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Chapter 1: Problem 4

Let \(Y\) be a nonsingular projective curve. Show that every nonconstant rational function \(f\) on \(Y\) defines a surjective morphism \(\varphi: Y \rightarrow \mathbf{P}^{1},\) and that for every \(P \in \mathbf{P}^{1}\) \(\varphi^{-1}(P)\) is a finite set of points.

Short Answer

Expert verified
The nonconstant rational function \(f\) on the nonsingular projective curve \(Y\) defines a surjective morphism \(\varphi: Y \rightarrow \mathbf{P}^{1}\) since each point in \(\mathbf{P}^{1}\) can be reached from \(Y\). For every point \(P \in \mathbf{P}^{1}\), the preimage \(\varphi^{-1}(P)\) is a finite set of points, as it corresponds to solving the equation \(f = P\), yielding a finite number of solutions.

Step by step solution

01

Define the morphism

Firstly, define the morphism \(\varphi: Y \rightarrow \mathbf{P}^{1}\) using the non-constant rational function \(f\). This can be done by associating each point on the curve \(Y\) with a point on the projective line \(\mathbf{P}^{1}\) through the function \(f\).
02

Proving surjectivity

This morphism is surjective by definition because for every point \(P\) in \(\mathbf{P}^{1}\), there exists a point in \(Y\) that maps to \(P\) through \(\varphi\), since we are given \(f\) is non-constant, and no point in \(\mathbf{P}^{1}\) is excluded as a target for \(\varphi\).
03

Proving the finiteness of the preimage

Given \(P \in \mathbf{P}^{1}\), \(\varphi^{-1}(P)\) corresponds to the solutions of the equation \(f = P\) (where \(f\) is considered as a function to the field). Since \(f\) is a rational function on a curve, and \(f\) is non-constant, the equation \(f = P\) defines a subvariety of \(Y\) of dimension 0, which corresponds to a set of points. So \(\varphi^{-1}(P)\) is a finite set of points.

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Key Concepts

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

Projective Curves
A projective curve can be thought of as a curve that is embedded in a projective space, such as a projective plane. The projective nature means we are thinking about geometric objects at infinity, which helps in considering solutions more abstractly and universally.

In the case of a nonsingular projective curve like \(Y\), the nonsingularity implies that locally, the curve behaves like a smooth manifold. This implies that the curve does not intersect itself and contains no cusp-like singular points.

Projective curves also allow us to consider complex points and abstract geometric relationships, which are essential in high-level algebraic geometry. They provide the structure needed for unified treatment of various geometric problems. Understanding the fundamentals of projective curves lets us explore relationships thoroughly and apply algebraic tools such as rational functions to understanding mappings from these curves.
Surjective Morphisms
A morphism is essentially a notion of a structure-preserving map from one mathematical object to another. Surjective morphisms are a specific type of these maps that cover every element of the target space. In the context of our exercise, it maintains that every point of the projective line \( \mathbf{P}^{1} \) must have at least one point from the curve \(Y\) mapping to it.

This ensures comprehensive coverage of \( \mathbf{P}^{1} \), and importantly, it tells us that the function \(f\) is nonconstant. A constant function could never cover all points of \( \mathbf{P}^{1} \). With a proper surjective morphism:
  • Each element in \( \mathbf{P}^{1} \) is accounted for in the mapping.
  • The morphism \( \varphi:Y \rightarrow \mathbf{P}^{1} \) via rational function \(f\) is truly covering \( \mathbf{P}^{1} \).
The key takeaway is that nonconstant rational functions can serve as surjective morphisms, forming essential mappings between algebraic structures.
Finite Preimage Sets
The concept of a finite preimage set simply means that for each point in the target space, there are only a limited number of points in the source space that map to it. In this scenario, the preimage \( \varphi^{-1}(P) \) corresponds to the solutions of \( f = P \).

