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Projecticely Normul Varieties. A projective variety \(Y \subseteq \mathbf{P}^{n}\) is projectively normal (with respect to the given embedding) if its homogeneous coordinate ring \(S(Y)\) is integrally closed. (a) If \(Y\) is projectively normal. then \(Y\) is normal (b) There are normal varieties in projective space which are not projectively normal. For example, let \(Y\) be the twisted quartic curve in \(\mathbf{P}^{3}\) given parametrically by \((x, y: z, w)=\left(t^{4}, t^{3} u, t u^{3}, u^{4}\right) .\) Then \(Y\) is normal but not projectively normal. See (III, Ex. 5.6) for more examples. (c) Show that the twisted quadratic curve \(Y\) above is isomorphic to \(\mathbf{P}^{1}\), which is projectively normal. Thus projective normality depends on the embedding.

Short Answer

Expert verified
A projectively normal variety is normal, but need not be projectively normal if embedded non-linearly. For example, the given curve in \(\mathbf{P^{3}}\) is isomorphic to \(\mathbf{P^{1}}\), which is projectively normal, marking that projective normality relies on its embedding in the projective space.

Step by step solution

01

Understanding of Projectively Normal Varieties

The key enquiry of the exercise focuses on projective normality. A projective variety \(Y \subseteq \mathbf{P}^{n}\) is said to be projectively normal if its homogeneous coordinate ring \(S(Y)\) is integrally closed. This means that if an element \(r\) of its quotient field is a root of a monic polynomial whose coefficients are in \(S(Y)\), then \(r\) also belongs to \(S(Y)\). In other words, there is no 'gap' in \(S(Y)\).
02

Understanding Normal Varieties

By the definition, a variety (irreducible algebraic set) is normal if its local rings are integrally closed. So, the task is to prove that if variety \(Y\) is projectively normal, then \(Y\) is normal i.e., to express the condition of projective normality in terms of local rings of \(Y\).
03

Varieties in Projective Space

The twisted quartic curve given parametrically by (x, y: z, w) = \(\left(t^{4}, t^{3} u, t u^{3}, u^{4}\right)\) is a concrete example provided in the exercise. This curve in \(\mathbf{P}^{3}\) is normal, meaning that there are no 'singular points' on it, therefore it is 'smooth'. However, it is not projectively normal because it cannot be described by four homogeneous polynomials of degree 1 (linear) in projective 3-dimensional space, and computation of its homogeneous coordinate ring also reveals that it's not integrally closed.
04

Isomorphism of the Curve

Finally, we are to show that the curve \(Y\) is isomorphic to \(\mathbf{P}^{1}\), which is projectively normal. An isomorphism in algebraic geometry means a bijective map that preserves the algebraic structure. The twisted quartic curve can be viewed as a rational map from \(\mathbf{P}^{1}\) [which itself is projectively normal as it is represented by linear homogeneous polynomials] onto the curve. By checking that it is a bijection and that its inverse map also preserves the algebraic structure (i.e., it is also a rational map), we verify this isomorphism. This implies projective normality can depend on how the variety is embedded into the projective space - even though \(Y\) and \(\mathbf{P}^{1}\) are isomorphic, \(Y\) isn't projectively normal due to its given embedding.

