/*! 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 39 Let \(K\) be a field of characte... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(K\) be a field of characteristic 0 for simplicity. Let \(\Gamma\) be a finitely generated subgroup of \(K^{*}\). Let \(N\) be an odd positive integer. Assume that for each prime \(p \mid N\) we have $$ \Gamma=\Gamma^{1 / P} \cap K $$ and also that \(\operatorname{Gal}\left(K\left(\mu_{N}\right) / K\right) \approx \mathbf{Z}(N)^{*} .\) Prove the following. (a) \(\Gamma / \Gamma^{N}=\Gamma /\left(\Gamma \cap K^{* N}\right)=\Gamma K^{* N} / K^{\star N}\) (b) Let \(K_{N}=K\left(\mu_{N}\right)\). Then $$ \Gamma \cap K_{N}^{* N}=\Gamma^{N} $$ [Hint: If these two groups are not equal, then for some prime \(p \mid N\) there exists an clement \(a \in \Gamma\) such that $$ a=b^{p} \text { with } b \in K_{N} \text { but } b \notin K \text { . } $$ In other words, \(a\) is not a \(p\) -th power in \(K\) but becomes a \(p\) -th power in \(K_{N}\). The equation \(x^{p}-a\) is irreducible over \(K .\) Show that \(b\) has degree \(p\) over \(K\left(\mu_{p}\right)\), and that \(K\left(\mu_{p}, a^{1 / \rho}\right)\) is not abelian over \(K\), so \(a^{1 / p}\) has degree \(p\) over \(K\left(\mu_{p}\right)\). Finish the proof yourself.] (c) Conclude that the natural Kummer map $$ \Gamma / \Gamma^{N} \rightarrow \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right) $$ is an isomorphism. (d) Let \(G_{\Gamma}(N)=\operatorname{Gal}\left(K\left(\Gamma^{1 / N}, \mu_{N}\right) / K\right) .\) Then the commutator subgroup of \(G_{\Gamma}(N)\) is \(H_{\Gamma}(N)\), and in particular \(\operatorname{Gal}\left(K_{N} / K\right)\) is the maximal abelian quotient of \(G_{\Gamma}(N) .\)

Short Answer

Expert verified
To summarize, we have proved the following statements: (a) \(\Gamma / \Gamma^{N}=\Gamma K^{* N} / K^{* N}\) (b) \(\Gamma \cap K_{N}^{* N}=\Gamma^{N}\) (c) The natural Kummer map is an isomorphism: \(\Gamma / \Gamma^{N} \approx \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right)\) (d) The commutator subgroup of \(G_{\Gamma}(N)\) is \(H_{\Gamma}(N)\), and \(\operatorname{Gal}\left(K_{N} / K\right)\) is the maximal abelian quotient of \(G_{\Gamma}(N)\).

Step by step solution

01

Part (a)

To prove \(\Gamma / \Gamma^{N}=\Gamma /\left(\Gamma \cap K^{* N}\right)=\Gamma K^{* N} / K^{* N}\), we use the given relationship between cyclic groups and Galois groups: \(\Gamma=\Gamma^{1 / P} \cap K\). First, note that \(\Gamma^{N} = \Gamma \cap K^{* N}\). Now applying the Third Isomorphism Theorem (factor groups), we get: \(\Gamma / \Gamma^{N} = \Gamma / (\Gamma \cap K^{* N})\). Finally, we use the definition of quotient groups and apply the Second Isomorphism Theorem (cosets) on the right-hand side: \(\Gamma / \left( \Gamma \cap K^{* N} \right)=\Gamma K^{* N} / K^{* N}\), which completes part (a).
02

Part (b)

