Section 26.4 (01HJ): Closed immersions of locally ringed spaces

26.4 Closed immersions of locally ringed spaces

We follow our conventions introduced in Modules, Definition 17.13.1.

Definition 26.4.1. Let $i : Z \to X$ be a morphism of locally ringed spaces. We say that $i$ is a closed immersion if:

The map $i$ is a homeomorphism of $Z$ onto a closed subset of $X$.
The map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective; let $\mathcal{I}$ denote the kernel.
The $\mathcal{O}_ X$-module $\mathcal{I}$ is locally generated by sections.

Lemma 26.4.2. Let $f : Z \to X$ be a morphism of locally ringed spaces. In order for $f$ to be a closed immersion it suffices that there exists an open covering $X = \bigcup U_ i$ such that each $f : f^{-1}U_ i \to U_ i$ is a closed immersion.

Proof. Omitted. $\square$

Example 26.4.3. Let $X$ be a locally ringed space. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals which is locally generated by sections as a sheaf of $\mathcal{O}_ X$-modules. Let $Z$ be the support of the sheaf of rings $\mathcal{O}_ X/\mathcal{I}$. This is a closed subset of $X$, by Modules, Lemma 17.5.3. Denote $i : Z \to X$ the inclusion map. By Modules, Lemma 17.6.1 there is a unique sheaf of rings $\mathcal{O}_ Z$ on $Z$ with $i_*\mathcal{O}_ Z = \mathcal{O}_ X/\mathcal{I}$. For any $z \in Z$ the stalk $\mathcal{O}_{Z, z}$ is equal to a quotient $\mathcal{O}_{X, i(z)}/\mathcal{I}_{i(z)}$ of a local ring and nonzero, hence a local ring. Thus $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ is a closed immersion of locally ringed spaces.

Definition 26.4.4. Let $X$ be a locally ringed space. Let $\mathcal{I}$ be a sheaf of ideals on $X$ which is locally generated by sections. The locally ringed space $(Z, \mathcal{O}_ Z)$ of Example 26.4.3 above is the closed subspace of $X$ associated to the sheaf of ideals $\mathcal{I}$.

Lemma 26.4.5. Let $f : X \to Y$ be a closed immersion of locally ringed spaces. Let $\mathcal{I}$ be the kernel of the map $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$. Let $i : Z \to Y$ be the closed subspace of $Y$ associated to $\mathcal{I}$. There is a unique isomorphism $f' : X \cong Z$ of locally ringed spaces such that $f = i \circ f'$.

Proof. Omitted. $\square$

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. $\square$

Lemma 26.4.7. Let $f : X \to Y$ be a morphism of locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals which is locally generated by sections. Let $i : Z \to Y$ be the closed subspace associated to the sheaf of ideals $\mathcal{I}$. Let $\mathcal{J}$ be the image of the map $f^*\mathcal{I} \to f^*\mathcal{O}_ Y = \mathcal{O}_ X$. Then this ideal is locally generated by sections. Moreover, let $i' : Z' \to X$ be the associated closed subspace of $X$. There exists a unique morphism of locally ringed spaces $f' : Z' \to Z$ such that the following diagram is a commutative square of locally ringed spaces

\[ \xymatrix{ Z' \ar[d]_{f'} \ar[r]_{i'} & X \ar[d]^ f \\ Z \ar[r]^{i} & Y } \]

Moreover, this diagram is a fibre square in the category of locally ringed spaces.

Proof. The ideal $\mathcal{J}$ is locally generated by sections by Modules, Lemma 17.8.2. The rest of the lemma follows from the characterization, in Lemma 26.4.6 above, of what it means for a morphism to factor through a closed subspace. $\square$

Comments (18)

Comment #404 by Keenan on December 23, 2013 at 20:14

In the proof of 01HP, $\mathscr{I}_y$ should be $\mathscr{I}_{f(y)}$ in a three places.

Comment #405 by Johan on December 25, 2013 at 01:59

Fixed here. Thanks!

Comment #434 by Leeroy on January 30, 2014 at 21:48

In example 25.4.3 it should be $\mathcal{O}_{X,z}/\mathcal{I}_z$ (or $\mathcal{O}_{X,i(z)}/\mathcal{I}_{i(z)}$ ) and not $\mathcal{O}_{X,x}/\mathcal{I}_x$ .

I was wondering, are you interested in people writing down those omitted simple proofs or do you not write them down just to make the thing more readable ? Obviously you would wish for people to contribute more meaningful stuff but since I'm learning I can't contribute anything interesting.

Comment #436 by Johan on January 31, 2014 at 01:25

Thanks for typo! Fixed here.

Yes, the goal is to fill in all the omitted proofs and any contribution like that is welcomed. If it becomes unreadable because of this (but I personally have never found this to be a problem), then we can use a technological solution to hide proofs (behind a link or something). Most of the omitted proofs should be easier or equivalent to things that are being proved in the same section or chapter. If you find that this is not the case, then please leave a comment saying so.

Comment #2582 by Dario Weißmann on May 28, 2017 at 18:37

Typo in Lemma 25.4.6: Let $X,Y$ be $a$ locally ringed spaces.

Comment #2617 by Johan on July 07, 2017 at 11:53

Thanks for the typo. Fixed here.

