/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 (Artin-Tate). Let \(G\) be a fin... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 6

(Artin-Tate). Let \(G\) be a finite group operating on a finite set \(S .\) For \(w \in S\), denote 1\. \(w\) by \([w]\), so that we have the direct sum $$ \left.\mathbf{Z}(S)=\sum_{w \in S} \mathbf{Z} \mid w\right] $$ Define an action of \(G\) on \(\mathbf{Z}(S)\) by defining \(\sigma[w]=[\sigma w]\) (for \(w \in S)\), and extending \(\sigma\) to \(\mathbf{Z}(S)\) by linearity. Let \(M\) be a subgroup of \(\mathbf{Z}(S)\) of rank # \([S]\). Show that \(M\) has a \(Z\) -basis \(\left\\{y_{w}\right\\}_{w e a}\) such that \(\sigma y_{w}=y_{\sigma w}\) for all \(w \in S .\) (Cf. my Algebraic Number Theory, Chapter IX, \$4. Theorem 1.)

Short Answer

Expert verified
We have constructed a subgroup \(M\) of \(\mathbf{Z}(S)\) with rank \(|S|\) and a \(\mathbb{Z}\)-basis \(\{y_w\}_{w \in S}\) such that \(\sigma y_w = y_{\sigma w}\) holds for all \(w \in S\) and \(\sigma \in G\). This basis is formed by defining \(y_w = [w] - [\sigma w]\) for all \(w \in S\) and extending it linearly. The elements of this basis are linearly independent, satisfying the required condition and completing the proof.

Step by step solution

01

Recap of Definitions and Notations

Let G be a finite group operating on a finite set S. For this exercise, we have to work with the direct sum \(\mathbf{Z}(S)=\sum_{w \in S} \mathbf{Z} \mid w]\). The elements of this direct sum have the form \(\sum a_w[w]\) for some integers \(a_w\). Define an action of \(G\) on \(\mathbf{Z}(S)\) by defining \(\sigma[w] = [\sigma w]\) (for \(w \in S\)), and extending \(\sigma\) to \(\mathbf{Z}(S)\) by linearity.
02

Construct the Basis

To construct a basis that satisfies the given properties, let's consider a basis for \(M\) with elements \(y_w\) for every \(w \in S\). Define \(y_w = [w] - [\sigma w]\) for all \(w \in S\), where \(\sigma \in G\). Notice that for any \(\sigma \in G\), \(\sigma y_w = \sigma([w] - [\sigma w]) = [\sigma w] - [\sigma^2 w] = y_{\sigma w}\), which shows that the \(\sigma y_w = y_{\sigma w}\) condition is satisfied.
03

Show that the constructed basis has rank S

We need to show that the constructed \(y_w\) elements form a basis for \(M\) of rank \(|S|\). Observe that this is the case since for any \(w \in S\), the elements \(y_w\) are in the submodule\(_{蟽\in G} y_w\), and by definition, \(y_w \ne 0\) for any \(w \in S\). Thus, since these elements are linearly independent, they form a basis for \(M\) of rank \(|S|\).
04

Conclusion

We have constructed a subgroup M of rank \(|S|\), with a \(\mathbb{Z}\)-basis \(\{y_{w}\}_{w \in S}\) such that the relation \(\sigma y_w = y_{\sigma w}\) holds for all \(w \in S\). This completes the proof and solves the exercise.

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Key Concepts

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

Finite groups
A finite group is a set endowed with an operation that satisfies the four fundamental properties of group theory: closure, associativity, the existence of an identity element, and the existence of inverse elements, with the crucial feature that the set contains a finite number of elements.
Understanding finite groups is important because they are foundational to many areas of mathematics and applied science. They model systems with symmetry and can be used to explore permutations, among other applications.
  • **Closure**: For any two elements \(a\) and \(b\) in the group \(G\), the result of the operation, say \(a \cdot b\), is also in \(G\).
  • **Associativity**: For any three elements \(a\), \(b\), and \(c\) in \(G\), the equation \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) holds.
  • **Identity Element**: There exists an element \(e\) in \(G\) such that for every element \(a\) in \(G\), the equation \(e \cdot a = a \cdot e = a\) holds.
  • **Inverse Element**: For each element \(a\) in \(G\), there exists an element \(b\) in \(G\) such that \(a \cdot b = b \cdot a = e\), where \(e\) is the identity element.
Sets with group actions
When a group acts on a set, it provides a way to symmetrically transform the elements of the set. The action of a group \(G\) on a set \(S\) is a rule that combines each element \(g\) of the group with each element \(s\) of the set to produce another element of the set \(g.s\). This interaction must satisfy two key properties:
It helps in breaking down complicated mathematical structures into simpler components, which can then be studied using group theory.
  • **Identity action**: The identity element \(e\) of \(G\) satisfies \(e.s = s\) for all \(s\) in \(S\).
  • **Compatibility**: For any two elements \(g, h\) in \(G\) and any element \(s\) in \(S\), the equation \((gh).s = g.(h.s)\) should hold true.
This action establishes a connection between the structure of the group and the symmetries within the set.
Direct sum and modules
The direct sum is a way to combine multiple algebraic structures into a single, larger structure. For sets and vector spaces, the direct sum creates an entity that holds elements made by taking pieces from each of the individual structures. In the case of modules, which are similar to vector spaces but more generalized, direct sums can manage elements from multiple modules at once.
This concept is crucial in comprehending the overall behavior of algebraic structures by observing the interplay between their individual components.In the context of this exercise, we consider the direct sum \(\mathbf{Z}(S)\) where each piece corresponds to an integer coefficient multiplied by an element of \(S\). Modules generalize the idea of linear algebra to sometimes more abstract algebraic systems. They allow algebraic operations similar to vector spaces over a ring, instead of just fields as vector spaces require.
Basis in linear algebra
In linear algebra, a basis is a set of vectors that are linearly independent and span a vector space. This means any vector in the space can be expressed as a unique combination of the vectors in the basis. The concept of a basis is critical as it simplifies complex vector spaces into comprehensible components, making operations like vector addition and scalar multiplication more understandable.
  • **Linear Independence**: Vectors in the basis are linearly independent, meaning no vector in the set is a linear combination of the others.
  • **Spanning the Space**: Any vector in the space can be represented as a linear combination of the basis vectors.
In this exercise, we constructed a \( \mathbb{Z} \)-basis for the submodule \(M\). This basis is tailored so that the group action respects the basis, showcasing a very elegant structured approach to understanding the symmetry behavior of vectors.

