Cuspidal cubic: delta invariant and normalization

The cubic is an integral curve, and its only singular point is the cusp $o$. On $Z=1$, the equation is $f=y^2-x^3$. Over $k(x)$ this monic quadratic is irreducible because $x^3$ has odd valuation at $x=0$ and is not a square; Gauss's lemma gives irreducibility in $k[x,y]$. The homogeneous cubic is not divisible by $Z$, hence is irreducible. 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 $A=k[x,y]\mathrm{/}(y^2-x^3)$ is a domain with fraction field $k(t)$ under $x=t^2$, $y=t^3$, where $t=y\mathrm{/}x$. On $D(Y)$ the ring is $k[x,z]\mathrm{/}(z-x^3)\cong k[x]$. 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 $-3x^2$ and $2y$, so a closed singular point must be $x=y=0$ because $\mathrm{char}k\mathrm{\ne }2,3$; this is $o$. On $Y=1$, the equation is $g=z-x^3$ and its partials $-3x^2$ and $1$ never vanish together. On $X=1$, the equation is $h=y^{2}z-1$ with partials $2yz$ and $y^2$; the equation forces $y\mathrm{\ne }0$, so they do not both vanish. Hence all other points are regular.

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

The finite projective normalization map. The homogeneous triple defining $\phi (T:S)=(T^{2}S:T^3:S^3)$ 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 $Y^{2}Z=T^6{S}^3=X^3$. On $D(Z)$ the source is $\mathrm{Spec}k[t]$ with $t=T\mathrm{/}S$, and the ring map $A=k[x,y]\mathrm{/}(y^2-x^3)\to k[t]$, $x\mapsto t^2$, $y\mapsto t^3$, is finite because $k[t]=A[t]$ and $t^2=x$; it is birational since $t=y\mathrm{/}x$ in $\mathrm{Frac}(A)$. On $D(Y)$ the inverse image is $\mathrm{Spec}k[s]$ with $s=S\mathrm{/}T$, and the target ring $k[x,z]\mathrm{/}(z-x^3)$ maps isomorphically to $k[s]$ by $x\mapsto s$, $z\mapsto s^3$. As $D(Z),D(Y)$ cover $X$, the projective map is finite. Its source 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 cusp. By step 2.2, $\phi$ is the normalization map. The only point of the affine source mapping to the cusp $o$ is $t=0$, so the stalk of ${\nu }_{\ast }{\mathcal{O}}_{{\mathbb{P}}^1}$ at $o$ is the local ring $k[t{]}_{(t)}$, and the image of ${\mathcal{O}}_{X,o}=A_{(x,y)}$ in it is the local ring $A_{(x,y)}=k[t^2,t^3{]}_{(t^2,t^3)}$. Writing $u=t^2$ one has $k[t]=k[u]\oplus tk[u]$ and $k[t^2,t^3]=k[u]\oplus t^{3}k[u]=k[u]\oplus tuk[u]$; localizing at $(t)$, which inverts no power of $u$, gives $k[t{]}_{(t)}=k[u{]}_{(u)}\oplus tk[u{]}_{(u)}$ and $k[t^2,t^3{]}_{(t^2,t^3)}=k[u{]}_{(u)}\oplus tuk[u{]}_{(u)}$. Hence the quotient is $tk[u{]}_{(u)}\mathrm{/}tuk[u{]}_{(u)}\cong k[u{]}_{(u)}\mathrm{/}uk[u{]}_{(u)}\cong k$, a one-dimensional $k$-vector space, and ${\delta }_o(X)=1$. Since $o$ is the only singular point by step 1.1 and $\delta$ vanishes at regular points, ${\sum }_x{\delta }_x(X)=1$.

Genus. By [F3] applied to the cubic $X$ (which has only isolated singularities by step 1.1) and steps 2.1 and 3.1, [F2, F3, step 2.1, step 3.1] $g(X^{\mathrm{n}\mathrm{u}})=\frac{(3-1)(3-2)}{2}-{\delta }_o(X)=1-1=0,$ so the geometric genus of $X$ is $0$.

The 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 $X^{\mathrm{n}\mathrm{u}}$. Hence the normalization of the cuspidal cubic is the projective line, and its geometric genus is zero by step 4.1.

Conclusion. Under the Axiom of Choice, for the cuspidal cubic $X=V_{+}(Y^{2}Z-X^3)$ over an algebraically closed field with $\mathrm{char}k\mathrm{\ne }2,3$: $p_a(X)=1$, the cusp $o$ is the unique singular point with ${\delta }_o(X)=1$, and the normalization is ${\mathbb{P}}_k^1$ via $t\mapsto (t^2:t^3:1)$ with geometric genus zero. 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.