Divisors and the Picard Group
Authors: Based on Shafarevich §III.1, Hartshorne §II.6–7, Fulton Ch. 8
Relations
Builds on: Local Properties of Varieties, Normalization and Resolution Extended by: Differential Forms, Riemann-Roch, and Hurwitz’s Formula Concepts used: Classical Algebraic Geometry
Throughout, \(k\) denotes an algebraically closed field and \(X\) is a smooth projective curve over \(k\). For background on DVRs and the local ring \(\mathcal{O}_{X,P}\), see Local Properties; for normalization and the smooth projective model, see Normalization and Resolution.
Table of Contents
- 1. Introduction
- 2. Weil Divisors on Curves
- 3. Rational Functions and Principal Divisors
- 4. The Picard Group
- 5. Line Bundles and the Divisor-Line Bundle Correspondence
- 6. Linear Systems and Maps to Projective Space
- References
1. Introduction 📐
A rational function \(f \in k(X)^\times\) on a projective curve \(X\) cannot be regular everywhere if it is non-constant: wherever it is non-zero and finite, it eventually vanishes somewhere and diverges somewhere else. The key insight of divisor theory is to track these zeros and poles globally, as a single algebraic object.
A divisor on \(X\) is a formal finite \(\mathbb{Z}\)-linear combination of closed points. Adding up the vanishing orders gives a map \(f \mapsto \mathrm{div}(f)\), and two divisors are declared linearly equivalent if they differ by such a principal divisor. The group of equivalence classes is the Picard group \(\mathrm{Pic}(X)\), the central invariant of this note.
The theory has three rewards:
- Classification of line bundles. There is a canonical isomorphism \(\mathrm{Pic}(X) \cong \{\text{line bundles on }X\}/{\cong}\).
- Maps to projective space. Every morphism \(X \to \mathbb{P}^r\) (up to linear equivalence) arises from a divisor \(D\) with \(h^0(D) = r+1\) and empty base locus.
- Group law on elliptic curves. For a smooth elliptic curve \(E\), the map \(P \mapsto [P] - [O]\) identifies \(E(k)\) with \(\mathrm{Pic}^0(E)\), realizing the mysterious chord-tangent law as group addition in the Picard group.
The notion of divisor on an algebraic curve is the algebraic analogue of the divisor of a meromorphic function on a compact Riemann surface. Abel’s theorem (1826) and Jacobi’s inversion theorem together characterize which divisors of degree zero are principal — a question that Riemann-Roch answers in the algebraic setting.
2. Weil Divisors on Curves 🔑
2.1 The DVR at a Smooth Point
Fix \(X\) a smooth projective curve over \(k = \bar{k}\). Recall from Local Properties that at each closed point \(P \in X\), the local ring \(\mathcal{O}_{X,P}\) is a discrete valuation ring (DVR): a local PID with maximal ideal \(\mathfrak{m}_P = (t_P)\) for some local parameter \(t_P\) (a uniformizer). The associated valuation
\[v_P : k(X)^\times \longrightarrow \mathbb{Z}\]
is defined by \(v_P(f) = n\) when \(f = t_P^n \cdot u\) with \(u \in \mathcal{O}_{X,P}^\times\). By convention \(v_P(0) = +\infty\).
Key properties:
- \(v_P(fg) = v_P(f) + v_P(g)\),
- \(v_P(f + g) \geq \min(v_P(f), v_P(g))\), with equality when \(v_P(f) \neq v_P(g)\),
- \(v_P(f) > 0 \Leftrightarrow P\) is a zero of \(f\); \(v_P(f) < 0 \Leftrightarrow P\) is a pole of \(f\); \(v_P(f) = 0 \Leftrightarrow f(P) \in k^\times\).
For any \(f \in k(X)^\times\), the set \(\{P \in X : v_P(f) \neq 0\}\) is finite. This follows because \(f\) defines a rational map \(X \dashrightarrow \mathbb{P}^1\) which, by the smoothness of \(X\), extends to a regular morphism \(X \to \mathbb{P}^1\); the preimage of \(0\) and \(\infty\) are each finite.
2.2 The Divisor Group
Definition (Divisor). A Weil divisor on \(X\) is a formal sum
\[D = \sum_{P \in X} n_P [P], \quad n_P \in \mathbb{Z},\]
where \(n_P = 0\) for all but finitely many \(P\). The divisor group is
\[\mathrm{Div}(X) = \bigoplus_{P \in X} \mathbb{Z} \cdot [P],\]
with the group law given by pointwise addition of coefficients: \((D + D')_P = n_P + n_P'\).
The coefficient \(n_P\) is called the multiplicity of \(D\) at \(P\). The support of \(D\) is \(\mathrm{supp}(D) = \{P : n_P \neq 0\}\).
2.3 Effective Divisors and the Degree Map
Definition (Effective divisor). A divisor \(D = \sum n_P [P]\) is effective, written \(D \geq 0\), if \(n_P \geq 0\) for all \(P\). We write \(D \geq D'\) if \(D - D' \geq 0\).
Definition (Degree). The degree map is the group homomorphism
\[\deg : \mathrm{Div}(X) \longrightarrow \mathbb{Z}, \quad \deg\!\left(\sum n_P [P]\right) = \sum_{P} n_P.\]
The kernel \(\mathrm{Div}^0(X) = \ker(\deg)\) is the subgroup of degree-zero divisors.
- The zero divisor \(0 \in \mathrm{Div}(X)\) has \(n_P = 0\) for all \(P\); \(\deg(0) = 0\). - A single point \([P]\) is an effective divisor of degree 1. - The divisor \(2[P] - [Q]\) has \(\deg = 1\), support \(\{P, Q\}\), and is not effective (since \(n_Q = -1\)). - On \(\mathbb{P}^1\), the “hyperplane class” is any \([P]\); all degree-\(n\) effective divisors on \(\mathbb{P}^1\) consist of \(n\) points (with multiplicity).
This problem establishes fluency with the basic operations on divisors.
Prerequisites: 2.3 Effective Divisors and the Degree Map
Let \(X = \mathbb{P}^1\). (a) Write down all effective divisors of degree 2 whose support has exactly one point. (b) Show that \(\mathrm{Div}^0(\mathbb{P}^1)\) is not finitely generated as an abelian group. (c) For an arbitrary smooth projective curve \(X\) of genus \(g\), how many points does a “general” effective divisor of degree \(d\) have in its support?
Key insight: Effective divisors of degree 2 on \(\mathbb{P}^1\) are either two distinct points or one point with multiplicity 2.
Sketch: (a) The only effective divisors of degree 2 with a single-point support are \(2[P]\) for any \(P \in \mathbb{P}^1\). There are \(|\mathbb{P}^1| = |k|\) such divisors (one per point).
\(\mathrm{Div}^0(\mathbb{P}^1) = \bigoplus_{P \in \mathbb{P}^1} \mathbb{Z} \cdot [P] / \langle \sum [P] = 0 \rangle\) (informally). More precisely, for any finite set \(S \subset \mathbb{P}^1\), the elements \(\{[P] - [Q] : P \in S\}\) are linearly independent in \(\mathrm{Div}^0\). Since \(|k| = \infty\) (as \(k = \bar{k}\)), no finite generating set suffices.
A general effective divisor of degree \(d\) on a curve has exactly \(d\) points in its support (all with multiplicity 1). Special divisors (with multiplicities) form a closed subset of the \(d\)-th symmetric product \(X^{(d)}\).