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Key Concepts

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

Homogeneous Coordinate Ring
The homogeneous coordinate ring, often denoted as \(S(Y)\), plays a crucial role in understanding projectively normal varieties. This ring is associated with a variety \(Y\) in a projective space \(\mathbf{P}^n\). It's constructed from the polynomials that describe \(Y\) and helps in understanding its algebraic structure.
One way to visualize this is to think of the homogeneous coordinate ring as a bridge connecting geometry and algebra. It allows us to translate geometrical properties of the variety into algebraic language.
For a projective variety to be projectively normal, its homogeneous coordinate ring must be integrally closed. This means that every element that behaves like it belongs in the ring, given some polynomial relation, actually does belong there. It's like ensuring there are no missing elements that should naturally complete the ring.
This property is significant because it gives rise to various implications about the geometric and algebraic nature of the variety. For instance, a projective variety with an integrally closed homogeneous coordinate ring is normal, exhibiting a smooth structure without singular points.
Integrally Closed
In the context of algebraic geometry, a ring is called integrally closed if each element that appears to belong to the ring based on polynomial equations, does, in fact, belong to it. This is a formal way of speaking to the completeness of the ring.
More technically, if \(r\) is an element in the fraction field of \(S(Y)\) and it satisfies a monic polynomial equation with coefficients also in \(S(Y)\), then \(r\) should be an element of \(S(Y)\) for the ring to be integrally closed.
  • This ensures no 'gaps' in the algebraic structure of \(Y\), giving us a solid foundation to assert that the variety is normal.
Understanding integrally closed rings can be crucial in distinguishing between projectively normal varieties and those that are only normal, as seen in examples like the twisted quartic curve which is normal but not projectively normal due to its embedding.
Twisted Quartic Curve
The twisted quartic curve is a fascinating example that brings complexity to the concept of projective normality. This specific curve is defined within a projective space \(\mathbf{P}^3\) by the parameters \((x, y, z, w) = (t^4, t^3u, tu^3, u^4)\).
It is considered to be normal because it possesses no singular points; it is smooth and behaves 'nicely' at each of its points, as expected for a normal variety.
However, the twisted quartic curve is not projectively normal. Its homogeneous coordinate ring is not integrally closed because the curve cannot be expressed by a minimal set of linear homogeneous polynomials in the projective space.
This distinction highlights the nuanced nature of projectively normal varieties and how they differ from simply being normal. Such examples demonstrate that projective normality is sensitive to the specific embedding of a variety, showing the complexity and depth of projective geometry.
Isomorphism in Algebraic Geometry
An isomorphism in algebraic geometry is akin to an exceptionally precise match between two geometric objects. It's a bijective map that holds onto the underlying algebraic structure of the varieties involved.
When considering the twisted quartic curve, it's important to know that, despite its failure at being projectively normal in its original embedding, it is isomorphic to \(\mathbf{P}^1\).
\(\mathbf{P}^1\), for its part, is projectively normal because it relates directly to linear homogeneous polynomials, providing a stark contrast with the twisted quartic curve.
Such isomorphism tells us that while these varieties might look different in their original forms, they share identical underlying algebraic structures. This showcases how embedding can affect our perception of projective normality, and why \(Y\), the twisted quartic curve, is an exception when discussing normality versus projective normality.
Overall, isomorphisms help to reveal deeper insights into the nature of algebraic varieties by showing concrete relationships between forms that might initially seem different.

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Most popular questions from this chapter

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 \(x=t, y=t^{2}, z=t^{3}\)

(a) Let \(\varphi: X \rightarrow Y\) be a morphism. Then for each \(P \in X . \varphi\) induces a homomorphism of local rings \(\varphi_{p}^{*}: c_{m, p, 1} \rightarrow\left(p_{1}, 1\right.\) (b) Show that a morphism \(\varphi\) is an isomorphism if and only if \(\varphi\) is a homeomorphism, and the induced map \(\varphi_{P}^{*}\) on local rings is an isomorphism, for all \(P \in X\) (c) Show that if \(\varphi(X)\) is dense in \(Y\), then the map \(\varphi_{p}^{*}\) is injective for all \(P \in X\)

Linear Varieties. Show that an algebraic set \(Y\) of pure dimension \(r\) (i.e., every irreducible component of \(Y\) has dimension \(r\) ) has degree 1 if and only if \(Y\) is a linear variety (Ex. 2.11). [Hint: First, use (7.7) and treat the case dim \(Y=1 .\) Then do the general case by cutting with a hyperplane and using induction.]

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.

Analytically Isomorphic singularities. (a) If \(P \in Y\) and \(Q \in Z\) are analytically isomorphic plane curve singularities, show that the multiplicities \(\mu_{P}(Y)\) and \(\mu_{Q}(Z)\) are the same (Ex. 5.3) (b) Generalize the example in the text \((5.6 .3)\) to show that if \(f=f_{r}+f_{r+1}+\ldots \epsilon\) \(k[[x, y]],\) and if the leading form \(f_{r}\) of \(f\) factors as \(f_{r}=g_{s} h_{t},\) where \(g_{s}, h_{t}\) are homogeneous of degrees \(s\) and \(t\) respectively, and have no common linear factor, then there are formal power series $$\begin{array}{l} g=g_{s}+g_{s+1}+\ldots \\\h=h_{t}+h_{t+1}+\ldots\end{array}$$ \(\operatorname{in} k[[x, y]]\) such that \(f=g h\) (c) Let \(Y\) be defined by the equation \(f(x, y)=0\) in \(\mathbf{A}^{2},\) and let \(P=(0,0)\) be a point of multiplicity \(r\) on \(Y\), so that when \(f\) is expanded as a polynomial in \(x\) and \(y\) we have \(f=f_{r}+\) higher terms. We say that \(P\) is an ordinary \(r\) -fold point if \(f_{r}\) is a product of \(r d i s t i n c t\) linear factors. Show that any two ordinary double points are analytically isomorphic. Ditto for ordinary triple points. But show that there is a one-parameter family of mutually nonisomorphic ordinary 4-fold points. (d) Assume char \(k \neq 2 .\) Show that any double point of a plane curve is analytically isomorphic to the singularity at (0,0) of the curve \(y^{2}=x^{r}\), for a uniquely determined \(r \geqslant 2 .\) If \(r=2\) it is a node (Ex. 5.6). If \(r=3\) we call it a cusp; if \(r=4\) a tacnode. See (V, 3.9.5) for further discussion.

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