Ramification of the double cover y^2=f(x)

The affine curve is smooth. Write $h=y^2-f(x)\in k[x,y]$; its partial derivatives are $2y$ and $-f^{\mathrm{\prime }}(x)$. At a common zero one has $y=0$ (characteristic $\mathrm{\ne }2$) and $f(x)=f^{\mathrm{\prime }}(x)=0$, so $x$ is a multiple root of $f$, contrary to squarefreeness. By the Jacobian criterion every closed point of $U$ is regular.

The plane model. The polynomial $G$ is homogeneous of degree $n$ and $Z\nmid G$, because $G(X,Y,0)=-F(X,0)=-a_n{X}^n$ with $a_n\mathrm{\ne }0$; its dehomogenization at $Z=1$ is $h=y^2-f(x)$, which is irreducible in $k[x,y]$ since ${\mathrm{deg}}_{y}h=2$ and $f$ is squarefree of degree at least three, hence not a square in $k[x]$. If $G=G_1{G}_2$ were a factorization into nonconstant forms, substituting $Z=1$ would express the irreducible $h$ as a product, so one factor would dehomogenize to a nonzero constant $c$; that factor is homogeneous of positive degree and equals $c$ on the hyperplane $Z=1$ at that point, hence is the form $c{Z}^{\mathrm{deg}{G}_i}$, forcing $Z\mid G$, a contradiction. Therefore $G$ is irreducible, $X$ is an integral plane curve of degree $n$, and $U=X\cap \{Z\mathrm{\ne }0\}$ is a dense open subscheme. By the plane-curve formula $H^0(X,{\mathcal{O}}_X)=k$ and $p_a(X)=\frac{(n-1)(n-2)}{2}$.

The only possible singular point. On the chart $Y\mathrm{\ne }0$ put $u=X\mathrm{/}Y$, $v=Z\mathrm{/}Y$, so that $X$ is cut out by $g(u,v)=v^{n-2}-F(u,v)$, where $F(u,v)={\sum }_{i=0}^n{a}_i{u}^i{v}^{n-i}$ is a binary form of degree $n$ with $a_n\mathrm{\ne }0$. The point $\mathrm{\infty }=(0:1:0)$ corresponds to $(u,v)=(0,0)$, and every term of $F$ and of its first partial derivatives has order at least $n-1\ge 2$ at the origin; hence the partials of $g$ at the origin are $-F_u=0$ and $(n-2)v^{n-3}-F_v$, which vanish at the origin for $n\ge 4$ and equal $(0,1)$ for $n=3$. Every other closed point of the chart lies either in $U$, which is regular by step 1.1, or has $Z=0$; but $Z=0$ on $X$ forces $X=0$, so the only such point is $\mathrm{\infty }$ itself. Hence $\mathrm{\infty }$ is the only possibly singular point of $X$, and it is singular exactly when $n\ge 4$.

A degree-two projection. The function field of $X$ is $k(X)=k(x,y)$ with $y^2=f(x)$ (step 1.2), and $y^2-f$ is irreducible over $k(x)$ because $f$ is squarefree of degree at least three and hence is not a square in $k(x)$; therefore $[k(X):k(x)]=2$. By [F5] the inclusion $k(x)\hookrightarrow k(C)=k(X)$ determines a dominant morphism $\pi :C\to {\mathbb{P}}_k^1$ with ${\pi }^{\ast }(x)=x$, and $\pi$ is finite and surjective of degree two.

