The canonical map: base-point-freeness and the hyperelliptic exception

(No canonical base point.) Suppose a geometric point $p$ were a base point of $\mathrm{\mid }{K}_C\mathrm{\mid }$. Then $h^0({\omega }_C(-p))=g$. By Serre duality [F1], $h^1({\mathcal{O}}_C(p))=g$, and Riemann--Roch [F2] gives $l(p)=1+1-g+g=2$. Since $l(0)=1$, there is a nonconstant function with poles bounded by $p$. A nonconstant function on a proper curve has a pole, so its pole divisor has degree one. By [F7] it gives a degree-one map to ${\mathbf{P}}_{\bar{k}}^1$ and hence an isomorphism, contradicting $g\ge 2$. Thus the canonical system is base-point-free over $\bar{k}$ and, by [F6], over $k$.

(Canonical map.) By [F3], the full canonical space has dimension $g$; by [F5] and step 1.1 its generating sections define ${\varphi }_K:C\to {\mathbf{P}}_k^{g-1}$ with ${\varphi }_K^{\ast }\mathcal{O}(1)\cong {\omega }_C$. The same construction identifies canonical divisors with hyperplane pullbacks. This proves assertion (1).

(Failure to separate points.) Over $\bar{k}$, let $p\mathrm{\ne }q$ be geometric points. If the canonical sections fail to separate them, their evaluation map to the two one-dimensional fibers has rank one: each component is nonzero by base-point-freeness. Its kernel is $H^0({\omega }_C(-p-q))$, so this space has dimension $g-1$. Duality [F1] gives $h^1({\mathcal{O}}_C(p+q))=g-1$, and Riemann--Roch [F2] gives $h^0({\mathcal{O}}_C(p+q))=2$.

(Failure to separate a tangent.) At a geometric point $p$, the first-jet evaluation has target the two-dimensional restriction of ${\omega }_C$ to $2p$. Base-point-freeness makes its value component nonzero. If the tangent direction is not separated, this map has rank one, and its kernel $H^0({\omega }_C(-2p))$ has dimension $g-1$. By [F1], $h^1({\mathcal{O}}_C(2p))=h^0({\omega }_C(-2p))=g-1$; Riemann--Roch [F2] then gives $h^0({\mathcal{O}}_C(2p))=2$.

(The resulting degree-two map.) In either step 2.2 or 2.3, the resulting two-dimensional Riemann--Roch space contains a nonconstant function. Its pole divisor is bounded by the effective divisor of degree two used there. It cannot have degree zero, and degree one would make $C_{\bar{k}}$ isomorphic to ${\mathbf{P}}_{\bar{k}}^1$ by [F7], contradicting $g\ge 2$. Its pole divisor therefore has degree two, and [F7] gives a degree-two map to ${\mathbf{P}}_{\bar{k}}^1$. It follows that a geometrically nonhyperelliptic curve separates every pair of geometric points and every tangent direction.

