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Chapter 3: Problem 4

. Let \(R\) be the ring of all real-valued continuous functions on the closed unit interval. If \(M\) is a maximal ideal of \(R\) prove that there exists a real number \(\gamma, 0 \leq \gamma \leq 1\), such that \(M=M_{\gamma}=[f(x) \in R \mid f(\gamma)=0\\}\).

Short Answer

Expert verified
For a maximal ideal \(M\) in \(R\), there exists \(\gamma \in [0,1]\) such that \(M = \{f \in R \mid f(\gamma) = 0\}\).

Step by step solution

01

Understand the Setup and Definitions

We are given a ring \(R\), which is the set of all real-valued continuous functions on the closed interval \([0,1]\), and \(M\), a maximal ideal of \(R\). Each function \(f\) in \(R\) is continuous over this interval.
02

Consider the Quotient Ring \(R/M\)

Since \(M\) is maximal, the quotient \(R/M\) is a field. We will prove that \(M\) takes a form of functions vanishing at a specific point, which makes \(R/M\) equivalent to \(\mathbb{R}\).
03

Establish a Connection to Points in the Interval

Consider evaluation at a point \(\gamma\). Define a homomorphism, \(\phi : R \to \mathbb{R} ; f \mapsto f(\gamma)\). The kernel of this homomorphism, \(\mathrm{ker}\, \phi = \{ f \in R \mid f(\gamma) = 0 \}\), is an ideal. If \(\gamma\) is fixed, this ideal, denoted as \(M_{\gamma}\), is formed by all functions that are zero at \(\gamma\).
04

Show that One Such \(M_{\gamma}\) is Maximal

Since \(M\) is maximal, \(R/M\) should be a field. Consider any \(f\) not in \(M\), implying \(f(\gamma) eq 0\) for some \(\gamma\) in \([0,1]\). We then have \(1/f(\gamma)\) exists in \(\mathbb{R}\) and \(\phi(f) eq 0\). Hence \(f\) can be mapped to a non-zero in \(\mathbb{R}\), confirming that \(M = M_{\gamma}\) for some \(\gamma \in [0,1]\).
05

Conclude with the Existence Statement

Using the above results, any maximal ideal \(M\) in \(R\) can be shown to be of the form \(M_{\gamma}\) for a specific \(\gamma\) in \([0, 1]\), proving our statement.

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Key Concepts

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

Ring of Continuous Functions
The concept of a ring of continuous functions revolves around the idea of collecting all possible continuous functions that map from a closed interval, such as \[0, 1\], to the real numbers. This forms a mathematical structure called a ring because it satisfies two main operations: addition and multiplication of functions. In this context:
  • Each function in the ring adds or multiplies with another function from the ring, remaining within the realm of continuous real-valued functions.
  • The identity elements are the zero function for addition and the constant function at 1 for multiplication, ensuring the axioms of a ring are met.
The elements themselves are not numbers but functions, each depicting smooth and consistent behavior over the interval. This idea plays a critical role in advanced mathematical concepts, such as ideal theory and topology, where understanding continuity becomes pivotal.
Quotient Ring
A quotient ring arises when we take a ring, in our case, the ring \(R\) of continuous functions, and form a new set by 'overlooking' a certain ideal, here, the maximal ideal \(M\). This is not just a simple set subtraction; it is a profound reorganization:
  • The components of the quotient ring are equivalence classes, where two functions are considered equivalent if their difference lies in the ideal \(M\).
  • In our example, because \(M\) is a maximal ideal, the resulting quotient ring \(R/M\) is a field, meaning every non-zero element has an inverse within this structure.
This transformation into a quotient ring simplifies problems, replacing complex function setups with simpler equations, permitting a broader range of mathematical operations and insights.
Homomorphism
A homomorphism in mathematics is essentially a function that respects the structure of the entities it maps between. In the process of proving the nature of the maximal ideal \(M\), a homomorphism \(\phi: R \to \mathbb{R}\) is employed. How this works here:
  • The homomorphism takes each function \(f\) in the ring \(R\) and evaluates it at a specific point \(\gamma\), thereby mapping it to a real number \(f(\gamma)\).
  • The kernel of this homomorphism, \text{ker } \phi\, identifies functions vanishing at \(\gamma\), forming \(M_\gamma\).
This mapping is critical not only in simplifying the function space by evaluating functions at precise points but also in understanding the behavior and nature of ideals involved.
Real-Valued Functions
Real-valued functions, as the name suggests, are functions whose output, or range, lies within the set of real numbers. In the context of our ring \(R\):
  • Each function provides a real number for each input in the closed interval \[0, 1\].
  • The continuous nature of these functions implies no abrupt changes or jumps, aligning smoothly across the domain.
This continuity ensures that all values between two given points are accounted for, making them highly suitable for mathematical analysis in calculus and real analysis. Real-valued functions form the backbone of function analysis, providing a real-number output for various input scenarios, and thereby highly relevant in mathematical applications of functions and equations.

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Most popular questions from this chapter

(a) Prove that \(x^{2}+1\) is irreducible over the field \(F\) of integers mod 11 and prove directly that \(F[x] /\left(x^{2}+1\right)\) is a field having 121 elements. (b) Prove that \(x^{2}+x+4\) is irreducible over \(F\), the field of integers mod 11 and prove directly that \(F[x] /\left(x^{2}+x+4\right)\) is a field having 121 elements. *(c) Prove that the fields of part (a) and part (b) are isomorphic.

In Problem 7 , if the least common multiple of \(a\) and \(b\) is denoted by \([a, b]\), prove that \([a, b]=\frac{a b}{(a, b)}\).

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