Since \(f\) is a nonconstant rational function on our nonsingular projective curve \(Y\), the equation \( f = P \) results in a subvariety of \(Y\) with dimension 0. Essentially, it's akin to solving a polynomial equation of one variable where solutions are restricted to discrete points rather than a continuous stretch. Therefore, the preimage \( \varphi^{-1}(P) \) is finite.

In simpler terms, when mapping a curve to a projective line, due to the properties of rational functions, you derive a limited number of points on the curve that correspond to each point on the \( \mathbf{P}^{1} \), enabling focused study of these relations in algebraic geometry.
  • For each \(P\) in \( \mathbf{P}^{1} \), there exists a limited number of preimages in \(Y\).
  • This finiteness is essential for in-depth analysis and applications in algebraic geometry.

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Most popular questions from this chapter

A variety \(Y\) is rational if it is birationally equivalent to \(\mathbf{P}^{n}\) for some \(n\) (or, equivalently by \((4.5),\) if \(K(Y)\) is a pure transcendental extension of \(k\) ). (a) Any conic in \(\mathbf{P}^{2}\) is a rational curve. (b) The cuspidal cubic \(y^{2}=x^{3}\) is a rational curve. (c) Let \(Y\) be the nodal cubic curve \(y^{2} z=x^{2}(x+z)\) in \(P^{2}\). Show that the projection \(\varphi\) from the point \(P=(0,0,1)\) to the line \(z=0\) (Ex. 3.14 ) induces a birational map from \(Y\) to \(\mathbf{P}^{1}\). Thus \(Y\) is a rational curve.

(a) Show that the following conditions are equivalent for a topological space \(X:\) (i) \(X\) is noetherian; (ii) every nonempty family of closed subsets has a minimal element; (iii) \(X\) satisfies the ascending chain condition for open subsets; (iv) every nonempty family of open subsets has a maximal element. (b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover. (c) Any subset of a noetherian topological space is noetherian in its induced topology. (d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

Let \(Y^{r} \subseteq \mathbf{P}^{n}\) be a variety of degree \(2 .\) Show that \(Y\) is contained in a linear subspace \(L\) of dimension \(r+1\) in \(\mathbf{P}^{n} .\) Thus \(Y\) is isomorphic to a quadric hypersurface in \(\mathbf{P}^{r+1}(\mathrm{E} \mathrm{x} .5 .12)\).

Given a curve \(Y\) of degree \(d\) in \(\mathbf{P}^{2}\), show that there is a nonempty open subset \(U\) of \(\left(\mathbf{P}^{2}\right)^{*}\) in its Zariski topology such that for each \(L \in U, L\) meets \(Y\) in exactly \(d\) points. \(\left[\text { Hint: Show that the set of lines in }\left(\mathbf{P}^{2}\right)^{*} \text { which are either tangent to } Y\) or pass \right. through a singular point of \(Y\) is contained in a proper closed subset.] This result shows that we could have defined the degree of \(Y\) to be the number \(d\) such that almost all lines in \(\mathbf{P}^{2}\) meet \(Y\) in \(d\) points, where "almost all" refers to a nonempty open set of the set of lines, when this set is identified with the dual projective space \(\left(\mathbf{P}^{2}\right)^{*}\).

Let \(Y \subseteq \mathbf{P}^{2}\) be a nonsingular plane curve of degree \(>1\), defined by the equation \(f(x, y, z)=0 .\) Let \(X \subseteq \mathbf{A}^{3}\) be the affine variety defined by \(f\) (this is the cone over \(Y ;\) see (Ex. 2.10) ). Let \(P\) be the point \((0,0,0),\) which is the vertex of the cone. Let \(\varphi: \tilde{X} \rightarrow X\) be the blowing-up of \(X\) at \(P\) (a) Show that \(X\) has just one singular point, namely \(P\) (b) Show that \(\tilde{X}\) is nonsingular (cover it with open affines). (c) Show that \(\varphi^{-1}(P)\) is isomorphic to \(Y\)
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