(Properness and the generic fiber.) Assume $g\ge 3$ and work over $\bar{k}$. The canonical map is nonconstant because $\mathrm{deg}{\omega }_C=2g-2>0$. Its target projective space is separated, so [F10] makes it proper. Let $Y$ be its scheme-theoretic image. The source and image are integral. Choose a standard projective chart containing the generic point of $Y$. Nonconstancy supplies a nonconstant affine coordinate ratio $h\in \bar{k}(Y)$ on that chart. The dominant map $C_{\bar{k}}\to Y$ embeds function fields by [F13], so regard $h$ as a nonconstant element of $\bar{k}(C)$. Since $\bar{k}$ is algebraically closed in $\bar{k}(C)$ by [F16], this nonconstant $h$ is transcendental over $\bar{k}$. Thus ${\mathrm{trdeg}}_{\bar{k}}\bar{k}(Y)=1$: it is at least one from $h$, and at most one from the inclusion $\bar{k}(Y)\hookrightarrow \bar{k}(C)$. The affine-domain dimension formula in [F16] on the chart shows that $Y$ is an integral curve. The function field $\bar{k}(C)$ is finitely generated of transcendence degree one, so $\bar{k}(C)\mathrm{/}\bar{k}(h)$ is algebraic and finitely generated, hence finite by [F17]. Since $\bar{k}(h)\subseteq \bar{k}(Y)\subseteq \bar{k}(C)$, the generic-fiber residue extension $\bar{k}(C)\mathrm{/}\bar{k}(Y)$ is finite. The generic point of $C$ maps to the generic point of $Y$. No closed point of $C$ maps to that generic point, since the morphism is proper and therefore sends the closed singleton of a closed point to a closed subset. Thus the generic fiber has just the source generic point, which is isolated, with the finite residue extension just proved. For a closed point $y\in Y$, the fiber is a proper closed subset of $C$: it cannot contain the generic point, which maps to the generic point of $Y$. By [F12] its support is a finite set of closed points. For each such point $p$, both $\kappa (p)$ and $\kappa (y)$ are finite over $\bar{k}$ by [F15], so $\kappa (p)\mathrm{/}\kappa (y)$ is finite; the points of this finite fiber are isolated. Hence every point is isolated in its fiber with finite residue extension, and the published fiber criterion [F11] makes the canonical map quasi-finite. Proper plus quasi-finite makes it finite.

(Canonical factorization.) Let $\varphi :C\to {\mathbf{P}}_k^1$ be a degree-two map and $L={\varphi }^{\ast }\mathcal{O}(1)$. The $g$ monomials in $H^0({\mathbf{P}}^1,\mathcal{O}(g-1))$ pull back to independent sections of $L^{\otimes (g-1)}$, so $h^0(L^{\otimes (g-1)})\ge g$. Its degree is $2g-2$. Riemann--Roch and Serre duality as in the proof of Hyperelliptic curves and hyperelliptic maps give $L^{\otimes (g-1)}\cong {\omega }_C$. The pulled-back monomials are therefore a basis of $H^0(C,{\omega }_C)$, and the canonical map is the Veronese factorization, up to projective coordinates. The same argument over $\bar{k}$ applies to a geometric degree-two map even when its quotient does not descend to ${\mathbf{P}}_k^1$.

