Coherent duality on a singular plane cubic

On $z=1$, the polynomial is $y^2-x^2(x+1)$. It is irreducible over $k(x)$ as a polynomial in $y$, since $x+1$ has odd valuation at $x=-1$ and hence is not a square; Gauss's lemma proves irreducibility in $k[x,y]$. The homogeneous polynomial is not divisible by $z$, and any homogeneous factorization would dehomogenize to a nontrivial factorization, so $C$ is integral and pure of dimension one. Its affine gradient vanishes at $(0,0)$ and its quadratic tangent cone is $y^2-x^2$, with two distinct lines. Thus $p$ is a node and $C$ is singular. In a regular ambient local ring along $C$, the nonzero hypersurface equation is regular and lowers dimension by one by [F3]. Lift a parameter tuple from the quotient and prepend the equation: [F3] makes this a parameter tuple of the ambient ring. The equation is therefore a regular parameter element, so [F1] makes $C$ CM.

The resolution $0\to {\mathcal{O}}_{{\mathbb{P}}^2}(-3)\stackrel{f}{\to }{\mathcal{O}}_{{\mathbb{P}}^2}\to i_{\ast }{\mathcal{O}}_C\to 0$ of [F1], dualized into ${\omega }_{{\mathbb{P}}^2}=\mathcal{O}(-3)$, has cokernel $i_{\ast }{\mathcal{O}}_C$ in degree one. Therefore [F2] gives ${\omega }_C={\mathcal{O}}_C$. The same resolution and its long exact sequence [F1] give $H^0(C,{\mathcal{O}}_C)=k$ and $H^1(C,{\mathcal{O}}_C)=H^2({\mathbb{P}}^2,\mathcal{O}(-3))=k$. The trace identifies the latter with $k$ and pairs it perfectly with the constants by [F2]. Finally $H^0(C,k_p)=k$ and $H^1(C,k_p)=0$, since a point sheaf has surjective restriction maps and is flasque, so [F1] applies. Apply [F2] with $d=1$ and $F=k_p$ to obtain the stated Ext and Hom groups. The example uses the A theorem for both pairings; singularity did not require a locally free hypothesis on $k_p$.