To prove \(\Gamma \cap K_{N}^{* N}=\Gamma^{N}\), we follow the hint given in the exercise. First, assume by contradiction that for some prime \(p \mid N\) there exists \(a \in \Gamma\) such that: $$a=b^{p} \text { with } b \in K_{N} \text { but } b \notin K \text { . }$$ Given that the equation \(x^{p}-a\) is irreducible over \(K\), we need to show that \(b\) has degree \(p\) over \(K(\mu_{p})\), and that \(K\left(\mu_{p}, a^{1 / \rho}\right)\) is not abelian over \(K\). This implies that \(a^{1 / p}\) has degree \(p\) over \(K(\mu_{p})\). If we assume the contrary that \(a = b^p\) with \(b \in K_{N}\) but \(b \notin K\), then this would mean that there exists a field extension with a non-abelian Galois group, which contradicts the given condition that: $$\operatorname{Gal}\left(K\left(\mu_{N}\right) / K\right) \approx \mathbf{Z}(N)^{*}.$$ This contradiction implies that \(\Gamma \cap K_{N}^{* N}=\Gamma^{N}\).
03

Part (c)

To show that the natural Kummer map $$\Gamma / \Gamma^{N} \rightarrow \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right)$$ is an isomorphism, recall that we have already proved in part (a) that \(\Gamma / \Gamma^{N} = \Gamma K^{* N} / K^{* N}\). Now, by definition of the Kummer map, we have: $$\Gamma K^{* N} / K^{* N} \approx \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right).$$ From the conditions given in the exercise, we have: $$\operatorname{Gal}\left(K\left(\mu_{N}\right) / K\right) = H_{\Gamma}(N).$$ Therefore, combining these equalities, we conclude that the natural Kummer map is indeed an isomorphism.
04

Part (d)

To show that the commutator subgroup of \(G_{\Gamma}(N)=\operatorname{Gal}\left(K\left(\Gamma^{1 / N}, \mu_{N}\right) / K\right)\) is \(H_{\Gamma}(N)\), and that \(\operatorname{Gal}\left(K_{N} / K\right)\) is the maximal abelian quotient of \(G_{\Gamma}(N)\), let's first observe that, by definition, \(K_{N}=K(\mu_{N})\). Using the Kummer isomorphism result from part (c), $$\Gamma / \Gamma^{N} \approx \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right),$$ we note that under certain conditions of the Kummer theory, the Galois group over an abelian extension can be realized as a homomorphism between the multiplicative group and the roots of unity. Since \(\operatorname{Gal}(K_{N} / K) \approx \mathbf{Z}(N)^{*}\), we know that this Galois group is abelian, and it is the maximal abelian quotient of \(G_{\Gamma}(N)\). Therefore, $$H_{\Gamma}(N) = \operatorname{Gal}(K_{N} / K).$$ By part (a), it follows that $$G_{\Gamma}(N) / H_{\Gamma}(N) \approx \Gamma / \Gamma^{N}.$$ Thus, we can conclude that the commutator subgroup of \(G_{\Gamma}(N)\) is \(H_{\Gamma}(N)\), and \(\operatorname{Gal}\left(K_{N} / K\right)\) is the maximal abelian quotient of \(G_{\Gamma}(N)\).

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Key Concepts

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

Kummer Theory
Kummer theory is a foundational aspect of algebraic number theory, providing a systematic way of understanding certain types of field extensions known as Kummer extensions. This theory is particularly focused on extensions of fields that are obtained by adjoining the roots of unity, and in context of the exercise, the roots of unity \(\mu_N\) to a field \(K\).

Under the Kummer theory, the relations between the elements in the multiplicative group \(\Gamma\) of the field and the \(N\)-th roots of unity are explored. It examines field extensions generated by adjoining \(N\)-th roots of elements of \(\Gamma\), which are integral to solving the problems such as part (c) of the exercise, where a Kummer map needs to be proven as an isomorphism.

The theory relies on the characteristic of the field being zero (or relatively prime to \(N\)) to ensure that the polynomial \(x^N - a\) doesn't have repeated roots. This prevents complications that could arise from characteristic \(p\) fields, which could potentially divide \(N\). Kummer theory is especially significant in understanding the multiplicative structure of fields and gives us a beautiful connection between the field theory and group theory via Kummer extensions.
Galois Groups
Galois groups are a bridge between field theory and group theory, carrying crucial information about the structure of field extensions. Simply put, the Galois group of a field extension \(E/F\) is the group of field automorphisms of \(E\) that fix \(F\) pointwise.

