In order to prevent bots from posting comments, we would like you to prove that you are human. Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.Īll contributions are licensed under the GNU Free Documentation License. A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). Your email address will not be published. Well since I think what it says in the text is correct. On Jat It refers to the discussion at the end of Section 26.10. In the proof of lemma 01IQ I think the sheaf is supported on, not. I'm not sure what is meant by "section", but immersions seem to be discussed rather in another chapter. Just before Remark 01IP is written "We will discuss locally closed subschemes and immersions at the end of this section". OK, I've tried to improve the exposition. The metric d : X × X R is just the function d. Then ( X, d ) is a metric space, which is said to be a subspace of ( M, d). We define a metric d on X by d ( x, y) d ( x, y) for x, y X. The fact that the two notions (open/closed) are defined separately for locally ringed spaces (and there is no definition of "immersion" or "subspace" for these, as far as I could see) makes it a bit hard to get a clear picture. The subset with that inherited metric is called a 'subspace.' Definition 2.1: Let ( M, d) be a metric space, and let X be a subset of M. The text should be more explicit about this (first of all by referring to that definition!). The point is that in 25.10.2, "closed subspace" is not meant in the topological sense but in the ringed space sense of Definition 01HN (aka 25.4.4). The terminology is a bit confusing, as the comments demonstrate. But multiple closed subschemes can correspond to the same closed subset. On Maat In 25.10.2 a closed subscheme of X is defined as a closed subset of X. Why the data of a closed subscheme consists of only a topological object? In the discussion following the proof of 01IQ, the first word of the sentence "If we define a." should be removed (or something should be added to the sentence). If it does, then of course $Z$ is identified with a unique locally closed subscheme of $Z'$, and so on.įor more information on immersions, we refer the reader to Morphisms, Section 29.3. To be explicit, if $Z \to X$ and $Z' \to X$ are two locally closed subschemes of $X$, then we say that $Z$ is contained in $Z'$ simply if the morphism $Z \to X$ factors through $Z'$. We may define a partial ordering on this set, which we call inclusion for obvious reasons. The interest of this is that the collection of locally closed subschemes of $X$ forms a set. The above then shows that any immersion $f : Y \to X$ factors uniquely as $Y \to Z \to X$ where $Z$ is a locally closed subspace of $X$ and $Y \to Z$ is an isomorphism. We usually just say “let $Z$ be a locally closed subscheme of $X$” since we may recover $U$ from the morphism $Z \to X$. Note that the notion of a quasi-coherent sheaf of $\mathcal \cup U = X$. Below we will prove that the same holds for a closed subspace of a scheme. In Lemma 26.9.2 we saw that any open subspace of a scheme is a scheme.
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