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Closed subschemes and closed immersions of schemes have been causing me a lot of confusion for a while now. I have a few questions that I think might clear things up. Please assume that when I use the term "ring" that I mean "commutative ring with identity" whose morphisms take $0 \mapsto 0$ and $1 \mapsto 1$. Sorry for the long question, but I feel it is necessary to spell out the definitions I am working with rather than expect people to chase them up.

My first exposure to this was Hartshorne page 85. There he makes the following definitions:

A closed immersion is a morphism of schemes $\iota: Y \longrightarrow X$ such that $\iota$ induces a homeomorphism of sp$(Y)$ onto a closed subset of sp$(X)$, and further that the induced map of sheaves, $\iota^{\#}: \mathcal{O}_{X} \longrightarrow \iota_{*}\mathcal{O}_{Y}$ is surjective.

A closed subscheme is an equivalence class of closed immersions, where we say that $\iota: Y \longrightarrow X$ and $\iota': Y' \longrightarrow X$ are equivalent if there is an isomorphism $\psi: Y' \longrightarrow Y$ satisfying $\iota' = \iota \circ \psi$.

After having a bit of confusion with the closed subscheme part, I consulted Görtz & Wedhorn where, on page 84 (Definition 3.41) they give their own definitions. Their definition for a closed immersion is identical, however their definition for a closed subscheme is as follows:

A closed subscheme of a scheme $X$ is given by a closed subset $Y \subseteq X$ (let $\iota: Y \hookrightarrow X$ be the inclusion) and a sheaf $\mathcal{O}_{Y}$ on $Y$ such that $(Y, \mathcal{O}_{Y})$ is a scheme, and such that the sheaf $\iota _{*}\mathcal{O}_{Y}$ is isomorphic to $\mathcal{O}_{X}/ \mathcal{I}$ for $\mathcal{I}$ a subsheaf of ideals of $\mathcal{O}_{X}$.

With that in place, my question is basically threefold:

1) Aside from the suggestive name, Hartshorne doesn't seem to suggest (at least not to me) that a closed subscheme is actually a scheme. Indeed, how does one even make sense of putting a scheme structure on an equivalence class of morphisms?

2) Görtz & Wedhorn seem to overcome this by simply defining it to be a scheme. How are their definitions equivalent (if they are)? One big problem I am having seeing this equivalence is relating the surjectivity of $\iota^{\#}$ and $\iota^{'\#}$ to the sheaf of ideals in Görtz & Wedhorn. The issue is, since the category of rings is not abelian (indeed, not even additive as far as I know), I can't expect to have kernels and cokernels, and so I can't expect to be able to take $\mathcal{I}$ to be the kernel of the surjective morphisms like I could if they were sheaves of abelian groups.

3) I tried to play around with the affine case in Hartshorne's definition to make sense of things. Let $A$ be a ring with ideals $\mathfrak{a}$ and $\mathfrak{b}$. These give morphisms of schemes Spec$(A /\mathfrak{a}) \longrightarrow$ Spec $A$ and Spec$(A /\mathfrak{b}) \longrightarrow$ Spec $A$ which induce homemorphisms onto the closed sets in the obvious way. My intuition suggests that these should be "equivalent" (for the purposes of defining a closed subscheme) precisely if $\mathfrak{a}$ and $\mathfrak{b}$ have the same radical, since then $V(\mathfrak{a})$ and $V(\mathfrak{b})$ would be the same. Indeed the radical is just the intersection of all primes containing them. However, by Hartshorne's definition, that would require that the isomorphism $i$ be induced by an isomorphism of rings $A / \mathfrak{a} \simeq A/ \mathfrak{b}$ whenever $\mathfrak{a}$ and $\mathfrak{b}$ have the same nilradical, which is obviously nonsense.

The long of the short is I am stumped and incredibly confused. Any advice, or references, or answers (partial or full) to all or any of these questions would be greatly appreciated.

Bonus question: Is there a way to fix my confusion $\textit{without}$ resorting to quasi-coherent sheaves?

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  • $\begingroup$ Hi Joe, I will answer your question in a moment, but in the meantime, I have edited your notation to get the Hartshorne definition to match the Gortz and Wedhorn definition. (Perhaps this edit already partially answers your question...) $\endgroup$ – Kenny Wong Aug 9 '17 at 15:24
  • $\begingroup$ Hartshorne's approach to the definition of closed subscheme resembles the notion of subobject in category theory. As between all the different equivalent closed immersion there is always (only) one who has as underlying set a closed subset of the original scheme and as structure sheave a quotient of the original structure sheaf. We can define a closed subscheme as this special representative. This is Gortz-Wedhorn definition. The objects between this two definition are obviously in a very natural bijection so there is no problem confusing this two definition in practice. $\endgroup$ – yamete kudasai Nov 30 '17 at 20:01
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I'll address your questions one by one, referring to the notation in my edit of your question.

