Maps to projective spaces -- Part 2

Again, let us consider a variety $X$ over a field $k.$ (One can relax this condition about $X$ in various places throughout the discussion, but we won't bother.) Last time, we have discussed that when we have scheme maps $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1}$ that do not vanish all together at any single point in $X,$ we can induce a $k$-scheme map $$\pi := [f_{0} : \cdots : f_{n}] : X \rightarrow \mathbb{P}^{n}$$ such that when $X$ is locally of finite type over $k,$ for any $x \in X(k)$ such that $f_{0}(x), \dots, f_{n}(x) \in \mathbb{A}^{1}(k) = k,$ we have $\pi(x) = [f_{0}(x) : \cdots f_{n}(x)] \in \mathbb{P}^{n}(k).$ We also recall that defined this map by the associated $k$-algebra maps $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{f_{i}}, \mathscr{O}_{X})$$ such that $(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (f_{0}/f_{i}, \dots, f_{n}/f_{i}).$

We now classify all $k$-scheme maps $\pi : X \rightarrow \mathbb{P}^{n}.$ We can consider the standard open cover $$\mathbb{P}^{n} = U_{0} \cup U_{1} \cup \cdots \cup U_{n},$$ where $U_{i} = D_{\mathbb{P}^{n}}(x_{i}).$ Writing $X_{i} := \pi^{-1}(U_{i}),$ the map $\pi$ is given by gluing the $k$-algebra maps $k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{i}, \mathscr{O}_{X_{i}}),$ for $i = 0, 1, \dots, n,$ each of which is determined by where we send $x_{0}/x_{i}, \dots, x_{n}/x_{i}.$ Recall that we can identify $$\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}} = \mathscr{O}_{U_{i}}$$ so that we have $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)) = k[x_{0}/x_{i}, \dots, x_{n}/x_{i}].$$ 

Review on $\mathscr{O}_{\mathbb{P}^{n}}(1)$. We have transition maps $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]_{x_{j}/x_{i}} \rightarrow k[x_{0}/x_{j}, \dots, x_{n}/x_{j}]_{x_{i}/x_{j}}$$ given by the multiplication of $x_{i}/x_{j}.$ An element $s$ of $H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1))$ is given by $$s = (f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}))_{0 \leq i \leq n}$$ such that $$x_{i}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = x_{j}f_{j}(x_{0}/x_{j}, \dots, x_{n}/x_{j})$$ for $0 \leq i, j \leq n.$ This implies that $$f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = a_{i,0}x_{0}/x_{i} + \cdots + a_{i,n}x_{n}/x_{i},$$ and $(a_{i,0}, \dots, a_{n,i}) = (a_{j,0}, \dots, a_{j,n})$ for $0 \leq i, j \leq n$ so that we can write $(a_{0}, \dots, a_{n}) = (a_{i,0}, \dots, a_{n,i})$ for $0 \leq i \leq n.$ Therefore, we have $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \simeq k x_{0} + \cdots + k x_{n}$$ given by $(a_{0}, \dots, a_{n}) \mapsto a_{0}x_{0} + \cdots + a_{n}x_{n},$ and we shall use the identification $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) = k x_{0} + \cdots + k x_{n}.$$ With respect to this identification, the restriction map $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \rightarrow H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1))$$ is given by $a_{0}x_{0} + \cdots + a_{n}x_{n} \mapsto a_{0}x_{0}/x_{i} + \cdots + a_{n}x_{n}/x_{i}.$

