Let \(X\) be the \(d\)-uple embedding (I, Ex. 2.12) of \(\PP^1\) in \(\PP^d\), for any \(d \geq 1\). We call \(X\) the rational normal curve of degree \(d\) in \(\PP^d\).
Show that \(X\) is projectively normal, and that its homogeneous ideal can be generated by forms of degree 2.
If \(X\) is any curve of degree \(d\) in \(\PP^n\), with \(d \leq n\), which is not contained in any \(\PP^{n-1}\), show that in fact \(d=n\), \(g(X)=0\), and \(X\) differs from the rational normal curve of degree \(d\) only by an automorphism of \(\PP^d\). Cf. (II, 7.8.5).
In particular, any curve of degree 2 in any \(\PP^n\) is a conic in some \(\PP^2\).
A curve of degree 3 in any \(\PP^n\) must be either a plane cubic curve, or the twisted cubic curve in \(\PP^3\).
Solution
Use homogeneous coordinates \[ [s:t] \] on \(\PP^1\) and \[ [x_0:\cdots:x_d] \] on \(\PP^d\). The \(d\)-uple embedding is \[ \nu_d([s:t]) = [s^d:s^{d-1}t:\cdots:st^{d-1}:t^d]. \]
1. For every \(m\ge0\), the restriction map\[ H^0(\PP^d,\mco_{\PP^d}(m)) \longrightarrow H^0(\PP^1,\mco_{\PP^1}(md)) \]is surjective.
The map sends \[ x_i\longmapsto s^{d-i}t^i. \] The target has basis \[ s^{md-r}t^r, \qquad 0\le r\le md. \] Write \[ r=qd+a, \qquad 0\le a<d. \] If \(q<m\), the target monomial is the image of \[ x_d^q x_a x_0^{m-q-1}; \] if \(q=m\), then \(a=0\) and it is the image of \(x_d^m\). Thus every basis monomial lies in the image.
2. The rational normal curve \(X=\nu_d(\PP^1)\) is projectively normal.
The line bundle cut out by hyperplanes on \(X\) pulls back to \[ \nu_d^*\mco_X(1)\cong\mco_{\PP^1}(d). \] The curve \(X\cong\PP^1\) is normal. Step 1 says that for every \(m\ge0\) the homogeneous degree-\(m\) polynomials on the ambient projective space surject onto \[ H^0(X,\mco_X(m)). \] By Exercise II.5.14(d), normality together with these surjective restriction maps in every degree is exactly the section criterion for projective normality. Hence \(X\) is projectively normal.
3. Let\[ Q_{ij}=x_i x_j-x_{i+1}x_{j-1} \qquad (0\le i<j-1\le d-1). \]These quadrics vanish on \(X\).
Under the parametrization, \[ x_i x_j \longmapsto s^{2d-i-j}t^{i+j} \] and \[ x_{i+1}x_{j-1} \longmapsto s^{2d-i-j}t^{i+j}. \] Their difference therefore restricts to zero on \(X\).
4. The quadrics \(Q_{ij}\) generate the homogeneous ideal of \(X\).
3.Consider a degree-\(m\) monomial\[ M=x_{i_1}\cdots x_{i_m}. \]Whenever two indices satisfy\[ b-a\ge2, \]the relation\[ x_a x_b\equiv x_{a+1}x_{b-1}\pmod J \]replaces them by two closer indices while preserving their sum.The sum of the squares of all indices strictly decreases, so repeatedstraightening terminates.
Consequently every degree-\(m\) monomial is congruent modulo \(J\) to one whose indices differ by at most \(1\). If the sum of its indices is \[ r=mq+a, \qquad 0\le a<m, \] the resulting monomial is \[ x_q^{m-a}x_{q+1}^{a}, \] with the evident interpretation \(x_d^m\) when \(r=md\). Thus \((R/J)_m\) is spanned by at most \[ md+1 \] monomials, one for each possible index sum \(0,\ldots,md\).
Their restrictions to \(X\) are the distinct monomials \[ s^{md-r}t^r, \qquad 0\le r\le md, \] which are linearly independent. Hence these \(md+1\) classes are a basis of \((R/J)_m\).
By step 1 the degree-\(m\) homogeneous coordinate ring of \(X\) also has dimension \(md+1\). Since \(J\) is contained in the homogeneous ideal \(I_X\), equality of the degree-\(m\) quotients forces \[ (I_X)_m=J_m \] for every \(m\). Therefore \[ I_X=J, \] so the homogeneous ideal is generated by quadrics. Together with step 2, this proves (a).
