[proofplan] A rational map $C \dashrightarrow \mathbb{P}^n_k$ is by definition a morphism on a Zariski-open dense subset $U \subset C$, and we must show it extends to a morphism defined at every point. Fix $p \in C$. We use the [order homomorphism](/theorems/2170) $\operatorname{ord}_p$ at the smooth point $p$ to renormalise the homogeneous coordinates of $\varphi$. Pick a local presentation $\varphi = [G_0 : \cdots : G_n]$ with $G_i \in \mathcal{O}_C(\eta)$ rational functions, not all zero. Let $m = \min_i \operatorname{ord}_p(G_i)$. Multiplying every coordinate by $t_p^{-m}$, where $t_p$ is a uniformiser at $p$, produces an equivalent presentation $\varphi = [t_p^{-m} G_0 : \cdots : t_p^{-m} G_n]$ in which every coordinate has $\operatorname{ord}_p \geq 0$ (so each lies in $\mathcal{O}_{C,p}$, i.e. is regular at $p$) and at least one coordinate has $\operatorname{ord}_p = 0$ (so it does not vanish at $p$). The renormalised tuple is therefore well-defined and not the zero tuple at $p$, defining $\varphi$ as a morphism on a neighbourhood of $p$. [/proofplan] [step:Reduce to showing $\varphi$ extends to a morphism at each $p \in C$] By definition, a rational map $\varphi: C \dashrightarrow \mathbb{P}^n_k$ is given by an equivalence class of pairs $(U, \varphi_U)$ where $U \subset C$ is a Zariski-open dense subset and $\varphi_U : U \to \mathbb{P}^n_k$ is a morphism, with two pairs equivalent if they agree on the intersection of their domains. The locus $\operatorname{dom}(\varphi) \subset C$ where $\varphi$ is regular — the union of all such $U$ — is a Zariski-open subset of $C$, and the morphism $\varphi: \operatorname{dom}(\varphi) \to \mathbb{P}^n_k$ is the maximal regular extension. To prove $\varphi$ is a morphism on all of $C$, it suffices to show that for every $p \in C$, there is a Zariski-open neighbourhood $V_p \ni p$ on which $\varphi$ is regular — for then the regular morphisms on the various $V_p$'s glue to a morphism on $\bigcup_p V_p = C$. So fix $p \in C$ throughout the rest of the proof. [/step] [step:Choose a presentation $\varphi = [G_0 : \cdots : G_n]$ with $G_i \in \mathcal{O}_C(\eta)$] Pick any nonempty open $U \subset \operatorname{dom}(\varphi)$ on which $\varphi$ is given by homogeneous polynomial coordinates. Working in the function field $\mathcal{O}_C(\eta)$ — which equals the field of fractions of $\mathcal{O}_{C,q}$ for every $q \in C$ (see [Function Field is Finitely Generated](/theorems/2142)) — we may divide all coordinates by any common nonzero coordinate to obtain a presentation \begin{align*} \varphi = [G_0 : G_1 : \cdots : G_n], \qquad G_i \in \mathcal{O}_C(\eta), \end{align*} not all zero, valid as rational maps; alternatively, we may keep the polynomial presentation directly — what matters is only that each homogeneous coordinate is a rational function, not necessarily regular at $p$. Since not all $G_i$ are zero (else $\varphi$ would not be a well-defined point of $\mathbb{P}^n_k$), at least one $G_j \in \mathcal{O}_C(\eta)^\times$ — that is, at least one coordinate is a nonzero element of the function field. The equivalence relation defining points of $\mathbb{P}^n_k$ is scaling by elements of $k^\times$ on the level of constants, but at the level of rational maps we may freely scale by any nonzero element $\lambda \in \mathcal{O}_C(\eta)^\times$ — for the maps $[G_0 : \cdots : G_n]$ and $[\lambda G_0 : \cdots : \lambda G_n]$ define the same rational map (they agree on the open subset where $\lambda$ is regular and nonzero, and