Review on pullbacks of quasi-coherent sheaves. Let $\pi : X \rightarrow Y$ be a scheme map and $\mathscr{G}$ a quasi-coherent sheaf on $Y.$ Then the pullback $\pi^{*}\mathscr{G}$ is a quasi-coherent sheaf on $X$ and on the restriction $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(B)$ of $\pi$ on affine opens, we have $$H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}) = H^{0}(\mathrm{Spec}(B), \mathscr{G}) \otimes_{B} A.$$ Given a map $\phi : \mathscr{G}_{1} \rightarrow \mathscr{G}_{2}$ of quasi-coherent sheaves on $Y,$ we have the induced map $\pi^{*}\phi : \pi^{*}\mathscr{G}_{1} \rightarrow \pi^{*}\mathscr{G}_{2}$ of quasi-coherent sheaves on $X$ such that the restriction $$H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}_{1}) \rightarrow H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}_{2})$$ is given by $$H^{0}(\mathrm{Spec}(B), \mathscr{G}_{1}) \otimes_{B} A \rightarrow H^{0}(\mathrm{Spec}(B), \mathscr{G}_{2}) \otimes_{B} A$$ with $s \otimes a \mapsto \phi(s) \otimes a.$

We can also have $\pi^{*} : H^{0}(Y, \mathscr{G}) \rightarrow H^{0}(X, \pi^{*}\mathscr{G})$ as follows (although this also depends on $\mathscr{G}$). Given $s \in H^{0}(Y, \mathscr{G}),$ we may consider the corresponding $\mathscr{O}_{Y} \rightarrow \mathscr{G}$, which we also call $s,$ given by $H^{0}(V, \mathscr{O}_{Y}) \rightarrow H^{0}(V, \mathscr{G})$ such that $1 \mapsto s|_{V}.$ Then we can consider its pullback $$\pi^{*}s : \mathscr{O}_{X} \simeq \pi^{*}\mathscr{O}_{Y} \rightarrow \pi^{*}\mathscr{G},$$ which corresponds to an element of $H^{0}(X, \pi^{*}\mathscr{G}).$ Restricting to the affine opens we fixed before, this gives rise to $$A \simeq B \otimes_{B} A \rightarrow H^{0}(\mathrm{Spec}(B), \mathscr{G})_{\otimes B} A = H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G})$$ given by $$b \otimes a \mapsto bs \otimes a.$$ Thus, the map $$\pi^{*} : H^{0}(Y, \mathscr{G}) \rightarrow H^{0}(X, \pi^{*}\mathscr{G})$$ must be given by gluing $$H^{0}(\mathrm{Spec}(B), \mathscr{G}) \rightarrow H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}) = H^{0}(\mathrm{Spec}(B), \pi^{*}\mathscr{G}) \otimes_{B} A$$ defined by $s \mapsto s \otimes 1.$ In particular, for any $s \in H^{0}(Y, \mathscr{G})$ and open $V \subset Y,$ we have $\pi^{*}|_{V} : H^{0}(V, \mathscr{G}) \rightarrow H^{0}(\pi^{-1}(V), \pi^{*}\mathscr{G})$ such that $s|_{V} \mapsto \pi^{*}(s)|_{\pi^{-1}(V)}.$ 

Going back to the previous discussion, considering the pullback map $$\pi^{*} : H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \rightarrow H^{0}(X, \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)),$$ since $\mathscr{O}_{\mathbb{P}^{n}}(1)$ is locally free, one can check that $$D_{X}(\pi^{*}x_{i}) = \pi^{-1}(D_{\mathbb{P}^{n}}(x_{i})) = \pi^{-1}(U_{i}) = X_{i}.$$ The global section $s_{i} := \pi^{*}x_{i}$ does not vanish anywhere in the open subset $X_{i}$ of $X,$ so the restriction of the line bundle $\pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)$ of $X$ on $X_{i}$ is trivial, but we will not use this trivialization yet. (We will use this later.)

