Let \(X_1 \subseteq \PP_k^4\) be the rational normal quartic curve, the \(4\dash\)uple embedding of \(\PP^1\) in \(\PP^4\). Let \(X_0 \subseteq \PP_k^3\) be a nonsingular rational quartic curve, such as the one in (I, Ex. 3.18b).
Use (9.8.3) to construct a flat family \(\ts{X_t}\) of curves in \(\PP^4\), parametrized by \(T = \AA^1\), with the given fibres \(X_1\) and \(X_0\) for \(t = 1\) and \(t = 0\).
Let \(\mci \subseteq \mco_{\PP^4 \times T}\) be the ideal sheaf of the total family \(X \subseteq \PP^4 \times T\). Show that \(\mci\) is flat over \(T\). Then show that \[ h^0(t, \mci) = \begin{cases} 0 & t \neq 0 \\ 1 & t = 0 \end{cases} \qquad h^1(t, \mci) = \begin{cases} 0 & t \neq 0 \\ 1 & t = 0 \end{cases} .\]
This gives another example of cohomology groups jumping at a special point.
(Erratum to the printed cohomology claim)
The two displayed jumps are not correct for the untwisted ideal sheaves \(\mci_t\). In fact, every fibre in the family below is a connected reduced rational curve, so \[ H^0(X_t,\mco_{X_t})=k. \] From \[ 0\to\mci_t\to\mco_{\PP^4}\to\mco_{X_t}\to0 \] and \(H^1(\PP^4,\mco_{\PP^4})=0\), one obtains \[ H^0(\PP^4,\mci_t)=H^1(\PP^4,\mci_t)=0 \] for every \(t\).
The advertised values are correct after replacing \(\mci_t\) by \(\mci_t(1)\). The solution proves the flat-family construction and flatness of \(\mci\), records the untwisted calculation, and then proves this corrected jump.
Solution
Use homogeneous coordinates \[ [x_0:x_1:x_2:x_3:x_4] \] on \(\PP^4\). The rational normal quartic is \[ \nu_1:\PP^1\longrightarrow\PP^4, \qquad [s:u]\longmapsto [s^4:s^3u:s^2u^2:su^3:u^4]. \]
1. Hartshorne’s projection degeneration (III.9.8.3), applied by scalingthe coordinate \(x_2\), gives a flat family\[ X\longrightarrow T=\AA^1 \]whose fibres for \(a\ne0\) are\[ X_a= \left\{ [s^4:s^3u:a s^2u^2:su^3:u^4] \right\}, \]and whose special fibre is the reduced twisted quartic\[ X_0= \left\{ [s^4:s^3u:0:su^3:u^4] \right\} \subseteq V(x_2)\cong\PP^3. \]
Let \[ P=[0:0:1:0:0]. \] This point does not lie on \(X_1\). For \(a\ne0\), let \[ \sigma_a([x_0:x_1:x_2:x_3:x_4]) = [x_0:x_1:a x_2:x_3:x_4]. \] Then \(X_a=\sigma_a(X_1)\), so over \(\GG_m\subset\AA^1\) the family is isomorphic to the product \(X_1\times\GG_m\). Example III.9.8.3 takes its flat projective closure over \(\AA^1\). Set-theoretically the special fibre is the projection of \(X_1\) from \(P\) to the hyperplane \(x_2=0\), namely the displayed twisted quartic.
Flatness makes the Hilbert polynomial of every fibre equal to that of the rational normal quartic: \[ P(m)=4m+1. \] The reduced projected curve is isomorphic to \(\PP^1\) by Exercise I.3.18, and its hyperplane bundle pulls back to \(\mco_{\PP^1}(4)\), so it has the same Hilbert polynomial \(4m+1\). Any nonzero nilpotent thickening supported on this curve would contribute a nonzero Hilbert polynomial. Hence the flat special fibre has no extra scheme structure and is exactly the reduced \(X_0\).
2. The ideal sheaf\[ \mci\subseteq\mco_{\PP^4\times T} \]of the total family is flat over \(T\).
There is an exact sequence \[ 0 \longrightarrow \mci \longrightarrow \mco_{\PP^4\times T} \longrightarrow \mco_X \longrightarrow0. \] The middle term is flat over \(T\), and \(\mco_X\) is flat over \(T\) because \(X\to T\) is flat. In an exact sequence, the kernel of a surjection between flat modules is flat. Therefore \(\mci\) is flat over \(T\).
