Question

Let \(D(X)\) be the group of divisors on \(X, D_{0}(X)\) the subgroup consisting of divisors of degree zero, and \(P(X)\) the subgroup of \(D_{0}(X)\) consisting of divisors of rational functions. Let \(C_{0}(X)=D_{0}(X) / P(X)\) be the quotient group. It is the divisor class group on \(X\). (a) If \(X=\mathbb{P}^{1}\), then \(C_{0}(X)=0\). (b) Let \(X=C\) be a nonsingular cubic. Pick \(P_{0} \in C\), defining \(\oplus\) on \(C\). Show that the map from \(C\) to \(C_{0}(X)\) that sends \(P\) to the residue class of the divisor \(P-P_{0}\) is an isomorphism from \((C, \oplus)\) onto \(C_{0}(X)\).

Short answer

The statement "if \(X = \mathbb{P}^1\), then \(C_0(X) = 0\)" means that the quotient group \(C_0(X)\) consisting of degree zero divisors modulo divisors of rational functions is trivial when \(X\) is the projective line. A nonsingular cubic curve \(C\) with the point \(P_0\) and the addition operation \(\oplus\) has a map to \(C_0(X)\) that sends point \(P\) to the residue class of the divisor \(P-P_0\), which is an isomorphism, meaning that the structure of the cubic curve is preserved by this map.

Step-by-step solution

Calculate \(D_0(X)\) for \(X=\mathbb{P}^1\).

For \(X=\mathbb{P}^1\), divisors are given by the degree of the polynomial. Since \(D_0(X)\) consists of divisors of degree zero, the only divisor in this category is the trivial one, a constant polynomial.

Calculate \(P(X)\) for \(X=\mathbb{P}^1\).

Recall that \(P(X)\) is the subgroup of \(D_0(X)\) consisting of divisors of rational functions. The rational functions on \(\mathbb{P}^1\) have the form \(\frac{f(x)}{g(x)}\) where \(f(x)\) and \(g(x)\) are polynomials. Since \(D_0(X)\) only has one element, the constant polynomial, the only possible rational functions are constants. Hence, \(P(X)\) is equal to \(D_0(X)\).

Calculate \(C_0(X)\) for \(X=\mathbb{P}^1\).

We have found that \(D_0(X)=P(X)\) for \(X=\mathbb{P}^1\). As a result, the quotient group \(C_0(X)=\frac{D_0(X)}{P(X)}\) is trivial and \(C_0(X)=0\). Thus, when \(X = \mathbb{P}^1\), we have shown that \(C_0(X) = 0\). Part (b):

Define the map from \(C\) to \(C_0(X)\).

Consider a map \(\phi:C\rightarrow C_0(X)\) defined by \(\phi(P) = (P-P_0)\mod P(X)\). Our goal is to show that this map is an isomorphism between \((C, \oplus)\) and \(C_0(X)\).

Show that the map is surjective.

Let \(D\) be any divisor class in \(C_0(X)\). Divisor classes are equivalence classes of divisors where two divisors are deemed equivalent if they differ by a divisor of a rational function. For any divisor \(D'\) in the class \(D\), we can express it in the form \(D' = (P-P_0)+P_1\) for some point \(P\in C\), where \(P_1\) is the divisor of a rational function. Then, \(\phi(P) = (P-P_0)\mod P(X) = D'\mod P(X) = D\). As we have shown that any element in \(C_0(X)\) has a preimage in \(C\), our map is surjective.

Show that the map is injective.

Let \(P\) and \(Q\) be two distinct points in \(C\) and suppose that \(\phi(P) = \phi(Q)\). This means that \((P-P_0)-(Q-P_0)\) is the divisor of a rational function. However, since this divisor has degree zero and the points are distinct, this difference cannot correspond to the divisor of a rational function as it involves canceling two distinct poles. This contradicts our assumption and establishes that \(\phi(P) \neq \phi(Q)\). Thus, the map is injective.

Show that the map respects the group operation.

Let \(P, Q, R \in C\) be points such that \(P \oplus Q = R\). We need to show that \(\phi(P\oplus Q) = \phi(P)\oplus_{C_0(X)}\phi(Q)\). By definition, \(\phi(P\oplus Q) = \phi(R) = (R-P_0)\mod P(X)\) and \(\phi(P)\oplus_{C_0(X)}\phi(Q) = (P-P_0)\mod P(X) \oplus_{C_0(X)}(Q-P_0)\mod P(X)\). Since the divisors of rational functions form an additive subgroup, the sum of two such divisors is also a divisor of a rational function. Therefore, \((P-P_0)\mod P(X) \oplus_{C_0(X)}(Q-P_0)\mod P(X) = (P-P_0 + Q-P_0)\mod P(X) = (R-P_0)\mod P(X)\). Hence, the map respects the group operation. Since we have established that our map is surjective, injective, and respects the group operation, it is an isomorphism from \((C, \oplus)\) onto \(C_0(X)\).

Key concepts

Algebraic Geometry

Algebraic geometry is a branch of mathematics that studies solutions to algebraic equations and their geometric properties. Imagine equations like \( x^2 + y^2 = 1 \) representing shapes in space. Algebraic geometry extends this idea to more complex equations.

It connects algebra, where we deal with polynomial equations, with geometry, which involves shapes.

• It explores solutions over different fields, like real numbers, complex numbers, and finite fields.
• It involves studying concepts like points of intersection and tangency.
• Provides tools to understand the structure and properties of algebraic varieties, the geometric objects defined by polynomial equations.

Algebraic geometry is foundational in understanding concepts like divisor class groups used in our exercise.

Complex Projective Line

The complex projective line, often denoted as \(\mathbb{P}^1\), is a fundamental concept in algebraic geometry. It can be visualized as a line extended far enough to include a point at infinity.

This is different from the usual complex plane because:

• It adds extra structure by including this 'point at infinity', making it a closed curve.
• Forms a surface akin to a sphere, often called the Riemann sphere.
• Every polynomial equation of degree \( n \) on \( \mathbb{P}^1 \) will have \( n \) solutions when counted with multiplicity.

This concept is widely used in algebraic geometry to analyze rational functions and divisors, as seen in the exercise.

Isomorphisms

In algebraic geometry, an isomorphism is a mapping between two structures that shows they are essentially the same. It's like matching puzzles of different shapes that fit perfectly together.

Isomorphisms preserve operations and properties, meaning:

• They map elements from one set to another without losing any structural information.
• They are both injective (one-to-one) and surjective (onto), ensuring no elements are missed or repeated.
• In our exercise, showing an isomorphism proves that two different looks at the divisors still represent the same underlying structure.

This demonstrates the concept of equivalence in the geometric structures we encounter.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They have the general form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials.

Key features include:

• The behavior of the function at the roots of \( q(x) \) (where it equals zero) can lead to vertical asymptotes or undefined points.
• They are defined everywhere except at the zeroes of the denominator.
• Rational functions can express complex ideas in algebraic geometry, such as those used to describe divisors and their equivalence classes.

In our exercise, understanding these functions is crucial to exploring how divisors act within the projective space and contributes to the divisor class group concepts.