This problem explores what goes wrong when the smoothness hypothesis is dropped.
Prerequisites: 2.2 The Divisor Group
Let \(C \subset \mathbb{P}^2\) be the nodal cubic \(y^2 z = x^2(x+z)\), which has a node at \(P = [0:0:1]\). The local ring \(\mathcal{O}_{C,P}\) is NOT a DVR. (a) Verify this by showing \(\mathfrak{m}_P = (x/z, y/z)\) requires two generators. (b) Explain why the definition of \(v_P(f)\) breaks down at \(P\). (c) The normalization \(\tilde{C} \cong \mathbb{P}^1\) separates the node into two smooth points \(\tilde{P}_1, \tilde{P}_2\). How should we define \(\mathrm{Div}(\tilde{C})\) versus \(\mathrm{Div}(C)\)?
Key insight: A DVR requires the maximal ideal to be principal; this forces the Zariski tangent space to be 1-dimensional. At a singular point the tangent space is strictly larger than \(\dim X\), so no single generator can exist.
Sketch: (a) In affine coordinates \((x,y) = (x/z, y/z)\), the nodal cubic becomes \(f = y^2 - x^2(x+1)\). At the origin, \(\nabla f|_{(0,0)} = (-2x(x+1) - x^2,\; 2y)|_{(0,0)} = (0, 0)\), so the Jacobian vanishes — \((0,0)\) is a singular point. The Zariski cotangent space \(\mathfrak{m}_P/\mathfrak{m}_P^2\) is spanned by the images of \(x\) and \(y\), giving \(\dim_k \mathfrak{m}_P/\mathfrak{m}_P^2 = 2\). By Nakayama’s lemma, \(\mathfrak{m}_P\) requires at least two generators, so \(\mathcal{O}_{C,P}\) is not a DVR. (Note: over the completion \(\hat{\mathcal{O}}_{C,P}\), the relation does factor as \((y - x\sqrt{x+1})(y + x\sqrt{x+1})\) since \(\sqrt{x+1}\) is a unit; that factoring reflects the two branches, not the failure of generation.)
Without a uniformizer, there is no canonical \(v_P : \mathcal{O}_{C,P}^\times \to \mathbb{Z}\) with the valuation property: the two branches through \(P\) give two distinct “orders of vanishing,” and there is no way to reconcile them into a single integer.
\(\mathrm{Div}(\tilde{C}) = \bigoplus_{Q \in \mathbb{P}^1} \mathbb{Z} \cdot [Q]\) is well-defined since \(\tilde{C}\) is smooth. \(\mathrm{Div}(C)\) can be defined using only the smooth points \(C \setminus \{P\}\), or by working on \(\tilde{C}\) and imposing the “compatibility at \(\tilde{P}_1, \tilde{P}_2\)” condition. The Picard group of \(C\) and \(\tilde{C}\) differ: \(\mathrm{Pic}^0(\tilde{C}) = 0\) while \(\mathrm{Pic}^0(C) \cong k^\times\).
3. Rational Functions and Principal Divisors 📐
3.1 Order of Vanishing
For a nonzero rational function \(f \in k(X)^\times\), the order of vanishing of \(f\) at \(P\) is \(v_P(f) \in \mathbb{Z}\). The finiteness property of Section 2.1 ensures that \(v_P(f) \neq 0\) for only finitely many \(P\), so we may form:
Definition (Principal divisor). The principal divisor of \(f \in k(X)^\times\) is
\[\mathrm{div}(f) = \sum_{P \in X} v_P(f)\, [P] \;\in\; \mathrm{Div}(X).\]
We decompose \(\mathrm{div}(f) = \mathrm{div}(f)_+ - \mathrm{div}(f)_-\) where
\[\mathrm{div}(f)_+ = \sum_{v_P(f) > 0} v_P(f)\,[P], \quad \mathrm{div}(f)_- = \sum_{v_P(f) < 0} (-v_P(f))\,[P]\]
are the zero divisor and pole divisor of \(f\), both effective.
The map \(\mathrm{div}: k(X)^\times \to \mathrm{Div}(X)\) is a group homomorphism (since \(v_P(fg) = v_P(f) + v_P(g)\)).
3.2 Principal Divisors and the Key Degree Theorem
Theorem (Degree of a principal divisor is zero). For any smooth projective curve \(X\) and any \(f \in k(X)^\times\),
\[\deg(\mathrm{div}(f)) = 0.\]
Proof sketch. The rational function \(f\) defines a regular morphism \(\varphi: X \to \mathbb{P}^1\) (extending the rational map, using smoothness). We must show \(|\varphi^{-1}(0)| = |\varphi^{-1}(\infty)|\) counted with multiplicity.
For a finite morphism \(\varphi: X \to \mathbb{P}^1\) of degree \(d\), every point \(Q \in \mathbb{P}^1\) has the property that
\[\sum_{P \in \varphi^{-1}(Q)} e_P = d,\]
where \(e_P = v_P(\varphi^*(t_Q))\) is the ramification index at \(P\) (here \(t_Q\) is a local parameter at \(Q\)). Applying this with \(Q = 0\) gives \(\deg(\mathrm{div}(f)_+) = d\), and with \(Q = \infty\) gives \(\deg(\mathrm{div}(f)_-) = d\). Hence
\[\deg(\mathrm{div}(f)) = \deg(\mathrm{div}(f)_+) - \deg(\mathrm{div}(f)_-) = d - d = 0. \quad \square\]
Corollary. The image of \(\mathrm{div}: k(X)^\times \to \mathrm{Div}(X)\) lies entirely within \(\mathrm{Div}^0(X)\).
The subgroup of principal divisors is
\[\mathrm{PDiv}(X) = \{\mathrm{div}(f) : f \in k(X)^\times\} \subset \mathrm{Div}^0(X).\]
The degree-zero statement fails for non-projective varieties. On \(\mathbb{A}^1\), the function \(f = x\) has \(\mathrm{div}(x) = [0]\), which has degree 1. Projectivity (compactness) is essential to ensure the pole at infinity is accounted for.
3.3 Worked Example on P^1
On \(\mathbb{P}^1\) with affine coordinate \(x\), the function field is \(k(x)\). Every nonzero \(f \in k(x)\) is a ratio of polynomials. Write
\[f = c \cdot \frac{\prod_i (x - a_i)^{m_i}}{\prod_j (x - b_j)^{n_j}}, \quad c \in k^\times.\]
Then: - \(v_{[a_i]}(f) = m_i\) (zero of order \(m_i\)), - \(v_{[b_j]}(f) = -n_j\) (pole of order \(n_j\)), - \(v_{[\infty]}(f) = \deg(\text{denominator}) - \deg(\text{numerator}) = \sum n_j - \sum m_i\).
So
\[\mathrm{div}(f) = \sum_i m_i [a_i] - \sum_j n_j [b_j] + \left(\sum_j n_j - \sum_i m_i\right)[\infty].\]
Two canonical examples:
- \(\mathrm{div}(x) = [0] - [\infty]\). Zero at \(0\), pole at \(\infty\), degree \(= 0\).
- \(\mathrm{div}\!\left(\dfrac{x-a}{x-b}\right) = [a] - [b]\) for \(a \neq b\). This shows \([a] - [b]\) is a principal divisor on \(\mathbb{P}^1\).
