Let \(C\) be the plane curve \(x^3 y+y^3 z+z^3 x=0\) over a field \(k\) of characteristic 3 (V, Ex. 2.4).
Show that the action of the \(k\)-linear Frobenius morphism \(f\) on \(H^1\left(C, \mathcal{O}_C\right)\) is identically 0 (Cf. (V, 4.21)).
Fix a point \(P \in C\), and show that there is a nonzero \(\xi \in H^1(\mathcal{L}(-P))\) such that \(f^* \xi=0\) in \(H^1(\mathcal{L}(-3 P))\).
Now let \(\mathcal{E}\) be defined by \(\xi\) as an extension \[ 0 \rightarrow \mathcal{O}_C \rightarrow \mathcal{E} \rightarrow \mathcal{L}(P) \rightarrow 0, \] and let \(X\) be the corresponding ruled surface over \(C\). Show that \(X\) contains a nonsingular curve \(Y \equiv 3 C_0-3 f\), such that \(\pi: Y \rightarrow C\) is purely inseparable.
Show that the divisor \(D=2 C_0\) satisfies the hypotheses of (2.21.b), but is not ample.
(Erratum to part (c))
The curve of class \[ 3C_0-3f \] constructed from the Frobenius-split extension is not nonsingular. The correct conclusion is that there is an integral curve \(Y\) of this class whose normalization is \(C\), and for which \[ \pi|_Y:Y\longrightarrow C \] is purely inseparable of degree \(3\).
Indeed, the proof below gives \(e=-1\) and \(g(C)=3\), so \[ K_X\equiv-2C_0+5f. \] Adjunction then gives \[ 2p_a(Y)-2 =(3C_0-3f)\cdot(C_0+2f) =6, \] hence \[ p_a(Y)=4. \] But the normalization is the genus-\(3\) curve \(C\). Thus \(Y\) has total \(\delta\)-invariant \(1\) and cannot be nonsingular.
Solution
As in Hartshorne’s notation, write \[ \mathcal L(A)=\OO_C(A) \] for the invertible sheaf associated to a divisor \(A\) on \(C\).
Write \[ F=x^3y+y^3z+z^3x. \] By P-AGH424FUNNYQUARTIC, the plane quartic \[ C=V(F)\subseteq\PP^2 \] is nonsingular; hence it has genus \(3\).
1. There is a natural identification\[ H^1(C,\OO_C)\cong H^2(\PP^2,\OO_{\PP^2}(-4)). \]
The quartic equation gives \[ 0 \longrightarrow \OO_{\PP^2}(-4) \xrightarrow{\cdot F} \OO_{\PP^2} \longrightarrow \OO_C \longrightarrow0. \] Since \[ H^1(\PP^2,\OO_{\PP^2})=0 \qquad\text{and}\qquad H^2(\PP^2,\OO_{\PP^2})=0, \] the connecting map in cohomology is the displayed isomorphism.
2. Under the identification in step 1, Frobenius on\(H^1(C,\OO_C)\) is represented by\[ \alpha\longmapsto F^2\alpha^3 \]on \(H^2(\PP^2,\OO_{\PP^2}(-4))\).
Pulling the quartic sequence of step 1 back by the characteristic-\(3\) Frobenius changes \[ \OO_{\PP^2}(-4) \] to \[ \OO_{\PP^2}(-12) \] and sends a cohomology class \(\alpha\) to \(\alpha^3\). The Frobenius-thickened quartic is cut out by \(F^3\), and \(C\) is a closed subscheme of that thickening. The induced restriction from the thickening to \(C\) is represented on the left terms by multiplication by \[ F^{3-1}=F^2. \] Naturality of the connecting homomorphism therefore gives exactly \[ \alpha\longmapsto F^2\alpha^3. \]
3. The Frobenius action on\[ H^1(C,\OO_C) \]is identically zero.
Using the standard Cech cover of \(\PP^2\), a basis of \[ H^2(\PP^2,\OO_{\PP^2}(-4)) \] is represented by \[ \frac1{x^2yz}, \qquad \frac1{xy^2z}, \qquad \frac1{xyz^2}. \] A Laurent monomial represents zero in this top Cech cohomology as soon as one of the exponents of \(x,y,z\) is nonnegative.