(Local closed-immersion argument over $\bar{k}$.) For the sufficient direction of assertion (2), assume $g\ge 3$ and that $C_{\bar{k}}$ is geometrically nonhyperelliptic. By Step 3.1 the canonical map separates geometric points and tangent directions; Step 3.2 proves that it is finite. Let $Y$ be the scheme-theoretic image of ${\varphi }_K$ in ${\mathbf{P}}_{\bar{k}}^{g-1}$. By Step 3.2, the map to the ambient projective space is finite. Cover $Y$ by opens $V=Y\cap W$, where $W=\mathrm{Spec}R$ ranges over the standard affine charts of the ambient projective space. Since $Y$ is closed, each $V=\mathrm{Spec}A$ is affine with $A$ a quotient of $R$. Since the map factors through the closed subscheme $Y$, its inverse image of $V$ is its inverse image of $W$, say $\mathrm{Spec}B$. Finiteness over $W$ gives finitely many $R$-module generators of $B$; the $R$-action factors through its quotient $A$, so these also generate $B$ over $A$. The scheme-theoretic-image property [F13] gives $A\hookrightarrow B$. At each closed point $y\in Y$, properness gives a point above it and point separation gives exactly one such point $p$. Let $B_y:=B{\otimes }_A{A}_y$. If $b_1,\dots ,b_n$ generate $B$ over $A$, their images generate $B_y$ over $A_y$, so $B_y$ is finite over $A_y$; by [F19] the ring map is integral. Every maximal ideal of $B_y$ contracts to the maximal ideal of the local ring $A_y$, again by [F19]. Maximal ideals of $B_y$ are precisely the points of the fiber over $y$: conversely, a prime over the maximal ideal is maximal because its quotient is an integral domain integral over the residue field $A_y\mathrm{/}{\mathfrak{m}}_{A_y}$, hence is a field. There is exactly one such point by point separation. Thus $B_y$ is local. If ${\mathfrak{q}}_y$ is its unique maximal ideal, then $B_y=(B_y{)}_{{\mathfrak{q}}_y}={\mathcal{O}}_{C_{\bar{k}},p}$; in particular this local ring is finite over $A_y$. The residue fields of $y$ and $p$ are $\bar{k}$. Tangent injectivity dualizes to a surjection ${\mathfrak{m}}_{A_y}\mathrm{/}{\mathfrak{m}}_{A_y}^2\to {\mathfrak{m}}_{B_y}\mathrm{/}{\mathfrak{m}}_{B_y}^2$. Hence ${\mathfrak{m}}_{B_y}={\mathfrak{m}}_{A_y}{B}_y+{\mathfrak{m}}_{B_y}^2$. The module ${\mathfrak{m}}_{B_y}\mathrm{/}{\mathfrak{m}}_{A_y}{B}_y$ is finite over the local ring $B_y$, so [F18] gives ${\mathfrak{m}}_{B_y}={\mathfrak{m}}_{A_y}{B}_y$. Equality of residue fields now gives $B_y=A_y+{\mathfrak{m}}_{A_y}{B}_y$. The finite $A_y$-module $B_y\mathrm{/}{A}_y$ is therefore equal to its product by ${\mathfrak{m}}_{A_y}$, so [F18] over $A_y$ gives $A_y=B_y$. For each affine $V=\mathrm{Spec}A$ in this cover, this proves that the cokernel of $A\hookrightarrow B$ vanishes after localization at every maximal ideal: maximal ideals are closed points because $A$ is of finite type over $\bar{k}$. A nonzero finite module has a maximal ideal in its support, so the cokernel is zero. Thus $A\to B$ is an isomorphism on every such affine, and $C_{\bar{k}}\to Y$ is an isomorphism. Since $Y$ is a closed subscheme, ${\varphi }_{K,\bar{k}}$ is a closed immersion. This proves the local criterion directly from point and tangent separation; it does not invoke a degree-threshold very-ampleness result.

(The factorization is not an embedding.) The Veronese map is a closed immersion. If its composite with $\varphi$ were a closed immersion, then $\varphi$ would be a closed immersion into its Veronese image: on affine charts this follows by factoring the surjection of coordinate rings for the composite through the coordinate ring of the Veronese image. But $\varphi$ is finite and surjective of degree two. A surjective closed immersion into ${\mathbf{P}}_{\bar{k}}^1$ is an isomorphism by [F20], since that target is reduced. This would force $\varphi$ to have degree one, a contradiction. Thus the canonical map is not a closed immersion whenever $C$ is geometrically hyperelliptic. Its generic degree onto the rational normal curve is two. This gives the necessary geometric condition in assertion (2) without any separability assumption.

(Generic geometric fibers.) For completeness, every degree-two map from $C_{\bar{k}}$ with $g\ge 2$ is separable, including in characteristic two. Only characteristic two needs proof. Put $\mathrm{\Omega }=\bar{k}$ and let $\mathrm{\Omega }(C)\mathrm{/}\mathrm{\Omega }(t)$ be the induced extension. If it were purely inseparable, choose $\alpha$ generating it with ${\alpha }^2=h(t)$ for $h\in \mathrm{\Omega }(t)$ nonsquare. Write $h=P(t)\mathrm{/}Q(t)$ with $P,Q\in \mathrm{\Omega }[t]$. Because $\mathrm{\Omega }$ is perfect, there are $P_0,Q_0\in \mathrm{\Omega }[z]$ such that $P(z^2)=P_0(z{)}^2$ and $Q(z^2)=Q_0(z{)}^2$. Under the embedding $\mathrm{\Omega }(t)\hookrightarrow \mathrm{\Omega }(z)$ given by $t\mapsto z^2$, the element $\alpha$ can be sent to $P_0(z)\mathrm{/}{Q}_0(z)$, whose square is $h(z^2)$. Irreducibility of $X^2-h$ makes this a field embedding $\mathrm{\Omega }(C)\hookrightarrow \mathrm{\Omega }(z)$. Both extensions over $\mathrm{\Omega }(t)$ have degree two, so the image is all of $\mathrm{\Omega }(z)$. Hence $C_{\bar{k}}$ is birational to ${\mathbf{P}}_{\bar{k}}^1$, and smooth proper birational curves are isomorphic; this contradicts $g\ge 2$. Thus the degree-two extension is separable. Its geometric generic fiber consists of two distinct points, which the canonical map identifies through the Veronese factorization.

