathcal{O}_ Y$ is zero by adjointness of $f_*$ and $f^*$. In other words, we obtain a morphism of sheaves of rings $\overline{f^\sharp } : \mathcal{O}_ X/\mathcal{I} \to f_*\mathcal{O}_ Y$. Note that $f_*\mathcal{O}_ Y = i_*g_*\mathcal{O}_ Y$ and that $\mathcal{O}_ X/\mathcal{I} = i_*\mathcal{O}_ Z$. By Sheaves, Lemma 6.32.4 we obtain a unique morphism of sheaves of rings $g^\sharp : \mathcal{O}_ Z \to g_*\mathcal{O}_ Y$ whose pushforward under $i$ is $\overline{f^\sharp }$. We omit the verification that $(g, g^\sharp )$ defines a morphism of locally ringed spaces and that $f = i \circ g$ as a morphism of locally ringed spaces. The uniqueness of $(g, g^\sharp )$ was pointed out above. $\square$


Comments (0)

There are also:

  • 8 comment(s) on Section 25.4: Closed immersions of locally ringed spaces

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01HP. Beware of the difference between the letter 'O' and the digit '0'.