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Chapter 1: Problem 2

The Twisted Cubic Curve. Let \(Y \subseteq \mathbf{A}^{3}\) be the set \(Y=\left\\{\left(t, t^{2}, t^{3}\right) | t \in k\right\\} .\) Show that \(Y\) is an affine variety of dimension 1. Find generators for the ideal \(I(Y) .\) Show that \(A(Y)\) is isomorphic to a polynomial ring in one variable over \(k .\) We say that \(Y\) is given by the parametric representation \(x=t, y=t^{2}, z=t^{3}\)

Short Answer

Expert verified
The algebraic set \(Y\) defined by the parametric equations \(x=t, y=t^{2}, z=t^{3}\) where \(t \in k\) is an affine variety of dimension 1. The ideal \(I(Y)\) is generated by the polynomials \((b^2 - a^3, ac - b^2, a^2 - b)\). The affine coordinate ring \(A(Y)\) is isomorphic to a polynomial ring \(k[T]\), a polynomial ring in one variable over \(k\).

Step by step solution

01

Showing that Y is an Affine Variety

To show that \(Y\) is an affine variety of dimension 1, we need to show that the zero set is non-empty, algebraic, irreducible, and that every chain of irreducible algebraic subsets of \(Y\) is of length 1. The given parametric representation \(x=t, y=t^{2}, z=t^{3}\) shows that \(Y\) is non-empty because the parameter \(t\) ranges over all of \(k\). Likewise, \(Y\) is an algebraic set as it is a subset of \(\mathbf{A}^{3}\) defined by a system of algebraic equations. It is irreducible as no nonempty algebraic subsets \(Y_1, Y_2\) of \(Y\)can cover \(Y\). The length of all chains of irreducible algebraic subsets of \(Y\) is 1 since \(Y\) itself is irreducible.
02

Finding Generators for the Ideal \(I(Y)\)

The ideal \(I(Y)\) is generated by the polynomials that \(Y\) vanishes on. We note that if \((a, b, c)\) is in \(Y\), then \((a, b, c) = (t, t^{2}, t^{3})\) for some \(t\). This implies that \(a = t, b = t^{2}\) and \(c = t^{3}\). Therefore \(b^2 - a^3 = 0, ac - b^2 = 0\) and \(a^2 - b = 0\). Thus, \(I(Y)\) is the ideal generated by \(f_1 = b^2 - a^3, f_2 = ac - b^2, f_3 = a^2 - b\).
03

Showing that \(A(Y)\) is Isomorphic to a Polynomial Ring in One Variable

The coordinate ring \(A(Y)\) is defined as \(k[X]/I(Y)\). We have to show that this is isomorphic to the polynomial ring \(k[T]\). Consider the map \(\phi: k[T] \rightarrow A(Y)\) defined by \(T \mapsto \pi (t)\) where \(\pi : k[T]\) is the canonical projection. This map is surjective because for any polynomial \(p(t) \in k[X]\) there exists a polynomial \(q(T) \in k[T]\) such that \( \pi (q(T)) = p(t)\). Moreover, all polynomials that get mapped to 0 making the map injective, therefore this map is an isomorphism.

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Key Concepts

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

Affine Variety
Understanding the concept of an Affine Variety is central to grasping the nature of the Twisted Cubic Curve. An affine variety is essentially a set of solutions to a system of algebraic equations over an algebraically closed field, often denoted as k. This variety must satisfy specific conditions鈥攊t must be algebraic, which means defined by polynomials, irreducible, meaning it cannot be divided into two non-trivial varieties, and non-empty.

A prime example is the Twisted Cubic Curve in our exercise. This curve, expressed as a set Y of ordered triples (t, t^2, t^3) where t is an element of the field k, showcases an affine variety. It is non-empty because there are an infinite number of values t can take in the field k. The curve is also irreducible since it cannot be split into simpler parts that are algebraic sets themselves, ensuring the criterion for a variety is met.
Polynomial Ring
A Polynomial Ring, typically denoted as k[x1, x2, ..., xn], represents the set of all polynomials with coefficients in a field k and indeterminates x1, x2, ..., xn. It is a fundamental concept in algebra that helps us understand how algebraic objects are constructed.

Through the lens of the Twisted Cubic Curve, we explore the significance of the polynomial ring in the context of the coordinate ring A(Y). The operation of the ring, such as addition and multiplication, is defined as it usually is for polynomials. The relation between the twisted cubic curve and polynomial rings is uncovered by mapping the coordinate ring A(Y) to a polynomial ring in one variable over k, which is illustrative of the curve's underlying simplicity despite its seemingly complex three-dimensional shape.
Ideal Generators
The concept of Ideal Generators pertains to the generators of an ideal in a ring, which are the elements of the ring that, through the operations of the ring, yield all the elements of the ideal.

Within our Twisted Cubic Curve problem, we need to identify polynomials that vanish at all points on the curve to find the ideal I(Y). When we speak of vanishing, we mean that if you replace the variables in the polynomial with corresponding values from a point on the curve, the result is zero. For instance, the generators for the ideal I(Y) of the twisted cubic are identified as f1 = b^2 - a^3, f2 = ac - b^2, f3 = a^2 - b. Every element of the ideal can be expressed as a combination of these generators, which indicates that these polynomials are foundational building blocks for the ideal associated with the curve.

