# Exercise 6.3.N. on Ravi Vakil's notes (on morphism from $S$ scheme to $\mathbb{P}_B^n$)

I am working on Exercise 6.3.N. on Ravi Vakil's notes (on morphism from $$S$$ scheme to $$\mathbb{P}_B^n$$) and I would like some assistance. The exercise states: Let $$B$$ be a ring. If $$X$$ is a $$B$$-scheme, and $$f_0, ..., f_n$$ are $$n+1$$ functions on $$X$$ with no common zeros, then show that $$[f_0, ..., f_n]$$ gives a morphism of $$B$$-schemes $$X \rightarrow \mathbb{P}_B^n$$.

I would appreciate any hint, comments, etc. Thank you very much!

Edit (Not by OP) (10/06/2021). In the 2017 version of the notes, this is exercise 6.3.M.

• Now in the 2022 Aug version, it's exercise 7.3.O Dec 16, 2022 at 3:15

If I denote the homogeneous coordinates on $\mathbf{P}^n_B$ by $x_0, \dots, x_n$ then $X_{f_i}$ should map into the affine open $D(x_i) \subset \mathbf{P}^n_B$. To specify this restriction you just need to give a $B$-homomorphism $B[x_0/x_i, \dots, x_n/x_i] \to \Gamma(X_{f_i}, \mathscr{O}_X)$. What's the right formula? Then check that these agree on overlaps.

• What does the notation $X_{f_i}$ mean here? Thank you!
– Oga
Apr 5, 2015 at 21:31
• @Oga The open set of points where $f_i$ does not vanish, in the sense that its image in the residue field is not zero. I'm sure he defines something like this at some point.
– Hoot
Apr 6, 2015 at 2:15
• Thank you for the answer. I figured it is the map that sends $x_l/x_i to f_l |_{X_{f_i}}$. Could you possibly explain me why the morphisms $[f_0, ..., f_n]$ and $[gf_0, ..., gf_n]$ to $\mathbb{P}_B^n$ are the same, if $g$ does not vanish anywhere on $X$? This is part b of the question. I can see that the map of topological spaces are the same, but they seem to have (a slightly) different maps of sheaves... Thank you!
– Oga
Apr 8, 2015 at 18:51
• @Oga I should have included more of an indication of what's going on. I think you want to send $x_l/x_i$ to $f_l/f_i$. This should make part (b) more obvious!
– Hoot
Apr 9, 2015 at 3:35
• oops, I guess I had a wrong map! Thank you for the clarification!
– Oga
Apr 9, 2015 at 20:34

I would suggest another solution. We have done most of the gluing work already in what is now (December 2022 version) exercise 7.3.F. There we construct a natural map $$\mathbb{A}_B^{n+1} \setminus \{0\} \to \mathbb{P}_B^n$$. (Technically we've only done it for fields $$k$$, but the construction of course works generally for rings $$B$$.) Hence it suffices to produce morphism $$X \to \mathbb{A}_B^n \setminus \{0\}$$, which also more or less directly solves subtask (b). But this is easily constructed: A morphism $$\varphi: X \to \mathbb{A}_B^n$$ is induced by the ring morphism \begin{align} B[x_0,...,x_n] &\to \Gamma(X,\mathcal{O}_X) \\ x_i &\mapsto f_i \end{align} and we are left to show that this morphism doesn't hit $$0 \in \mathbb{A}_B^n$$. Note that the pullback of the function $$x_i \in \Gamma(\mathbb{A}_B^n, \mathcal{O}_{\mathbb{A}_B^n})$$ is $$f_i \in \Gamma(X,\mathcal{O}_X)$$. Since we have a morphism of locally ringed spaces, $$x_i$$ vanishing at $$p \in \mathbb{A}_B^n$$ implies $$f_i$$ vanishes at all $$q \in \varphi^{-1}(p) \subseteq X$$. But at $$0 \in \mathbb{A}_B^n$$, all $$x_i$$ vanish. Thus all $$f_i$$ vanish at all $$q \in \varphi^{-1}(0)$$. By assumption such a point $$q$$ can't exist, showing that $$\operatorname{im} \varphi \subseteq \mathbb{A}_B^n \setminus \{0\}$$.