Let \(X\) be a curve in \(\PP^3\), which is not contained in any plane.
If \(O \notin X\) is a point, such that the projection from \(O\) induces a birational morphism \(\varphi\) from \(X\) to its image in \(\PP^2\), show that \(\varphi(X)\) must be singular. Hint: Calculate \(\dim H^0(X, \OO_X(1))\) two ways.
If \(X\) has degree \(d\) and genus \(g\), conclude that \(g<\frac{1}{2}(d-1)(d-2)\). (Use (Ex. 1.8).)
Now let \(\ts{X_t}\) be the flat family of curves induced by the projection (III, 9.8.3) whose fibre over \(t=1\) is \(X\), and whose fibre \(X_0\) over \(t=0\) is a scheme with support \(\varphi(X)\). Show that \(X_0\) always has nilpotent elements. Thus the example (III, 9.8.4) is typical.
Solution
Let \[ Y=\varphi(X)\subseteq\PP^2. \] Projection from \(O\) is defined by the three-dimensional space of hyperplanes through \(O\), so \[ \varphi^*\mco_Y(1)\cong\mco_X(1). \]
1. Since \(X\subseteq\PP^3\) is not contained in a plane,\[ h^0(X,\mco_X(1))\ge4. \]
Restriction gives a linear map \[ H^0(\PP^3,\mco_{\PP^3}(1)) \longrightarrow H^0(X,\mco_X(1)). \] Its kernel consists of linear forms vanishing identically on \(X\), equivalently planes containing \(X\). By hypothesis there is no such plane, so the map is injective. The source has dimension \(4\), proving the claim.
2. If \(Y\) were nonsingular, then\[ h^0(X,\mco_X(1))=3. \]
The morphism \[ \varphi:X\longrightarrow Y \] is birational by hypothesis. If \(Y\) were nonsingular, then both \(X\) and \(Y\) would be smooth projective curves. A birational morphism between smooth projective curves is an isomorphism.
Therefore \[ H^0(X,\mco_X(1)) \cong H^0(Y,\mco_Y(1)). \] The integral plane curve \(Y\) is not a line: if it were, then \(X\), being isomorphic to \(Y\cong\PP^1\), would have \[ \deg\mco_X(1) = \deg\varphi^*\mco_Y(1) =1, \] and hence would be a line in \(\PP^3\), contrary to nondegeneracy.
Thus the defining equation of \(Y\) has degree at least \(2\). Twisting its ideal sequence by \(\mco_{\PP^2}(1)\) gives \[ 0 \longrightarrow \mco_{\PP^2}(1-\deg Y) \longrightarrow \mco_{\PP^2}(1) \longrightarrow \mco_Y(1) \longrightarrow0. \] The first sheaf has no global sections, and \[ H^1(\PP^2,\mco(q))=0 \] for every \(q\). Hence restriction gives \[ H^0(\PP^2,\mco(1)) \xrightarrow{\sim} H^0(Y,\mco_Y(1)). \] The left side has dimension \(3\), proving the claim.
3. The plane image \(Y\) is singular.
If \(Y\) were nonsingular, steps 1 and 2 would give simultaneously \[ h^0(X,\mco_X(1))\ge4 \] and \[ h^0(X,\mco_X(1))=3, \] a contradiction. Hence \[ \boxed{Y\text{ is singular}.} \] This proves (a).
4. If \(\deg X=d\), then\[ \deg Y=d. \]
Because \[ \varphi^*\mco_Y(1)\cong\mco_X(1) \] and \(\varphi\) is birational, it has degree one on function fields. A nonconstant morphism of projective curves is finite, so the usual degree formula for pullback of a line bundle applies. Therefore \[ \deg\mco_X(1) = \deg\varphi\cdot\deg\mco_Y(1) = \deg\mco_Y(1). \] The two sides are respectively \(\deg X\) and \(\deg Y\), so both equal \(d\).
5. The arithmetic genus of the singular plane image is\[ p_a(Y)=\frac12(d-1)(d-2)>g(X). \]
The plane-curve genus formula gives \[ p_a(Y) = \frac12(d-1)(d-2). \]
Since \(\varphi:X\to Y\) is a birational morphism from the nonsingular curve \(X\), it is the normalization of \(Y\). By Exercise IV.1.8, \[ p_a(Y) = g(X) + \sum_{P\in Y}\delta_P. \] Step 3 says that \(Y\) is singular, so at least one local ring is not normal and the corresponding normalization quotient has positive length. Hence \[ \sum_{P\in Y}\delta_P>0. \] Therefore \[ \boxed{ g(X) < \frac12(d-1)(d-2) }. \] This proves (b).
6. In the flat projection family, the special fibre satisfies\[ p_a(X_0)=g(X). \]
The family in Hartshorne III.9.8.3 is flat and projective over the parameter curve. Flatness preserves the Hilbert polynomial, hence the arithmetic genus of the one-dimensional fibres. Since the fibre at \(t=1\) is the nonsingular curve \(X\), \[ p_a(X_0)=p_a(X)=g(X). \]
7. The special fibre \(X_0\) is not reduced.
By construction the support of \(X_0\) is the plane curve \(Y\). Suppose \(X_0\) were reduced. Since \(Y\) is integral, \[ X_0=Y \] as schemes. Step 6 would then give \[ p_a(Y)=g(X). \] But step 5 gives the strict inequality \[ p_a(Y)>g(X), \] a contradiction. Thus \(X_0\) is nonreduced.
8. The structure sheaf of \(X_0\) has nonzero nilpotent elements.
A scheme is reduced exactly when its structure sheaf has zero nilradical. By step 7, \(X_0\) is not reduced, so its nilradical is nonzero. Hence \[ \boxed{X_0\text{ has nilpotent elements}.} \] This proves (c).
9. Q.E.D.
Step 3 proves (a), step 5 proves (b), and steps <1>6–<1>8 prove (c).