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Chapter 14: Problem 3

If \(A\) or \(B\) is flat, prove that \(\operatorname{Tor}_{n}(A, B)=0\) for all \(n \geq 1\)

Short Answer

Expert verified
If either module \( A \) or \( B \) is flat, then \( \operatorname{Tor}_n(A, B) = 0 \) for all \( n \geq 1 \).

Step by step solution

01

Understanding Tor functors

The Tor functor, denoted by \( \operatorname{Tor}_n(A, B) \), measures the 'non-flatness' of two modules \( A \) and \( B \) over a commutative ring \( R \). Specifically, if \( A \) or \( B \) is flat, \( \operatorname{Tor}_n(A, B) \) is zero for all \( n \geq 1 \). This concept is an extension of the tensor product of modules.
02

Definition of Flatness

A module \( M \) over a ring \( R \) is flat if the functor \( - \otimes_R M \) is exact, meaning it preserves the exactness of any short exact sequence of \( R \)-modules. In simpler terms, this means that any sequence of \( R \)-modules that is exact remains exact after tensoring with \( M \).
03

Applying the definition

Consider if \( A \) is a flat module. Then for any short exact sequence of \( R \)-modules \( 0 \to X \to Y \to Z \to 0 \), the tensor product sequence \( 0 \to X \otimes_R A \to Y \otimes_R A \to Z \otimes_R A \to 0 \) is also exact due to the flatness of \( A \). This implies that the derived functors \( \operatorname{Tor}_n(A, B) \) for \( n \geq 1 \) are zero.
04

Considering flatness of B

Similarly, if \( B \) is flat instead of \( A \), then for any short exact sequence \( 0 \to X \to Y \to Z \to 0 \) of \( R \)-modules, the sequence \( 0 \to X \otimes_R B \to Y \otimes_R B \to Z \otimes_R B \to 0 \) is exact, thereby making \( \operatorname{Tor}_n(A, B) = 0 \) for all \( n \geq 1 \).
05

Conclusion with generality

Regardless of which module is flat (either \( A \) or \( B \)), the property of exactness holds and the higher Tor functors \( \operatorname{Tor}_n(A, B) \) vanish for all \( n \geq 1 \). Either module flatness provides the same result due to symmetric property of tensor product in the argument of Tor.

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Key Concepts

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

Tor functor
The Tor functor, denoted as \( \operatorname{Tor}_n(A, B) \), plays an essential role in algebra, particularly in module theory and homological algebra. This functor quantifies the failure of flatness when dealing with modules over a ring. It essentially measures how far a tensor product of two modules \( A \) and \( B \) deviates from being exact.
  • If either module \( A \) or \( B \) is flat, \( \operatorname{Tor}_n(A, B) \) becomes trivial for all \( n \geq 1 \).
  • The functor is related to the process of deriving tensor products, which yields a series of abelian groups that are used to understand the interaction of modules.
Understanding Tor functors helps in gaining insight into the nature of module extensions and the behavior of modules, both in theory and in practical applications like algebraic topology.
Flat module
Flatness is a foundational concept in module theory. A module \( M \) over a ring \( R \) is described as flat if the tensoring functor \( - \otimes_R M \) preserves exactness of sequences.
What does this mean practically? Consider a short exact sequence of \( R \)-modules:
  • \( 0 \to X \to Y \to Z \to 0 \).
When \( M \) is flat, tensoring this sequence with \( M \) results in another exact sequence:
  • \( 0 \to X \otimes_R M \to Y \otimes_R M \to Z \otimes_R M \to 0 \).
Flat modules are crucial because they simplify computations and ensure consistency in various algebraic frameworks. They are widely used in algebraic geometry and commutative algebra.
Tensor product
The tensor product is a pivotal construction in algebra, allowing the combination of modules over a ring in a manner that generalizes the outer product for vector spaces. Given two \( R \)-modules \( A \) and \( B \), their tensor product \( A \otimes_R B \) forms a new module.
  • The construction adheres to bilinearity, meaning it respects the operations of module addition and scalar multiplication.
  • Because of its properties, the tensor product is a versatile tool for forming more complex structures from simpler ones.
The usefulness of the tensor product spans various mathematical disciplines such as multilinear algebra, geometry, and even physics, where it is used to construct objects like vector spaces of tensors.
Exact sequence
In the realm of algebra, exact sequences help describe the structure-preserving maps between algebraic objects like groups or modules. A sequence of modules and homomorphisms
  • \( \cdots \to A \xrightarrow{f} B \xrightarrow{g} C \to \cdots \)
is exact at a module \( B \) if the image of the homomorphism \( f \) equals the kernel of \( g \).
Exact sequences succinctly capture key features of module transformations and relationships:
  • Short exact sequences contain only three nontrivial modules and represent fundamental algebraic concepts like extensions and morphisms.
  • They are applied in a broad range of areas including homological algebra, topology, and complex analysis to understand the properties and interactions of different algebraic structures.
Exact sequences are a powerful tool for simplifying complex relationships into more manageable forms.

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Most popular questions from this chapter

Prove that \(\operatorname{Tor}_{0}(A, B) \cong A \otimes B\) and that $$ \operatorname{Tor}_{n}(A, B \oplus C) \cong \operatorname{Tor}_{n}(A, B) \oplus \operatorname{Tor}_{n}(A, C) $$

Let \(R \subseteq S\) be rings with \(_{R} S\) flat. Let \(A\) be a right \(R\)-module and let \(B\) be a left \(S\)-module. Note that \(B\) is a left \(R\)-module by restriction. Prove that $$ \operatorname{Tor}_{n}^{S}\left(A \otimes_{R} S, B\right) \cong \operatorname{Tor}_{n}^{R}(A,{R} B) $$ To this end, start with a projective resolution for \(A\) and tensor it with the flat module \(_{R} S\) to obtain a projective resolution for \(A \otimes_{R} S\). Then use these sequences to compute the appropriate Tor groups. Now let \(R\) be a right Noetherian ring and use all the notation of this chapter. In particular, let \(S=R[x ; \sigma]\) be graded with \(S_{n}=R x^{n}\) and let \(T=S[y]\) be graded by total degree. Furthermore, view \(R\) as the left \(S\)-module \(R \cong S / S x\) and as the left \(T\)-module \(R \cong T / I=\) \(T /(T x+T y)\)

If \(\operatorname{pd} A=k\) or \(\operatorname{pd} B=k\), show that \(\operatorname{Tor}_{n}(A, B)=0\) for all \(n \geq k+1\).
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