The two affine charts. Finiteness of $\pi$ (step 2.2) makes the preimages $U^{\mathrm{\prime }}={\pi }^{-1}(\mathrm{Spec}k[x])$ and $C_t={\pi }^{-1}(\mathrm{Spec}k[t])$ of the two standard affine charts of ${\mathbb{P}}_k^1$ affine, with coordinate rings $A^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}$ module-finite over $k[x]$ and over $k[t]$ respectively (Finite morphisms of schemes), and $\mathrm{Frac}{A}^{\mathrm{\prime }}=\mathrm{Frac}{B}^{\mathrm{\prime }}=k(C)$. Put $t=1\mathrm{/}x$, $w=y{t}^{g+1}$, and $\psi (t)=t^{n}f(1\mathrm{/}t)$ when $n=2g+2$, $\psi (t)=t^{n+1}f(1\mathrm{/}t)$ when $n=2g+1$; then $w^2=\psi (t)$ in $k(C)$. The subrings $A=k[x,y]\mathrm{/}(y^2-f)\subseteq A^{\mathrm{\prime }}$ and $B=k[t,w]\mathrm{/}(w^2-\psi )\subseteq B^{\mathrm{\prime }}$ have $\mathrm{Frac}A=k(x,y)=k(C)$ and $\mathrm{Frac}B=k(t,w)=k(C)$, the latter because $[k(C):k(t)]=[k(C):k(x)]=2$ by step 2.2 while $w\mathrm{\notin }k(t)$. The polynomial $\psi$ is squarefree: its roots are the inverses of the nonzero roots of $f$, all simple because $f$ is squarefree, together with $t=0$ in the odd case, where $\psi =t\phi$ with $\phi (t)=t^{n}f(1\mathrm{/}t)$ of constant term $a_n\mathrm{\ne }0$, so that root is simple as well. Since $\mathrm{char}k\mathrm{\ne }2$, the Jacobian criterion [F1] shows that the localizations of $A$ (step 1.1) and of $B$ at maximal ideals are regular local rings, their localizations at the zero ideal are fraction fields, and so $A$ and $B$ are regular Noetherian domains, hence integrally closed [F11]; the same applies to $A^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}$, which are coordinate rings of affine opens of the smooth curve $C$. Now $x\in A^{\mathrm{\prime }}$ and $y^2=f(x)\in A^{\mathrm{\prime }}$ with $y\in k(C)=\mathrm{Frac}{A}^{\mathrm{\prime }}$, so $y\in A^{\mathrm{\prime }}$ and $A\subseteq A^{\mathrm{\prime }}$; dually $t\in B^{\mathrm{\prime }}$ and $w^2=\psi (t)\in B^{\mathrm{\prime }}$ give $w\in B^{\mathrm{\prime }}$ and $B\subseteq B^{\mathrm{\prime }}$. Every element of $A^{\mathrm{\prime }}$ is integral over $k[x]\subseteq A$ and every element of $B^{\mathrm{\prime }}$ is integral over $k[t]\subseteq B$, so integrally closedness gives $A^{\mathrm{\prime }}=A$ and $B^{\mathrm{\prime }}=B$. Consequently the closed points of the finite chart are the maximal ideals of $A$, the points of $C$ over $\mathrm{\infty }$ are the maximal ideals of $B$ lying over $(t)$, and the local rings of $C$ at these points are the corresponding localizations.

Ramification over finite points. Fix $x_0\in k$ and let $p\in C$ lie over $x_0$, so that $p$ corresponds to a maximal ideal of $A$ (step 3.1), while the pullback of the uniformizer $x-x_0$ of ${\mathbb{P}}_k^1$ at $x_0$ is the function $x-x_0$, whence $e_p={\mathrm{ord}}_p(x-x_0)$. If $f(x_0)\mathrm{\ne }0$, choose $y_0\in k^{\times }$ with $y_0^2=f(x_0)$ and write $f(x)-f(x_0)=(x-x_0)u(x)$, without requiring $u(x_0)\mathrm{\ne }0$. At $p=(x_0,y_0)$ the factor $y+y_0$ is a unit, so $(y-y_0)=(x-x_0)u(x)\mathrm{/}(y+y_0)$ in the local ring and its maximal ideal $(x-x_0,y-y_0)$ is $(x-x_0)$. At $p^{\mathrm{\prime }}=(x_0,-y_0)$ the factor $y-y_0$ is a unit, and the same identity gives $(y+y_0)=(x-x_0)u(x)\mathrm{/}(y-y_0)$, so the maximal ideal $(x-x_0,y+y_0)$ is again $(x-x_0)$. Thus $x-x_0$ is a uniformizer at both points and $e_p=1$, regardless of whether $x_0$ is a critical point of $f$; these are exactly the two points over $x_0$. If $f(x_0)=0$, write $f=(x-x_0)u(x)$ with $u(x_0)\mathrm{\ne }0$; then $x-x_0=y^{2}u(x{)}^{-1}$ in $k(C)$, so $x-x_0$ lies in the square of the maximal ideal at the unique point $p=(x_0,0)$ of $U$ and generates its square, whence the maximal ideal is generated by $y$, making $y$ a uniformizer with ${\mathrm{ord}}_p(x-x_0)=2$, $e_p=2$, and the fibre is $\{p\}$. In both cases the fibre-degree formula of [F6] shows that no further point lies over $x_0$, because the displayed contributions already sum to $\mathrm{deg}\pi =2$ with trivial residue extensions.