Any divisor \(D = \sum n_P [P]\) of degree zero on \(\mathbb{P}^1\) is \(\mathrm{div}(f)\) for \(f = \prod_P (x - P)^{n_P}\) (where we set \(x - \infty = 1\), and the \([\infty]\) contribution balances automatically by the degree condition). Hence \(\mathrm{Pic}^0(\mathbb{P}^1) = 0\).
This problem develops the mechanical skill of reading off zeros and poles from a rational function.
Prerequisites: 3.1 Order of Vanishing
Compute \(\mathrm{div}(f)\) for each of the following on \(\mathbb{P}^1\) with coordinate \(x\), and verify in each case that \(\deg(\mathrm{div}(f)) = 0\): (a) \(f = x^2 - 1\), (b) \(f = x^3/(x^2+1)\) (assume \(\mathrm{char}(k) \neq 2\)), (c) \(f = (x^2-1)/(x^2-4)\).
Key insight: Factor the numerator and denominator, read off zeros/poles in \(\mathbb{A}^1\), then compute the order at \(\infty\) from the degree difference.
Sketch: (a) \(f = (x-1)(x+1)\). Zeros: \([1]\) and \([-1]\), each multiplicity 1. Pole at \(\infty\) of order 2 (numerator degree 2 > denominator degree 0). So \(\mathrm{div}(f) = [1] + [-1] - 2[\infty]\). Degree: \(1+1-2=0\). ✓
\(f = x^3/(x^2+1)\). Zeros: \([0]\) with multiplicity 3. Poles: \([i]\) and \([-i]\) each multiplicity 1, and \([\infty]\) with order \(3-2=1\). So \(\mathrm{div}(f) = 3[0] - [i] - [-i] - [\infty]\). Degree: \(3-1-1-1=0\). ✓
\(f = (x-1)(x+1)/((x-2)(x+2))\). \(\mathrm{div}(f) = [1]+[-1]-[2]-[-2]\). Order at \(\infty\): equal degrees (\(2-2=0\)), so no zero or pole there. Degree: \(1+1-1-1=0\). ✓
This problem gives a direct proof of \(\deg(\mathrm{div}(f))=0\) in the simplest non-trivial case.
Prerequisites: 3.2 Principal Divisors and the Key Degree Theorem
Let \(f = (ax+b)/(cx+d) \in k(x)\) with \(ad - bc \neq 0\) (a Möbius transformation). (a) Explicitly compute \(\mathrm{div}(f)\) as a divisor on \(\mathbb{P}^1\). (b) Verify \(\deg(\mathrm{div}(f)) = 0\) directly, without invoking the general theorem. (c) The corresponding morphism \(\varphi: \mathbb{P}^1 \to \mathbb{P}^1\) has degree 1; reconcile this with the proof sketch in §3.2.
Key insight: A degree-1 morphism \(\mathbb{P}^1 \to \mathbb{P}^1\) is an isomorphism; every point has exactly one preimage with ramification index 1.
Sketch: (a) \(f = (ax+b)/(cx+d)\). The unique zero of the numerator is \(x_0 = -b/a\) (if \(a \neq 0\)), the unique zero of the denominator is \(x_1 = -d/c\) (if \(c \neq 0\)). Order at \(\infty\): degrees are equal, so \(v_\infty(f) = 1 - 1 = 0\). Hence \(\mathrm{div}(f) = [x_0] - [x_1]\), which has degree \(1 - 1 = 0\).
Directly: \(\deg(\mathrm{div}(f)) = 1 + (-1) = 0\).
Since \(\varphi\) has degree 1, every point \(Q \in \mathbb{P}^1\) has \(\sum_{P \in \varphi^{-1}(Q)} e_P = 1\), i.e., a unique preimage with ramification index 1. The proof gives \(\deg(\mathrm{div}(f)_+) = 1 = \deg(\mathrm{div}(f)_-)\).
4. The Picard Group 🔑
4.1 Linear Equivalence
Definition (Linear equivalence). Two divisors \(D, D' \in \mathrm{Div}(X)\) are linearly equivalent, written \(D \sim D'\), if \(D - D' = \mathrm{div}(f)\) for some \(f \in k(X)^\times\).
Linear equivalence is an equivalence relation on \(\mathrm{Div}(X)\) (reflexivity: \(\mathrm{div}(1) = 0\); symmetry: \(\mathrm{div}(f^{-1}) = -\mathrm{div}(f)\); transitivity: \(\mathrm{div}(fg) = \mathrm{div}(f) + \mathrm{div}(g)\)).
\(D \sim D'\) means: \(D\) and \(D'\) are “cut out by the same geometric data up to a change of rational function.” On \(\mathbb{P}^1\), any two points are linearly equivalent (since \([P] - [Q] = \mathrm{div}((x-P)/(x-Q))\)), so there is essentially one “degree-1 class.”
4.2 Definition and Basic Structure
Definition (Picard group). The Picard group of \(X\) is the quotient
\[\mathrm{Pic}(X) = \mathrm{Div}(X) / \mathrm{PDiv}(X).\]
The degree-zero Picard group is
\[\mathrm{Pic}^0(X) = \mathrm{Div}^0(X) / \mathrm{PDiv}(X).\]
Since \(\mathrm{PDiv}(X) \subset \mathrm{Div}^0(X)\) (by the degree theorem), the degree map descends to a well-defined surjective homomorphism \(\deg: \mathrm{Pic}(X) \to \mathbb{Z}\), giving the short exact sequence
\[0 \longrightarrow \mathrm{Pic}^0(X) \longrightarrow \mathrm{Pic}(X) \xrightarrow{\;\deg\;} \mathbb{Z} \longrightarrow 0.\]
This sequence is split: any choice of base point \(P_0 \in X\) gives a section \(\mathbb{Z} \to \mathrm{Pic}(X)\) by \(n \mapsto n[P_0]\). Hence
\[\mathrm{Pic}(X) \cong \mathbb{Z} \oplus \mathrm{Pic}^0(X).\]
The factor \(\mathbb{Z}\) records the degree; the factor \(\mathrm{Pic}^0(X)\) records the “finer” geometry of the curve.
The splitting \(\mathrm{Pic}(X) \cong \mathbb{Z} \oplus \mathrm{Pic}^0(X)\) depends on the choice of \(P_0\). Changing \(P_0\) to \(P_0'\) changes the section by \([P_0'] - [P_0] \in \mathrm{Pic}^0(X)\), giving a different (but isomorphic) splitting.
4.3 Computations: P^1, A^1, and Elliptic Curves
Proposition. \(\mathrm{Pic}(\mathbb{P}^1) \cong \mathbb{Z}\).
Proof. By the example in §3.3, every degree-zero divisor on \(\mathbb{P}^1\) is principal, so \(\mathrm{Pic}^0(\mathbb{P}^1) = 0\). The exact sequence gives \(\mathrm{Pic}(\mathbb{P}^1) \cong \mathbb{Z}\), generated by the class of any point \([P]\). \(\square\)
Proposition. \(\mathrm{Pic}(\mathbb{A}^1) = 0\).
Proof. Every divisor on \(\mathbb{A}^1\) has the form \(D = \sum_{i=1}^r n_i [a_i]\) for distinct \(a_i \in k\). Set \(f = \prod_{i=1}^r (x - a_i)^{n_i} \in k[x] \subset k(x)^\times\). This is a polynomial (regular on \(\mathbb{A}^1\)), and \(\mathrm{div}(f) = D\) since there is no “point at infinity” on \(\mathbb{A}^1\). Every divisor is principal, so \(\mathrm{Pic}(\mathbb{A}^1) = 0\). \(\square\)
For a smooth elliptic curve \(E\) with a chosen base point \(O \in E\), there is a group isomorphism \(\mathrm{Pic}^0(E) \cong E(k)\) given by \[P \longmapsto [P] - [O].\] The group law on \(E\) (the chord-tangent construction) corresponds exactly to addition in \(\mathrm{Pic}^0(E)\). This identification is the content of Abel’s theorem for elliptic curves and will be proved in the Riemann-Roch note using Riemann-Roch’s theorem and its corollaries.
