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Chapter 18: Problem 9

Now assume that \(R\) is a commutative domain and let \(V\) be a torsion free \(R\)-module. If \(T\) is a multiplicatively closed subset of \(R\), verify all the steps in the construction of the module \(V T^{-1}\). Furthermore, if \(S \supseteq T\) is also a multiplicatively closed subset of \(R\), prove that \(\left(V T^{-1}\right) S^{-1} \cong V S^{-1}\) as \(R S^{-1}\)-modules.

Short Answer

Expert verified
Both module constructions are verified: \((V T^{-1}) S^{-1} \cong V S^{-1}\).

Step by step solution

01

Define Torsion-Free and Module Localization

A module \(V\) over a ring \(R\) is torsion-free if for any non-zero \(r \in R\) and \(v \in V\), \(rv = 0\) implies \(v = 0\). Localization involves constructing \(V T^{-1}\), where the elements are equivalence classes \( \frac{v}{t} \) with \(v \in V\) and \(t \in T\). This allows division by elements of \(T\) in the module.
02

Constructing VT^{-1}

For the module \(V T^{-1}\), define the relation \( (v, t) \sim (v', t') \) if there exists \(s \in T\) such that \(s(tv' - t'v) = 0\). The equivalence class \( [v, t] \) is represented as \( \frac{v}{t} \). Verifying the equivalence relation ensures this forms a module over \(R T^{-1}\).
03

Verify VT^{-1} is an R T^{-1}-Module

To verify \(V T^{-1}\) is an \(R T^{-1}\)-module, define addition \( \frac{v}{t} + \frac{v'}{t'} = \frac{vt' + v't}{tt'} \) and scalar multiplication \( r \cdot \frac{v}{t} = \frac{rv}{t} \) for \(r \in R\). Check these operations satisfy associativity, commutativity, and distributivity, confirming module structure.
04

Consider Sufficiently Large Multiplicatively Closed Set

Suppose \(S \supseteq T\) is another multiplicatively closed subset of \(R\). Consider both the module \((V T^{-1}) S^{-1}\) and \(V S^{-1}\). The aim is to show these are isomorphic as \(R S^{-1}\)-modules.
05

Show Isomorphism Between (VT^{-1}) S^{-1} and VS^{-1}

Construct a map \( \phi : (V T^{-1}) S^{-1} \to V S^{-1} \) by \( \phi(\frac{[v, t]}{s}) = \frac{v}{ts} \). Show \(\phi\) is well-defined, bijective, and respects module operations of \(R S^{-1}\), thus establishing an isomorphism.

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Key Concepts

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

Torsion-Free Modules
Torsion-free modules are an essential concept in module theory. Imagine a module, which is a generalization of vector spaces, over a ring \( R \). When we say this module is torsion-free, it means that multiplying a non-zero element from the ring by a non-zero element of the module cannot result in zero. In other words, if \( r \cdot v = 0 \) for some \( r \in R \) and \( v \in V \), then it must be that \( v = 0 \). This property is vital because it ensures that there are no 'hidden' zero-equivalents within the module when considered over the ring.
  • It preserves multiplicative injectivity.
  • Facilitates simpler operations and calculations.
  • Provides a foundation for further module constructions.
Understanding torsion-free modules sets the base for exploring more complex structures like localized modules.
Localization of Modules
Localization is a process that transforms a module \( V \) by focusing it around a particular set of elements, allowing for the division by those elements. This technique is particularly beneficial when dealing with rings and modules where division is not naturally possible.
To construct the localization of a module \( V \) with respect to a multiplicatively closed set \( T \) of a commutative domain \( R \), we create the module \( V T^{-1} \). Here, elements of \( V T^{-1} \) are fractions of the form \( \frac{v}{t} \), where \( v \in V \) and \( t \in T \).
This construction:
  • Allows the module to have elements akin to rational numbers, extending its 'divisibility' properties.
  • Simplifies scenarios by effectively "ignoring" complicating factors present in \( R \) not captured by \( T \).
  • Creates a bridge to further study and manipulations involving modules and their relationships with the ring elements.
By localizing modules, we gain powerful tools for algebraic manipulation and theory advancement.
Commutative Domains
A commutative domain, in the context of algebra, is a type of ring with two critical properties:
  • Commutativity: For any two elements \( a \) and \( b \), the product is independent of order: \( a \cdot b = b \cdot a \).
  • No zero divisors: If \( a \cdot b = 0 \), then either \( a = 0 \) or \( b = 0 \).
These properties mean that every element behaves similarly to numbers under multiplication, making algebraic manipulation more natural and intuitive.
Commutative domains serve as a natural setting for module theory because:
  • They allow the extension of many properties and solutions from numbers to more abstract structures.
  • The absence of zero divisors ensures more predictable behavior in algebraic computations.
  • They form the backbone for defining many meaningful algebraic constructs, including polynomial rings and more.
Delving into modules over a commutative domain opens doors to robust algebraic explorations and applications.
Multiplicatively Closed Sets
A multiplicatively closed set within a ring \( R \) is a set \( T \) where, if you take any two elements of the set and multiply them together, the result remains in the set.
In mathematical terms, this means for any elements \( t_1, t_2 \in T \), the product \( t_1 \cdot t_2 \in T \). Such sets play a crucial role in localization because they form the collection of 'denominators' that we use to extend modules and rings.
Key aspects of multiplicatively closed sets include:
  • Always include the element 1, because it's the multiplicative identity.
  • Are used to construct various algebraic structures, such as localized rings or modules, by enlarging perspectives on divisibility and elimination.
  • Help in proving significant propositions in mathematics, like the isomorphism mentioned earlier.
Using multiplicatively closed sets, mathematicians can extend the rings and modules, providing a broader scope for algebraic study and applications.

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Most popular questions from this chapter

Find an example of a nonzero finitely generated projective \(R\)-module that is not a progenerator. If \(R\) is a simple ring and \(I\) is a nonzero right ideal of \(R\), prove that \(I_{R}\) is a generator.

Let \(P\) be a prime ideal of the commutative integral domain \(R\). If \(R+P R_{P}=R_{P}\), prove that \(P\) is a maximal ideal of \(R\). In particular, conclude that Lemma \(18.5\) fails for nonmaximal ideals.

Suppose \(R\) is a right Noetherian ring and \(P\) is a finitely generated projective \(R\)-module. Prove that \(P\) is a progenerator if and only if \(p^{\infty} \cong R^{\infty}\)

Suppose \(P \cong \oplus \sum_{i=1}^{\infty} P_{i}\) is an infinite direct sum of countably generated projective \(R\)-modules \(P_{i}\) with \(T\left(P_{i}\right)=R .\) If \(R\) is right Noetherian, show that there exists an epimorphism \(\mu: P \rightarrow R\). Conclude from the Eilenberg Trick that \(P\) is free.
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