The Stacks project

Lemma 26.4.6. 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 (2)

Comment #8525 by on

I think it could be interesting to expand the statement in the following way:

Let , be locally ringed spaces. Let be a sheaf of ideals locally generated by sections. Let be the associated closed subspace. Let be a morphism of locally ringed spaces. The following are equivalent: 1. The morphism factors through , 2. The map is zero, 3. is contained in the kernel of , and 4. is contained in the kernel of .

Moreover, if this is the case, then the morphism such that is unique.

The already existing proof does 12 and, implicitly, 23. The proof of 34 is easy by the adjointness (the same strategy proves 43), so it is left to show 42, and this is straightforward.

Comment #8529 by on

OK, I think we can leave it to the reader to see the equivalences of 2, 3, and 4. Presumably this is already silently used in some places.

There are also:

  • 18 comment(s) on Section 26.4: Closed immersions of locally ringed spaces

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