Generalize the statement that “not every secant is a multisecant” as follows. If \(X\) is a curve in \(\PP^n\), not contained in any \(\PP^{n-1}\), and if \(\characteristic k=0\), show that for almost all choices of \(n-1\) points \(P_1, \ldots, P_{n-1}\) on \(X\), the linear space \(L^{n-2}\) spanned by the \(P_i\) does not contain any further points of \(X\).
Solution
Put \[ d=\deg X. \] The curve is nondegenerate by hypothesis.
1. One has\[ d\ge n. \]
If \(d\le n\), then Exercise IV.3.4(b) applied to the nondegenerate curve \(X\subseteq\PP^n\) gives \[ d=n. \] Thus in all cases \(d\ge n\).
2. Choose a general hyperplane\[ H\subseteq\PP^n. \]Then\[ \Gamma=X\cap H \]is a reduced set of \(d\) points in uniform position in\[ H\cong\PP^{n-1}. \]
In characteristic zero, the Uniform Position Lemma says that a general hyperplane section of an integral nondegenerate projective curve is reduced and in uniform position: any two subsets of the same cardinality have the same Hilbert function. This is Harris’s Uniform Position Lemma (J. Harris, Galois groups of enumerative problems); see also the curve-notes exposition, §16.1. The characteristic-zero hypothesis in the exercise is exactly the hypothesis under which this monodromy statement is used.
3. The \(d\) points of \(\Gamma\) span the whole hyperplane \(H\).
Suppose instead that \[ \Gamma\subseteq L \] for some hyperplane \[ L\subsetneq H. \] Since \(L\) has codimension \(2\) in \(\PP^n\), choose a hyperplane \[ H'\ne H \] with \[ H\cap H'=L. \] Nondegeneracy of \(X\) implies that neither \(H\) nor \(H'\) contains \(X\).
The hyperplane divisor \[ D_H=H|_X \] is the reduced divisor consisting of the \(d\) points of \(\Gamma\). Since every point of \(\Gamma\) lies in \(H'\), one has \[ D_{H'}\ge D_H. \] Both divisors have degree \(d\), so \[ D_{H'}=D_H. \]
Let \(s,s'\) be the two corresponding nonzero sections of \(\mco_X(1)\). Their zero divisors agree, so \[ \frac{s'}s \] is a nowhere-vanishing regular function on the integral projective curve \(X\). It is therefore a scalar \[ c\in k^\times. \] Thus \[ s'-cs=0 \] on \(X\). But \(s'-cs\) is a nonzero linear form because \(H'\ne H\), so this would place \(X\) in a hyperplane, a contradiction. Hence \(\Gamma\) spans \(H\).
4. Every set of \(n\) distinct points of \(\Gamma\) spans \(H\).
By step 3, the finite set \(\Gamma\) spans the \((n-1)\)-dimensional projective space \(H\). Hence some \(n\) points of \(\Gamma\) form a projective basis of \(H\).
For a finite set \[ Z\subseteq H, \] the value of its Hilbert function in degree \(1\) is the rank of the evaluation map \[ H^0(H,\mco_H(1)) \longrightarrow H^0(Z,\mco_Z). \] For \(n\) points this rank is \(n\) exactly when no hyperplane of \(H\) contains all of them, equivalently exactly when they span \(H\).
Step 2 says that all \(n\)-point subsets of \(\Gamma\) have the same Hilbert function. Since one such subset spans \(H\), every such subset has degree-one Hilbert function equal to \(n\) and therefore spans \(H\).
5. Every set of \(n-1\) distinct points\[ P_1,\ldots,P_{n-1}\in\Gamma \]is linearly independent and spans an \((n-2)\)-plane\[ L=\langle P_1,\ldots,P_{n-1}\rangle. \]
1.Therefore the \(n-1\) points are linearly independent and their span hasdimension \(n-2\).
6. For the points in step 5,\[ L\cap X = \{P_1,\ldots,P_{n-1}\} \]scheme-theoretically.
4.Thus there is no additional set-theoretic intersection point.
2.At each \(P_i\), a local equation of \(H\) is therefore a uniformizer in theregular local ring \(\mco_{X,P_i}\). Since the ideal of\[ L\cap X \]contains that local equation, its local intersection length at \(P_i\) is atmost \(1\), and it is nonzero because \(P_i\in L\). Hence that length is exactly\(1\).Therefore the scheme-theoretic intersection consists exactly of the\(n-1\) reduced chosen points.
7. Let\[ U\subseteq X^{n-1} \]be the open set of ordered tuples of distinct linearly independent points.There is a finite morphism\[ q:\mathcal Z\longrightarrow U \]whose fibre over\[ \mathbf P=(P_1,\ldots,P_{n-1}) \]is\[ X\cap\langle P_1,\ldots,P_{n-1}\rangle. \]
Linear independence is an open rank condition, and pairwise distinctness is the complement of the diagonals, so \(U\) is open. It is nonempty by step 5.
On \(U\), taking the linear span defines a morphism \[ \lambda: U \longrightarrow \operatorname{Gr}(n-2,n), \] where the Grassmannian parametrizes \((n-2)\)-planes in \(\PP^n\). Pull back the universal \((n-2)\)-plane and intersect it with \[ U\times X. \] This gives a closed subscheme \[ \mathcal Z \subseteq U\times X \] with the stated fibres.
The projection \[ q:\mathcal Z\to U \] is projective. Every fibre is finite: if an \((n-2)\)-plane contained the integral curve \(X\), then \(X\) would lie in a hyperplane, contrary to nondegeneracy. Hence \(q\) is projective and quasi-finite, therefore finite.
8. The locus\[ B = \left\{ \mathbf P\in U: \operatorname{length} \bigl( X\cap\langle P_1,\ldots,P_{n-1}\rangle \bigr) \ge n \right\} \]is a proper closed subset of \(U\).
Since \(q\) is finite, the coherent sheaf \[ q_*\mco_{\mathcal Z} \] has fibre dimension \[ \dim_{\kappa(\mathbf P)} \left( q_*\mco_{\mathcal Z}\otimes\kappa(\mathbf P) \right) = \operatorname{length}(\mathcal Z_{\mathbf P}). \] Upper semicontinuity of fibre dimension for a coherent sheaf makes the locus where this number is at least \(n\) closed.
Step 6 gives a tuple for which the fibre consists of exactly \(n-1\) reduced points, hence has length \(n-1\). Therefore that tuple is not in \(B\), so \(B\) is proper.
9. For every tuple\[ \mathbf P\in U\setminus B, \]the span\[ \langle P_1,\ldots,P_{n-1}\rangle \]contains no further point of \(X\).
The fibre in step 7 always contains the \(n-1\) distinct chosen points, so its length is at least \(n-1\). For \[ \mathbf P\notin B, \] step 8 gives length at most \(n-1\). Hence the length is exactly \(n-1\). There can therefore be no additional point of \(X\) in the span.
10. Q.E.D.
8. Hence it is adense open subset of \(X^{n-1}\).
By step 9, every tuple in this dense open set has the required property. Thus for almost all choices of \[ P_1,\ldots,P_{n-1}\in X, \] their \((n-2)\)-plane contains no further point of \(X\).