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This question is an exercise 2.4 p.96 from Qing Liu's book "Algebraic Geometry and Arithmetic Curves".

Let $X$, $Y$ be schemes over a locally Noetherian scheme $S$, with $Y$ of finite type over $S$. Let $x\in X$. Show that for any morphism of $S$-schemes $f_x:\text{Spec}(\mathcal{O}_{X,x})\to Y$, there exist an open subset $U\ni x$ of $X$ and a morphism of $S$-schemes $f:U\to Y$ such that $f_x=f\cdot i_x$, where $i_x:\text{Spec}(\mathcal{O}_{X,x})\to U$ is the canonical morphism (in other words, the morphism $f_x$ extends to an open neighborhood of $x$).

What is the idea of the proof?

vitaliy
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1 Answers1

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First, you should convince yourself that the question whether a given morphism $\operatorname{Spec} \mathcal{O}_{X,x} \to Y$ extends to an open neighborhood of $x$ comes down to the following question in commutative algebra:

Let $A$ and $B$ be two rings and let $\mathfrak{p}$ be a prime ideal of $A$. Does a given ring homomorphism $B \to A_\mathfrak{p}$ factor as $B \to A_f \to A_\mathfrak{p}$ for some $f \in A \setminus \mathfrak{p}$?

Note that there is no reason to hope that such a factorization always exists. Indeed, the answer to the above question might be "no".

But what if $B$ is a finitely generated $R$-algebra with $R$ a Noetherian ring, $A$ is an $R$-algebra and the given morphism is a morphism of $R$-algebras? This is the case you have to deal with in solving your exercise (I'll assume you are able to work out why - if you have trouble, just drop a comment). Then the above question is guaranteed to have a positive answer. Let's prove this. Write $B = R[T_1,\dotsc,T_n] / (g_1,\dotsc,g_m)$, denote the composite of the canonical projection $R[T_1,\dotsc,T_n] \twoheadrightarrow B$ with the given morphism $B \to A_\mathfrak{p}$ by $\varphi$ and write

$$\varphi(T_i) = \frac{a_i}{f_i}, \quad i=1,\dotsc,n \, ,$$ $$\varphi(g_j) = \frac{a_j'}{h_j}, \quad j=1,\dotsc,m \, .$$

Now, for each $j \in \lbrace 1,\dotsc,m \rbrace$, choose some $f_j' \in A \setminus \mathfrak{p}$ such that $f_j' a_j' = 0$ in $A$ (which exists since $\varphi(g_j) = 0 \in A_\mathfrak{p}$). I now claim that $f = \prod_{i=1}^n f_i \prod_{j=1}^m f_j'$ has the desired property, i.e. that the given homomorphism factors through $A_f \to A_\mathfrak{p}$. To see why, we first make use of the fact that, by construction, $f$ is a common denominator of all the $\varphi(T_i)$ and note that

$$ T_i \mapsto \frac{a_i \prod_{k \neq i} f_k \prod_{j=1}^m f_j'}{f}, \quad i=1,\dotsc,n$$

defines a morphism of $R$-algebras $R[T_1,\dotsc,T_n] \to A_f$ whose composite with $A_f \to A_\mathfrak{p}$ agrees with $\varphi$. In addition, we have, by construction, $a_j' f = 0$ for all $j \in \lbrace 1,\dotsc,m \rbrace$; thus, the morphism just defined maps each $g_j$ to $0$ and, as a consequence, factors over $B$.

