Coherent Duality on Projective Cohen-Macaulay Schemes — Examples

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 $k_p$.

The three affine charts are polynomial planes, so $X$ is smooth, projective and pure of dimension two. The smooth specialization in [F1] gives ${\omega }_X=\mathcal{O}(-3)$. A basis of $H^0(\mathcal{O}(m))$ consists of $x_0^{a_0}{x}_1^{a_1}{x}_2^{a_2}$ with nonnegative exponents summing to $m$. By [F2], the dual basis of $H^2(\mathcal{O}(-m-3))$ is $x_0^{-a_0-1}{x}_1^{-a_1-1}{x}_2^{-a_2-1}$. Multiplication followed by the coefficient of $(x_0{x}_1{x}_2{)}^{-1}$ gives the Kronecker pairing. This is the A theorem's $i=0$ pairing for $F=\mathcal{O}(m)$ under the locally free Ext identification.

The point sheaf has $H^0=k$ and no higher cohomology, since it is flasque and [F2] applies. Applying the A theorem [F1] with $F=k_p$ in degrees $i=0,1,2$ gives respectively the stated Ext degree two, degree one, and degree zero groups. Thus the same surface theorem handles a coherent sheaf which is not locally free, as well as the twisting bundles. AC is inherited through [F1]–[F2].

The scheme is smooth and pure of dimension one; its local rings are regular and hence CM. It is not proper: after base change to ${\mathbb{A}}_x^1$, the closed subscheme $V(xt-1)\subset {\mathbb{A}}_x^1\times {\mathbb{A}}_t^1$ has image $D(x)$ under projection, which is not closed. Thus the structure map is not universally closed.

By [F1], $H^1(X,F)=0$, whereas ${\mathrm{Ext}}_X^0(F,{\omega }_X)=\mathrm{\Gamma }(X,{\omega }_X)=k[t]dt\mathrm{\ne }0$. The asserted formula in degree $i=1$ would equate these nonzero and zero spaces. This refutes it while retaining smoothness and CM. On an affine nonproper scheme a local dualizing complex still exists; it does not carry the proper global trace-duality representation of the A theorem. The counterexample uses the precise pairing refuted, with ordinary cohomology rather than a compact-support replacement.