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Chapter 4: Problem 3

If \(X\) is a curve of genus \(\geqslant 2\) which is a complete intersection (II, Ex. 8.4 ) in some \(\mathbf{P}^{n},\) show that the canonical divisor \(K\) is very ample. Conclude that a curve of genus 2 can never be a complete intersection in any \(\mathbf{P}^{n}\). Cf. (Ex. 5.1 ).

Short Answer

Expert verified
Given a curve \(X\) of genus \(g\geq 2\) as a complete intersection, the canonical divisor \(K_{X}\) is very ample. However, a curve with genus 2 cannot be a complete intersection in any \(\mathbb{P}^{n}\) because this contradicts with the degree of canonical bundle.

Step by step solution

01

Show that the canonical divisor is very ample

If \(X\) is a complete intersection, then it is embedded in \(\mathbf{P}^{n}\) by a very ample divisor. The canonical divisor \(K_{X}\) is a divisor from the ambient space \(\mathbf{P}^{n}\). Since the restriction of a very ample divisor is very ample, we can conclude that the canonical divisor is very ample.
02

Rules out genus 2 as complete intersection

Consider a curve of genus 2. The degree of the canonical bundle \(K_{X}\) of the curve \(X\), which is a complete intersection, is equal to \(2g-2 = 2\). A very ample divisor on a projective space curve of degree \(d\) embeds the curve into \(\mathbb{P}^d\). But for the canonical divisor on a genus 2 curve, we have \(d=2\), and so our curve would be embedded into \(\mathbb{P}^2\). However, every curve in \(\mathbb{P}^2\) is a plane curve and so is of genus 0 or 1, a contradiction because we assumed our curve \(X\) has genus 2. Thus, a curve of genus 2 cannot be a complete intersection in any \(\mathbf{P}^{n}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

genus of a curve
The genus of a curve is an important concept in algebraic geometry. It essentially measures the complexity of the curve. You can think of it as a shape equivalent to the number of 'holes' in a surface. For example, a sphere has genus 0, while a torus (the surface of a donut) has genus 1.

The genus can be calculated using various formulas. For curves, if you know the number of points where two surfaces intersect, called the degree, it can help in finding the genus. For a smooth projective curve, the genus is related to the degree by the formula: \[ g = \frac{{(d-1)(d-2)}}{2} \]where \(d\) is the degree of the curve.

In the context of our exercise, understanding that a curve has genus 2 helps us make certain conclusions about its properties in projective space. It determines that such a curve cannot be embedded as a complete intersection in any \(\mathbf{P}^{n}\). This is because, according to the Riemann-Roch theorem, a curve of genus 2 contradicts the required conditions for complete intersection embedding, thereby ruling out that possibility.
complete intersection
In algebraic geometry, a complete intersection is a type of variety generated by the simultaneous zeroes of several polynomials. When a curve is described as a complete intersection, it implies that the curve arises from the intersection of hypersurfaces.

The concept becomes particularly rich when considered in projective space, \(\mathbf{P}^{n}\). Here, a complete intersection is achieved when the number of equations describing the variety is exactly what is needed to specify the dimension. For example, in \(\mathbf{P}^{3}\), we need two equations to define a curve, because each equation reduces the dimension from 3 to 2, and then to 1.

When a curve has certain genus, such as genus 2, it influences its ability to be a complete intersection. The explicit exercise shows that a curve of genus 2 cannot be a complete intersection in any \(\mathbf{P}^{n}\). This is strictly due to the fact that the conditions required for being a plane curve, based on the resultant degree, lead to a contradiction when they do not match with the given genus.
very ample divisor
A very ample divisor is a concept that helps to define how a variety, like a curve, can be embedded into a projective space. In simple terms, it's a tool that enables us to see the curve inside the projective space as a tightly and nicely fitted image.

When we say a divisor is very ample, it means that the sections of the line bundle it represents can embed the variety into a projective space in such a way that the variety retains its structure. The divisor can be thought of as both describing a 'shape' and 'spacing' necessary to help visualize the curve's presentation in its ambient space.

In our specific problem from the exercise, because the curve is a complete intersection, it's embedded in \(\mathbf{P}^{n}\) using a very ample divisor. The canonical divisor, which is central to defining the geometry of a curve, retains the very ample property when restricted to the curve, confirming that the curve can be embedded fully and robustly 鈥� but only if it is not of genus 2, highlighting the exception drawn in the solution.

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Most popular questions from this chapter