c_c_chaos
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  • Dear c_c_chaos, your answer is perfect and very pedagogical: +1. May I be so bold as to ask you how you learned algebraic geometry: did you follow courses or are you self-taught? – Georges Elencwajg May 01 '15 at 10:17
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    Dear @GeorgesElencwajg, I both followed courses and am self-taught. To be more precise, in the last three years, I attented quite a number of courses on algebraic geometry - covering foundational topics such as the language of schemes, (Zarisiki) sheaf cohomology or étale cohomology as well as more specialised topics such as toric or abelian varieties. However, I felt that I understood only a tiny part of the lectures properly and had considerable trouble solving exercises. As a consequence of my growing frustration, I essentially decided to go back to the very beginning and began... – c_c_chaos May 01 '15 at 19:42
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    ... reading the book "Algebraic Geometry 1" by Görtz and Wedhorn and doing most of the exercises. This is where I feel that most of my "working knowledge" comes from although my previous exposure to most of the material surely had some effect as well. – c_c_chaos May 01 '15 at 19:45
  • Dear @c_c_chaos, your answer to my comment is incredibly informative. I have argued elsewhere that for mastering a topic in mathematics it is not sufficient to perfectly understand a book or lecture but that one has to dirty one's hands with concrete little problems. This is exactly what you seem to be doing and I predict that you will thus become (or already are!) a very competent algebraic geometer. Since you use a pseudonym I suppose it is inappropriate to ask you what courses you have taken, but I'll try my luck ... – Georges Elencwajg May 01 '15 at 20:02
  • By the way I'm sure I'm not the only one to prefer communication with a user known by real name rather than by pseudonym. Unless you have a compelling reason not to, I strongly encourage you to use your actual name here, especially since your answers are nothing to be shy about ! It is easy to ask a moderator to transfer your reputation to your new username. I did it myself a long time ago: my pseudonym was the fake Hispanic and deliberately transparent El Georges :-) – Georges Elencwajg May 01 '15 at 20:14
  • Dear @c_c_chaos, thank you for your answer - it is great! Though I have some experience in commutative algebra it is quite hard sometimes for me to understand that certain question from geometry is actually local. And this is the case. – vitaliy May 01 '15 at 23:58
  • Dear @vitaliy, does this mean that you would appreciate some elaboration on how to carry out the reduction from your question to the affine case my answer deals with? – c_c_chaos May 02 '15 at 14:03
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    Dear @GeorgesElencwajg, your question is not inappropriate at all, although I am not sure whether I understand the question and its intention correctly. In case you are simply asking for a list of courses on algebraic geometry I have taken, here it is: 1. Lectures: Algebraic Geometry 1 (roughly Hartshorne I.1-II.6), Algebraic Geometry 2 (roughly Hartshorne II.7-8 and III up to Serre duality; I cannot remember what, if anything, was covered after Serre duality), Etale Cohomology, Abelian Varieties 1 (after a book by Mumford; I remember next to nothing about the covered material). ... – c_c_chaos May 02 '15 at 14:12
  • Dear c_c_chaos, thank you for your detailed answer. – Georges Elencwajg May 02 '15 at 14:17
  • ... 2. Seminars: Algebraic Curves (after the script by Fulton), Elliptic Curves (up to the Mordell-Weil theorem, after the more elementary of the two books by Silverman), Toric Varieties (again after a book by Fulton). I did neither include courses on more basic topics like commutative algebra and homological algebra nor such on neighboring fields such as algebraic number theory. (It might by worth noting that my overall background is, I guess, more "arithmetic" than "geometric".) The list is not chronological. – c_c_chaos May 02 '15 at 14:23
  • Dear @c_c_chaos, I totally understand your answer so there is no need to give the detailed explanation. In my comment I should have written "And this was the case". – vitaliy May 03 '15 at 00:52
  • Dear c_c_chaos, I don't see how noetherianity of $R$ plays a role. This is a completely unimportant nitpick, but I mention it to check whether something escaped me in the proof. – Georges Elencwajg May 29 '15 at 07:30
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    Dear @GeorgesElencwajg, if I'm not mistaken, the proof makes use of the fact that there are only finitely many $g_i$ - i.e., of the fact that the finitely generated $R$-algebra $B$ can be written as $R[T_1,\dotsc,T_n]/\mathfrak{a}$ where $\mathfrak{a}$ is a finitely generated ideal. By Hilbert's basis theorem, this is automatic if $R$ is Noetherian. However, if $R$ is not required to be Noetherian, one must actually assume that $B$ is of this form - that is, replace "of finite type" by "of finite presentation" in the original question. – c_c_chaos May 29 '15 at 15:28
  • Dear c_c_chaos: you are absolutely right and indeed that point had escaped me. Bravo, but I can't upvote you now because I already did long ago :-) – Georges Elencwajg May 29 '15 at 17:03
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    Can you elaborate on the reduction at the very beginning of the solution? Specifically, I don't understand how you reduce to the case where $Y$ is affine. – Gal Porat Mar 13 '17 at 22:49
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    @GalPorat Your question is already 3 years old but I'll answer it if anyone reading this is interested. Essentially, the reduction is due to the general fact that if you have a map $\text{Spec }R\to Y$ with $(R,P)$ any local ring, the image is always contained inside any open (affine) set that contains the image of $P$. – George Sep 28 '20 at 12:25