Maps to projective spaces -- Part 1

Let $X$ be a variety over a field $k.$ Given any set-theoretic functions $f_{0}, \dots, f_{n} : X(k) \rightarrow k$ that has no common zeros, we may define a set map $X(k) \rightarrow \mathbb{P}^{n}(k)$ given by $$x \mapsto [f_{0}(x) : \cdots : f_{n}(x)].$$

Let's think about a scheme-theoretic version of this map. First, we assume that $f_{0}, \dots, f_{n}$ are elements of $H^{0}(X, \mathscr{O}_{X}).$ Note that an element $f \in H^{0}(X, \mathscr{O}_{X})$ corresponds with a $k$-scheme map $X \rightarrow \mathbb{A}^{1}$ that corresponds to the $k$-algebra map $k[t] \rightarrow H^{0}(X, \mathscr{O}_{X})$ defined by $t \mapsto f,$ so we are fixing $k$-scheme maps $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1},$ analogous to the previous situation.

A zero of $f$ is a point of $X$ that is sent to the maximal ideal $(t)$ in $\mathbb{A}^{1} = \mathrm{Spec}(k[t]).$ One can easily check that for any $x \in X,$ the following are equivalent:

1. $[f] \in \mathscr{O}_{X,x}$ sits inside $\mathfrak{m}_{X,x}$;
2. $x \in X$ is zero of $f.$

Example. Consider the case when $X = \mathrm{Spec}(k[x_{1}, \dots, x_{m}])$ and $x = (x_{1} - a_{1}, \dots, x_{m} - a_{m}).$ In this case, we can observe that $x$ is a zero of $f \in k[x_{1}, \dots, x_{m}]$ if and only if $f(a_{1}, \dots, a_{m}) = 0.$

Hence, we may now require that $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1}$ have no common zeros. (The data can be thought as $f_{0}, \dots, f_{n} \in H^{0}(X, \mathscr{O}_{X})$ as well.)

Special case: $n = 1$. We first consider the case $n = 1,$ where we see how $f_{0}, f_{1} \in H^{0}(X, \mathscr{O}_{X})$ that have no common zeros may induce a $k$-scheme map $X \rightarrow \mathbb{P}^{1}.$ Due to our hypothesis, we have the following open cover: $X = X_{f_{0}} \cup X_{f_{1}},$ where $$X_{f} := \{x \in X : f \text{ does not vanish at } x\}.$$ It is important to note that (the restriction of) $f$ is invertible in $H^{0}(X_{f}, \mathscr{O}_{X})$ because for any affine open $U = \mathrm{Spec}(R) \subset X,$ we have $$X_{f} \cap U = D_{R}(f|_{U}) \simeq \mathrm{Spec}(R_{f|_{U}}).$$

We now construct two $k$-scheme maps $X_{f_{i}} \rightarrow U_{i} = \mathrm{Spec}(k[x_{0}/x_{i}, x_{1}/x_{i}])$ for $i = 0, 1$ and glue them together. For the construction for each $i,$ it is enough to construct a $k$-algebra map $k[x_{0}/x_{i}, x_{1}/x_{i}] \rightarrow H^{0}(X_{f_{i}}, \mathscr{O}_{X}).$ We do this by the following assignments:

• $x_{0}/x_{1} \mapsto f_{0}/f_{1} := f_{0}f_{1}^{-1}$;
• $x_{1}/x_{0} \mapsto f_{1}/f_{0} := f_{1}f_{0}^{-1}.$

The $k$-scheme maps $X_{f_{0}} \rightarrow U_{0}$ and $X_{f_{1}} \rightarrow U_{1}$ agree on the intersections (which can be checked with their $k$-algebra maps), so they glue to give a $k$-scheme map $X \rightarrow \mathbb{P}^{1}.$ Let $x \in X$ be a $k$-point, and suppose that $f_{0}(x), f_{1}(x)$ are $k$-points of $\mathbb{A}^{1}$ (which always happens when $k$ is algebraically closed).

Claim. If $X$ is locally of finite type over $k,$ then the map $\pi : X \rightarrow \mathbb{P}^{1}$ we constructed sends $x$ to $[f_{0}(x) : f_{1}(x)] \in \mathbb{P}^{1}(k).$

Proof. We have two cases: either $x \in X_{f_{0}}$ or $x \in X_{f_{1}}.$ Suppose we are in the first case. We may take an affine open neighborhood $U \subset X_{f_{0}}$ of $x$ such that $W = \mathrm{Spec}(k[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{r}))$ and write $$x = (y_{1} - b_{1}, \dots, y_{m} - b_{m})$$ in $k[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{r})$ for some $b_{1}, \dots, b_{m} \in k.$ Consider the following restriction of $\pi$: $$W \hookrightarrow X_{f_{0}} \rightarrow U_{0} = \mathrm{Spec}(k[x_{1}/x_{0}]).$$ We have $x \in W,$ and we would like to compute what $\pi(x) \in W$ is. The $k$-algebra map corresponding to the above composition is $$\phi : k[x_{1}/x_{0}] \rightarrow H^{0}(X_{f_{0}}, \mathscr{O}_{X}) \rightarrow H^{0}(W, \mathscr{O}_{X}) = \frac{k[y_{1}, \dots, y_{m}]}{(g_{1}, \dots, g_{r})},$$ which is defined by $x_{1}/x_{0} \mapsto f_{1}/f_{0},$ where now we can think of $f_{0}, f_{1}$ with polynomial expressions: $f_{i} = f_{i}(y_{1}, \dots, y_{m}).$ With this notation, we have $$\begin{align*}\pi(x) &= \phi^{-1}((y_{1}-b_{1}, \dots, y_{m}-b_{m})) \\ &= \{h(x_{1}/x_{0}) \in k[x_{1}/x_{0}] : h(f_{1}/f_{0}) \in x\} \\ &= \{h(x_{1}/x_{0}) \in k[x_{1}/x_{0}] : h(f_{1}(b_{1}, \dots, b_{m})/f_{0}(b_{1}, \dots, b_{m}) = 0\} \\ &= (x_{1}/x_{0} - f_{1}(b_{1}, \dots, b_{m})/f_{0}(b_{1}, \dots, b_{m})) \\ &= [f_{0}(b_{1}, \dots, b_{m})) : f_{1}(b_{1}, \dots, b_{m})] \\ &= [f_{0}(x) : f_{1}(x)],\end{align*}$$ as desired. The other case where $x \in X_{f_{1}}$ can be proven by an almost identical argument. $\Box$

General case. One may notice that the arguments for the special case $n = 1$ works in general without any difficulty. We shall now write $\pi := [f_{0} : \cdots : f_{n}]$ for the reasons described above.