For any curve \(X,\) the algebraic fundamental group \(\pi_{1}(X)\) is defined as \(\lim \operatorname{Gal}\left(K^{\prime} / K\right),\) where \(K\) is the function field of \(X,\) and \(K^{\prime}\) runs over all Galois extensions of \(K\) such that the corresponding curve \(X^{\prime}\) is 茅tale over \(X(\mathrm{III}, \mathrm{Ex} .10 .3)\) Thus, for example, \(\pi_{1}\left(\mathbf{P}^{1}\right)=1(2.5 .3) .\) Show that for an elliptic curve \(X\) \(\pi_{1}(X)=\prod_{\text {prime }} \mathbf{Z}_{l} \times \mathbf{Z}_{l} \quad \text { if char } k=0\) \(\pi_{1}(X)=\prod_{l \neq p} \mathbf{Z}_{l} \times \mathbf{Z}_{l}\) if char \(k=p\) and Hasse \(X=0\) \(\pi_{1}(X)=\mathbf{Z}_{p} \times \prod_{l \neq p} \mathbf{Z}_{l} \times \mathbf{Z}_{l} \quad\) if char \(k=p\) and Hasse \(X \neq 0\) where \(\mathbf{Z}_{l}=\lim \mathbf{Z} / l^{n}\) is the \(l\) -adic integers. [Hints: Any Galois 茅tale cover \(X^{\text {' }}\) of an elliptic curve is again an elliptic curve If the degree of \(X\) ' over \(X\) is relatively prime to \(p\), then \(X\) ' can be dominated by the cover \(n_{X}: X \rightarrow X\) for some integer \(n\) with \((n, p)=1 .\) The Galois group of the covering \(n_{X}\) is \(\mathbf{Z}\) in \(\times \mathbf{Z}\),n. Etale covers of degree divisible by \(p\) can occur only if the Hasse invariant of \(X\) is not zero. Note: More generally, Grothendieck has shown [SGA 1, X, 2.6, p. 272] that the algebraic fundamental group of any curve of genus \(g\) is isomorphic to a quotient of the completion, with respect to subgroups of finite index, of the ordinary topological fundamental group of a compact Riemann surface of genus \(g,\) i.e., a group with \(2 g\) generators \(a_{1}, \ldots, a_{q}, b_{1}, \ldots, b_{q}\) and the relation \(\left(a_{1} b_{1} a_{1}^{-1} b_{1}^{-1}\right) \cdots\) \(\left(a_{q} b_{q} a_{q}^{-1} b_{q}^{-1}\right)=1.\)

Let \(X\) be an elliptic curve over a field \(k\) of characteristic \(p .\) Let \(F: X_{p} \rightarrow X\) be the \(k\) -linear Frobenius morphism \((2.4 .1) .\) Use \((4.10 .7)\) to show that the dual morphism \(\hat{F}^{\prime}: X \rightarrow X_{p}\) is separable if and only if the Hasse invariant of \(X\) is 1 Now use (Ex. 4.7 ) to show that if the Hasse invariant is 1 , then the subgroup of points of order \(p\) on \(X\) is isomorphic to \(\mathbf{Z} / p ;\) if the Hasse invariant is \(0,\) it is 0.

Again let \(X\) be an elliptic curve over \(k\) of characteristic \(p,\) and suppose \(X\) is defined over the field \(\mathbf{F}_{q}\) of \(q=p^{r}\) elements, i.e., \(X \subseteq \mathbf{P}^{2}\) can be defined by an equation with coefficients in \(\mathbf{F}_{q}\). Assume also that \(X\) has a rational point over \(\mathbf{F}_{q} .\) Let \(F^{\prime}: X_{q} \rightarrow X\) be the \(k\) -linear Frobenius with respect to \(q\) (a) Show that \(X_{q} \cong X\) as schemes over \(k\), and that under this identification, \(F^{\prime}: X \rightarrow X\) is the map obtained by the \(q\) th-power map on the coordinates of points of \(X,\) embedded in \(\mathbf{P}^{2}\) (b) Show that \(1_{X}-F^{\prime}\) is a separable morphism and its kernel is just the set \(X\left(\mathbf{F}_{q}\right)\) of points of \(X\) with coordinates in \(\mathbf{F}_{q}\) (c) Using (Ex. 4.7 ), show that \(F^{\prime}+\hat{F}^{\prime}=a_{x}\) for some integer \(a\), and that \(N=\) \(q-a+1,\) where \(N=\\# X\left(\mathbf{F}_{q}\right)\) (d) Use the fact that \(\operatorname{deg}\left(m+n F^{\prime}\right)>0\) for all \(m, n \in \mathbf{Z}\) to show that \(|a| \leqslant 2 \vee q\) This is Hasse's proof of the analogue of the Riemann hypothesis for elliptic curves (App. C. Ex. 5.6). (e) Now assume \(y=p,\) and show that the Hasse invariant of \(X\) is 0 if and only if \(a \equiv 0(\bmod p) .\) Conclude for \(p \geqslant 5\) that \(X\) has Hasse invariant 0 if and only if \(N=p+1.\)

If \(X \rightarrow \mathbf{A}_{\mathrm{C}}^{1}\) is a family of elliptic curves having a section, show that the family is trivial. \([\)Hints: Use the section to fix the group structure on the fibres. Show that the points of order 2 on the fibres form an 茅tale cover of \(\mathbf{A}_{\mathbf{C}}^{1},\) which must be trivial, since \(\mathbf{A}_{\mathbf{C}}^{1}\) is simply connected. This implies that \(\lambda\) can be defined on the family, so it gives a \(\operatorname{map} \mathbf{A}_{\mathbf{C}}^{1} \rightarrow \mathbf{A}_{\mathbf{C}}^{1}-\\{0,1\\} .\) Any such map is constant, so \(i\) is constant, so the family is trivial.

A curve of degree 5 and genus 2 in \(\mathbf{P}^{3}\) is contained in a unique quadric surface \(Q\) .Show that for any abstract curve \(X\) of genus \(2,\) there exist embeddings of degree 5 in \(\mathbf{P}^{3}\) for which \(Q\) is nonsingular, and there exist other embeddings of degree 5 for which \(Q\) is singular.
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