This problem generalizes the computation \(\mathrm{Pic}(\mathbb{A}^1)=0\) to higher dimensions.
Prerequisites: 4.3 Computations: P^1, A^1, and Elliptic Curves
Show that \(\mathrm{Pic}(\mathbb{A}^n) = 0\) for all \(n \geq 1\). (Hint: \(k[\mathbb{A}^n] = k[x_1,\ldots,x_n]\) is a UFD; use the fact that the Picard group of a UFD domain is trivial.)
Key insight: The Picard group of a normal variety equals the class group, and for a UFD, every Weil divisor is principal.
Sketch: The coordinate ring of \(\mathbb{A}^n\) is \(k[x_1,\ldots,x_n]\), a UFD. Every irreducible polynomial \(p \in k[x_1,\ldots,x_n]\) defines a prime Weil divisor \(V(p)\). Since the ring is a UFD, every Weil divisor (codimension-1 irreducible subvariety) is the zero locus of a single irreducible polynomial, hence is \(\mathrm{div}(p)\) for that polynomial \(p\). Thus every Weil divisor is principal, so \(\mathrm{Cl}(\mathbb{A}^n) = 0\). For smooth varieties, \(\mathrm{Pic} = \mathrm{Cl}\), so \(\mathrm{Pic}(\mathbb{A}^n) = 0\).
This problem establishes \(\mathrm{Pic}(\mathbb{P}^n) \cong \mathbb{Z}\), generated by a hyperplane.
Prerequisites: 4.2 Definition and Basic Structure
- Show that every Weil divisor on \(\mathbb{P}^n\) is linearly equivalent to \(d \cdot [H]\) for some integer \(d\) and any hyperplane \(H \cong \mathbb{P}^{n-1}\). (Hint: Use homogeneous polynomials and the degree.) (b) Show that \([H] \not\sim 0\), i.e., no hyperplane is principal. Conclude that \(\mathrm{Pic}(\mathbb{P}^n) \cong \mathbb{Z}\).
Key insight: Rational functions on \(\mathbb{P}^n\) are ratios of homogeneous polynomials of the same degree; their divisors always have degree zero.
Sketch: (a) A prime divisor on \(\mathbb{P}^n\) is an irreducible hypersurface \(V(F)\) for a homogeneous polynomial \(F\) of some degree \(d\). For any hyperplane \(H = V(L)\) (degree 1), we have \(\mathrm{div}(F/L^d) = [V(F)] - d[H]\), so \([V(F)] \sim d[H]\). By linearity, every Weil divisor is equivalent to \(d[H]\) for some \(d\).
- If \([H] = \mathrm{div}(f)\) for some \(f \in k(\mathbb{P}^n)^\times\), then \(f = G/H'\) for homogeneous \(G, H'\) of the same degree. But \(\mathrm{div}(G/H') = [V(G)] - [V(H')]\), which has degree \(\deg G - \deg H' = 0\) as a divisor, while \([H]\) has degree 1. Contradiction. Hence the map \(\deg: \mathrm{Pic}(\mathbb{P}^n) \to \mathbb{Z}\) is an isomorphism.
This problem identifies specific torsion elements in the Picard group of an elliptic curve.
Prerequisites: 4.3 Computations: P^1, A^1, and Elliptic Curves
Let \(E: y^2 = (x-e_1)(x-e_2)(x-e_3)\) be an elliptic curve over \(k\) with \(\mathrm{char}(k) \neq 2\), and let \(O = [\infty]\) be the base point. Let \(P_i = (e_i, 0)\) be the three 2-torsion points. (a) Show that \(2[P_i] \sim 2[O]\), so \([P_i] - [O]\) has order dividing 2 in \(\mathrm{Pic}^0(E)\). (b) Show that \([P_i] \not\sim [O]\) (i.e., \(P_i \neq O\) in \(E(k)\)). (c) Conclude that \(\{[O], P_1, P_2, P_3\}\) contribute distinct elements of order exactly 2 in \(\mathrm{Pic}^0(E)\), so \((\mathbb{Z}/2\mathbb{Z})^2 \hookrightarrow \mathrm{Pic}^0(E)[2]\).
Key insight: The function \(x - e_i\) has a double zero at \(P_i\) and a double pole at \(O\) (since \(O\) is a flex point at infinity with local parameter \(1/x\)).
Sketch: (a) The function \(f_i = x - e_i \in k(E)^\times\) has \(\mathrm{div}(f_i) = 2[P_i] - 2[O]\) (double zero at \(P_i\), double pole at \(O\) counted by the Weierstrass model). So \(2[P_i] \sim 2[O]\), i.e., \(2([P_i]-[O]) = 0\) in \(\mathrm{Pic}^0(E)\).
If \([P_i] \sim [O]\), there would be a rational function \(g: E \to \mathbb{P}^1\) with a simple zero at \(P_i\) and a simple pole at \(O\) (i.e., \(\mathrm{div}(g) = [P_i]-[O]\)). Such \(g\) has degree 1, making \(E \cong \mathbb{P}^1\). But \(E\) has genus 1 \(\neq\) 0, contradiction.
The three differences \([P_i]-[O]\) are distinct nonzero elements of order 2. Together with 0, they form a subgroup isomorphic to \((\mathbb{Z}/2\mathbb{Z})^2\) inside \(\mathrm{Pic}^0(E)[2] \cong E[2]\).
5. Line Bundles and the Divisor-Line Bundle Correspondence 🔑
5.1 Rank-1 Locally Free Sheaves
A line bundle on \(X\) (equivalently, a rank-1 locally free sheaf or invertible sheaf) is a sheaf \(\mathcal{L}\) of \(\mathcal{O}_X\)-modules such that there exists an open cover \(\{U_i\}\) of \(X\) with \(\mathcal{L}|_{U_i} \cong \mathcal{O}_{U_i}\) for each \(i\).
The set of isomorphism classes of line bundles on \(X\) forms an abelian group under tensor product:
- Multiplication: \([\mathcal{L}] \cdot [\mathcal{L}'] = [\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{L}']\)
- Identity: \([\mathcal{O}_X]\) (the structure sheaf)
- Inverse: \([\mathcal{L}]^{-1} = [\mathcal{L}^{-1}] = [\mathcal{H}om(\mathcal{L}, \mathcal{O}_X)]\) (the dual sheaf)
This group is also denoted \(\mathrm{Pic}(X)\), and the fundamental theorem below justifies this notation.
5.2 The Sheaf O(D)
Definition (Sheaf of a divisor). For a divisor \(D \in \mathrm{Div}(X)\), define the sheaf \(\mathcal{O}(D)\) by
\[\mathcal{O}(D)(U) = \{f \in k(X)^\times : \mathrm{div}(f)|_U + D|_U \geq 0\} \cup \{0\}\]
for each open \(U \subset X\). Concretely, \(f\) belongs to \(\mathcal{O}(D)(U)\) if and only if, for every \(P \in U\), \(v_P(f) \geq -n_P\) (where \(D = \sum n_P [P]\)).