Remark. What's important for us is the notion of the "inverse" of $s \in H^{0}(X, \mathscr{L})$ when $s$ is nowhere vanishing. Let $X = \bigcup_{\alpha \in I}U_{\alpha}$ be any trivializing open cover for $\mathscr{L}.$ Fix $p \in X,$ and pick any $U_{\alpha} \ni p.$ We have $\mathscr{O}_{U_{\alpha}} \simeq \mathscr{L}|_{U_{\alpha}},$ which gives $\mathscr{O}_{X,p} \simeq \mathscr{L}_{p}.$ Denote by $s'_{p}$ the element of $\mathscr{O}_{X,p}$ corresponding to $s_{p} \in \mathscr{L}_{p},$ the image of $s \in H^{0}(X, \mathscr{L}).$ We have $s_{p} \notin \mathfrak{m}_{X,p}\mathscr{L}_{p}$ (i.e., $s$ does not vanish at $p$), so $s'_{p} \in \mathfrak{m}_{X,p},$ which means that $s'_{p}$ is a unit in $\mathscr{O}_{X,p}.$ Thus, refining our trivializing open cover if necessary, we may assume that there is $w_{\alpha} \in H^{0}(U_{\alpha}, \mathscr{O}_{X})$ such that $s'|_{U_{\alpha}}w_{\alpha} = 1 \in H^{0}(U_{\alpha}, \mathscr{O}_{X}).$ These $w_{\alpha}$'s agree on the intersections of $U_{\alpha}$'s, so we may glue them to construct $w \in H^{0}(X, \mathscr{O}_{X}),$ which we refer to as the inverse of $s.$ (I am not sure whether this is a standard terminology.) Note that even though we needed a trivialization of $\mathscr{L}$ to construct this element, it does not depend on the choice of such a trivialization. 

Moreover, note that any element (not necessarily nonvanishing one) of $H^{0}(X, \mathscr{L})$ has a unique corresponding element in $H^{0}(X, \mathscr{O}_{X}),$ given by gluing the ones defined on open subsets in a trivialization of $\mathscr{L}.$ (Note that this element does not depend on the choice of a trivialization.)

Going back to our discussion, now, it makes sense for us to consider the elements $s_{0}/s_{i}, \dots, s_{n}/s_{i} \in H^{0}(X_{i}, \mathscr{O}_{X}),$ namely the elements corresponding to $s_{0}|_{X_{i}}, \dots, s_{n}|_{X_{i}} \in H^{0}(X_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1))$ multiplied by the inverse of $s_{i}|_{X_{i}} \in H^{0}(X_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)).$ Note that $\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}} \simeq \mathscr{O}_{U_{i}},$ so $$\pi^{*}(\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}}) \simeq \pi^{*}\mathscr{O}_{U_{i}} \simeq \mathscr{O}_{X_{i}}.$$ Now, the upshot is that (one can check that) the map $$H^{0}(U_{i}, \mathscr{O}_{U_{i}}) \rightarrow H^{0}(X_{i}, \mathscr{O}_{X_{i}})$$ given by $\pi : X \rightarrow \mathbb{P}^{n}$ coincides with the one corresponding to the pullback map $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}}) \rightarrow H^{0}(X_{i}, \pi^{*}(\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}})).$$ That is, such a map is given by $$(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (s_{0}/s_{i}, \dots, s_{n}/s_{i}).$$ So far, we have shown the following:

Theorem 1. Any $k$-scheme map $\pi : X \rightarrow \mathbb{P}^{n}$ is given by $k$-algebra maps $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}) \rightarrow H^{0}(\pi^{-1}(U_{i}), \mathscr{O}_{X})$$ such that $x_{0}/x_{i}, \dots, x_{n}/x_{i} \mapsto s_{0}/s_{i}, \dots, s_{n}/s_{i},$ where $s_{i}$ corresponds to the restriction of $\pi^{*}x_{i}$ on $\pi^{-1}(U_{i}).$

More on construction. If $\mathscr{L}$ is any line bundle on $X$ with global sections $s_{0}, s_{1}, \dots, s_{n} \in H^{0}(X, \mathscr{L})$ that do not vanish at a single point in $X$ altogether, then we have an open cover $X = X_{0} \cup \cdots \cup X_{n},$ where $X_{i} := D_{X}(s_{i}).$ Refining these open sets further and gluing back, we can construct a $k$-scheme map $\pi : X \rightarrow \mathbb{P}^{n}$ given by $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{i}, \mathscr{O}_{X})$$ such that $(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (s_{0}/s_{i}, \dots, s_{n}/s_{i})$ for $i = 0, 1, \dots, n.$