(1) In Hartshorne's definition of a closed subscheme, $Y$ is the thing that is actually a scheme. It corresponds directly to Gortz & Wedhorn's $Y$.

Hartshorne's $\iota$ is the inclusion map of $Y$ into $X$. This corresponds to Gortz & Wedhorn's $\iota$.

The reason why Hartshorne talks about equivalence classes is that we need some way of determining whether two closed subschemes of $X$ are essentially the same. For example, suppose $$X = {\rm Spec \ } k[X,Y], \ \ \ \ Y = {\rm Spec \ } k[U], \ \ \ \ Y' = {\rm Spec \ } k[T]$$ and suppose the inclusion morphisms $$ \iota : Y \to X, \ \ \ \ \ \iota ' : Y' \to X$$ are defined as the scheme morphisms associated to the ring morphisms $$ k[X,Y] \to k[U], \ \ \ \ \ X \mapsto U, \ \ \ \ \ Y \mapsto 0, $$ and $$ k[X,Y] \to k[T], \ \ \ \ \ X \mapsto T, \ \ \ \ \ Y \mapsto 0, $$ respectively.

For all intents and purposes, $Y$ embedded by $\iota$ is the same thing as $Y'$ embedded by $\iota '$. There is no difference between $Y$ and $Y'$ apart from my choice of notation. So we should consider them equivalent. Indeed, there exists an isomorphism $\psi : Y' \to Y$ such that $\psi \circ \iota = \iota '$: this $\psi$ is the morphism of schemes associated to the ring morphism sending $U \mapsto T$. So $Y$ and $Y'$ with their respective inclusion maps are considered equivalent by Hartshorne's definition.

The scheme structures are well-defined on equivalence classes because if $\iota : Y \to X$ and $\iota ' : Y ' \to X$ are two equivalent closed subschemes with their respective immersions, then $Y$ and $Y'$ are isomorphic as schemes (via $\psi$) by definition, and therefore, they have the same scheme structures.

(2) Suppose the scheme $Y$ together with the embedding $\iota : Y \to X$ is a closed subscheme by Hartshorne's definition. How can we see that the quasi-coherent sheaf $\iota_\star \mathcal O_Y$ is of the form $\mathcal O_X / \mathcal I$ for some quasi-coherent sheaf $\mathcal I$ that is a sheaf of ideals of $\mathcal O_X$? We would like to do this in order to match up with Gortz & Wedhorn's definition.

Well, we have a surjective morphism of sheaves: $$ \mathcal O_X \twoheadrightarrow \iota_\star \mathcal O_Y$$ (This morphism of sheaves is the $\iota^{\#}$ in Hartshorne's definition.)

Since the category of quasi-coherent sheaves on $X$ is an abelian category, there exists a quasi-coherent sheaf $\mathcal I$ that acts as the kernel for the above sheaf morphism, giving a short exact sequence: $$ 0 \to \mathcal I \hookrightarrow \mathcal O_X \twoheadrightarrow \iota_\star \mathcal O_Y \to 0$$ But then, $\mathcal I $ is a sub-$\mathcal O_X$-module of $\mathcal O_X$, which is the same as saying that $\mathcal I$ is a sheaf of ideals of $\mathcal O_X$, and furthermore, $\iota_\star \mathcal O_Y \cong \mathcal O_X / \mathcal I$ as sheaves, so we have constructed everything that is required in Gortz & Wedhorn's definition.

Note that we are working in the category of quasi-coherent sheaves on $X$, and not in the category of rings. Your comment about the category of rings not being an additive category is unimportant. (Here is a local version of my statement: on an affine scheme, specifying a quasi-coherent sheaf is equivalent to specifying a module over a fixed ring. The category of modules over a fixed ring is certainly an abelian category, even if the category of rings is not.)

It should also possible to go backwards from Gortz & Wedhorn's definition to Hartshorne's definition. The annoying thing is that Hartshorne wants $\iota : Y \to X$ to be a morphism of schemes, whereas the $\iota : Y \to X$ provided by Gortz & Wedhorn is only an inclusion of topological spaces. So we need to specify a sheaf morphism $\iota^\# : \mathcal O_X \to \iota_\star \mathcal O_Y$ such that $(\iota, \iota^\#) : (Y, \mathcal O_Y) \to (X, \mathcal O_X)$ defines a genuine morphism of schemes. It is clear that we ought to define $\iota^\#$ to be the composition $ \mathcal O_X \twoheadrightarrow \mathcal O_X / \mathcal I \cong \mathcal O_Y$ in order to ensure that the current construction really is the inverse of the previous one. But unless I'm missing something obvious, I can't any simple way of showing that $(\iota, \iota^\#) : (Y, \mathcal O_Y) \to (X, \mathcal O_X)$ is a morphism of schemes without doing a lot of work. Personally, I would start by arguing that the only way that $\iota_\star \mathcal O_Y$ can be isomorphic to $\mathcal O_X / \mathcal I $ is if $Y$ is the support of the sheaf $\mathcal O_X / \mathcal I$. Since the question is of a local nature, I would now focus my attention on an open affine $U \cong {\rm Spec \ } A \subset X$. Writing $\mathcal I|_U$ as $\widetilde{I}$ for some ideal $I \subset A$, I would show that the support of $\mathcal O_X / I$ is the set $V(I) \subset {\rm Spec \ } A$, which is homeomorphic to ${\rm Spec \ }(A/I)$ as a topological space. Finally, by examining sections on basic open sets and restriction maps between basic open sets, I would try to show that $(Y, \mathcal O_Y)$ has the structure of ${\rm Spec \ } (A / I)$ as a scheme, and that the map $(\iota, \iota^\#) : (Y, \mathcal O_Y) \to (X, \mathcal O_X)$ is the morphism of schemes associated to the natural ring morphism $A \twoheadrightarrow A / I$.