Comment #3779 by qiao on November 06, 2018 at 16:45

Typo in Lemma 01HP (Lemma 25.4.6) $f^*\mathcal{I} \to f^*\mathcal{O}_ X = \mathcal{O}_ Y$ perhaps not "=" here

Comment #3780 by qiao on November 06, 2018 at 16:51

Typo in Lemma 01HP (Lemma 25.4.6) perhaps not "=" here $\begin{equation} f^* \mathcal{I} \to f^* \mathcal{O}_ X = \mathcal{O}_ Y \end{equation}$

Comment #3909 by Johan on January 22, 2019 at 11:09

@#3779 and #3780: I am not going to change that for now. If more people complain then I will.

Comment #5037 by Erica on April 18, 2020 at 23:14

In lemma 25.4.6, maybe the map should be $f^{*}\mathcal{I}\rightarrow f^{*}\mathcal{O}_{X}\rightarrow \mathcal{O}_{Y}$ ? This typo is not so important

Comment #5172 by Andrés on June 09, 2020 at 11:53

Why do we ask for the ideal sheaf to be locally generated by sections? If you omit this from the definition, then you'd still have a correspondance between ideal sheafs and closed immersions. In the case of schemes, then one gets a correspondance between coherent ideal sheafs and closed immersions of schemes.

Comment #5173 by Johan on June 09, 2020 at 12:57

Please see the discussion in Section 17.13.

Comment #5564 by minsom on October 31, 2020 at 13:40

I think in the proot Lemma 26.4.6 (Lemma 01HP) " ~ since $f_y^{\#}(1)=1$ we see that $1∉\mathcal{I}_{f(y)}$ ~ " is not true because if $O_{Y,y} = 0$ then, it can be $1 \in \mathcal{I}_{f(y)}$ . How can we fix this?

Comment #5565 by minsom on October 31, 2020 at 21:13

Sorry, I see know the stalk ring is a local ring and the local ring cannot be zero ring.

Comment #7079 by Tim on February 25, 2022 at 05:40

In the lemma 26.4.6, by adjointness, why don't we use the condition $I\rightarrow f_*O_Y$ is zero instead which seems to be easier to work.

Comment #7164 by Hussein on April 04, 2022 at 07:05

Typo in Lemmas 01HP & 01HQ we have $f^*O_Y=O_X$ ? I believe it should be $f^{*}\mathcal{I}\rightarrow f^{*}\mathcal{O}_{X}\rightarrow \mathcal{O}_{Y}$

Comment #7256 by Johan on May 01, 2022 at 18:30

@#7079: of course since it is equivalent you may do so, but I think the lemma is used in the form it is stated so I do not want to change the statement. @#7164: for any map of ringed spaces $f : Y \to X$ we have $f^*\mathcal{O}_X = \mathcal{O}_Y$ . Note that in Lemma 26.4.7 the roles of $X$ and $Y$ are swapped.

Comment #8456 by Elías Guisado on June 04, 2023 at 14:57

I think the following folklore (that #5172 hints at) could be an interesting remark to add to this section: Given a locally ringed space $X$ , there is a one-to-one correspondence between ideal sheaves over $X$ locally generated by sections and closed immersions of l.r.s. into $X$ , modulo isomorphism over $X$ . The map to the right sends $\mathcal{I}\subset\mathcal{O}_X$ to the closed subspace of $X$ associated to this sheaf of ideals (Definition 26.4.4), whereas the map to the left sends a closed immersion $f:Z\to X$ to $\operatorname{Ker}(\mathcal{O}_X\to f_*\mathcal{O}_Z)$ . The composite "first to the right, then to the left" is the identity by 26.4.3, whereas the composite "first to the left, then to the right" is the identity by 26.4.5. Moreover, this one-to-one correspondence promotes to a contravariant isomorphism of posets when the ideal sheaves are ordered by inclusion and the order in the closed subspaces of $X$ is given by the existence of a map $Z\to Z'$ over $X$ (by monicity of closed immersions, there is at most one such map). On the one hand, if we have ideal sheaves $\mathcal{I}'\subset\mathcal{I}\subset\mathcal{O}_X$ , then we get a map $Z\to Z'$ over $X$ by 26.4.6, where $Z$ and $Z'$ are the closed subspaces associated to $\mathcal{I}$ and $\mathcal{I}'$ , respectively. On the other hand, if we have closed subspaces $f:Z\to X$ and $f':Z'\to X$ and a map $g:Z\to Z'$ over $X$ , then $\mathcal{I}$ equals the kernel of $\mathcal{O}_X\to f_*\mathcal{O}_Z$ , i.e., of $\mathcal{O}_X\to f'_*\mathcal{O}_{Z'}\to f_*'g_*\mathcal{O}_X$ . But this contains $\mathcal{I}'$ .

Also, at the beginning of the section, maybe one could inform the reader that the condition (3) of Definition 26.4.1 will not be ever used throughout the section? Everything stated in this section is still true if we were to drop (3) from this definition, and the proofs don't change (condition (3) is never invoked). I am aware of the discussion in Modules, Section 17.13, and I don't disagree with it. I just think it would be informative to have some note in this section regarding this issue.

The Stacks project

26.4 Closed immersions of locally ringed spaces

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