rational maps are determined by their restriction to any open dense set). [/step] [step:Apply the order homomorphism to renormalise the coordinates] At the smooth point $p \in C$, by the [order homomorphism](/theorems/2170), there is a group homomorphism \begin{align*} \operatorname{ord}_p : \mathcal{O}_C(\eta)^\times \to \mathbb{Z}, \end{align*} and $f \in \mathcal{O}_{C,p} \iff \operatorname{ord}_p(f) \geq 0$ (with the convention that $0 \in \mathcal{O}_{C,p}$). Let $t = t_p \in \mathfrak{m}_p$ be a uniformiser at $p$, so $\operatorname{ord}_p(t) = 1$ and $(t) = \mathfrak{m}_p$. Define \begin{align*} m := \min\{\operatorname{ord}_p(G_i) : G_i \neq 0\} \in \mathbb{Z}. \end{align*} This minimum is over a nonempty finite set (at least one $G_j \neq 0$, by Step 2), so $m$ is a well-defined integer. (We extend $\operatorname{ord}_p$ to $0$ by the convention $\operatorname{ord}_p(0) = +\infty$, so the minimum picks up only the nonzero $G_i$.) Multiply every coordinate by $t^{-m} \in \mathcal{O}_C(\eta)^\times$: \begin{align*} \varphi = [t^{-m} G_0 : t^{-m} G_1 : \cdots : t^{-m} G_n]. \end{align*} This is the same rational map as $[G_0 : \cdots : G_n]$ — they differ only by scaling by the nonzero rational function $t^{-m}$, which is allowed by the projective coordinate equivalence as discussed in Step 2. [guided] Why is multiplying by $t^{-m}$ the right move? In affine algebraic geometry, regularity of a rational function $G$ at $p$ is the condition $G \in \mathcal{O}_{C,p}$, i.e. $\operatorname{ord}_p(G) \geq 0$. In **projective** algebraic geometry, the homogeneous coordinates carry an extra freedom — they can be scaled by any nonzero rational function without changing the underlying map. So we can adjust orders globally, not pin them down absolutely. The minimum order $m = \min_i \operatorname{ord}_p(G_i)$ is the obstruction to all coordinates being regular at $p$: it tells us how badly the worst coordinate diverges (or how cautiously the most-vanishing coordinate vanishes). By scaling the entire tuple by $t^{-m}$, we shift every order down by $m$: \begin{align*} \operatorname{ord}_p(t^{-m} G_i) = \operatorname{ord}_p(t^{-m}) + \operatorname{ord}_p(G_i) = -m + \operatorname{ord}_p(G_i), \end{align*} using that $\operatorname{ord}_p$ is a group homomorphism (Theorem 2170). After shifting: \begin{itemize} \item Every coordinate has order $\geq -m + m = 0$, i.e. lies in $\mathcal{O}_{C,p}$. So every coordinate is regular at $p$. \item At least one coordinate — the one(s) achieving the minimum $m$ — has order exactly $-m + m = 0$, so it lies in $\mathcal{O}_{C,p} \setminus \mathfrak{m}_p$. By the second part of the order-homomorphism characterisation, this means it does not vanish at $p$ (i.e. its residue in $\mathcal{O}_{C,p} / \mathfrak{m}_p$ is nonzero). \end{itemize} The combination — all coordinates regular, at least one nonvanishing — is precisely what is needed for the homogeneous tuple to define a point of $\mathbb{P}^n_k$ at $p$. The single-variable nature of $\mathfrak{m}_p$ — which is principal, generated by $t_p$ — is essential: it lets us encode "vanishing to order $r$" as multiplication by $t^r$, so that scaling the tuple is a single uniform operation. On a higher-dimensional smooth variety, the maximal ideal at a point is no longer principal, and rational maps from smooth varieties to projective space can genuinely fail to extend (e.g. the rational map $\mathbb{A}^2 \dashrightarrow \mathbb{P}^1$, $(x,y) \mapsto [x:y]$ fails to extend at the origin). [/guided] [/step] [step:Check the renormalised coordinates define a morphism on a neighbourhood of $p$] Set $\widetilde G_i := t^{-m} G_i \in \mathcal{O}_C(\eta)$ for $i = 0, \ldots, n$. We verify two properties. **Every $\widetilde G_i$ lies in $\mathcal{O}_{C,p}$.