Ramification over infinity. The points of $C$ over $\mathrm{\infty }$ are the maximal ideals of $B$ over $(t)$, that is, the maximal ideals of $B\mathrm{/}(t)=k[w]\mathrm{/}(w^2-\psi (0))$ (step 3.1). If $n$ is even, then $\psi (0)=a_n\mathrm{\ne }0$ and $k[w]\mathrm{/}(w^2-a_n)\cong k\times k$, so there are exactly two points $p_{\pm }$ over $\mathrm{\infty }$, and $w$ is a unit at each of them. At $p_{+}$ with $c=\sqrt{a_n}$, the identity $(w-c)(w+c)=\psi (t)-\psi (0)=tg(t)$, where $g=(\psi -\psi (0))\mathrm{/}t\in k[t]$, shows that $w-c=tg(t)(w+c{)}^{-1}\in (t)$ because $w+c$ has nonzero residue $2c$ and is a unit; hence the maximal ideal $(t,w-c)$ equals $(t)$ and $t$ is a uniformizer at $p_{+}$, and the same computation with $-c$ at $p_{-}$ gives a uniformizer $t$ there too. If $n$ is odd, then $\psi (0)=0$ and $B\mathrm{/}(t)=k[w]\mathrm{/}(w^2)$ is local with maximal ideal $(w)$, so there is exactly one point $p_{\mathrm{\infty }}$ over $\mathrm{\infty }$; since $\psi =t\phi$ with $\phi (0)=a_n\mathrm{\ne }0$, the element $\phi (t)$ is a unit at $p_{\mathrm{\infty }}$, so $t=w^2\phi (t{)}^{-1}\in (w{)}^2$. In the local ring $B_{(t,w)}$, one has $B_{(t,w)}\mathrm{/}(w)\cong k[t{]}_{(t)}\mathrm{/}(t\phi (t))\cong k$, so its maximal ideal $(t,w)$ is $(w)$; hence $w$ is a uniformizer at $p_{\mathrm{\infty }}$ and ${\mathrm{ord}}_{p_{\mathrm{\infty }}}(t)=2$. In both cases the pullback of the uniformizer $t$ of ${\mathbb{P}}_k^1$ at $\mathrm{\infty }$ is the function $t$ itself, so the orders just computed are the ramification indices: $e_{p_{+}}=e_{p_{-}}=1$ for even $n$, and $e_{p_{\mathrm{\infty }}}=2$ for odd $n$.

Genus and projectivity. The proper plane model $X$ and finite normalization make $C$ proper of finite type by [F3, F10]. The finite projection gives the ample sheaf $L={\pi }^{\ast }\mathcal{O}(1)$ by [F10]. A positive power of $L$ is closed H-very ample over the Noetherian base field by [F10], so $C$ is projective. It is smooth of dimension one by [F3]. The two affine charts in step 3.1 are the inverse images of the standard cover of ${\mathbb{P}}_k^1$; their intersection is obtained by inverting $x$ or $t$, with $t=x^{-1}$ and $w=x^{-(g+1)}y$. Properness supplies separatedness. Thus [F9] applies and gives $h^1(C,{\mathcal{O}}_C)=g$, with classes $x^{-1}y,\dots ,x^{-g}y$. The genus definition in [F3] now gives $g(C)=g$.

