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I’m reading Qing Liu’s Algebraic Geometry and Arithmetic Curves. I’m confusing my double definitions of projective morphism over a ring $A$. In definition 2.3.42, projective scheme over $A$ is an $A$-scheme that is isomorphic to a closed subscheme of $\operatorname{Proj} A[x_0,...,x_n]$. In definition 3.1.11, $X \rightarrow \operatorname{Spec A}$ is projective if it factors in to a closed immersion $X \rightarrow \operatorname{Proj} A[x_0,...,x_n]$ followed by the canonical morphism $\operatorname{Proj} A[x_0,...,x_n] \rightarrow \operatorname{Spec} A$. So I think A scheme $X$ can be said projective in definition 3.1.11 sense. So I think this two definitions should be coincide.

I can understand if $X$ is projective in the sense of definition 3.1.11, $X$ is projective in the sense of 2.3.42. But I can’t understand the converse, when $X$ is projective in the sense of 2.3.42, why does $f:X \rightarrow \operatorname{Spec} A$ have a decomposition two morphisms.

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    $\begingroup$ Please put all of your math inside math mode and use \operatorname{Spec} to format $\operatorname{Spec}$ and the like. I've made this upgrade for you this time - you can check out what I did via the edit button. $\endgroup$
    – KReiser
    Commented Nov 15, 2020 at 6:37
  • $\begingroup$ Thank you for your advice. Sorry, I’m not used to using stack exchange. I should not do the same mistakes hereafter. $\endgroup$
    – masui
    Commented Nov 15, 2020 at 14:21

1 Answer 1

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Given a scheme $S$, an $S$-scheme is a scheme $X$ together with a morphism $X \to S$ (often referred to as the structure map), and a morphism of $S$-schemes is a morphism of schemes $X \to Y$ such that the structure map $X \to S$ factors as $X \to Y \to S$. So, phrased in this language, the second definition literally just says that $X$ can be realized as a closed $A$-subscheme of some projective space over $A$. The factorization of $X \to \operatorname{Spec} A$ as a composition $X \to \operatorname{Proj} A[x_0, \dots, x_n] \to \operatorname{Spec} A$ is encoded in the fact that a projective scheme over $A$ is (by definition) an $A$-scheme, and the isomorphism to a closed subscheme of projective space is a morphism of $A$-schemes.

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  • $\begingroup$ Here $S$ is the base scheme and the $S$-scheme $X$ can be thought of as $moduli$ space. Am I right ? $\endgroup$
    – MAS
    Commented Nov 15, 2020 at 5:25
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    $\begingroup$ Yeah, in your context, $S = \operatorname{Spec} A$, I just stated it in general since there's no need to restrict to affine schemes in that definition. There are lots of ways to think of a morphism of schemes $X \to S$; one useful interpretation is that it's a family of schemes, with members of the family corresponding to fibers of the morphism. I'm not sure "moduli space" is quite the right word for it, since that generally refers to a space parametrizing some objects up to isomorphism. $\endgroup$ Commented Nov 15, 2020 at 5:30
  • $\begingroup$ Yes, it is appropriate to use the term $\text{family of schemes}$ instead of $\text{moduli space}$. I agree $\endgroup$
    – MAS
    Commented Nov 15, 2020 at 5:35
  • $\begingroup$ Thank you for your answer. But I can’t understand why the following morphism is A-scheme, “the isomorphism to a closed subscheme of projective space is a morphism of A-schemes” . In the first definition, dose that isomorphism mean A-scheme isomorphism? $\endgroup$
    – masui
    Commented Nov 15, 2020 at 5:56
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    $\begingroup$ @masui: Yes, it does mean isomorphic as an $A$-scheme, not just as an abstract scheme (without the structure map). $\endgroup$ Commented Nov 15, 2020 at 6:57

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