Notation. We write $\pi := [s_{0} : \cdots : s_{n}].$

Theorem 2. We have $\mathscr{L} \simeq \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)$ such that $s_{i} \leftrightarrow \pi^{*}x_{i}$ on $X.$

Remark. For any line bundle $\mathscr{L}$ on a scheme $X.$ If $s \in H^{0}(X, \mathscr{L})$ is nowhere vanishing, then we have $\mathscr{L} \simeq \mathscr{O}_{X}$ given as follows. Consider the map $\mathscr{O}_{X} \rightarrow \mathscr{L}$ given by $$H^{0}(U, \mathscr{O}_{X}) \rightarrow H^{0}(U, \mathscr{L})$$ such that $1 \mapsto s|_{U}.$ This defines an isomorphism of $\mathscr{O}_{X}$-modules for the following reason. Fixing any $p \in X,$ if we consider the map $\mathscr{O}_{X,p} \rightarrow \mathscr{L}_{p}$ on stalks at $p,$ it is an $\mathscr{O}_{X,p}$-linear isomorphism because $s_{p} \notin \mathfrak{m}_{X,p}\mathscr{L}_{p},$ which ensures that $s_{p}$ corresponds to a unit in the local ring $\mathscr{O}_{X,p}$ under the $\mathscr{O}_{X,p}$-linear isomorphism $\mathscr{L}_{p} \simeq \mathscr{O}_{X,p}$ given by the hypothesis that $\mathscr{L}$ is a line bundle.

Trivialization of a line bundle from its sections. Given any line bundle $\mathscr{L}$ on $X$ and $s_{0}, s_{1}, \dots, s_{n} \in H^{0}(X, \mathscr{L})$ that do not vanish altogether at any point in $X,$ the maps $\phi_{i} : \mathscr{O}_{X_{i}} \overset{\sim}{\longrightarrow} \mathscr{L}|_{X_{i}}$ given by $$H^{0}(U, \mathscr{O}_{X_{i}}) \simeq H^{0}(U, \mathscr{L}|_{X_{i}})$$ such that $1 \mapsto s_{i}|_{U}$ is a trivialization of the line bundle $\mathscr{L}.$ That is, any other trivialization of $\mathscr{L}$ is compatible with this one.

Proof of Theorem 2. We have $\phi_{i} : \mathscr{O}_{X_{i}} \overset{\sim}{\longrightarrow} \mathscr{L}|_{X_{i}}$ given by $$H^{0}(U, \mathscr{O}_{X_{i}}) \simeq H^{0}(U, \mathscr{L}|_{X_{i}})$$ such that $1 \mapsto s_{i}|_{U}.$ We can consider $\mathscr{L}$ as a line bundle defined by gluing $\mathscr{O}_{X_{i}}$'s with the transition functions $$\phi_{ij} := \phi_{j}^{-1}\phi_{i} : \mathscr{O}_{X_{ij}} \rightarrow \mathscr{L}|_{X_{ij}} \rightarrow \mathscr{O}_{X_{ij}},$$ where $X_{ij} = X_{i} \cap X_{j}.$ Note that these transition functions are given by $1 \mapsto s_{i} \mapsto s_{i}/s_{j}.$ Now, note that $\pi^{-1}(U_{i}) = D_{X}(s_{i}) = X_{i}$ and for any open $U \subset X_{i},$ we have $$H^{0}(U, \mathscr{L}) \simeq H^{0}(U, \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1))$$ given by $s_{i} \mapsto \pi^{*}(x_{i}).$ This map takes $s_{i}/s_{j}$ to $\pi^{*}x_{i}/\pi^{*}x_{j},$ which finishes the proof. $\Box$

Reference. Hartshorne's book and Vakil's notes

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.