The branch locus. Combining step 4.1 and step 4.2: over a point $x_0\in k$ with $f(x_0)\mathrm{\ne }0$ there are exactly two points of $C$, both with $e=1$; over each of the $n$ roots of $f$ there is exactly one point, with $e=2$; and over $\mathrm{\infty }$ there are two points with $e=1$ when $n$ is even and one point with $e=2$ when $n$ is odd. Since $k$ is algebraically closed, every residue field extension at these closed points is trivial, so the fibre-degree formula ${\sum }_{p\in {\pi }^{-1}(q)}{e}_p[\kappa (p):\kappa (q)]=\mathrm{deg}\pi =2$ holds at every closed point $q$ of ${\mathbb{P}}_k^1$, consistently with the counts. The ramification locus of $\pi$ is therefore the set of $n+[n\text{ odd}]=2g+2$ points over the roots of $f$ and, for odd $n$, over infinity, all of index two, and the branch locus is exactly the set of the $n$ roots of $f$ together with $\mathrm{\infty }$ when $n$ is odd, again $2g+2$ points; over each branch point lies exactly one ramification point, and over every other closed point of ${\mathbb{P}}_k^1$ lie two points with $e=1$.

The differential ${\omega }_0=\mathrm{d}x\mathrm{/}y$ and its divisor. The form ${\omega }_0=\mathrm{d}x\mathrm{/}y$ is a nonzero rational differential on $C$ [F8]: $y\mathrm{\ne }0$ in the field $k(C)$, and $\mathrm{d}x\mathrm{\ne }0$ because $k(C)\mathrm{/}k(x)$ is separable of degree two in characteristic $\mathrm{\ne }2$. At a point over $x_0$ with $f(x_0)\mathrm{\ne }0$, step 4.1 provides the uniformizer $x-x_0$ and $y$ is a unit, so $\mathrm{ord}(\mathrm{d}x\mathrm{/}y)=\mathrm{ord}(y^{-1}\mathrm{d}(x-x_0))=0$; at the point over a root $x_0$ of $f$, writing $f=(x-x_0)u$ with $u(x_0)\mathrm{\ne }0$ gives $x-x_0=y^{2}u(x{)}^{-1}$ and therefore $\mathrm{d}x\mathrm{/}y=2u(x{)}^{-1}(1+y^{2}u(x{)}^{-2}{u}^{\mathrm{\prime }}(x){)}^{-1}\mathrm{d}y\mathrm{.}$ The coefficient is a unit at that point, so $\mathrm{d}x\mathrm{/}y$ is a unit multiple of $\mathrm{d}y$ and has order $0$; this formula shows that $\mathrm{d}x$ itself is not a unit multiple of $\mathrm{d}y$ there. Over infinity, $x=1\mathrm{/}t$, $\mathrm{d}x=-t^{-2}\mathrm{d}t$ and $y=w{t}^{-(g+1)}$ give $\mathrm{d}x\mathrm{/}y=-t^{g-1}\mathrm{d}t\mathrm{/}w$. In the even case $t$ is a uniformizer and $w$ a unit at $p_{\pm }$ (step 4.2), so ${\mathrm{ord}}_{p_{\pm }}(\mathrm{d}x\mathrm{/}y)=g-1$; in the odd case $t=w^2\phi (t{)}^{-1}$ gives $\mathrm{d}t=2w\phi (t{)}^{-1}(1+w^2\phi (t{)}^{-2}{\phi }^{\mathrm{\prime }}(t){)}^{-1}\mathrm{d}w$, so that $\mathrm{d}t\mathrm{/}w$ is a unit multiple of $\mathrm{d}w$, and $t^{g-1}=w^{2g-2}\phi (t{)}^{-(g-1)}$ gives ${\mathrm{ord}}_{p_{\mathrm{\infty }}}(\mathrm{d}x\mathrm{/}y)=2g-2$. Hence $\mathrm{div}(\mathrm{d}x\mathrm{/}y)=(g-1)(p_{+}+p_{-})$ in the even case and $\mathrm{div}(\mathrm{d}x\mathrm{/}y)=(2g-2)p_{\mathrm{\infty }}$ in the odd case; both divisors have degree $2g-2$, and each is a canonical divisor on the smooth proper geometrically integral curve $C$.