** For each $i$, either $G_i = 0$ (so $\widetilde G_i = 0 \in \mathcal{O}_{C,p}$ immediately), or $G_i \neq 0$, in which case \begin{align*} \operatorname{ord}_p(\widetilde G_i) = \operatorname{ord}_p(t^{-m}) + \operatorname{ord}_p(G_i) = -m + \operatorname{ord}_p(G_i) \geq -m + m = 0, \end{align*} using that $\operatorname{ord}_p : \mathcal{O}_C(\eta)^\times \to \mathbb{Z}$ is a group homomorphism (Theorem 2170, Step 4) and the definition of $m$. By the [order homomorphism](/theorems/2170) characterisation, $\operatorname{ord}_p(\widetilde G_i) \geq 0$ implies $\widetilde G_i \in \mathcal{O}_{C,p}$. **At least one $\widetilde G_i$ does not vanish at $p$.** Pick $j$ with $\operatorname{ord}_p(G_j) = m$ (such $j$ exists by definition of the minimum). Then \begin{align*} \operatorname{ord}_p(\widetilde G_j) = -m + m = 0. \end{align*} Again by the [order homomorphism](/theorems/2170) characterisation, $\operatorname{ord}_p(\widetilde G_j) = 0$ means $\widetilde G_j \in \mathcal{O}_{C,p} \setminus \mathfrak{m}_p = \mathcal{O}_{C,p}^\times$, and in particular the residue of $\widetilde G_j$ in $k = \mathcal{O}_{C,p}/\mathfrak{m}_p$ is nonzero — that is, $\widetilde G_j(p) \neq 0$. **Spread to a neighbourhood.** Since $\widetilde G_0, \ldots, \widetilde G_n \in \mathcal{O}_{C,p}$, each is regular on some Zariski-open neighbourhood of $p$ in $C$. The intersection $V_p$ of these $n+1$ neighbourhoods, intersected further with the locus $\{\widetilde G_j \neq 0\}$ (which is Zariski-open since $\widetilde G_j$ is regular at $p$ and nonvanishing there), is a Zariski-open neighbourhood of $p$ on which all $\widetilde G_i$ are regular and $\widetilde G_j$ is nonvanishing. On $V_p$, the map \begin{align*} V_p &\to \mathbb{P}^n_k \\ q &\mapsto [\widetilde G_0(q) : \cdots : \widetilde G_n(q)] \end{align*} is well-defined (the tuple is never identically zero, since $\widetilde G_j(q) \neq 0$ throughout $V_p$) and is a morphism (by the standard criterion that a tuple of regular functions, not all vanishing, defines a morphism into projective space). It agrees with $\varphi$ on $V_p \cap \operatorname{dom}(\varphi)$, since both are presentations of the same rational map. [/step] [step:Glue the local extensions to obtain a morphism on $C$] By Step 4, for every $p \in C$ there is a Zariski-open neighbourhood $V_p \ni p$ on which $\varphi$ extends to a morphism $\varphi_p : V_p \to \mathbb{P}^n_k$. The collection $\{V_p\}_{p \in C}$ covers $C$. On any overlap $V_p \cap V_q$, the two extensions $\varphi_p$ and $\varphi_q$ both extend the same rational map $\varphi$, so they agree on the open dense subset $V_p \cap V_q \cap \operatorname{dom}(\varphi)$. Two morphisms from a variety to a separated variety (such as $\mathbb{P}^n_k$) that agree on a dense subset must agree on the entire intersection of their domains: this is the [identity principle for morphisms into separated varieties](/theorems/???), valid because the equaliser of two morphisms into a separated target is closed. The morphisms $\varphi_p$ therefore glue uniquely to a single morphism $\Phi: C \to \mathbb{P}^n_k$, which extends $\varphi$. Since $\varphi$ was a rational map and $\Phi$ extends it everywhere, $\varphi$ itself is a morphism. This completes the proof. [/step]