# Extending a morphism of schemes

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?

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 $B$.

• 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 1 '15 at 10:17
• 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 1 '15 at 19:42
• ... 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 1 '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 1 '15 at 20:02
• 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