For example, cubing the first basis vector gives \[ x^{-6}y^{-3}z^{-3}. \] Every monomial of \[ F^2=(x^3y+y^3z+z^3x)^2 \] makes at least one exponent nonnegative after multiplication by this Laurent monomial. Explicitly the six monomial types are \[ x^6y^2, \quad y^6z^2, \quad z^6x^2, \quad x^3y^4z, \quad x^4yz^3, \quad xy^3z^4, \] and multiplication gives respectively an \(x\)-, \(y\)-, \(z\)-, \(y\)-, \(z\)-, or \(y\)-exponent at least zero. Hence \[ F^2(x^{-6}y^{-3}z^{-3})=0 \] in \(H^2(\PP^2,\OO(-4))\).
The other two basis vectors are obtained by cyclic permutation, and \(F\) is cyclically symmetric. Their images vanish as well. Step 2 therefore shows that Frobenius acts as zero on \(H^1(C,\OO_C)\). This proves part (a).
4. For every point \(P\in C\), the natural map\[ H^1(C,\OO_C(-P)) \longrightarrow H^1(C,\OO_C) \]is an isomorphism, while\[ \ker\left( H^1(C,\OO_C(-3P)) \longrightarrow H^1(C,\OO_C) \right) \]has dimension \(2\).
From \[ 0\to\OO_C(-P)\to\OO_C\to\OO_P\to0, \] the map \[ H^0(C,\OO_C)\longrightarrow H^0(P,\OO_P) \] is an isomorphism. Hence the induced map on \(H^1\) is an isomorphism.
For \(3P\), Riemann–Roch gives \[ h^1(C,\OO_C(-3P))=5, \] because the line bundle has negative degree and genus \(3\). Also \[ h^1(C,\OO_C)=3. \] The exact sequence \[ 0\to\OO_C(-3P)\to\OO_C\to\OO_{3P}\to0 \] shows that the map on \(H^1\) is surjective. Its kernel therefore has dimension \[ 5-3=2. \]
5. There exists\[ 0\ne\xi\in H^1(C,\OO_C(-P)) \]such that\[ f^*\xi=0 \quad\text{in}\quad H^1(C,\OO_C(-3P)). \]
3. Thus the image of the leftvertical arrow is contained in the two-dimensional kernel identified instep
4.
But the source has dimension \[ h^1(C,\OO_C(-P))=h^1(C,\OO_C)=3. \] Hence the left vertical map has nonzero kernel. Choose \[ 0\ne\xi \] in that kernel. This proves part (b).
6. Let \(\mathcal E\) be the extension determined by \(\xi\):\[ 0 \longrightarrow \OO_C \longrightarrow \mathcal E \longrightarrow \OO_C(P) \longrightarrow0. \]Then \(\mathcal E\) is normalized and the ruled surface\[ X=\PP(\mathcal E) \]has invariant\[ \boxed{e=-1}. \]
The inclusion of \(\OO_C\) gives a nonzero global section of \(\mathcal E\). Let \(M\) be a line bundle of negative degree.
If \(\deg M\le-2\), then both \(M\) and \(M(P)\) have negative degree, so the twisted extension has no global sections.
If \(\deg M=-1\) and \(M\not\cong\OO_C(-P)\), then \(M(P)\) is a nontrivial degree-zero line bundle and again has no global sections. Finally, for \[ M=\OO_C(-P), \] the connecting map \[ H^0(C,\OO_C)\longrightarrow H^1(C,\OO_C(-P)) \] in the twisted extension sends \(1\) to the extension class \(\xi\), which is nonzero. Thus the quotient section of the twisted sequence does not lift, and \[ H^0(C,\mathcal E(-P))=0. \] Therefore \(\mathcal E\) is normalized.
Since \[ \det\mathcal E\cong\OO_C(P), \] one has \[ e=-\deg\det\mathcal E=-1. \] The quotient \[ \mathcal E\twoheadrightarrow\OO_C(P) \] defines the normalized section \(C_0\), and \[ C_0^2=-e=1. \]
7. Frobenius pullback splits the extension:\[ f^*\mathcal E \cong \OO_C\oplus\OO_C(3P). \]
5. Hence the sequence splits.
8. A splitting in step 7 produces a morphism\[ \varphi:C\longrightarrow X \]such that\[ \pi\circ\varphi=f \]and\[ \varphi^*\OO_X(C_0)\cong\OO_C. \]
Choose the quotient supplied by the splitting, \[ f^*\mathcal E\twoheadrightarrow\OO_C. \] By the universal property of the projective bundle of one-dimensional quotients, this quotient is equivalent to a morphism \[ \varphi:C\longrightarrow\PP(\mathcal E)=X \] lying over the base map \(f\).
