04.04.E1 · algebraic-geometry / curves

Curves exercise pack (Hartshorne Ch. IV supplement)

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Formal definition of the pack Intermediate

Hartshorne Chapter IV is the worked-example chapter of the book: smooth projective curves over an algebraically closed field $k$ , studied through Riemann-Roch and Serre duality. It opens with Riemann-Roch 04.04.01 and the basic vanishing/special-divisor dictionary, proves Hurwitz's ramification formula 04.04.02, develops the criteria for a divisor to embed a curve in projective space, treats elliptic curves 04.04.03 with their group law and $j$ -invariant, and closes with the canonical embedding and the resulting classification of curves by genus. Many exercises braid several of these threads at once.

This pack collects nine exercises — two easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The exercises group loosely by Hartshorne section: numerical Riemann-Roch warm-ups (easy), Hurwitz and embedding computations (medium), and elliptic-curve and canonical-embedding arguments (hard).

The conventions throughout are Hartshorne's: $X$ is a smooth projective curve of genus $g$ over $k = \overset{ˉ}{k}$ ; $K_{X}$ a canonical divisor with $de g K_{X} = 2 g - 2$ and $ℓ (K_{X}) = g$ ; $ℓ (D) = h^{0} (O_{X} (D))$ ; Riemann-Roch reads $ℓ (D) - ℓ (K_{X} - D) = de g D + 1 - g$ . A divisor $D$ is very ample when $O_{X} (D)$ embeds $X$ in $P^{ℓ (D) - 1}$ ; the canonical divisor is very ample exactly when $X$ is non-hyperelliptic of genus $\geq 3$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Prove that a divisor $D$ of degree $\geq 2 g + 1$ on a smooth projective curve $X$ of genus $g$ is very ample.

Solution. A complete linear system $∣ D ∣$ is very ample iff it separates points and separates tangent vectors: for all points $P, Q \in X$ (including $P = Q$ ),

ℓ (D - P - Q) = ℓ (D) - 2.

Separating points ( $P \neq = Q$ ) means no hyperplane through the image of $P$ is forced through $Q$ ; separating tangents ( $P = Q$ ) means the embedding is an immersion at $P$ . Both conditions are captured by the displayed equality (Hartshorne IV.3.1).

Apply Riemann-Roch 04.04.01 to $D$ and to $D - P - Q$ . Since $de g D \geq 2 g + 1$ , also $de g D \geq 2 g - 1$ , so $de g (K_{X} - D) < 0$ and $ℓ (K_{X} - D) = 0$ ; thus $ℓ (D) = de g D + 1 - g$ . Likewise $de g (D - P - Q) = de g D - 2 \geq 2 g - 1$ , so $ℓ (K_{X} - D + P + Q) = 0$ and

ℓ (D - P - Q) = (de g D - 2) + 1 - g = ℓ (D) - 2.

The equality holds for every pair $(P, Q)$ , so $∣ D ∣$ separates points and tangents, hence $D$ is very ample. $□$

This is Hartshorne IV.3.2(b), the everyday embedding criterion. Two consequences route through it: every curve of genus $g$ embeds in $P^{3}$ (take a generic $D$ of degree $2 g + 1$ , then project), and any line bundle of large enough degree is automatically very ample, which is why ampleness and very-ampleness coincide asymptotically 04.05.05.

Exercises Intermediate

Exercise 2 (easy, numeric). Riemann-Roch for a generic divisor.

On a curve of genus $4$ , compute $ℓ (D)$ for a divisor $D$ of degree $10$ , and for a divisor $D^{'}$ of degree $5$ in general position.

Hint

When $de g D > 2 g - 2$ , the index $ℓ (K_{X} - D)$ vanishes and Riemann-Roch is exact. For $de g D^{'} = 5$ between $g - 1$ and $2 g - 2$ , "general position" means $ℓ (K_{X} - D^{'}) = 0$ as well.

Answer

Here $g = 4$ , so $de g K_{X} = 2 g - 2 = 6$ .

For $de g D = 10 > 6$ : $de g (K_{X} - D) = 6 - 10 = - 4 < 0$ , so $ℓ (K_{X} - D) = 0$ and

ℓ (D) = de g D + 1 - g = 10 + 1 - 4 = 7.

