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

Question

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.

Answer 1

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.

Answer 2

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 $\operatorname{im} \varphi \subseteq \mathbb{A}_B^n \setminus \{0\}$.