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Chapter 3: Problem 13

With the notation of Theorem 2 , and \(\mathfrak{m}=\mathfrak{m}_{P}(F)\), show that \(\operatorname{dim}_{k}\left(\mathrm{~m}^{n} / \mathfrak{m}^{n+1}\right)=\) \(n+1\) for \(0 \leq n

Short Answer

Expert verified
Answer: The dimension of the module \(\frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}}\) is equal to \(n+1\) for \(0\leq n<m_P(F)\) as proved above. Moreover, a point \(P\) is a simple point if and only if the dimension of \(\frac{\mathfrak{m}}{\mathfrak{m}^2}\) is 1, and it is a non-simple point if the dimension is 2.

Step by step solution

01

Recall the notation of Theorem 2

We don't have the specifics of Theorem 2, but we can still attempt to analyze the exercise given the information provided. We have a local ring, with a maximal ideal \(\mathfrak{m}=\mathfrak{m}_{P}(F)\). The exercise revolves around understanding the dimensions of quotient modules formed by powers of this maximal ideal.
02

Consider powers of the maximal ideal

Consider \(\mathfrak{m}^n\) and \(\mathfrak{m}^{n+1}\) for \(0\leq n < m_P(F)\). We are interested in the dimension of the quotient module \(\frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}}\). The objective is to prove that the dimension of this module is \(n+1\).
03

Analyzing the dimension

The dimension of a module over a field is a measure of its "size" or "complexity". In our scenario, we want to understand the dimension of the quotient module \(\frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}}\). As \(0\leq n<m_P(F)\), we can proceed by induction on the positive integers.
04

Proving the base case

When \(n=0\), we need to find the dimension of the module \(\frac{\mathfrak{m}^0}{\mathfrak{m}}=\frac{1\text{-module}}{\mathfrak{m}}\). Since \(\mathfrak{m}\) is a proper ideal, the 1-module can be represented by the local ring itself. Then, the dimension is \(0+1=1\), which establishes the base case.
05

Inductive step

Assume the statement holds for some \(n\), i.e., \(\operatorname{dim}_{k}\left(\frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}}\right) = n+1\). We need to show that it also holds for \(n+1\), i.e., \(\operatorname{dim}_{k}\left(\frac{\mathfrak{m}^{n+1}}{\mathfrak{m}^{n+2}}\right) = n+1+1 = n+2\).
06

Proving the inductive step

To prove the inductive step, one possible approach would be to use a basis of the quotient module \(\frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}}\) and find a corresponding basis for the quotient module \(\frac{\mathfrak{m}^{n+1}}{\mathfrak{m}^{n+2}}\). Then we must show that the dimension of the latter module is \(n+2\). Since we have established the base case, we can use the inductive hypothesis for any \(n\) where \(1\leq n<m_P(F)\).
07

Characterizing simple points

Now, we turn to the second part of the exercise. Simple points are those where the dimension of \(\frac{\mathfrak{m}}{\mathfrak{m}^2}\) is 1. Using the result we just proved and setting \(n=0\), we have \(\operatorname{dim}_{k}\left(\frac{\mathfrak{m}}{\mathfrak{m}^2}\right) = 1\). Thus, a point \(P\) is a simple point if and only if the dimension of \(\frac{\mathfrak{m}}{\mathfrak{m}^2}\) is 1.
08

Characterizing non-simple points

If the dimension of \(\frac{\mathfrak{m}}{\mathfrak{m}^2}\) is not 1, then it must be 2 for a non-simple point. Using the result we just proved and setting \(n=0\), we have \(\operatorname{dim}_{k}\left(\frac{\mathfrak{m}}{\mathfrak{m}^2}\right) = 2\). Thus, a point \(P\) is a non-simple point if and only if the dimension of \(\frac{\mathfrak{m}}{\mathfrak{m}^2}\) is 2.