In words: \(\mathcal{O}(D)\) consists of rational functions that are “allowed to have poles only where \(D\) has positive multiplicity, and only up to the order prescribed by \(D\).”
Near any point \(P \in X\), choose an open \(U_P \ni P\) and a rational function \(g_P\) with \(\mathrm{div}(g_P)|_{U_P} = -D|_{U_P}\) (possible since \(D\) has integer coefficients and \(X\) is smooth). Then \(\mathcal{O}(D)|_{U_P} \cong \mathcal{O}_{U_P}\) via \(f \mapsto f/g_P\). This confirms \(\mathcal{O}(D)\) is an invertible sheaf.
5.3 Global Sections and the Identification Pic(X) = Line Bundles
The space of global sections of \(\mathcal{O}(D)\) is
\[H^0(X, \mathcal{O}(D)) = \{f \in k(X)^\times : \mathrm{div}(f) + D \geq 0\} \cup \{0\}.\]
This is a finite-dimensional \(k\)-vector space. We write \(h^0(D) = \dim_k H^0(X, \mathcal{O}(D))\).
Key structural facts:
- \(\mathcal{O}(D) \otimes \mathcal{O}(D') \cong \mathcal{O}(D + D')\).
- \(\mathcal{O}(D)^{-1} \cong \mathcal{O}(-D)\).
- \(\mathcal{O}(0) = \mathcal{O}_X\) (sections of \(\mathcal{O}(0)\) are regular functions; on a projective variety, these are constants).
Theorem (Divisor–line bundle correspondence). The map
\[\mathrm{Div}(X) \longrightarrow \{\text{line bundles on }X\}, \quad D \longmapsto \mathcal{O}(D),\]
is a surjective group homomorphism with kernel \(\mathrm{PDiv}(X)\). Hence it induces an isomorphism
\[\mathrm{Pic}(X) \xrightarrow{\;\sim\;} \{\text{isomorphism classes of line bundles on }X\}.\]
Proof sketch. Surjectivity: any line bundle \(\mathcal{L}\) embeds into the constant sheaf \(k(X)\) (since \(X\) is integral), giving a rational section \(s\); then \(\mathcal{L} \cong \mathcal{O}(D)\) where \(D = -\mathrm{div}(s)\). The kernel calculation: \(\mathcal{O}(D) \cong \mathcal{O}(D')\) iff they share a rational section differing by a unit in \(k(X)^\times\), iff \(D - D' = \mathrm{div}(f)\). \(\square\)
Example: \(\mathcal{O}(n)\) on \(\mathbb{P}^1\). Take \(D = n[\infty]\) on \(\mathbb{P}^1\). Then
\[H^0(\mathbb{P}^1, \mathcal{O}(n)) = \{f \in k(x) : v_\infty(f) \geq -n,\text{ no other poles}\}.\]
Since \(v_\infty(x^j) = -j\), the sections are exactly polynomials \(a_0 + a_1 x + \cdots + a_n x^n\) of degree at most \(n\). So \(h^0(\mathcal{O}(n)) = n + 1\).
\(H^0(\mathbb{P}^1, \mathcal{O}(n)) = 0\) for \(n < 0\) (no rational function can satisfy \(v_\infty(f) \geq -n > 0\) without being zero).
This problem builds fluency with the global sections formula.
Prerequisites: 5.3 Global Sections and the Identification Pic(X) = Line Bundles
On \(\mathbb{P}^1\) with coordinate \(x\), compute \(H^0(\mathbb{P}^1, \mathcal{O}(D))\) and its dimension for: (a) \(D = 2[0] + [1] - [\infty]\), (b) \(D = -[0] + 3[\infty]\), (c) \(D = [0] - [1]\) (degree 0).
Key insight: \(f \in H^0(\mathbb{P}^1, \mathcal{O}(D))\) iff \(v_P(f) \geq -n_P\) for all \(P\), i.e., \(f\) has at most a pole of order \(n_P\) at each \(P\) where \(n_P > 0\) and is required to vanish to order \(|n_P|\) where \(n_P < 0\).
Sketch: (a) \(D = 2[0]+[1]-[\infty]\): \(f\) must have \(v_0(f)\geq -2\), \(v_1(f)\geq -1\), \(v_\infty(f)\geq 1\), and be regular elsewhere. In terms of \(x\): \(f = p(x)/(x^2(x-1))\) where \(p\) is a polynomial, and \(v_\infty(f)\geq 1\) forces \(\deg p \leq 1\). So \(f = (ax+b)/(x^2(x-1))\) — but we also need \(v_\infty(f)\geq 1\), meaning \(\deg(\text{denom}) - \deg(\text{num}) \geq 1\), i.e., \(3 - \deg p \geq 1\), so \(\deg p \leq 2\). But also checking: \(f=(ax^2+bx+c)/(x^2(x-1))\) has \(v_\infty = 3-2=1\geq 1\). So \(H^0 = \{(ax^2+bx+c)/(x^2(x-1))\}\), dimension 3.
\(D = -[0]+3[\infty]\): \(v_0(f)\geq 1\) (zero at 0), \(v_\infty(f)\geq -3\), regular elsewhere. So \(f = x \cdot q(x)\) for \(q\) a polynomial with \(\deg(xq)\leq 3\), giving \(f \in \{ax, bx^2, cx^3\}\), dimension 3.
\(D=[0]-[1]\): \(v_0(f)\geq -1\), \(v_1(f)\geq 1\), regular elsewhere, \(v_\infty(f)\geq 0\). So \(f\) has a zero at 1, at most a simple pole at 0, and no other poles. \(f = a(x-1)/x\) for \(a\in k\). Dimension 1.
This problem verifies the tensor product formula \(\mathcal{O}(D)\otimes\mathcal{O}(D')\cong\mathcal{O}(D+D')\) directly.
Prerequisites: 5.2 The Sheaf O(D)
- Show directly from the definition that if \(f \in \mathcal{O}(D)(U)\) and \(g \in \mathcal{O}(D')(U)\), then \(fg \in \mathcal{O}(D+D')(U)\). (b) Conclude that there is a natural map \(\mathcal{O}(D) \otimes \mathcal{O}(D') \to \mathcal{O}(D+D')\), and verify it is an isomorphism.
Key insight: Valuations are additive: \(v_P(fg) = v_P(f) + v_P(g)\).
Sketch: (a) If \(f\in\mathcal{O}(D)(U)\) then \(v_P(f)\geq -n_P\) for all \(P\in U\); if \(g\in\mathcal{O}(D')(U)\) then \(v_P(g)\geq -n_P'\). Since \(v_P(fg) = v_P(f)+v_P(g) \geq -n_P - n_P' = -(n_P+n_P')\), we have \(fg\in\mathcal{O}(D+D')(U)\).
- The natural map \(\mathcal{O}(D)(U)\times\mathcal{O}(D')(U)\to\mathcal{O}(D+D')(U)\), \((f,g)\mapsto fg\), is bilinear and factors through the tensor product. To see it’s an isomorphism, work locally: on a sufficiently small open \(U\) containing \(P\), choose a local uniformizer \(t\) at \(P\), and write \(f = t^a u\), \(g = t^b v\) with \(u,v\) units; then \(fg = t^{a+b}(uv)\), and the map is \(t^a\mathcal{O}_U \otimes t^b\mathcal{O}_U \xrightarrow{\sim} t^{a+b}\mathcal{O}_U\) via multiplication, an isomorphism.