For $de g D^{'} = 5$ : now $de g (K_{X} - D^{'}) = 6 - 5 = 1 \geq 0$ , so $ℓ (K_{X} - D^{'})$ could be nonzero. For a general effective divisor of degree $5$ , the special divisor $K_{X} - D^{'}$ of degree $1$ is non-effective (a general degree- $1$ class has $ℓ = 0$ unless $g = 0$ ), so $ℓ (K_{X} - D^{'}) = 0$ and

ℓ (D^{'}) = de g D^{'} + 1 - g = 5 + 1 - 4 = 2.

(For a special $D^{'}$ — e.g. if $K_{X} - D^{'}$ happens to be an effective point — $ℓ (D^{'})$ could jump to $3$ ; the genus- $4$ curve is the canonical curve in $P^{3}$ and its $g_{3}^{1}$ is the special system.)

Exercise 3 (medium, numeric). Hurwitz for a hyperelliptic curve.

A hyperelliptic curve $X$ of genus $g$ admits a degree- $2$ map $f : X \to P^{1}$ . Use the Riemann-Hurwitz formula to determine the number of branch points.

Hint

Riemann-Hurwitz 04.04.02: $2 g - 2 = de g f \cdot (2 \cdot 0 - 2) + \sum_{P} (e_{P} - 1)$ . A double cover in characteristic $\neq = 2$ has only simple ( $e_{P} = 2$ ) ramification.

Answer

The target $P^{1}$ has genus $0$ . With $de g f = 2$ , Riemann-Hurwitz reads

2 g - 2 = 2 (2 \cdot 0 - 2) + P \sum (e_{P} - 1) = - 4 + P \sum (e_{P} - 1) .

In characteristic $0$ (or $\neq = 2$ ) a degree- $2$ map is tamely ramified, and every ramification point has $e_{P} = 2$ (the fiber over a branch point is a single point counted twice), contributing $e_{P} - 1 = 1$ each. If there are $r$ branch points,

2 g - 2 = - 4 + r ⟹ r = 2 g + 2.

So a hyperelliptic curve of genus $g$ is a double cover of $P^{1}$ branched over exactly $2 g + 2$ points — the Weierstrass points. For $g = 1$ this gives $4$ branch points, matching the four roots (including $\infty$ ) of a Legendre-form elliptic curve 04.04.03; for $g = 2$ , six branch points.

Exercise 4 (medium, symbolic). The $j$ -invariant of an elliptic curve.

For the Weierstrass curve $y^{2} = x^{3} + a x + b$ over $k$ ( $char k \neq = 2, 3$ ), recall the discriminant $Δ = - 16 (4 a^{3} + 27 b^{2})$ and the $j$ -invariant $j = - 1728 (4 a)^{3} /Δ$ . Show that two such curves are isomorphic iff they have the same $j$ , and compute $j$ for $y^{2} = x^{3} + x$ and $y^{2} = x^{3} + 1$ .

Hint

The only coordinate changes preserving Weierstrass form are $x \mapsto u^{2} x, y \mapsto u^{3} y$ , scaling $(a, b) \mapsto (u^{4} a, u^{6} b)$ . Check $j$ is invariant and that it pins the pair $(a, b)$ up to this scaling.

Answer

An admissible change of variable preserving short Weierstrass form is $x = u^{2} x^{'}$ , $y = u^{3} y^{'}$ for $u \in k^{\times}$ , sending $(a, b) \mapsto (u^{4} a, u^{6} b)$ and $Δ \mapsto u^{12} Δ$ . Under this, $(4 a)^{3} /Δ \mapsto u^{12} (4 a)^{3} / (u^{12} Δ)$ is unchanged, so $j$ is an isomorphism invariant.

Conversely, given equal $j$ , the ratio $a^{3} : b^{2}$ is determined (since $j$ is a function of $4 a^{3} / (4 a^{3} + 27 b^{2})$ ). If both $a, b \neq = 0$ , solve $u^{4} = a / a^{'}$ and check the same $u$ gives $u^{6} = b / b^{'}$ using equal $j$ ; the degenerate cases $a = 0$ ( $j = 0$ ) and $b = 0$ ( $j = 1728$ ) are handled by the corresponding scalings. So equal $j$ implies isomorphic curves over $\overset{ˉ}{k}$ .

Compute. For $y^{2} = x^{3} + x$ : $a = 1, b = 0$ , so $Δ = - 16 \cdot 4 = - 64$ and

j = - 1728 \cdot \frac{( 4 ) ^{3}}{- 64} = - 1728 \cdot \frac{64}{- 64} = 1728.