3. For the untwisted ideal sheaves, the printed jump does not occur:\[ H^0(\PP^4,\mci_t)=H^1(\PP^4,\mci_t)=0 \]for every \(t\in T\).
Since \(\mco_X\) is flat over \(T\), restricting the exact sequence of step 2 to a fibre remains exact: \[ 0 \longrightarrow \mci_t \longrightarrow \mco_{\PP^4} \longrightarrow \mco_{X_t} \longrightarrow0. \] Every \(X_t\) is isomorphic to \(\PP^1\), hence is connected and proper, so \[ H^0(X_t,\mco_{X_t})=k. \] The restriction map \[ H^0(\PP^4,\mco_{\PP^4})=k \longrightarrow H^0(X_t,\mco_{X_t})=k \] is the identity on constants and therefore an isomorphism. Since \[ H^1(\PP^4,\mco_{\PP^4})=0, \] the long exact sequence gives \[ H^0(\PP^4,\mci_t)=0 \qquad\text{and}\qquad H^1(\PP^4,\mci_t)=0. \] Thus the two displayed functions in the printed statement are both identically zero.
4. After twisting by \(\mco_{\PP^4}(1)\), for every \(a\ne0\) one has\[ H^0(\PP^4,\mci_a(1)) = H^1(\PP^4,\mci_a(1)) = 0. \]
Twisting the fibre sequence gives \[ 0 \longrightarrow \mci_a(1) \longrightarrow \mco_{\PP^4}(1) \longrightarrow \mco_{X_a}(1) \longrightarrow0. \] For \(a\ne0\), the embedding \(X_a\cong\PP^1\hookrightarrow\PP^4\) is the complete linear series \(|\mco_{\PP^1}(4)|\). Therefore the restriction map \[ H^0(\PP^4,\mco(1)) \longrightarrow H^0(X_a,\mco_{X_a}(1)) \] identifies two five-dimensional spaces and sends the five coordinate linear forms to a basis of \(H^0(\PP^1,\mco(4))\). Hence it is an isomorphism. Since \(H^1(\PP^4,\mco(1))=0\), the long exact sequence gives the claimed vanishing.
5. For the special fibre,\[ h^0(\PP^4,\mci_0(1)) = h^1(\PP^4,\mci_0(1)) = 1. \]
The pullback of \(\mco_{X_0}(1)\) to \(\PP^1\) is again \(\mco_{\PP^1}(4)\), so \[ h^0(X_0,\mco_{X_0}(1))=5. \] Under the parametrization \[ [s:u]\longmapsto[s^4:s^3u:0:su^3:u^4], \] the restriction map \[ H^0(\PP^4,\mco(1)) \longrightarrow H^0(\PP^1,\mco(4)) \] sends the coordinate basis to \[ s^4,\ s^3u,\ 0,\ su^3,\ u^4. \] Its kernel is therefore the one-dimensional span of \(x_2\), and its image has codimension one, missing the section \(s^2u^2\).
The long exact sequence of \[ 0 \to \mci_0(1) \to \mco_{\PP^4}(1) \to \mco_{X_0}(1) \to0 \] and the vanishing \(H^1(\PP^4,\mco(1))=0\) identify \[ H^0(\PP^4,\mci_0(1)) \] with that kernel and \[ H^1(\PP^4,\mci_0(1)) \] with that cokernel. Both have dimension one.
6. Thus the intended semicontinuity example is\[ h^0(t,\mci(1)) = h^1(t,\mci(1)) = \begin{cases} 0,&t\ne0,\\ 1,&t=0. \end{cases} \]
Steps 4 and 5 give the displayed values. Step 3 proves that the same assertion with \(\mci\) in place of \(\mci(1)\) is false.
7. Q.E.D. for the corrected statement.
Steps <1>1–<1>2 construct the required flat family and prove flatness of its ideal sheaf. Step 3 resolves the printed cohomology claim, and steps <1>4–<1>6 prove the missing-twist correction that has the advertised jump.