This problem uses the Riemann-Roch theorem (stated without proof) to compute h^0 for explicit divisors on curves of low genus.
Prerequisites: 5.3 Global Sections and the Identification Pic(X) = Line Bundles
Riemann-Roch states: for a smooth projective curve of genus \(g\) and a divisor \(D\), \(h^0(D) - h^0(K-D) = \deg(D) + 1 - g\), where \(K\) is the canonical divisor. Using this (without proof), compute \(h^0(D)\) for: (a) \(D = n[P]\) on \(\mathbb{P}^1\) (genus 0), (b) \(D = n[P]\) on an elliptic curve \(E\) (genus 1) for \(n = 0, 1, 2, 3\).
Key insight: On \(\mathbb{P}^1\), the canonical divisor has degree \(-2\) (genus 0), so \(h^0(K-D)=0\) for \(\deg D>-2\). On an elliptic curve, \(K\sim 0\), so Riemann-Roch becomes \(h^0(D)-h^0(-D) = \deg D\).
Sketch: (a) \(g=0\), \(\deg K = -2\). For \(D=n[P]\) with \(n\geq 0\): \(\deg(K-D) = -2-n < 0\), so \(h^0(K-D)=0\). Riemann-Roch: \(h^0(n[P]) = n+1\). Matches our direct computation.
- \(g=1\), \(K\sim 0\). Riemann-Roch: \(h^0(D)-h^0(-D)=\deg D\). For \(n=0\): \(D=0\), \(h^0(0)=1\) (constants). For \(n=1\): \(h^0([P])-h^0(-[P])=1\); \(h^0(-[P])=0\) (degree \(-1\)), so \(h^0([P])=1\). For \(n=2\): \(h^0(2[P])=2\). For \(n=3\): \(h^0(3[P])=3\).
This problem establishes when the space of global sections is trivial.
Prerequisites: 5.3 Global Sections and the Identification Pic(X) = Line Bundles
Let \(X\) be a smooth projective curve. (a) Show that if \(\deg(D) < 0\) then \(H^0(X, \mathcal{O}(D)) = 0\). (b) Show that if \(D \sim D'\) then \(h^0(D) = h^0(D')\). (c) Show that \(h^0(0) = 1\) (the only global sections of \(\mathcal{O}_X\) are constants).
Key insight: A nonzero section \(f \in H^0(X,\mathcal{O}(D))\) satisfies \(\mathrm{div}(f)+D \geq 0\), so \(\deg(\mathrm{div}(f)+D) = \deg(D) \geq 0\) (since \(\deg(\mathrm{div}(f))=0\)).
Sketch: (a) If \(f \neq 0\) were in \(H^0(X,\mathcal{O}(D))\), then \(\deg(\mathrm{div}(f)+D) \geq 0\) (as an effective divisor). But \(\deg(\mathrm{div}(f)+D) = \deg D < 0\), contradiction.
If \(D \sim D'\), then \(D' = D + \mathrm{div}(g)\) for some \(g\in k(X)^\times\). The map \(f \mapsto f/g\) is an isomorphism \(H^0(X,\mathcal{O}(D)) \xrightarrow{\sim} H^0(X,\mathcal{O}(D'))\): indeed, \(\mathrm{div}(f)+D\geq 0\) iff \(\mathrm{div}(f/g)+D'\geq 0\).
$H^0(X,_X) = {fk(X)^: (f)}{0} = $ regular functions on \(X\). By projectivity of \(X\), any regular function is constant. So \(H^0(X,\mathcal{O}_X) = k\), dimension 1.
This problem derives the duality \(\mathcal{O}(D)^\vee \cong \mathcal{O}(-D)\) from first principles.
Prerequisites: 5.2 The Sheaf O(D)
For a divisor \(D\) on a smooth projective curve \(X\): (a) Show that \(\mathcal{H}om(\mathcal{O}(D), \mathcal{O}_X) \cong \mathcal{O}(-D)\) as \(\mathcal{O}_X\)-modules. (b) Deduce that \(\mathrm{Pic}(X)\) is indeed a group under \(\otimes\) (i.e., every element has an inverse).
Key insight: Multiplication by \(f \in \mathcal{O}(D)(U)\) maps into \(\mathcal{O}_X(U)\) precisely when \(f \in \mathcal{O}(-D)(U)\).
Sketch: (a) A morphism of \(\mathcal{O}_X\)-modules \(\phi: \mathcal{O}(D) \to \mathcal{O}_X\) over an open \(U\) is determined by where the “canonical section” \(1 \in k(X)^\times\) (locally trivializing \(\mathcal{O}(D)\)) maps to. Say \(\phi_U(1) = h \in \mathcal{O}_X(U)\). For \(\phi\) to be a well-defined map \(\mathcal{O}(D)(U) \to \mathcal{O}_X(U)\), we need \(f \cdot h \in \mathcal{O}_X(U)\) for all \(f\in\mathcal{O}(D)(U)\). This forces \(v_P(h) \geq -v_P(f) \geq n_P = v_P(D)\) for all \(P\), i.e., \(h\in\mathcal{O}(-D)(U)\). So \(\mathcal{H}om(\mathcal{O}(D),\mathcal{O}_X)\cong\mathcal{O}(-D)\).
- From (a): \(\mathcal{O}(D) \otimes \mathcal{O}(-D) \cong \mathcal{O}(D) \otimes \mathcal{O}(D)^\vee \cong \mathcal{O}_X\) (using the natural evaluation pairing for rank-1 locally free sheaves). So \([\mathcal{O}(-D)]\) is the inverse of \([\mathcal{O}(D)]\) in \(\mathrm{Pic}(X)\).
6. Linear Systems and Maps to Projective Space 📐
6.1 Complete Linear Systems
Definition (Linear system). A linear system on \(X\) is a set of the form \(\mathfrak{d} = \{D' \in \mathrm{Div}(X) : D' \geq 0,\, D' \sim D\}\) for some divisor \(D\), i.e., a set of effective divisors linearly equivalent to \(D\).
The complete linear system associated to \(D\) is
\[|D| = \{D' \in \mathrm{Div}(X) : D' \geq 0,\, D' \sim D\}.\]
There is a canonical bijection
\[|D| \;\longleftrightarrow\; \mathbb{P}(H^0(X, \mathcal{O}(D))),\]
given by \(\mathrm{div}(f) + D \leftrightarrow [f]\) for \(f \in H^0(X, \mathcal{O}(D)) \setminus \{0\}\). The dimension of \(|D|\) is
\[\dim|D| = h^0(D) - 1.\]
The bijection \(|D| \leftrightarrow \mathbb{P}(H^0(X,\mathcal{O}(D)))\) comes from the fact that two nonzero sections \(f, f' \in H^0(X,\mathcal{O}(D))\) give the same effective divisor (i.e., \(\mathrm{div}(f)+D = \mathrm{div}(f')+D\)) iff \(\mathrm{div}(f/f')=0\) iff \(f/f' \in k^\times\) (the only global regular functions on a projective variety are constants). So divisors correspond to lines in \(H^0(X,\mathcal{O}(D))\).
6.2 Base Locus and the Map phi_D
Definition (Base locus). The base locus of \(|D|\) is
\[\mathrm{Bs}(|D|) = \bigcap_{D' \in |D|} \mathrm{supp}(D').\]
Equivalently, \(P \in \mathrm{Bs}(|D|)\) iff every nonzero section \(s \in H^0(X, \mathcal{O}(D))\) vanishes at \(P\).