For the normalized section \(C_0\), the kernel of \[ \mathcal E\twoheadrightarrow\OO_C(P) \] is \(\OO_C\), so the standard section-divisor formula gives \[ \OO_X(C_0)\cong\OO_X(1). \] The pullback of the tautological quotient bundle along \(\varphi\) is the chosen quotient line bundle \(\OO_C\). Hence \[ \varphi^*\OO_X(C_0)\cong\OO_C. \]
9. Let \(Y\subset X\) be the scheme-theoretic image of \(\varphi\). Then\(Y\) is integral, \(\varphi:C\to Y\) is its normalization, and\[ \boxed{Y\equiv3C_0-3f}. \]
Because \(C\) is proper and \[ \pi\circ\varphi=f \] is finite, \(\varphi\) is proper and quasi-finite, hence finite. Its scheme-theoretic image \(Y\) is therefore an integral projective curve.
Let \[ m=\deg(C\to Y), \] and write \[ Y\equiv aC_0+bf. \] Since \[ f=(\pi|_Y)\circ\varphi \] has degree \(3\), \[ 3=ma. \] Also step 8 gives \[ 0=\deg\varphi^*\OO_X(C_0)=m(C_0\cdot Y). \] Hence \[ C_0\cdot Y=0. \] As \(C_0^2=1\), this says \[ a+b=0. \]
Thus either \[ (m,a,b)=(1,3,-3) \] or \[ (m,a,b)=(3,1,-1). \] The second possibility is impossible: an integral curve of numerical class \[ C_0-f \] would have degree one over \(C\), hence would be a section, whereas the normalized-bundle argument of P-AGH5214IRRCURVESCHARP, step 2, shows that every section \(C_0+bf\) has \(b\ge0\).
Therefore \(m=1\) and \[ Y\equiv3C_0-3f. \] Since \(C\) is nonsingular and the finite map \(C\to Y\) is birational, it is the normalization of \(Y\).
10. The morphism\[ \pi|_Y:Y\longrightarrow C \]is purely inseparable of degree \(3\).
Step 9 identifies the function field of \(Y\) with that of its normalization \(C\). Under this identification, the induced extension of function fields for \(\pi|_Y\) is exactly the extension induced by \[ f:C\longrightarrow C. \] The characteristic-\(3\) Frobenius is purely inseparable of degree \(3\). Therefore so is \(\pi|_Y\).
11. The curve \(Y\) is singular; more precisely,\[ p_a(Y)=4, \qquad g(\widetilde Y)=3. \]
The base curve is a nonsingular plane quartic, so \(g=3\). Step 6 gives \(e=-1\), hence the canonical divisor of the ruled surface is numerically \[ K_X\equiv-2C_0+(2g-2-e)f=-2C_0+5f. \] Using step 9 and adjunction, \[ \begin{aligned} 2p_a(Y)-2 &=Y\cdot(Y+K_X)\\ &=(3C_0-3f)\cdot(C_0+2f)\\ &=6. \end{aligned} \] Thus \[ p_a(Y)=4. \] Its normalization is \(C\) by step 9, so its geometric genus is \(3\). Consequently \(Y\) is singular, proving the correction recorded above.
12. The divisor\[ D=2C_0 \]satisfies the numerical hypotheses referred to in (2.21.b), namely\[ a>0, \qquad b>\frac12ae, \]but \(D\) is not ample.
For \[ D=2C_0 \] one has \[ a=2, \qquad b=0, \qquad e=-1. \] Thus \[ a>0 \] and \[ b=0>\frac12(2)(-1)=-1. \] So the numerical hypotheses of (2.21.b) are satisfied.
On the other hand, step 9 gives the integral curve \[ Y\equiv3C_0-3f. \] Hence \[ D\cdot Y = 2C_0\cdot(3C_0-3f) = 6-6 =0. \] An ample divisor has strictly positive intersection with every irreducible curve by Nakai–Moishezon T-SRFNAKAI. Therefore \[ \boxed{D\text{ is not ample}.} \] This is exactly the characteristic-\(3\) failure that the exercise exhibits.
13. Q.E.D. for the corrected statement.
Steps <1>1–<1>3 prove part (a), steps <1>4–<1>5 prove part (b), and steps <1>6–<1>12 prove the corrected form of part (c) and the non-ampleness counterexample.