Question

Let \(C\) be an irreducible projective plane curve, \(P_{1}, \ldots, P_{n}\) simple points on \(C\), \(m_{1}, \ldots, m_{n}\) integers. Show that there is a \(z \in k(C)\) with ord \(_{P_{l}}(z)=m_{i}\) for \(i=1, \ldots, n\). (Hint: Take lines \(L_{i}\) as in Problem \(5.14\) for \(P_{i}\), and a line \(L_{0}\) not through any \(P_{j}\), and let \(\left.z=\Pi L_{i}^{m_{i}} L_{0}-{\Sigma m_{i}} .\right)\)

Short answer

#Question# Show that for a given irreducible projective plane curve C and simple points P1, P2, ..., Pn on C, as well as given integers m1, m2, ..., mn, there exists a z ∈ k(C) with ord_Pi(z) = mi for i=1, 2, ..., n. #Answer# We first define lines L_i for each point Pi, and a line L_0 that does not pass through any of the points P1, P2,..., Pn. We then define the function z = Π L_i^m_i L_0^{-Σ m_i}. Since z is a rational function on the curve C, z ∈ k(C). By analyzing the function z over each of the simple points Pi, we observe that ord_Pi(z) = mi for i=1, 2, ..., n. Thus, there exists a z ∈ k(C) with ord_Pi(z) = mi for the given integers and points on the curve C.

Step-by-step solution

L[i] and L0 lines definition

For every Pi (where i ranges from 1 to n), let L_i be the line such that P_i belongs to L_i, as in Problem 5.14. Now, choose a line L_0 that does not pass through any of the points P1, P2,..., Pn.

Define the function z

Define z as follows: z = Π L_i^m_i L_0^{-Σ m_i}. Here Π represents the product of the lines L_i raised to the power of m_i, and L_0 raised to the power of the negative sum of m_i (i.e., L_0^{-Σ m_i}). It is important to note that z ∈ k(C) as z is a rational function on the curve C.

Show that ord_Pi(z) = mi

To show that ord_Pi(z) = mi, we will analyze the function z over each of the simple points Pi. Recall that ord_Pi(z) denotes the order of a function z at point Pi (i.e., the multiplicity of the point Pi in the divisor of z). For each i, we will have ord_Pi(L_i) = 1 as L_i is through Pi and Pi is a simple point. Also, ord_Pi(L_0) = 0 since L_0 does not pass through any of the points P1, P2,..., Pn. Now, let's consider ord_Pi(Π L_i^m_i L_0^{-Σ m_i}). Since the order of a function is additive for multiplication and scalar multiples, we get: ord_Pi(z) = ord_Pi(Π L_i^m_i L_0^{-Σ m_i}) = Σ (m_j * ord_Pi(L_j)) - (Σ m_j) * ord_Pi(L_0) For each j ≠ i, ord_Pi(L_j) = 0 as L_j does not pass through Pi. Therefore, we get: ord_Pi(z) = m_i * ord_Pi(L_i) - (Σ m_j) * 0 = m_i We have proven that ord_Pi(z) = mi for i=1, 2, ..., n.

Key concepts

Projective Plane Curve

A projective plane curve is essentially a curve located within the projective plane, which is a mathematical construct used in algebraic geometry. It extends the concept of the standard coordinate plane by adding "points at infinity" to account for parallel lines intersecting. These curves are defined by polynomial equations.
In our context, the curve denoted by \(C\) is irreducible, meaning it cannot be factored into simpler polynomial components. This quality is important because it means the curve has a certain kind of mathematical "wholeness"—it is not a product of simpler, overlapping parts.
The projective plane helps in studying geometric properties that remain constant under transformations such as rotations and translations, making it a powerful tool for mathematicians.

Rational Functions

Rational functions are functions presented as fractions, where both the numerator and the denominator are polynomials. An example of a rational function is \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).
For a projective plane curve \(C\), we represent rational functions in a similar way, but they are defined over the entire projective plane, implying they work well with the points at infinity.
In the exercise, the function \(z\) constructed by combining lines \(L_i\) and \(L_0\) is an example of such a rational function on \(C\). It integrates the idea of complex polynomial calculations that adjust according to the points \(P_i\) and the coefficients \(m_i\). This functionality is essential in expressing divisors and computing orders at specific points.

Divisors

Divisors in algebraic geometry are formal sums of points on a curve. They are important for describing properties of functions and curves by identifying the zeros and poles of functions. Generally, a divisor can be written as \(D = m_1P_1 + m_2P_2 + \ldots + m_nP_n\), where \(P_i\) are points on the curve and \(m_i\) are integers.
The divisor of a rational function captures where the function equals zero (zeros) or tends to infinity (poles).

• A positive \(m_i\) indicates a zero of order \(m_i\) at \(P_i\).
• A negative \(m_i\) indicates a pole of order \(|m_i|\) at \(P_i\).

In our exercise, we manipulate the lines \(L_i\) to ensure that the function \(z\) has a divisor that correctly interprets the specified integers \(m_i\) at the corresponding simple points \(P_i\).

Order of a Function

The order of a function at a given point in algebraic geometry describes how the function behaves near that point. Specifically, it tells us about the multiplicity of zeros or poles at the point.
For a function \(f\) and a point \(P\), the order, denoted \(\text{ord}_P(f)\), is:

• Positive if \(f\) has a zero at \(P\).
• Negative if \(f\) has a pole at \(P\).
• Zero if \(f\) neither vanishes nor becomes infinite at \(P\).

In the given problem, constructing \(z\) with the desired order \( \text{ord}_{P_i}(z) = m_i \) is critical for validating that the specified condition is satisfied at each point \(P_i\). By defining the function \(z\) with consideration for lines like \(L_i\) and \(L_0\), we can systematically control the order at each point of interest within the projective curve.