Definition (Map to projective space). Choose a basis \(s_0, \ldots, s_r \in H^0(X, \mathcal{O}(D))\). Define a (possibly rational) map
\[\phi_D : X \dashrightarrow \mathbb{P}^r, \quad P \longmapsto [s_0(P) : s_1(P) : \cdots : s_r(P)].\]
More precisely, after choosing a local trivialization of \(\mathcal{O}(D)\) near \(P\), the \(s_i(P)\) are elements of \(k\); the ratio \([s_0(P):\cdots:s_r(P)]\) is well-defined as an element of \(\mathbb{P}^r\) as long as not all \(s_i\) vanish at \(P\).
Theorem. \(\phi_D\) extends to a regular morphism \(X \to \mathbb{P}^r\) if and only if \(\mathrm{Bs}(|D|) = \emptyset\).
If \(P \in \mathrm{Bs}(|D|)\), then \(s_i(P) = 0\) for all \(i\), and \(\phi_D\) is undefined at \(P\). Removing base points (e.g., by passing to a subsystem or a different \(D'\)) is often necessary before constructing a morphism.
6.3 Amplitude and Very Ampleness
Definition (Very ample). A divisor \(D\) is very ample if:
- \(\mathrm{Bs}(|D|) = \emptyset\) (so \(\phi_D\) is a morphism), and
- \(\phi_D : X \hookrightarrow \mathbb{P}^r\) is a closed immersion (i.e., an embedding).
Condition 2 means \(\phi_D\) is injective on points and on tangent spaces: \(\phi_D(P) \neq \phi_D(Q)\) for \(P \neq Q\) (separates points), and \(d\phi_D|_P: T_{X,P} \hookrightarrow T_{\mathbb{P}^r, \phi_D(P)}\) is injective (separates tangent vectors).
Definition (Ample). A divisor \(D\) is ample if \(nD\) is very ample for some \(n \gg 0\).
On a projective variety, \(D\) is ample if and only if \(H^i(X, \mathcal{F} \otimes \mathcal{O}(nD)) = 0\) for all \(i > 0\), all coherent sheaves \(\mathcal{F}\), and all sufficiently large \(n\) (depending on \(\mathcal{F}\)). This is the cohomological characterization of ampleness.
6.4 Examples: Veronese and Conics
Example 1: \(\mathcal{O}(1)\) on \(\mathbb{P}^1\). Take \(D = [\infty]\), so \(\mathcal{O}(D) = \mathcal{O}(1)\). A basis for \(H^0(\mathbb{P}^1, \mathcal{O}(1))\) is \(\{1, x\}\) (constants and linear polynomials, dimension 2). The map \(\phi_D: \mathbb{P}^1 \to \mathbb{P}^1\) sends \([s:t] \mapsto [s:t]\), which is the identity. This is a closed immersion, so \(\mathcal{O}(1)\) is very ample on \(\mathbb{P}^1\).
Example 2: The degree-\(n\) Veronese embedding. Take \(D = n[\infty]\) on \(\mathbb{P}^1\), so \(\mathcal{O}(D) = \mathcal{O}(n)\). A basis for \(H^0(\mathbb{P}^1, \mathcal{O}(n))\) is \(\{1, x, x^2, \ldots, x^n\}\) (dimension \(n+1\)). In homogeneous coordinates \([s:t]\):
\[\nu_n : \mathbb{P}^1 \hookrightarrow \mathbb{P}^n, \quad [s:t] \longmapsto [s^n : s^{n-1}t : s^{n-2}t^2 : \cdots : t^n].\]
This is the degree-\(n\) Veronese embedding. It is a closed immersion for all \(n \geq 1\), so \(\mathcal{O}(n)\) is very ample on \(\mathbb{P}^1\).
Why is \(\nu_n\) a closed immersion? Injectivity on points: \([s:t] \neq [s':t']\) implies some ratio \(s/t \neq s'/t'\), hence \((s/t)^n \neq (s'/t')^n\) or \((s/t)^{n-1} \neq (s'/t')^{n-1}\), so the images in \(\mathbb{P}^n\) differ. Injectivity on tangent vectors: the differential \(d\nu_n\) at \([1:0]\) maps \(\partial/\partial(t/s)|_{[1:0]}\) to \((0, 1, 0, \ldots, 0)\), which is nonzero.
Example 3: A conic as a Veronese image. Let \(C \subset \mathbb{P}^2\) be a smooth conic (say \(xz = y^2\)). By the classification of conics, \(C \cong \mathbb{P}^1\). Under this isomorphism, the hyperplane class on \(C\) (i.e., \(\mathcal{O}_C(1) = \mathcal{O}_{\mathbb{P}^2}(1)|_C\)) corresponds to \(\mathcal{O}_{\mathbb{P}^1}(2)\). The complete linear system \(|2[P]|\) on \(\mathbb{P}^1\) has \(h^0 = 3\) and gives exactly the Veronese \(\nu_2: \mathbb{P}^1 \hookrightarrow \mathbb{P}^2\), whose image is \(C\). So \(C\) is “the degree-2 Veronese of \(\mathbb{P}^1\).”
This problem computes the base locus for an explicit subsystem.
Prerequisites: 6.2 Base Locus and the Map phi_D
On \(\mathbb{P}^1\), consider the linear system \(\mathfrak{d} = \{D' \in |2[\infty]| : D' \geq [0]\}\) (the subsystem of \(|2[\infty]|\) consisting of effective divisors that contain \([0]\) with multiplicity at least 1). (a) Identify \(\mathfrak{d}\) with a projective subspace of \(\mathbb{P}(H^0(\mathbb{P}^1,\mathcal{O}(2)))\). (b) Compute \(\mathrm{Bs}(\mathfrak{d})\). (c) What morphism \(\phi: \mathbb{P}^1 \to \mathbb{P}^1\) does \(\mathfrak{d}\) define?
Key insight: The condition \(D' \geq [0]\) forces every section to vanish at 0, cutting out a hyperplane in the space of sections.
Sketch: (a) \(H^0(\mathbb{P}^1,\mathcal{O}(2)) = \mathrm{span}\{1, x, x^2\}\). The condition \(D' \geq [0]\) means the section \(f\) with \(\mathrm{div}(f)+2[\infty] = D' \geq [0]\) must have \(v_0(f) \geq 1\), so \(f\) vanishes at \(x=0\). The sections in question are \(f \in \{ax + bx^2 : a, b \in k\} = x \cdot k[x]_{\leq 1}\). So \(\mathfrak{d} \leftrightarrow \mathbb{P}(\mathrm{span}\{x, x^2\}) \cong \mathbb{P}^1\).
The sections in \(\mathfrak{d}\) are \(ax + bx^2 = x(a+bx)\), all vanishing at \(x=0\) (i.e., \([0]\)) and nowhere else simultaneously (the other zero is \(-a/b\), varying). So \(\mathrm{Bs}(\mathfrak{d}) = \{[0]\}\).
With basis \(\{x, x^2\}\), the map is \([s:t] \mapsto [s \cdot t: s^2 \cdot t^2]\)… more carefully in affine coordinates, \(x \mapsto [x : x^2] = [1 : x]\) (dividing by \(x\)), giving the identity map \(\mathbb{P}^1 \to \mathbb{P}^1\), \([s:t]\mapsto [s:t]\), well-defined away from \([0]\). Since \(\mathrm{Bs}\neq\emptyset\), this is only a rational map.