The delta invariant at infinity. By [F4] one has $p_a(X)=g(X^{\mathrm{n}\mathrm{u}})+{\sum }_x{\delta }_x(X)$, the sum over the closed points of $X$; since ${\delta }_x$ vanishes at regular points and $\mathrm{\infty }$ is the only possibly singular point of $X$ (step 2.1), the sum reduces to ${\delta }_{\mathrm{\infty }}$, and it is empty when $n=3$, where $\mathrm{\infty }$ is regular and ${\delta }_{\mathrm{\infty }}=0$. With $p_a(X)=\frac{(n-1)(n-2)}{2}$ (step 1.2) and $g(X^{\mathrm{n}\mathrm{u}})=g(C)=g$ (step 4.3), ${\delta }_{\mathrm{\infty }}=\frac{(n-1)(n-2)}{2}-g,$ which is $2g^2$ for $n=2g+2$, $2g(g-1)$ for $n=2g+1$, and $0$ for $(n,g)=(3,1)$; in particular it is a nonnegative integer in each family.

Regular differentials. Let $\omega$ be a global section of the canonical bundle ${\omega }_C={\mathrm{\Omega }}_{C\mathrm{/}k}^1$, that is, a regular differential on $C$ [F8]; the zero section is the case $R=0$ below. If $\omega \mathrm{\ne }0$, then $\omega =h{\omega }_0$ for a unique $h\in k(C{)}^{\times }$ [F8]. Since ${\omega }_0$ has order $0$ at every point of the finite chart and generates the free rank-one module ${\omega }_{C,p}$ there (step 5.2), regularity of $\omega$ forces $h\in {\mathcal{O}}_C(U^{\mathrm{\prime }})=A=k[x,y]\mathrm{/}(y^2-f)$ (step 3.1), so $h=R(x)+S(x)y$ for unique $R,S\in k[x]$, the elements $1,y$ forming a $k(x)$-basis of $k(C)$ (step 2.2, step 3.1). Regularity on $U^{\mathrm{\prime }}$ is then automatic, and by [F8] it remains to impose at the points over infinity the condition ${\mathrm{ord}}_p(h)\ge -{\mathrm{ord}}_p({\omega }_0)$ coming from $\mathrm{div}(\omega )=\mathrm{div}(h)+\mathrm{div}({\omega }_0)$. In the even case, at $p_{\pm }$ one has $\mathrm{ord}(R(x))=-\mathrm{deg}R$ and $\mathrm{ord}(S(x)y)=-(g+1)-\mathrm{deg}S$ for nonzero $R$ and $S$, using $y=w{t}^{-(g+1)}$ with $w$ a unit, while ${\mathrm{ord}}_{p_{\pm }}({\omega }_0)=g-1$; hence ${\mathrm{ord}}_{p_{\pm }}(h)\ge -(g-1)$ is required. For $S=0$ this is exactly $\mathrm{deg}R\le g-1$. For $S\mathrm{\ne }0$: if the two orders differ, then $\mathrm{ord}(h)$ is the smaller one, and since $-(g+1)-\mathrm{deg}S\le -(g+1)<-(g-1)$ this is impossible; if the two orders are equal, so that $\mathrm{deg}R=g+1+\mathrm{deg}S$, then the coefficient of $t^{-\mathrm{deg}R}$ in $h$ at $p_{\pm }$ is $r+w(p_{\pm })s$ with leading coefficients $r,s\mathrm{\ne }0$, and regularity at both $p_{+}$ and $p_{-}$ would force $r+\sqrt{a_n}s=r-\sqrt{a_n}s=0$, impossible in characteristic $\mathrm{\ne }2$ with $a_n\mathrm{\ne }0$. Hence $S=0$ and $\mathrm{deg}R\le g-1$. In the odd case, at $p_{\mathrm{\infty }}$ one has $\mathrm{ord}(R(x))=-2\mathrm{deg}R$ and $\mathrm{ord}(S(x)y)=-(2g+1)-2\mathrm{deg}S$ for nonzero $R$ and $S$, while ${\mathrm{ord}}_{p_{\mathrm{\infty }}}({\omega }_0)=2g-2$; if $S\mathrm{\ne }0$, then $-(2g+1)-2\mathrm{deg}S\le -2g-1<-(2g-2)$, so if the orders of the two summands differ the order of $h$ is too small, and they cannot be equal because $-2\mathrm{deg}R$ is even while $-(2g+1)-2\mathrm{deg}S$ is odd. Hence again $S=0$, and $-2\mathrm{deg}R\ge -(2g-2)$ gives $\mathrm{deg}R\le g-1$. Conversely, every $R\in k[x]$ with $\mathrm{deg}R\le g-1$ yields a regular differential $R(x)\mathrm{d}x\mathrm{/}y$: it is regular on $U^{\mathrm{\prime }}$, and at infinity its order is $g-1-\mathrm{deg}R\ge 0$ in the even case and $2(g-1-\mathrm{deg}R)\ge 0$ in the odd case (step 5.2, step 4.2). Therefore $H^0(C,{\omega }_C)=\{R(x)\mathrm{d}x\mathrm{/}y:R\in k[x],\mathrm{deg}R\le g-1\},$ the differentials $\mathrm{d}x\mathrm{/}y,x\mathrm{d}x\mathrm{/}y,\dots ,x^{g-1}\mathrm{d}x\mathrm{/}y$ are linearly independent over $k$, and $h^0(C,{\omega }_C)=g$.

