Problem 12
Question
Let \(A\) be a Noetherian ring and \(M\) a finite \(A\) -module. Let \(I\) be an ideal of \(A\) such that \(\operatorname{supp}(M) \subset \mathscr{I}(t) .\) Then \(I^{n} M=0\) for some \(n>0 .\)
Step-by-Step Solution
Verified Answer
Given a Noetherian ring \(A\) and a finite \(A\)-module \(M\) with \(\operatorname{supp}(M) \subseteq \mathscr{I}(t)\), we have shown that there exists a non-negative integer \(n\) such that \(I^nM=0\), using the properties of Noetherian rings and finite \(A\)-modules. Thus, \(I^nM=0\) for some \(n>0\).
1Step 1: Understand supp(M) and I(t)
The support of an A-module M, denoted by supp(M), is defined as the set of all prime ideals P in A such that the localization of M at P, denoted by M_P, is non-zero. In other words, it is the set of prime ideals where M does not vanish: \[supp(M) = \{P \in Spec(A) \mid M_P \neq 0\}.\]
I(t) is defined as the set of all prime ideals that contain I to the power of t: \[\mathscr{I}(t) = \{P \in Spec(A) \mid I^t \subseteq P\}.\]
Given that supp(M) ⊆ I(t), this means that all the prime ideals in which M doesn't vanish also contain I raised to the power t.
2Step 2: Use the properties of a Noetherian ring and a finite A-module
Since A is a Noetherian ring and M is a finite A-module, there exists a non-negative integer n such that annihilator of M, denoted Ann(M), contains I^n. In other words: \[I^n \subseteq Ann(M).\]
For any m ∈ M, since I^n ⊆ Ann(M), we have I^n * m = 0. Therefore, I^n * M = 0.
3Step 3: Conclude the proof
We have shown that there exists an n such that I^n * M = 0. Thus, for the given Noetherian ring A and finite A-module M with supp(M) ⊆ I(t), it is true that I^n * M = 0 for some n > 0.
Key Concepts
Support of a ModuleLocalization of ModulesAnnihilator of a Module
Support of a Module
The support of a module is a fundamental concept in the study of ring and module theory in abstract algebra. When dealing with a module M over a ring A, the support gives us a way to understand where the module is 'active'. Specifically, the support of M, denoted as
Understanding the support is vital when attempting to comprehend the behavior of a module with respect to certain substructures of the ring, like ideals. For instance, if an ideal
supp(M), consists of all the prime ideals P of A such that the localization of M at P, M_P, is not zero. This essentially means that the module has a 'presence' at these prime ideals. In practical terms, if you're working with an A-module and you take its support, you're listing out all the spots in the ring where the module has some non-zero elements after localization.Understanding the support is vital when attempting to comprehend the behavior of a module with respect to certain substructures of the ring, like ideals. For instance, if an ideal
I is such that supp(M) is contained within I(t), the particular ideal's influence extends over all the areas where the module is non-zero. It's like knowing which regions a certain policy has an effect on when studying governmental policies.Localization of Modules
Localization is a process akin to zooming into a specific portion of a module in relation to a chosen prime ideal. More technically, for a module M over a ring A and a prime ideal P, the localization of M at P, denoted as M_P, is constructed to reflect the behavior of M 'near' P. The local perspective obtained through localization is useful for studying modules locally rather than globally.
In the context of localization, we use the idea of inverting elements not in P to better understand how M behaves when those elements are allowed to be 'units' or invertible. The process of localizing a module is paramount for understanding modules and rings because it reveals intricacies that are not visible at the 'global' level. For instance, properties that hold true for all localizations of a module might lead to global conclusions, making localization both a powerful and an indispensable tool in algebraic geometry and commutative algebra.
In the context of localization, we use the idea of inverting elements not in P to better understand how M behaves when those elements are allowed to be 'units' or invertible. The process of localizing a module is paramount for understanding modules and rings because it reveals intricacies that are not visible at the 'global' level. For instance, properties that hold true for all localizations of a module might lead to global conclusions, making localization both a powerful and an indispensable tool in algebraic geometry and commutative algebra.
Annihilator of a Module
In the context of modules over a ring, the annihilator offers a snapshot of the elements in the ring that 'annihilate' the module. The annihilator Annn(M) of a module M over a ring A consists of all elements a in A such that a*m = 0 for every element m in M. This means that the annihilator is the set of 'kill switches' for M within A. In essence, any element of the ring that's part of the annihilator wipes out the entire module by multiplying with it.
The concept of an annihilator is significant because it gives insights into how a module interacts with its ring. If every element of a certain ideal annihilates the module, it points to a strong relationship between the module's structure and that ideal. In the context of a Noetherian ring, as in the exercise provided, the existence of an annihilator containing a power of an ideal
The concept of an annihilator is significant because it gives insights into how a module interacts with its ring. If every element of a certain ideal annihilates the module, it points to a strong relationship between the module's structure and that ideal. In the context of a Noetherian ring, as in the exercise provided, the existence of an annihilator containing a power of an ideal
I^n tells us that the module M is in some sense 'controlled' by the ideal I, as the ideal's elements can annihilate the module's elements. This property gives us a concrete criterion to work with when examining when a module might actually be 'zero' after being multiplied by powers of an ideal.Other exercises in this chapter
Problem 10
Let \(A, B\) be local nings with maximal ideals \(m_{A}, m_{s}\), respectively. Let \(f: A \rightarrow B\) be a homomorphism. We say that \(f\) is local if \(f^
View solution Problem 11
Let \(A\) be a commutative ring and \(M\) an \(A\) -module. Define the support of \(M\) by $$ \operatorname{supp}(M)=\left\\{p \in \operatorname{spec}(A): M_{w}
View solution Problem 13
Let \(A\) be any commutative ring, and \(M, N\) modules over \(A\). If \(M\) is finitely presented, and \(S\) is a multiplicative subset of \(A\), show that $$
View solution Problem 18
Let \(A\) be a Noetherian local ring. Let \(K\) be a finite complex, as follows: $$ 0 \rightarrow K^{0} \rightarrow \cdots \rightarrow K^{n} \rightarrow 0 $$ su
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