This problem verifies that a degree-3 divisor on an elliptic curve embeds it in P^2.
Prerequisites: 6.3 Amplitude and Very Ampleness, 5.3 Global Sections and the Identification Pic(X) = Line Bundles
Let \(E\) be a smooth elliptic curve (genus 1) with base point \(O\). (a) Using Riemann-Roch (from Exercise 10), show \(h^0(3[O]) = 3\). (b) Show \(\mathrm{Bs}(|3[O]|) = \emptyset\). (c) The resulting map \(\phi_{3[O]}: E \to \mathbb{P}^2\) is a closed immersion; what is the degree of its image?
Key insight: A degree-\(d\) divisor on a curve of genus \(g\) is very ample if \(d \geq 2g+1\) (by the Riemann-Roch theorem). For \(g=1\), this means \(d \geq 3\).
Sketch: (a) From Exercise 10(b): \(h^0(3[O]) = 3\) by Riemann-Roch.
Base-point-freeness: for any \(P \in E\), we want a section \(s \in H^0(E,\mathcal{O}(3[O]))\) that doesn’t vanish at \(P\). The divisor \(3[O]-[P]\) has degree 2, and \(h^0(3[O]-[P]) = h^0(2[O]+(O-P)) = 2\) (by Riemann-Roch, \(h^0 = \deg+1-g+h^0(K-D) = 2+1-1+0=2\)). So the evaluation map \(H^0(\mathcal{O}(3[O])) \to k_P\) is not identically zero, meaning some \(s\) doesn’t vanish at \(P\).
The image \(\phi_{3[O]}(E) \subset \mathbb{P}^2\) is a smooth plane curve of degree 3 (since \(\mathcal{O}(3[O])\) corresponds to the pullback of \(\mathcal{O}_{\mathbb{P}^2}(1)\) restricted to the image, and the degree of the image equals \(\deg(3[O]) = 3\)). A smooth plane cubic has genus 1, consistent with \(E\) having genus 1.
This problem establishes the formula \(\dim|D| = \deg(D)\) for \(\deg(D) \geq 0\) on \(\mathbb{P}^1\).
Prerequisites: 6.1 Complete Linear Systems, 5.3 Global Sections and the Identification Pic(X) = Line Bundles
Let \(X = \mathbb{P}^1\). (a) Show that for any divisor \(D\) on \(\mathbb{P}^1\) with \(\deg(D) = d \geq 0\), we have \(h^0(D) = d+1\). (b) Conclude that \(\dim|D| = d\) for \(d \geq 0\). (c) Give a geometric interpretation: what does a general element of \(|D|\) look like?
Key insight: On \(\mathbb{P}^1\), every divisor of degree \(d\geq 0\) is linearly equivalent to \(d[\infty]\), and \(H^0(\mathbb{P}^1,\mathcal{O}(d[\infty]))\) is spanned by \(\{1,x,\ldots,x^d\}\).
Sketch: (a) By the example in §3.3, every degree-0 divisor on \(\mathbb{P}^1\) is principal. So \(D \sim d[\infty]\) for any \(D\) with \(\deg D = d\). By Exercise 11(b), \(h^0(D) = h^0(d[\infty])\). A section of \(\mathcal{O}(d[\infty])\) is a rational function \(f\) with \(v_\infty(f) \geq -d\) and no other poles — i.e., a polynomial of degree \(\leq d\). This vector space has dimension \(d+1\).
\(\dim|D| = h^0(D)-1 = d\).
A general element of \(|D|\) is \(\mathrm{div}(f) + d[\infty]\) for a degree-\(d\) polynomial \(f = c\prod_{i=1}^d (x-a_i)\) with distinct roots \(a_i\). Geometrically, it is a set of \(d\) distinct points in \(\mathbb{A}^1 \subset \mathbb{P}^1\) (plus possibly a contribution at \(\infty\) if \(d[\infty]\) has extra multiplicity).
This problem analyzes a 1-dimensional linear system (a pencil) and the associated map.
Prerequisites: 6.2 Base Locus and the Map phi_D
A pencil is a 1-dimensional linear system, i.e., \(\dim|D| = 1\) so \(h^0(D) = 2\). (a) On \(\mathbb{P}^1\), identify all pencils: what are the possible divisors \(D\) that give a pencil? (b) For \(X\) a smooth projective curve of genus \(g \geq 1\) and a pencil \(|D|\) with \(\deg D = 1\), show the map \(\phi_D: X \to \mathbb{P}^1\) is an isomorphism (so \(X \cong \mathbb{P}^1\), hence \(g = 0\), contradiction). Conclude that no curve of genus \(\geq 1\) has a degree-1 pencil. (c) What is the minimum degree of a divisor on an elliptic curve \(E\) (genus 1) that gives a morphism \(E \to \mathbb{P}^1\)?
Key insight: A degree-1 pencil defines a degree-1 map to \(\mathbb{P}^1\), hence an isomorphism, which forces genus 0.
Sketch: (a) By Exercise 15, \(\dim|D| = \deg D\) on \(\mathbb{P}^1\), so \(\dim|D|=1\) iff \(\deg D=1\). Any \(D\) of degree 1 gives a pencil; all are equivalent to \([\infty]\).
If \(h^0(D)=2\) and \(\deg D=1\), the map \(\phi_D: X\to\mathbb{P}^1\) has degree \(\deg D=1\) (the degree of a map equals \(\deg D\) for a base-point-free system). A degree-1 morphism of curves is an isomorphism. But \(\mathbb{P}^1\) has genus 0, while \(X\) has genus \(g\geq 1\); isomorphic curves have the same genus. Contradiction.
For \(E\) of genus 1, Riemann-Roch gives \(h^0(D)=\deg D\) for \(\deg D\geq 1\). A pencil requires \(h^0(D)=2\), so \(\deg D=2\). The map \(\phi_D: E\to\mathbb{P}^1\) has degree 2 (a degree-2 map or “hyperelliptic involution”). So the minimum degree is 2.
References
| Reference Name | Brief Summary | Link to Reference |
|---|---|---|
| Shafarevich, Basic Algebraic Geometry Vol. 1, §III.1 | Primary source; develops divisors on curves, linear systems, and Picard groups in the classical algebraic geometry setting | SpringerLink |
| Hartshorne, Algebraic Geometry, §II.6–7 | Definitive modern treatment of Weil divisors, Cartier divisors, and invertible sheaves; proves the divisor–line bundle correspondence in full generality | Springer |
| Fulton, Algebraic Curves | Accessible introduction to divisors on curves with emphasis on the Riemann-Roch theorem; free PDF available | Fulton’s website |
| Silverman, Arithmetic of Elliptic Curves, Ch. II §1–3 | Divisors on curves in the context of elliptic curves; explains the Picard group and its identification with the curve’s group law | Springer |
| Milne, Algebraic Geometry lecture notes, Ch. 12 | Freely available lecture notes covering divisors, intersection theory, and Picard groups; very complete and up-to-date | jmilne.org |
| Reid, Undergraduate Algebraic Geometry, Ch. 9 | Elementary treatment of linear systems and maps to projective space, with worked examples | Cambridge |
| Wikipedia, Divisor (algebraic geometry) | Concise summary of Weil vs. Cartier divisors, Picard group, and the class group | Wikipedia |
| Vakil, Foundations of Algebraic Geometry, Ch. 27–32 | Stanford lecture notes with a thorough treatment of divisors, line bundles, and linear systems in scheme-theoretic language | math.stanford.edu |