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Most popular questions from this chapter

Products of Affine Varieties. Let \(X \subseteq \mathbf{A}^{n}\) and \(Y \subseteq \mathbf{A}^{m}\) be affine varieties. (a) Show that \(X \times Y \subseteq \mathbf{A}^{n+m}\) with its induced topology is irreducible. [Hint: Suppose that \(X \times Y\) is a union of two closed subsets \(Z_{1} \cup Z_{2}\). Let \(X_{1}=\) \(\left\\{x \in X | x \times Y \subseteq Z_{1}\right\\}, i=1,2 .\) Show that \(X=X_{1} \cup X_{2}\) and \(X_{1}, X_{2}\) are closed. Then \(\left.X=X_{1} \text { or } X_{2} \text { so } X \times Y=Z_{1} \text { or } Z_{2} .\right]\) The affine variety \(X \times Y\) is called the product of \(X\) and \(Y\). Note that its topology is in general not equal to the product topology (Ex. 1.4 ). (d) Show that \(\operatorname{dim} X \times Y=\operatorname{dim} X+\operatorname{dim} Y\) (b) Show that \(A(X \times Y) \cong A(X) \otimes_{k} A(Y)\) (c) Show that \(X \times Y\) is a product in the category of varieties, i.e., show (i) the projections \(X \times Y \rightarrow X\) and \(X \times Y \rightarrow Y\) are morphisms, and (ii) given a variety \(Z,\) and the morphisms \(Z \rightarrow X, Z \rightarrow Y,\) there is a unique morphism \(Z \rightarrow X \times Y\) making a commutative diagram

The d-Uple Embedding. For given \(n, d>0,\) let \(M_{0}, M_{1}, \ldots, M_{\mathrm{Y}}\) be all the monomials of degree \(d\) in the \(n+1\) variables \(x_{0}, \ldots, x_{n},\) where \(N=\left(\begin{array}{c}n+d \\\ n\end{array}\right)-1 . \mathrm{Wc}\) define a mapping \(\rho_{d}: \mathbf{P}^{n} \rightarrow \mathbf{P}^{\prime}\) by sending the point \(P=\left(a_{0}, \ldots, a_{n}\right)\) to the point \(\rho_{d}(P)=\left(M_{0}(a), \ldots, M_{\mathrm{v}}(a)\right)\) obtained by substituting the \(a_{i}\) in the monomials \(M_{j}\) This is called the \(d\) -uple embedding of \(\mathbf{P}^{n}\) in \(\mathbf{P}^{N}\). For example, if \(n=1, d=2\), then \(N=2,\) and the image \(Y\) of the 2 -uple cmbedding of \(\mathbf{P}^{1}\) in \(\mathbf{P}^{2}\) is a conic. (a) Let \(\theta: k\left[y_{0}, \ldots, y_{v}\right] \rightarrow k\left[x_{0}, \ldots, x_{n}\right]\) be the homomorphism defined by sending \(y_{i}\) to \(M_{t},\) and let a be the kernel of \(0 .\) Then a is a homogencous prime ideal, and so \(Z(a)\) is a projective variety in \(\mathbf{P}^{\prime}\) (b) Show that the image of \(\rho_{d}\) is exactly \(Z\) (a). (One inclusion is casy. The other will require some calculation.) (c) Now show that \(\rho_{d}\) is a homeomorphism of \(\mathbf{P}^{n}\) onto the projective variety \(Z\) (a). (d) Show that the twisted cubic curve in \(\mathbf{P}^{3}\) (Ex. 2.9 ) is equal to the 3 -uple embed\(\operatorname{ding}\) of \(\mathbf{P}^{1}\) in \(\mathbf{P}^{3},\) for suitable choice of coordinates.

There are quasi-affine varieties which are not affine. For example, show that \(\mathrm{N}=\mathbf{A}^{2}-\\{(0,0)\\}\) is not affine. \([\text {Hint}: \text { Show that }((X) \cong h[x, 1] \text { and use }(3.5)\) See (III, Ex. 4.3 for another proof.

Vormal Varielies. A variety \(Y\) is normal all a point \(P \in Y\) if \((p\) is an integrally closed ring. \(Y\) is normal if it is normal at every point (a) Show that every conic in \(\mathbf{P}^{2}\) is normal. (b) Show that the quadric surfaces \(Q_{1}, Q_{2}\) in \(\mathbf{P}^{3}\) given by equations \(Q_{1}: x y=z w\) \(Q_{2}: x y=z^{2}\) are normal (cf. (II. Ex. 6.4) for the latter.) (c) Show that the cuspidal cubic \(y^{2}=x^{3}\) in \(A^{2}\) is not normal (d) If \(Y\) is affine. then \(Y\) is normal \(\Leftrightarrow A(Y)\) is integrally closed. (e) Let \(Y\) be an affine variety. Show that there is a normal affine variety \(\tilde{Y}\), and a morphism \(\pi: \tilde{Y} \rightarrow Y\), with the property that whenever \(Z\) is a normal variety, and \(\varphi: Z \rightarrow Y\) is a domincint morphism (i.e., \(\varphi(Z)\) is dense in \(Y\) ), then there is a unique morphism \(\theta: Z \rightarrow \tilde{Y}\) such that \(\varphi=\pi\) 0. \(\tilde{Y}\) is called the normalization of \(Y\). You will need \((3.9 \mathrm{A})\) above.

If we identify \(\mathbf{A}^{2}\) with \(\mathbf{A}^{1} \times \mathbf{A}^{1}\) in the natural way, show that the Zariski topology on \(\mathbf{A}^{2}\) is not the product topology of the Zariski topologies on the two copies of \(\mathbf{A}^{1}\)
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