5. If \(\mcl\) is a line bundle of degree \(e\ge0\) on a smooth projectivecurve, then\[ h^0(\mcl)\le e+1. \]
A degree-\(0\) line bundle has at most one nonzero section: a nonzero section has an effective zero divisor of degree \(0\), so it is nowhere vanishing and trivializes the bundle.
For \(e>0\), choose any point \(P\). The exact sequence \[ 0 \longrightarrow \mcl(-P) \longrightarrow \mcl \longrightarrow \mcl|_P \longrightarrow0 \] shows \[ h^0(\mcl)\le h^0(\mcl(-P))+1. \] Induction on the degree gives the stated bound.
6. Let \(X\subseteq\PP^n\) be nondegenerate of degree \(d\le n\). Then\[ d=n \qquad\text{and}\qquad h^0(X,\mco_X(1))=n+1. \]
Nondegeneracy means that no nonzero linear form vanishes identically on \(X\). Hence restriction is injective: \[ H^0(\PP^n,\mco(1)) \hookrightarrow H^0(X,\mco_X(1)). \] Therefore \[ n+1 \le h^0(X,\mco_X(1)). \] The hyperplane bundle has degree \(d\), so step 5 gives \[ h^0(X,\mco_X(1)) \le d+1. \] Using \(d\le n\), \[ n+1 \le h^0(X,\mco_X(1)) \le d+1 \le n+1. \] All inequalities are equalities, proving the claim.
7. Under the hypotheses of step 6, one has\[ X\cong\PP^1. \]
Put \[ \mcl=\mco_X(1). \] By step 6, \[ \deg\mcl=d, \qquad h^0(\mcl)=d+1. \] Choose an effective hyperplane divisor \[ H=P_1+\cdots+P_d \] counting multiplicities. Set \[ \mce=\mcl(-P_1-\cdots-P_{d-1}). \] Then \(\deg\mce=1\). Repeatedly applying the one-point exact sequence used in step 5 gives \[ h^0(\mce) \ge h^0(\mcl)-(d-1) =2. \] The degree-one case of step 5 gives the opposite inequality, hence \[ h^0(\mce)=2. \]
5.Thus \(|\mce|\) defines a nonconstant morphism\[ X\longrightarrow\PP^1 \]of degree \(\deg\mce=1\). A degree-one morphism between smooth projectivecurves is an isomorphism. Therefore\[ X\cong\PP^1, \qquad g(X)=0. \]
8. The embedding of step 6 is, up to an automorphism of \(\PP^d\), therational normal curve.
By steps <1>6–<1>7, \[ d=n, \qquad X\cong\PP^1, \qquad \mco_X(1)\cong\mco_{\PP^1}(d). \] The restriction map on hyperplane sections is injective between vector spaces of the same dimension \(d+1\), hence is an isomorphism. Thus the given embedding is defined by the complete linear system \[ |\mco_{\PP^1}(d)|. \] Choosing a different basis of this \((d+1)\)-dimensional space changes the map only by an automorphism of \(\PP^d\). Therefore the given curve differs from the standard \(d\)-uple embedding only by an automorphism of \(\PP^d\). This proves (b).
9. Every curve of degree \(2\) in projective space is a plane conic.
Let \(\PP^r\) be the linear span of the curve. Then the curve is nondegenerate in \(\PP^r\). It cannot have \(r=1\), since the only integral projective curve whose span is \(\PP^1\) is \(\PP^1\) itself, of degree \(1\). Thus \(r\ge2\), and \[ 2\le r. \] Applying part (b) in the span gives \[ r=2 \] and identifies the curve with the degree-two rational normal curve, namely a conic in \(\PP^2\). This proves (c).
10. Every curve of degree \(3\) is either a plane cubic or the twistedcubic in \(\PP^3\).
Again let \(\PP^r\) be the linear span. The case \(r=1\) is impossible. If \(r=2\), the curve is a plane curve of degree \(3\), hence a plane cubic.
If \(r\ge3\), then \[ 3\le r, \] so part (b) applied inside the span gives \[ r=3 \] and says that the curve is the degree-three rational normal curve. By Exercise I.2.12(d), this is the twisted cubic. This proves (d).
11. Q.E.D.
Steps <1>1–<1>4 prove (a), steps <1>5–<1>8 prove (b), step 9 proves (c), and step 10 proves (d).