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Most popular questions from this chapter

(a) Let \(M\) be a projective finite module over the Dedekind ring o. Show that there exist free modules \(F\) and \(F^{\prime}\) such that \(F \supset M \supset F^{\prime}\), and \(F, F^{\prime}\) have the same rank, which is called the rank of \(M\). (b) Prove that there exists a basis \(\left\\{e_{1}, \ldots, e_{n}\right\\}\) of \(F\) and ideals \(a_{1}, \ldots, a_{n}\) such that \(M=a_{1} e_{1}+\cdots+a_{n} e_{n}\), or in other words, \(M \approx \oplus a_{i}\) (c) Prove that \(M \approx 0^{n-1} \oplus\) a for some ideal \(a\), and that the association \(M \mapsto a\) induces an isomorphism of \(K_{0}(0)\) with the group of ideal classes Pic(o). (The group \(K_{0}(0)\) is the group of equivalence classes of projective modules defined at the end of \(\$ 4\).)

(a) Let \(A\) be a commutative ring and let \(M\) be an \(A\) -module. Let \(S\) be a multiplicative subset of \(A\). Define \(S^{-1} M\) in a manner analogous to the one we used to define \(S^{-1} A\), and show that \(S^{-1} M\) is an \(S^{-1} A\) -module. (b) If \(0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{*} \rightarrow 0\) is an exact sequence, show that the sequence \(0 \rightarrow S^{-1} M^{\prime} \rightarrow S^{-1} M \rightarrow S^{-1} M^{\prime \prime} \rightarrow 0\) is exact.

Consider the multiplicative group \(Q^{*}\) of non-zero rational numbers. For a non-zero rational number \(x=a / b\) with \(a, b \in \mathbf{Z}\) and \((a, b)=1\), define the height $$ h(x)=\log \max (|a|,|b|) $$ (a) Show that \(h\) defines a seminorm on \(Q^{*}\), whose kernel consists of \(\pm 1\) (the torsion group). (b) Let \(M_{1}\) be a finitely generated subgroup of \(\mathbf{Q}^{*}\), generated by rational numbers \(x_{1}, \ldots, x_{m}\). Let \(M\) be the subgroup of \(Q^{*}\) consisting of those elements \(x\) such that \(x^{t} \in M_{1}\) for some positive integer \(s .\) Show that \(M\) is finitely generated, and using Exercise 7 , find a bound for the seminorm of a set of generators of \(M\) in terms of the seminorms of \(x_{1} \ldots \ldots x_{m}\). Note. The above two exercises are applied in questions of diophantine approximation. See my Diophantine approximation on toruses, Am.J. Math. \(86(1964)\), pp. \(521-533\), and the discussion and references I give in Ency. clopedia of Mathematical Sciences, Number Theory III, Springer Verlag, 1991, Pp. \(240-243\).

(a) Let \(n\) range over the positive integers and let \(p\) be a prime number. Show that the abelian groups \(A_{n}=\mathbf{Z} / p^{n} \mathbf{Z}\) form a projective system under the canonical homomorphism if \(n \geqq m .\) Let \(\mathbf{Z}_{p}\) be its inverse limit. Show that \(\mathbf{Z}_{p}\) maps surjectively on each \(\mathbf{Z} / p^{n} \mathbf{Z} ;\) that \(\mathbf{Z}_{p}\) has no divisors of 0, and has a unique maximal ideal generated by \(p\). Show that \(\mathbf{Z}_{p}\) is factorial, with only one primc, namely \(p\) itself. (b) Next consider all ideals of \(\mathbf{Z}\) as forming a directed system, by divisibility. Prove that $$ \frac{\varliminf_{(\boldsymbol{a})}} \mathbf{Z} /(a)=\prod_{p} \mathbf{z}_{p} $$ where the limit is taken over all ideals \((a)\), and the product is taken over all primes \(p\)

Let \(E\) be a module over a ring. Let \(\left\\{M_{i}\right\\}\) be a directed family of modules. If \(E\) is finitely generated, show that the natural homomorphism is injective. If \(E\) is finitely presented, show that this homomorphism is an isomorphism. Hint: First prove the statements when \(E\) is free with finite basis. Then, say \(E\) is finitely presented by an exact sequence \(F_{1} \rightarrow F_{0} \rightarrow E \rightarrow 0\). Consider the diagram:
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