For $y^{2} = x^{3} + 1$ : $a = 0, b = 1$ , so $(4 a)^{3} = 0$ and $j = 0$ .

Thus $y^{2} = x^{3} + x$ has $j = 1728$ (the curve with extra automorphism $i$ , of order $4$ ) and $y^{2} = x^{3} + 1$ has $j = 0$ (the curve with the order- $6$ automorphism). These are exactly the two curves with automorphism groups larger than ${\pm 1}$ , Hartshorne IV.4.

Exercise 5 (medium, numeric). Genus of a curve of bidegree $(a, b)$ on a quadric.

Let $X \subset P^{1} \times P^{1}$ be a smooth curve of bidegree $(a, b)$ (class $a {pt} \times P^{1} + b P^{1} \times {pt}$ ). Compute its genus.

Hint

Adjunction on the surface $Q = P^{1} \times P^{1}$ : $2 g - 2 = X \cdot (X + K_{Q})$ , with $K_{Q}$ of bidegree $(- 2, - 2)$ and the intersection product on $Q$ given by $(a, b) \cdot (c, d) = a d + b c$ .

Answer

On $Q = P^{1} \times P^{1}$ the intersection pairing on divisor classes is $(a, b) \cdot (c, d) = a d + b c$ , and the canonical class is $K_{Q} = (- 2, - 2)$ 04.08.02. The adjunction formula gives $2 g - 2 = X \cdot (X + K_{Q})$ for the smooth curve $X$ of class $(a, b)$ .

Compute $X + K_{Q} = (a - 2, b - 2)$ , so

X \cdot (X + K_{Q}) = (a, b) \cdot (a - 2, b - 2) = a (b - 2) + b (a - 2) = 2 ab - 2 a - 2 b .

Hence $2 g - 2 = 2 ab - 2 a - 2 b$ , i.e.

g = ab - a - b + 1 = (a - 1) (b - 1) .

Checks: bidegree $(1, 1)$ (a "diagonal" conic, rational) gives $g = 0$ ; bidegree $(2, 2)$ gives $g = 1$ (an elliptic curve as a $(2, 2)$ on the quadric); bidegree $(2, 3)$ gives $g = 2$ . This is Hartshorne IV.1 Example / Exercise, the quadric-surface analogue of the plane genus-degree formula [from 04.03.E1 Exercise 3].

Exercise 6 (hard, proof). The canonical embedding for non-hyperelliptic curves.

Let $X$ be a smooth projective curve of genus $g \geq 3$ . Prove that the canonical divisor $K_{X}$ is very ample iff $X$ is not hyperelliptic, embedding non-hyperelliptic $X$ as a curve of degree $2 g - 2$ in $P^{g - 1}$ .

Hint

By the point-and-tangent criterion (lead exercise), $K_{X}$ fails to separate $P, Q$ iff $ℓ (K_{X} - P - Q) = ℓ (K_{X}) - 1 = g - 1$ . By Riemann-Roch this is equivalent to $ℓ (P + Q) = 2$ , i.e. a $g_{2}^{1}$ , i.e. hyperelliptic.

Answer

The complete canonical system $∣ K_{X} ∣$ has $ℓ (K_{X}) = g$ , so it maps $X \to P^{g - 1}$ . By the very-ampleness criterion (lead exercise) $∣ K_{X} ∣$ is very ample iff for all points $P, Q$ ,

ℓ (K_{X} - P - Q) = ℓ (K_{X}) - 2 = g - 2.

Apply Riemann-Roch 04.04.01 to $D = P + Q$ (degree $2$ ):

ℓ (P + Q) - ℓ (K_{X} - P - Q) = de g (P + Q) + 1 - g = 2 + 1 - g = 3 - g .

Since $ℓ (K_{X}) = g$ , the failure $ℓ (K_{X} - P - Q) = g - 1$ (one bigger than the very-ample value $g - 2$ ) is equivalent, by the displayed equation, to

ℓ (P + Q) = (g - 1) + (3 - g) = 2.

But $ℓ (P + Q) = 2$ means the degree- $2$ divisor $P + Q$ moves in a pencil — a $g_{2}^{1}$ — which by definition makes $X$ hyperelliptic (the pencil gives a degree- $2$ map $X \to P^{1}$ ).

