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Chapter 25: Problem 3

Suppose \(R\) is a domain. Show that \(R\) is an Ore domain if and only if every two nonzero right ideals of \(R\) have a nonzero intersection. If \(R\) is right Noetherian, prove that \(\mathbf{Q}_{\mathrm{cl}}(R)\) exists. For the latter, let \(0 \neq r, t \in R\) and observe that the right ideal \(\sum_{k=0}^{\infty} t^{k} r R\) must be finitely generated.

Short Answer

Expert verified
An Ore domain's nonzero ideals intersect nontrivially. A right Noetherian domain has a well-defined classical quotient ring.

Step by step solution

01

Understanding Ore Domains

An Ore domain is a ring where for any two elements \( r \) and \( s \), with \( s eq 0 \), there exist nonzero elements \( a \) and \( b \) such that \( ar = bs \). This property ensures that any two nonzero right ideals of \( R \) will have a nonzero intersection.
02

The Necessity Direction

Assume \( R \) is an Ore domain. Consider two nonzero right ideals \( I \) and \( J \). Since \( R \) is an Ore domain, for any \( i \in I \) and \( j \in J \), there exist nonzero elements in \( R \) such that their product lies in both \( I \) and \( J \). Thus, \( I \cap Jeq \{0\}.\)
03

The Sufficiency Direction

Now, assume each pair of nonzero right ideals has a nonzero intersection. If \( r, s eq 0 \in R \), generate principal ideals \( rR \) and \( sR \). Their nonzero intersection contains a nonzero element that can be expressed as \( ar = bs\), verifying the Ore condition.
04

Existence of Classical Quotient Ring

For a right Noetherian domain \( R \), every right ideal is finitely generated. Consider the series \( \sum_{k=0}^\infty t^k rR \), which, being a right ideal, must be finitely generated by Noetherian property.
05

Construction of Classical Quotient

The finitely generated series implies that there is a limit beyond which the coefficients stabilize, ensuring that we can construct the classical quotient ring \( \mathbf{Q}_{\mathrm{cl}}(R) \), consisting of fractions with numerator in \( R \) and denominators being nonzero divisors.

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Key Concepts

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

Ore Domain
An Ore domain is an important concept in ring theory. For a ring to qualify as an Ore domain, it must satisfy a specific condition involving its elements. If you take any two elements, say \( r \) and \( s \) (with \( s eq 0 \)), you should be able to find nonzero elements \( a \) and \( b \) such that \( ar = bs \). This implies that the ring must allow commutative-like behavior to create equivalences, even within non-commutative structures.

The condition indicates that even if \( R \) is non-commutative, a form of compatibility with fractions exists, facilitating their formation.
  • This property ensures that every pair of nonzero right ideals intersects non-trivially.
  • The Ore condition is crucial for constructing fractions in non-commutative rings, which paves the way for forming more complex structures like quotient rings.
Right Noetherian Domain
A right Noetherian domain is a type of ring that has a neat and organized structure. Specifically, it is a domain in which every right ideal is finitely generated. This means that for any set of elements that you define as a right ideal, only a finite number of them will generate the entire ideal.

The Noetherian property is significant because it prevents rings from being "too large" or unwieldy, by limiting the complexity of ideals.
  • All ascending chains of right ideals must eventually stabilize, which is its defining feature.
  • This property directly assists in the demonstration that the classical quotient ring \( \mathbf{Q}_{\mathrm{cl}}(R) \) can be constructed, ensuring the finiteness needed for the construction process.
Classical Quotient Ring
The concept of a classical quotient ring \( \mathbf{Q}_{\mathrm{cl}}(R) \) is essentially the generalization of forming a quotient ring from an Ore domain. For a right Noetherian domain, this ring consists of elements which are typically represented as fractions, where the numerator is from the domain \( R \) and the denominator is a nonzero divisor.

The existence of \( \mathbf{Q}_{\mathrm{cl}}(R) \) stems from the ability to create a field-like environment where divisions are possible.
  • In a right Noetherian domain, the existence of a classical quotient ring is guaranteed, providing more structural flexibility.
  • The finitely generated condition of the series \( \sum_{k=0}^\infty t^k rR \) ensures stability within this new ring structure.
Ideals in Ring Theory
Ideals form a foundational concept in ring theory, appearing similarly to the way numbers are divisible in arithmetic. In a ring \( R \), an ideal is a special subset that is closed under addition and compatible with multiplication by any element from \( R \). For ideals, there are two main types鈥攔ight ideals and left ideals.

In this context, nonzero right ideals are considered; their properties are pivotal in the definitions of structures like Noetherian and Ore domains.
  • The intersection of any two nonzero right ideals in an Ore domain is nonzero, solidifying their importance in structural organization.
  • Finitely generated ideals in a Noetherian domain ensure that any complex structure can be simplified or collapsed to a manageable form, making them critical in advanced ring constructions.

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Most popular questions from this chapter

Consider the collection of formal fractions with denominators in \(T\) and define \(r a^{-1} \sim s b^{-1}\) if and only if there exist \(x, y \in T\) with \(a x=b y\) and \(r x=s y .\) Prove that \(\sim\) is an equivalence relation and then let \(\left[\mathrm{ra}^{-1}\right]\) denote the class of \(\mathrm{ra}^{-1}\).

Let \(T\) be a right denominator set in \(R\). Characterize those right ideals \(X\) of \(R\) with \(\left(X T^{-1}\right) \cap R=X .\) If \(R\) is right Noetherian, prove that the same is true of \(R T^{-1}\).
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