(Finiteness over $k$ and descent.) Under the same hypotheses, the canonical map over $k$ is proper, and the following fiber argument proves it quasi-finite, hence finite. Its scheme-theoretic image $Y$ is integral. Choose a standard projective chart containing its generic point. Nonconstancy supplies a nonconstant affine coordinate ratio $h\in k(Y)$. The dominant map $C\to Y$ embeds function fields by [F13], so regard $h$ as a nonconstant element of $k(C)$. Geometric integrality makes $h$ transcendental over $k$ by [F16]. The inclusion $k(Y)\hookrightarrow k(C)$ and the fact that $k(C)$ has transcendence degree one show that $k(Y)$ has transcendence degree one, so [F16] makes $Y$ an integral curve. Finite generation of $k(C)$ of transcendence degree one makes $k(C)\mathrm{/}k(h)$, and therefore $k(C)\mathrm{/}k(Y)$, finite. Thus the generic fiber is the single isolated generic point with finite residue extension. The generic point of $C$ maps to the generic point of $Y$, and properness keeps closed source points from mapping there: a proper map sends the closed singleton of a closed source point to a closed subset of the projective target, whereas the generic point of the integral curve $Y$ is not closed. Each closed fiber is a proper closed subset of $C$ with finite support by [F12], and its residue extensions are finite by [F15]. The fiber criterion and properness then give finiteness by [F11]. On each standard affine chart $U=\mathrm{Spec}A$ of ${\mathbf{P}}_k^{g-1}$, write ${\varphi }_K^{-1}(U)=\mathrm{Spec}B$; finiteness makes $B$ a finite $A$-module. Its base change to $\bar{k}$ is the chart map of the closed immersion proved above, so $A{\otimes }_k\bar{k}\to B{\otimes }_k\bar{k}$ is surjective. The cokernel of $A\to B$ tensors to zero; since $\bar{k}\mathrm{/}k$ is faithfully flat, that cokernel is zero. Therefore every chart map is surjective and ${\varphi }_K$ is a closed immersion over $k$ by [F14]. This proves the sufficient direction of assertion (2).

(The converse.) If $g=2$, the canonical map has target ${\mathbf{P}}_k^1$ and is nonconstant because $\mathrm{deg}{\omega }_C=2$. A closed immersion of a proper integral curve into ${\mathbf{P}}_k^1$ would have one-dimensional closed image, hence image all of ${\mathbf{P}}_k^1$, and would make $C\cong {\mathbf{P}}_k^1$, contradicting $g=2$. If $C_{\bar{k}}$ is geometrically hyperelliptic, step 4.2 shows that the base-changed canonical map is not a closed immersion; closed immersions remain closed after field extension. Thus a closed immersion forces $g\ge 3$ and geometric nonhyperellipticity. Together with steps 3.1--3.2, 4.1--4.2, and 5.1, this proves assertion (2).

(Conclusion.) Assertions (1)--(3) follow from steps 1.1--2.1, 3.1--3.3, 4.1--4.3, and 5.1--6.1. The Axiom of Choice enters through the cited duality, properness, finite-morphism, Nakayama, scheme-image, and descent suppliers; no additional choice is made in the point, tangent, or local-ring computations.