So $∣ K_{X} ∣$ fails to separate some pair $P, Q$ iff $X$ is hyperelliptic. Contrapositive: $X$ non-hyperelliptic $\Rightarrow K_{X}$ very ample, giving an embedding $X ↪ P^{g - 1}$ of degree $de g K_{X} = 2 g - 2$ . This is the canonical embedding (Hartshorne IV.5.2). For $g = 3$ it realizes a non-hyperelliptic curve as a smooth plane quartic; for $g = 4$ , as the complete intersection of a quadric and a cubic in $P^{3}$ .

Exercise 7 (hard, proof). Every genus-1 curve with a point is a plane cubic.

Let $(X, P_{0})$ be a smooth projective curve of genus $1$ with a chosen point $P_{0}$ . Show that the linear system $∣3 P_{0} ∣$ embeds $X$ as a smooth cubic curve in $P^{2}$ , and that $P_{0}$ becomes an inflection point.

Hint

Compute $ℓ (n P_{0})$ by Riemann-Roch for small $n$ . Degree $3 = 2 g + 1$ on a genus- $1$ curve is very ample (lead exercise). The image has degree $3$ in $P^{2}$ .

Answer

For genus $g = 1$ , $de g K_{X} = 0$ and $ℓ (K_{X}) = 1$ , so $K_{X} \sim 0$ (the canonical class is the zero class, $O_{X}$ ). Riemann-Roch 04.04.01 applied to $D = n P_{0}$ ( $n \geq 1$ ) gives $de g (K_{X} - n P_{0}) = - n < 0$ , so $ℓ (K_{X} - n P_{0}) = 0$ and

ℓ (n P_{0}) = n + 1 - g = n (n \geq 1) .

Thus $ℓ (P_{0}) = 1$ , $ℓ (2 P_{0}) = 2$ , $ℓ (3 P_{0}) = 3$ . The degree $de g (3 P_{0}) = 3 = 2 g + 1$ , so $∣3 P_{0} ∣$ is very ample by the lead exercise, embedding $X ↪ P^{ℓ (3 P_{0}) - 1} = P^{2}$ . The image has degree equal to $de g (3 P_{0}) = 3$ , a smooth plane cubic.

A basis of $L (3 P_{0})$ is ${1, x, y}$ with $x \in L (2 P_{0}) ∖ L (P_{0})$ (a pole of order $2$ at $P_{0}$ ) and $y \in L (3 P_{0}) ∖ L (2 P_{0})$ (pole of order $3$ ). The seven functions $1, x, y, x^{2}, x y, y^{2}, x^{3}$ lie in the $6$ -dimensional $L (6 P_{0})$ , so satisfy one linear relation — a Weierstrass cubic $y^{2} + \dots = x^{3} + \dots$ after scaling. The point $P_{0}$ maps to the unique point at infinity $[0 : 1 : 0]$ , where the tangent line ${Z = 0}$ meets the cubic with multiplicity $3$ : $P_{0}$ is an inflection (flex) point. This is exactly the construction of the group law 04.04.03: $P_{0}$ is the identity, and collinearity of three points means their sum is $0$ (Hartshorne IV.4.6).

Exercise 8 (hard, proof). Clifford-type bound and the genus-2 classification.

Show that every smooth projective curve of genus $2$ is hyperelliptic, and that its canonical map is the degree- $2$ map to $P^{1}$ .

Hint

For $g = 2$ , $de g K_{X} = 2$ and $ℓ (K_{X}) = 2$ , so $∣ K_{X} ∣$ is a $g_{2}^{1}$ . A base-point-free $g_{2}^{1}$ is a degree- $2$ map.

Answer

For genus $g = 2$ : $de g K_{X} = 2 g - 2 = 2$ and $ℓ (K_{X}) = g = 2$ . So the canonical system $∣ K_{X} ∣$ is a linear system of dimension $ℓ (K_{X}) - 1 = 1$ and degree $2$ — a $g_{2}^{1}$ .

First, $∣ K_{X} ∣$ is base-point free. If $P$ were a base point, then $ℓ (K_{X} - P) = ℓ (K_{X}) = 2$ ; but $de g (K_{X} - P) = 1$ , and Riemann-Roch gives $ℓ (K_{X} - P) - ℓ (P) = 1 + 1 - 2 = 0$ , so $ℓ (K_{X} - P) = ℓ (P)$ . On a curve of genus $\geq 1$ , $ℓ (P) = 1$ (no nonconstant function with a single simple pole, else $X ≅ P^{1}$ ). Hence $ℓ (K_{X} - P) = 1 \neq = 2$ , contradiction. So $∣ K_{X} ∣$ has no base points.

