Problem 17
Question
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.
Step-by-Step Solution
VerifiedKey Concepts
Conic in Projective Plane
- Conics include familiar shapes like circles, ellipses, parabolas, and hyperbolas, all of which can be smoothly transformed into one another.
- The absence of singular points (like cusps or nodes) implies the integral closure of the corresponding ring, affirming the normality.
Quadric Surfaces in Projective Space
- Each quadric in \( \mathbf{P}^{3} \) adheres to algebraic properties that ensure no singularities, covering the entire space smoothly.
- The equations describe surfaces like hyperboloids or ellipsoids, each of which is naturally devoid of cusps.
Cuspidal Cubic
- The function \( 1/x \) can exist in the function field of the coordinate ring for this cubic, fulfilling integral conditions yet absent from the ring itself.
- This reveals issues with integrality at the cusp, where the ring isn't closed, directly attacking normality.
Affine Variety Normalization
- A process known as normalization allows transforming \( Y \) into a normal variety \( \tilde{Y} \).
- The morphism \( \pi: \tilde{Y} \rightarrow Y \) ensures any dense morphism from a normal \( Z \) adjusts through \( \tilde{Y} \).