Riemann-Hurwitz for a tame double cover with 2r branch points

For every closed point $q$ of ${\mathbb{P}}_k^1$ and every closed point $p$ of $C$ with $f(p)=q$, [F6] gives ${\mathrm{deg}}_k{f}^{\ast }[q]=2{\mathrm{deg}}_{k}q$ and $f^{\ast }[q]={\sum }_{p\in f^{-1}(q)}{e}_p[p]$, hence ${\sum }_{p\in f^{-1}(q)}{e}_p{f}_p=2$ with $f_p=[\kappa (p):\kappa (q)]$; consequently $e_p{f}_p\le 2$, so $e_p\in \{1,2\}$, $f_p\le 2$, every residue extension is separable by [F1], and all ramification is tame.

In case (ii), $h$ has an irreducible factor $g$ of multiplicity one. The $g$-adic valuation of $h$ in $k(x)$ is therefore $1$, so $h$ is not a square in $k(x)$ and $y^2-h(x)$ is irreducible over $k(x)$. Thus $K=k(x)[y]\mathrm{/}(y^2-h(x))$ is a field, finitely generated of transcendence degree one over $k$.

The field $k$ is relatively algebraically closed in $K$. Let $A=k[x,y]\mathrm{/}(y^2-h(x))$, so $K=\mathrm{Frac}(A)$. Over $\bar{k}$, $h$ remains squarefree and has a simple root; its valuation there is odd, so it is not a square in $\bar{k}(x)$. Hence $A{\otimes }_k\bar{k}\cong \bar{k}[x,y]\mathrm{/}(y^2-h(x))$ is a domain. Since $\bar{k}\mathrm{/}k$ is flat, $A\hookrightarrow A{\otimes }_k\bar{k}$ is injective and $K{\otimes }_k\bar{k}\cong S^{-1}(A{\otimes }_k\bar{k})$, where $S=A\setminus \{0\}$, is a domain. If $\alpha \in K$ is algebraic over $k$ but not in $k$, then $L=k(\alpha )$ is a finite extension of degree greater than one. Since $k$ is perfect, $L\mathrm{/}k$ is separable; therefore $L{\otimes }_k\bar{k}$ is a product of $[L:k]>1$ copies of $\bar{k}$, not a domain. Flatness makes $L{\otimes }_k\bar{k}\hookrightarrow K{\otimes }_k\bar{k}$, a contradiction. Thus $k$ is relatively algebraically closed in $K$.

Construct the model explicitly. Write $h=x^{2r}+a_1{x}^{2r-1}+\cdots +a_{2r}$ and define $w(s)=s^{2r}h(s^{-1})=1+a_{1}s+\cdots +a_{2r}{s}^{2r}$. Glue $U_0=\mathrm{Spec}{B}_0$, $B_0=k[x,y]\mathrm{/}(y^2-h(x))$, to $U_{\mathrm{\infty }}=\mathrm{Spec}{B}_{\mathrm{\infty }}$, $B_{\mathrm{\infty }}=k[s,z]\mathrm{/}(z^2-w(s))$, on $D(x)$ and $D(s)$ by $s=x^{-1}$ and $z=y\mathrm{/}{x}^r$. The equations agree on this overlap. Each chart is a hypersurface, hence finitely presented over $k$; the resulting map $\pi :C_h\to {\mathbb{P}}_k^1$ has inverse images $U_0,U_{\mathrm{\infty }}$ over the standard affine charts, and each chart ring is free of rank two over its base coordinate ring, so $\pi$ is finite of degree two. The polynomial $w$ is squarefree: its roots are the reciprocals of the nonzero roots of $h$, and $w(0)=1$; there is at least one nonzero root since $h$ has $2r\ge 2$ distinct roots and at most one is zero. On either chart, at a prime containing $y$ (respectively $z$), the derivative $h^{\mathrm{\prime }}(x)$ (respectively $w^{\mathrm{\prime }}(s)$) is a unit because the polynomial is squarefree; away from those primes, $2y$ (respectively $2z$) is a unit. The relative Jacobian criterion therefore makes both charts smooth over $k$. Over $\bar{k}$, each chart ring is a domain because its squarefree polynomial has a simple root and is not a square in the rational function field. The charts meet in a nonempty open, so their gluing remains integral after base change. Their common function field is $K$, and both chart rings have dimension one by [F18], so this is a curve; it is smooth and geometrically integral. The finite map to the proper curve ${\mathbb{P}}_k^1$ makes it proper by [F17]. By [F12] it is the smooth proper geometrically integral model of $K$, unique up to the isomorphism compatible with the specified identification of its function field. The projective-embedding result in [F17] makes this model projective. The Given Axiom of Choice supplies Dependent Choice by [F19] for the Cartier-to-Weil divisor suppliers used below.

For every closed point $p$ of $C_h$ with $q=\pi (p)$, the divisor identity ${\pi }^{\ast }\mathrm{div}(h)=2\mathrm{div}(y)$ and the additivity and closed-point formula of [F6] give $2{\mathrm{ord}}_p(y)=e_p{\mathrm{ord}}_q(h)$, where ${\mathrm{ord}}_q(h)$ is the order of the rational function $h$ at $q$; since $h$ is monic of degree $2r$, its only pole is at $\mathrm{\infty }$ and there ${\mathrm{ord}}_{\mathrm{\infty }}(h)=-2r$, while at a finite point $q=\mathrm{V}(g)$ one has ${\mathrm{ord}}_q(h)=1$ if $g\mid h$ (multiplicity of the irreducible factor, $h$ squarefree) and ${\mathrm{ord}}_q(h)=0$ otherwise.