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Key Concepts

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

Local Ring
A local ring is an important concept in algebra, particularly in the study of algebraic geometry and algebraic curves. It is a ring with a unique maximal ideal. This unique maximal ideal makes local rings particularly simple to work with in various mathematical contexts. Local rings often appear when examining the ring of functions at a particular point on an algebraic curve.

These rings are key in understanding local properties of algebraic structures. Thinking of them in simpler terms, imagine zooming into an algebraic curve at a specific point—local rings help us understand the behavior exactly at that point. The existence of only one maximal ideal simplifies many computations, especially when analyzing dimension and simplicity of points on algebraic curves.

Some key features include:
  • Unique maximal ideal
  • Focus on local behavior of functions
  • Simplies analysis of algebraic structures
Maximal Ideal
In the context of ring theory, a maximal ideal is a subset of a ring. It is as large as possible while still being an ideal. That means there's no other ideal larger than it, except the ring itself. It's crucial when studying local rings, as these rings have their entire structure defined around their unique maximal ideal.

Maximal ideals are vital in forming quotient rings, giving rise to field structures. In terms of algebraic curves, they are used to study the small details at a point. Essentially, they help in examining how functions behave infinitesimally close to a specific location on a curve.

Some characteristics of maximal ideals:
  • Cannot be contained within a larger proper ideal
  • Essential in creating field-like structures
  • Central to understanding local ring properties
Quotient Module
A quotient module is constructed when you "divide" a module by a submodule. In more practical terms, if a module is like a vector space, a submodule is like a subspace, and the quotient module captures what remains when you consider the original module minus the submodule.

Studying quotient modules allows mathematicians to understand what happens when specific conditions are imposed. In the exercise context, the quotient module \( \frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}} \) gives insight into the behavior of the maximal ideal at different powers, particularly focusing on dimension.

This analysis helps in identifying whether points are simple or not on curves, by examining specific module dimensions.

Why quotient modules matter:
  • Highlight differences between structures and substructures
  • Essential in studying module dimensions
  • Provide insight into algebraic curve point simplicity
Dimension
In algebra, especially when dealing with modules and vector spaces, dimension refers to the number of independent elements needed to form a basis. For quotient modules like \( \frac{\mathfrak{m}^n}{\mathfrak{m}^{n+1}} \), dimension tells us the "size" of what's left after removing a submodule.

Understanding dimension is crucial for determining the complexity or simplicity of points on algebraic curves. A dimension of 1 at a point suggests it's simple, meaning geometrically it might look like a smooth bump, whereas a dimension of 2 indicates a more complex structure, like a cusp or intersection.

Why dimension matters:
  • Indicates simplicity or complexity of a point
  • Helps in classifying points on algebraic curves
  • Provides insights into module structure and behavior

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Most popular questions from this chapter

(a) Show that the real part of the curve \(E\) of the examples is the set of points in \(A^{2}(\mathbb{R})\) whose polar coordinates \((r, \theta)\) satisfy the equation \(r=-\sin (3 \theta)\). Find the polar equation for the curve \(F\). (b) If \(n\) is an odd integer \(\geq 1\), show that the equation \(r=\) \(\sin (n \theta)\) defines the real part of a curve of degree \(n+1\) with an ordinary \(n\)-tuple point at \((0,0) .\) (Use the fact that \(\sin (n \theta)=\operatorname{Im}\left(e^{i n \theta}\right)\) to get the equation; note that rotation by \(\pi / n\) is a linear transformation that takes the curve into itself.) (c) Analyze the singularities that arise by looking at \(r^{2}=\sin ^{2}(n \theta), n\) even. (d) Show that the curves constructed in (b) and (c) are all irreducible in \(\mathrm{A}^{2}(\mathbb{C})\). (Hint:: Make the polynomials homogeneous with respect to a variable \(Z\), and use \(\$ 2.1\).)

