Nodal cubic: arithmetic genus one, delta one, geometric genus zero

The cubic is an integral curve, and its only singular point is the ordinary node $o$. On $Z=1$, let $f=y^2-x^3-x^2=y^2-x^2(x+1)$. Over $k(x)$ the monic quadratic $f$ is irreducible: $x^2(x+1)$ has valuation one at $x+1$, so it is not a square; Gauss's lemma gives irreducibility in $k[x,y]$. The homogeneous cubic is not divisible by $Z$, hence is irreducible as well. Since $k$ is algebraically closed, $X$ is geometrically integral. The charts $D(Z)$ and $D(Y)$ cover $X$, because $Y=Z=0$ in its equation forces $X=0$. On $D(Z)$ the coordinate ring is the domain $A=k[x,y]\mathrm{/}(y^2-x^3-x^2)$ with fraction field $k(t)$ under $x=t^2-1$, $y=t(t^2-1)$; here $t=y\mathrm{/}x$ in the fraction field. On $D(Y)$ the coordinate ring is $R=k[x,z]\mathrm{/}((1-x^2)z-x^3)$. The identity $1=(1-x^2)(1+x^2+xz)$ in $R$ makes $1-x^2$ invertible, and solving for $z$ gives $R\cong k[x,(1-x^2{)}^{-1}]$. Both chart rings are finite-type domains and hence Noetherian; [F7] gives their dimension one and the chain dimension one of their finite open cover $X$. The projective cubic is a closed subscheme of ${\mathbb{P}}_k^2$, hence proper by [F8]; it is also separated and finite type. Thus it is an integral proper curve, and its generic point is regular because its local ring is the function field. Every other point is closed. On $Z=1$ the partials of $f$ are $-3x^2-2x=-x(3x+2)$ and $2y$. A closed singular point must have $y=0$, so the curve equation gives $x=0$ or $x=-1$; at $x=-1$ the first partial is $-1$, in every characteristic. Thus the only singular point on this chart is $o$. On $Y=1$ the equation $g=z-x^3-x^{2}z$ has partials $-3x^2-2xz$ and $1-x^2$; if the second vanishes, then $x^2=1$ and $g=-x^3\mathrm{\ne }0$, so there is no singular point there. On $X=1$ the equation $h=y^{2}z-1-z$ has partials $2yz$ and $y^2-1$; the equation $z(y^2-1)=1$ forces the second partial to be nonzero. The quadratic tangent cone at $o$ is $y^2-x^2=(y-x)(y+x)$, with distinct tangent lines because $\mathrm{char}k\mathrm{\ne }2$. Hence $o$ is an ordinary node and all other points are regular.

Arithmetic genus. By step 1.1, $X$ is an integral proper plane curve of degree three, so [F1] gives $p_a(X)=\frac{(3-1)(3-2)}{2}=1$.

Explicit normalization and finite projective map. Put $q=T^2-S^2$. The homogeneous triple defining $\psi (T:S)=(Sq:Tq:S^3)$ has no common zero: if $S=0$, then $Y=T^3\mathrm{\ne }0$, while if $S\mathrm{\ne }0$, then $Z=S^3\mathrm{\ne }0$. It lands on $X$ because $Y^{2}Z=T^2{q}^2{S}^3=S^3{q}^2(q+S^2)=X^3+X^{2}Z$. On $D(Z)$ the source chart is $\mathrm{Spec}k[t]$ with $t=T\mathrm{/}S$, and its ring map is $A=k[x,y]\mathrm{/}(y^2-x^3-x^2)\to k[t]$, $x\mapsto t^2-1$, $y\mapsto t(t^2-1)$. It is finite because $k[t]=A[t]$ and $t^2=x+1$; it is birational since $t=y\mathrm{/}x$ in $\mathrm{Frac}(A)$. On $D(Y)$ the inverse image is $\mathrm{Spec}k[u,(1-u^2{)}^{-1}]$, where $u=S\mathrm{/}T$, and the chart map $R\to k[u,(1-u^2{)}^{-1}]$ sends $x\mapsto u$, $z\mapsto u^3\mathrm{/}(1-u^2)$; it is an isomorphism by the description of $R$ in step 1.1. As $D(Z)$ and $D(Y)$ cover $X$, these chart maps show that $\psi$ is finite. Its source ${\mathbb{P}}_k^1$ is normal and integral by [F6], so the finite birational map identifies it with the normalization by the initial property in [F4].

Delta at the node. By step 2.2, $\psi$ is the normalization map. The two points $t=1$ and $t=-1$ of the affine source map to $o$ and no other point does, so the stalk of ${\nu }_{\ast }{\mathcal{O}}_{{\mathbb{P}}^1}$ at $o$ is the semilocalization $k[t{]}_S$ with $S=\{h\in k[t]:h(1)\mathrm{\ne }0,h(-1)\mathrm{\ne }0\}$. Put $u=t^2-1$ and identify the coordinate ring of $Z=1$ with the subring $A=k[u,tu]=k[u]\oplus tuk[u]\subseteq k[t]$; also $k[t]=k[u]\oplus tk[u]$. The semilocalization equals $k[u{]}_{(u)}\oplus tk[u{]}_{(u)}$: if $h(t)=a(u)+tb(u)$ is nonzero at both $1$ and $-1$, its norm $h(t)h(-t)=a(u{)}^2-(u+1)b(u{)}^2$ is nonzero at $u=0$, so $h$ is invertible after localizing over $k[u{]}_{(u)}$, and every element of $k[u]\setminus (u)$ is nonzero at both points. The image of ${\mathcal{O}}_{X,o}=A_{(u,tu)}$ in this semilocalization is $k[u{]}_{(u)}\oplus tuk[u{]}_{(u)}$. The quotient is $tk[u{]}_{(u)}\mathrm{/}tuk[u{]}_{(u)}\cong k[u{]}_{(u)}\mathrm{/}uk[u{]}_{(u)}\cong k$, of dimension one. Hence ${\delta }_o(X)=1$, and since $o$ is the only singular point by step 1.1 and delta vanishes off the singular locus, ${\sum }_x{\delta }_x(X)=1$.

Genus. By [F3] and steps 1.1 and 3.1, $g(X^{\mathrm{n}\mathrm{u}})=\frac{(3-1)(3-2)}{2}-{\sum }_x{\delta }_x(X)=1-1=0$; the geometric genus of $X$ is $0$.

Normalization is the line. By step 2.2, the displayed map is a normal integral finite birational model of $X$; the uniqueness clause in [F4] identifies it with $X^{\mathrm{n}\mathrm{u}}$. Thus the normalization is the projective line, and the geometric genus computed in step 4.1 is zero.

Conclusion. Under the Axiom of Choice, for the nodal cubic $X=V_{+}(Y^{2}Z-X^3-X^{2}Z)$ over an algebraically closed field with $\mathrm{char}k\mathrm{\ne }2$, the arithmetic genus is $1$, the unique singular point is the ordinary node $o$ with delta invariant $1$, and the normalization is the projective line; the normalization has geometric genus $0$. Choice is used in the normalization and normality interfaces [F4, F6] and the delta/genus interfaces [F2, F3], in steps 2.2, 3.1, 4.1 and 5.1.