If ${\mathrm{ord}}_p(y)=0$, Step 1.5 shows that $q\mathrm{\ne }\mathrm{\infty }$ and that $h$ has order zero at $q$. Thus $p$ lies on the finite chart, where the actual relative-differentials presentation is ${\mathrm{\Omega }}_{B_0\mathrm{/}k[x]}\cong (B_0\mathrm{/}(2y))dy$ by [F16]. The element $y$ is a unit at $p$, so this localized module is zero; hence ${\mathrm{\ell }}_p=0$ and $p$ is unramified with separable residue extension, that is $e_p=1$ by [F4].

In case (i), the branch locus is exactly $\{q_1,\dots ,q_{2r}\}$ and every point $p$ over $q_i$ has $e_p=2$; by Step 1.1 there is exactly one such $p=p_i$, with $f_{p_i}=1$, so $\kappa (p_i)=\kappa (q_i)=k$ and $e_{p_i}=2$, while every point over a non-branch point has $e_p=1$.

If ${\mathrm{ord}}_p(y)=m>0$, then Step 1.5 shows that $q=\mathrm{V}(g)$ for an irreducible factor $g$ of $h$, with ${\mathrm{ord}}_q(h)=1$, and $e_p=2m\ge 2$. The fibre of the actual finite chart over $q$ is $\kappa (q)[y]\mathrm{/}(y^2)$, so it has a unique point with residue field $\kappa (q)$. Locally write $h=gu$ with $u$ a unit. The maximal ideal at that point is $(g,y)$ and $g=y^2\mathrm{/}u$, hence it is $(y)$; by [F9], $y$ is a uniformizer. Therefore $e_p={\mathrm{ord}}_p(g)=2$ and $m=1$. The point is tame and has ${\mathrm{\ell }}_p=e_p-1=1$ by [F4].

If ${\mathrm{ord}}_p(y)<0$, then Step 1.5 forces $q=\mathrm{\infty }$. On the infinity chart the coordinates are $s=1\mathrm{/}x$ and $z=y\mathrm{/}{x}^r$, with $z^2=w(s)$. The fibre over $s=0$ is $k[z]\mathrm{/}(z^2-1)\cong k\times k$, so it consists of two distinct $k$-rational points. At each, $z$ is a unit and [F16] gives ${\mathrm{\Omega }}_{B_{\mathrm{\infty }}\mathrm{/}k[s]}\cong (B_{\mathrm{\infty }}\mathrm{/}(2z))dz=0$ locally; each is unramified, hence has $e_p=1$. These are the two rational points at infinity.

In case (i), the ramified points $p_i$ are tame with separable residue extension, so [F4] gives ${\mathrm{\ell }}_{p_i}=e_{p_i}-1=1$, and every unramified point has ${\mathrm{\ell }}_p=0$ by the criterion ${\mathrm{\ell }}_p=0\iff e_p=1$ and separable residue; hence $R_f={\sum }_{i=1}^{2r}[p_i]$ with ${\mathrm{deg}}_k{R}_f={\sum }_i{\mathrm{deg}}_k(p_i)=2r$.

Combining Steps 2.1, 2.3 and 2.4, the ramified points of $\pi$ are exactly the points $p_g$ over the closed points $\mathrm{V}(g)$ for the irreducible factors $g$ of $h$, each unique with $e_{p_g}=2$, $f_{p_g}=1$, hence $\kappa (p_g)=k[x]\mathrm{/}(g)$ and ${\mathrm{\ell }}_{p_g}=1$; every other point is unramified with ${\mathrm{\ell }}_p=0$; therefore $R_{\pi }={\sum }_{g\mid h}[p_g]$ and ${\mathrm{deg}}_k{R}_{\pi }={\sum }_{g\mid h}\mathrm{deg}(g)=\mathrm{deg}(h)=2r$.

In case (i), $k(C)\mathrm{/}k({\mathbb{P}}_k^1)$ has degree $2$ and is separable by [F1], so Riemann-Hurwitz applies: $2g(C)-2=2(2\cdot 0-2)+{\mathrm{deg}}_k{R}_f=-4+2r$, and therefore $g(C)=r-1$ with $g$ as in [F10].

In case (ii) the extension $k(C_h)\mathrm{/}k({\mathbb{P}}_k^1)$ is separable by [F1], so Riemann-Hurwitz gives $2g(C_h)-2=2(2\cdot 0-2)+{\mathrm{deg}}_k{R}_{\pi }=-4+2r$, that is $g(C_h)=r-1$, with $g$ as in [F10]; when $h$ splits over $k$ with distinct roots $c_1,\dots ,c_{2r}\in k$ the branch locus is exactly those $2r$ distinct $k$-rational points, so $C_h\to {\mathbb{P}}_k^1$ is a cover of case (i) and the two computations agree.

The small cases: for $r=1$ the model of $y^2=x^2-1$ is the smooth plane conic $Y^2=X^2-Z^2$ with rational point $[1:0:1]$; its partial derivatives $-2X,2Y,2Z$ cannot vanish simultaneously at a projective point. Step 4.2 gives genus zero, and the rational point defines a degree-one divisor, so [F20] gives $C_h\cong {\mathbb{P}}_k^1$. The projection identification is explicit: $[U:V]⟼[U^2+V^2:U^2-V^2:2UV]\mathrm{.}$ On $X+Y\mathrm{\ne }0$ its inverse is $[X+Y:Z]$, and on $X-Y\mathrm{\ne }0$ it is $[Z:X-Y]$; the formulas agree on the overlap. For $r=2$ and $c\in k\setminus \{0,1\}$ the polynomial $(x^2-1)(x^2-c)$ is monic squarefree of degree $4$, so the model gives a quartic family of genus $1$. For $r=3$ every squarefree sextic gives a model of genus $2$. Each case is consistent with $g(C_h)=r-1$.