A base-point-free $g_{2}^{1}$ defines a morphism $φ_{∣ K_{X} ∣} : X \to P^{1}$ of degree $2$ (the degree equals $de g K_{X} = 2$ ). A degree- $2$ map to $P^{1}$ is exactly a hyperelliptic structure. Therefore every genus- $2$ curve is hyperelliptic, and its canonical map is the hyperelliptic double cover.

This also shows the canonical map is not an embedding for $g = 2$ (it is $2$ -to- $1$ ), consistent with Exercise 6: the canonical embedding requires non-hyperelliptic, which forces $g \geq 3$ . By Riemann-Hurwitz 04.04.02 (Exercise 3) this double cover branches over $2 g + 2 = 6$ points, so every genus- $2$ curve is $y^{2} = f_{6} (x)$ for a sextic $f_{6}$ with distinct roots. Hartshorne IV.5.

Exercise 9 (medium, proof). Curves of genus 0 are projective lines.

Let $X$ be a smooth projective curve of genus $0$ over $k = \overset{ˉ}{k}$ with a $k$ -rational point $P$ . Show $X ≅ P^{1}$ .

Hint

Compute $ℓ (P)$ by Riemann-Roch. A degree- $1$ divisor with $ℓ = 2$ gives a degree- $1$ map to $P^{1}$ .

Answer

For genus $g = 0$ , $de g K_{X} = - 2$ . Apply Riemann-Roch 04.04.01 to $D = P$ (degree $1$ ): $de g (K_{X} - P) = - 3 < 0$ , so $ℓ (K_{X} - P) = 0$ and

ℓ (P) = de g P + 1 - g = 1 + 1 - 0 = 2.

So $L (P)$ is $2$ -dimensional, spanned by $1$ and a nonconstant function $f$ with a single simple pole at $P$ . The map $f : X \to P^{1}$ (sending $P \mapsto \infty$ ) has degree equal to the number of poles counted with multiplicity, namely $1$ . A degree- $1$ morphism of smooth projective curves is an isomorphism (it is finite, birational, and the target is normal). Hence $X ≅ P^{1}$ .

This completes the bottom of the classification by genus: $g = 0$ gives $P^{1}$ ; $g = 1$ gives the plane cubics / elliptic curves (Exercise 7); $g = 2$ the hyperelliptic sextic double covers (Exercise 8); $g \geq 3$ splits into hyperelliptic and canonically-embedded non-hyperelliptic curves (Exercise 6). Over $k = \overset{ˉ}{k}$ every genus- $0$ curve has a rational point, so the hypothesis is automatic.

Exercise pack EP1 for 04-algebraic-geometry/04-curves. Hartshorne Chapter IV supplement: Riemann-Roch for curves, Hurwitz and ramification, embeddings in projective space, elliptic curves and the $j$ -invariant, canonical embedding and classification by genus (§IV.1–§IV.5).

Prerequisites

04.04.01 pending
04.04.02 pending
04.04.03 pending
04.08.02 pending
04.08.03 pending
04.05.05 pending

Tier anchors

beginner
intermediate: Hartshorne Ch. IV exercises (§IV.1 Riemann-Roch, §IV.2 Hurwitz and ramification, §IV.3 embeddings in projective space, §IV.4 elliptic curves and the $j$-invariant, §IV.5 canonical embedding and the classification by genus)
master

References

Hartshorne — Algebraic Geometry (GTM 52) · Ch. IV exercises across §IV.1 (Riemann-Roch, special divisors, genus of curves on $\mathbb{P}^1\times\mathbb{P}^1$), §IV.2 (Hurwitz, the Riemann-Hurwitz count, the $j$-line cover), §IV.3 (very ampleness of $\deg\geq 2g+1$ divisors, projective normality), §IV.4 (elliptic curves, group law, $j$-invariant, automorphisms), §IV.5 (canonical embedding, hyperelliptic curves, classification by genus)
Silverman — The Arithmetic of Elliptic Curves (GTM 106) · Ch. III — Weierstrass equations, the group law, the $j$-invariant, automorphisms
Arbarello-Cornalba-Griffiths-Harris — Geometry of Algebraic Curves I · Ch. I–III — special divisors, the canonical curve, Brill-Noether theory

Estimated time

intermediate: 6h