Conclusion. Step 1.1 shows that the affine curve $y^2=f(x)$ is smooth, and step 1.2 and step 2.1 identify the plane model $X=V_{+}(G)$ as an integral plane curve of degree $n$ with $p_a(X)=\frac{(n-1)(n-2)}{2}$ whose only possibly singular point is $\mathrm{\infty }$, singular exactly when $n\ge 4$. Step 2.2 exhibits the degree-two projection $\pi :C\to {\mathbb{P}}_k^1$ from the normalization, step 3.1 identifies the two standard affine charts with the explicit rings $A=k[x,y]\mathrm{/}(y^2-f)$ and $B=k[t,w]\mathrm{/}(w^2-\psi )$, and step 4.1 and step 4.2 compute the ramification over the finite points and over infinity. Step 5.1 shows that $\pi$ is finite and surjective of degree two, branched exactly at the $n$ roots of $f$ and, when $n$ is odd, at infinity — that is, at $2g+2$ branch points — with exactly one ramification point of index two over each of them and two points with $e=1$ over every other closed point of ${\mathbb{P}}_k^1$. Step 5.2 and step 6.1 compute $\mathrm{div}(\mathrm{d}x\mathrm{/}y)$ and identify $H^0(C,{\omega }_C)$ with the $g$-dimensional space of differentials $R(x)\mathrm{d}x\mathrm{/}y$ with $\mathrm{deg}R\le g-1$, and step 4.3 computes $H^1(C,{\mathcal{O}}_C)$ from the two affine charts and concludes $g(C)=g$. Finally step 5.3 computes the delta invariant ${\delta }_{\mathrm{\infty }}=\frac{(n-1)(n-2)}{2}-g$, equal to $2g^2$ for $n=2g+2$, to $2g(g-1)$ for $n=2g+1$ and to $0$ for $(n,g)=(3,1)$, so that the geometric genus of $X$ is $g$. The Axiom of Choice is used through the normalization and curve/function-field/finiteness routes [F3], [F5], [F6], the differential and Čech-cohomology routes [F8], [F9], and the properness and ampleness suppliers [F10], at the steps where those inputs are applied.