[Apologies for earlier edits of this post where I misread your statement of Gortz & Wedhorn!]

(3) I'm afraid that your hypothesis that ${\rm Spec}(A / I)$ and ${\rm Spec}(A / I')$ are equivalent when $\sqrt{I} = \sqrt{I'}$ is not true. For example, consider: $$ X = {\rm Spec \ } k[T], \ \ \ \ \ Y = {\rm Spec \ } k[T] / (T), \ \ \ \ \ \ \ Y' = {\rm Spec \ } k[T] / (T^2).$$ $Y$ can be thought of as a closed subvariety of $X$: it is the point at the origin. But $Y'$ is a totally different thing - it isn't a variety at all. $Y'$ should be thought of as a "double point" at the origin.

From Hartshorne's perspective, $Y$ and $Y'$ cannot possibly be isomorphic because $k[T]/(T)$ is a reduced ring whereas $k[T]/(T^2)$ is non-reduced, so it's impossible to construct a scheme isomorphism between $Y$ and $Y'$.

And how are $Y$ and $Y'$ different from Gortz & Wedhorn's point of view? After all, the underlying topological spaces are the same, and the inclusion maps are also the same when viewed as continuous maps between topological spaces. The difference between $Y$ and $Y'$ is that their structures sheaves are different. $\iota_\star \mathcal O_Y$ is isomorphic to $\mathcal O_X / \mathcal I$ with $\mathcal I = \widetilde{(T)}$, whereas $\iota_\star \mathcal O_{Y'}$ is isomorphic to $\mathcal O_X / \mathcal I'$ with $\mathcal I' = \widetilde{(T^2)}$.

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    $\begingroup$ This is a great answer. I felt like these were all things I should already know, and yet it still left me feeling like I understood them better. $\endgroup$ – Tobias Kildetoft Aug 10 '17 at 7:40
  • $\begingroup$ Thank you for your answer Kenny. Sorry I didn't get back to you the other day, I thought I would go back and have a look at a few things before replying. This has definitely cleared things up for me. One thing that struck me about what you said, and actually opened up a few more questions, was your statement "Note that we are working in the category of quasi-coherent sheaves on X, and not in the category of rings". How is this so? Why are we ignoring the ring structure that the sheaves are taking their value in? $\endgroup$ – Joe Aug 11 '17 at 23:12
  • $\begingroup$ To elaborate, the sheaf $\mathcal{O}_{X}$ on the basic open set $D(h)$ is the $\textit{ring}$ $A_{h}$, isn't it? Not just the localization as a module? I ask this because I have had a lot of confusion about when I am able to apply exactness results on sheaves, and I have run into problems because certain exactness results hold for modules over a ring, but not rings themselves. $\endgroup$ – Joe Aug 11 '17 at 23:20
  • $\begingroup$ Hi Joe! These are great questions. Fix an open $U \subset X$. I think the point is that $\mathcal I (U)$ and $(\mathcal O_X / \mathcal I)(U)$ are not rings in the first place! $\mathcal I(U)$ is an ideal of $\mathcal O_X(U)$; if $U \cap Y \neq \emptyset$, then it doesn't contain $1$. Meanwhile, $(\mathcal O_X / \mathcal I)(U) \cong \iota_\star O_Y(U)$ is trivial if $U \subset X \ \backslash \ Y$, whereas you have defined a ring to be something where $0 \neq 1$. We should instead think of both $\mathcal I (U)$ and $\mathcal O_X / \mathcal I (U)$ as modules over the ring $\mathcal O_X(U)$. $\endgroup$ – Kenny Wong Aug 11 '17 at 23:39
  • $\begingroup$ And yes, $\mathcal O_X(D(h))$ is the ring $A_h$. Similarly, $\mathcal I(D(h))$ is $I_h$, which is a module over the ring $A_h$, and $(\mathcal O_X / \mathcal I)(D(h))$ is $(A/I)_h$, which is again a module over the ring $A_h$. $\endgroup$ – Kenny Wong Aug 11 '17 at 23:41

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