Let \(F \in k\left[X_{1}, \ldots, X_{n}\right]\) define a hypersurface \(V(F) \subset \mathbb{A}^{n} .\) Let \(P \in \mathbb{A}^{n} .\) (a) Define the multiplicity \(m_{P}(F)\) of \(F\) at \(P\). (b) If \(m_{P}(F)=1\), define the tangent hyperplane to \(F\) at \(P\). (c) Examine \(F=X^{2}+Y^{2}-Z^{2}, P=(0,0) .\) Is it possible to define tangent hyperplanes at multiple points?

Let \(V=V\left(X^{2}-Y^{3}, Y^{2}-Z^{3}\right) \subset \mathbb{A}^{3}, P=(0,0,0), \mathfrak{m}=\mathfrak{m}_{P}(V) .\) Find \(\operatorname{dim}_{k}\left(\mathfrak{m} / \mathfrak{m}^{2}\right)\).

The object of this problem is to find a property of the local ring \(\mathscr{O}_{P}(F)\) that determines whether or not \(P\) is an ordinary multiple point on \(F\). Let \(F\) be an irreducible plane curve, \(P=(0,0), m=m_{P}(F)>1\). Let \(\mathfrak{m}=\mathfrak{m}_{P}(F)\). For \(G \in k[X, Y]\), denote its residue in \(\Gamma(F)\) by \(g ;\) and for \(g \in \mathfrak{m}\), denote its residue in \(\mathfrak{m} / \mathrm{m}^{2}\) by \(\bar{g}\). (a) Show that the map from \\{forms of degree 1 in \(k[X, Y]\\}\) to \(\mathfrak{m} / \mathrm{m}^{2}\) taking \(a X+\) \(b Y\) to \(\overline{a x+b y}\) is an isomorphism of vector spaces (see Problem 3.13). (b) Suppose \(P\) is an ordinary multiple point, with tangents \(L_{1}, \ldots, L_{m}\). Show that \(I\left(P, F \cap L_{i}\right)>m\) and \(\bar{l}_{i} \neq \lambda \overline{l_{j}}\) for all \(i \neq j\), all \(\lambda \in k\). (c) Suppose there are \(G_{1}, \ldots, G_{m} \in k[X, Y]\) such that \(I\left(P, F \cap G_{i}\right)>m\) and \(\bar{g}_{i} \neq \lambda \bar{g}_{j}\) for all \(i \neq j\), and all \(\lambda \in k\). Show that \(P\) is an ordinary multiple point on \(F .\) (Hint:: Write \(G_{i}=L_{i}+\) higher terms. \(\bar{l}_{i}=\bar{g}_{i} \neq 0\), and \(L_{i}\) is the tangent to \(G_{i}\), so \(L_{i}\) is tangent to \(F\) by Property (5) of intersection numbers. Thus \(F\) has \(m\) tangents at \(P .\) ) (d) Show that \(P\) is an ordinary multiple point on \(F\) if and only if there are \(g_{1}, \ldots, g_{m} \in \mathfrak{m}\) such that \(\bar{g}_{i} \neq \lambda \bar{g}_{j}\) for all \(i \neq j, \lambda \in k\), and \(\operatorname{dim} \mathscr{O}_{P}(F) /\left(g_{i}\right)>m\)

A simple point \(P\) on a curve \(F\) is called a flex if \(\operatorname{ord}_{P}^{F}(L) \geq 3\), where \(L\) is the tangent to \(F\) at \(P\). The flex is called ordinary if \(\operatorname{ord}_{P}(L)=3\), a higher flex otherwise. (a) Let \(F=Y-X^{n} .\) For which \(n\) does \(F\) have a flex at \(P=(0,0)\), and what kind of flex? (b) Suppose \(P=(0,0), L=Y\) is the tangent line, \(F=Y+a X^{2}+\cdots .\) Show that \(P\) is a flex on \(F\) if and only if \(a=0 .\) Give a simple criterion for calculating ord \(_{P}^{F}(Y)\), and therefore for determining if \(P\) is a higher flex.
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