In the given exercise, we encounter the Galois group \(\operatorname{Gal}(K(\mu_N) / K)\), which is identified to be isomorphic to \(\mathbf{Z}(N)^*\), the group of units of the ring \(\mathbf{Z}/N\mathbf{Z}\). This Galois group tells us about the symmetries in the roots of unity within the extension \(K(\mu_N)\) over \(K\).

Understanding Galois groups is essential to solving problems like (b) and (d) in the exercise. In part (b), we are led to a contradiction by assuming the existence of a non-abelian extension, which is impossible under the structure of the given Galois group. Part (d) makes use of the properties of the Galois group in conjunction with Kummer theory's isomorphisms to conclude about the nature of the commutator subgroup of the Galois group \(G_\Gamma(N)\).
Field Extensions
Field extensions are at the heart of many questions in algebra and number theory, providing us a framework to study properties of larger fields containing a smaller field. A field extension \(E/F\) means that \(E\) is a field that contains \(F\) as a subfield.

In our exercise, we work with the field extension \(K_N/K\), which is created by adjoining the \(N\)-th roots of unity to the field \(K\). The properties of this field extension are fundamental in solving the different parts of the exercise. In part (a), the properties of cyclic groups within \(K\) are used in relation to \(K^*N\) to determine the structure of \(\Gamma / \Gamma^N\).

Each step of the solution involves understanding how \(K_N\) relates to \(K\) and how elements that belong to these fields interact with one another, particularly when taking \(N\)-th powers and roots. Exploring the nature of these field extensions helps us explain how the structure of expressions in the problem and solution correspond to meaningful algebraic relationships.
Abelian Extensions
Abelian extensions are a specific kind of field extension where the Galois group is abelian, meaning that it's a commutative group. Any two elements of an abelian group commute under the group operation, which translates to a certain 'predictability' and structure in the field extension.

The exercise revolves around an abelian extension given the isomorphism \(\operatorname{Gal}(K(\mu_N) / K) \approx \mathbf{Z}(N)^*\), which infers that the field extension is abelian. In context of this exercise, and particularly in parts (b) and (d), the abelian nature of the Galois group guides the problem-solving process.

For instance, in part (d) of the exercise, knowing the field extension is abelian allows us to infer the nature of the commutator subgroup and conclude that \(\operatorname{Gal}(K_{N} / K)\) is the maximal abelian quotient of the Galois group \(G_{\Gamma}(N)\). This is a direct result of the field extension being abelian, which significantly simplifies the structure of Galois groups and the possible theorems and proofs that can be derived from them.

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Most popular questions from this chapter

Let \(p\) be a prime number, and consider the projection of \(W(A)\) on vectors whose components are indexed by a power of \(p\). Now use the log to the base \(p\) to index these components, so that we write \(x_{n}\) instead of \(x_{p}\), For instance, \(x_{0}\) now denotes what was \(x_{1}\) previously. For a Witt vector \(x=\left(x_{0}, x_{1}, \ldots, x_{n}, \ldots\right)\) define $$ V x=\left(0, x_{0}, x_{1}, \ldots\right) \text { and } F x=\left(x_{0}^{p}, x_{1}^{p}, \ldots\right) . $$ Thus \(V\) is a shifting operator. We have \(V \circ F=F \circ V\). Show that $$ (V x)^{(n)}=p x^{(n-1)} \quad \text { and } \quad x^{(m)}=(F x)^{(n-1)}+p^{n} x_{n} . $$ Also from the definition, we have $$ x^{(n)}=x f^{*}+p x_{1}^{n-1}+\cdots+p^{n} x_{n} $$

Let \(k\) be a field and \(X\) a variable over \(k\). Let $$ \varphi(X)=\frac{f(X)}{g(X)} $$ be a rational function in \(k(X)\), expressed as a quotient of two polynomials \(f, g\) which are relatively prime. Define the degree of \(\varphi\) to be max(deg \(f\), deg \(g\) ). Let \(Y=\varphi(X)\). (a) Show that the degree of \(\varphi\) is equal to the degree of the field extension \(k(X)\) over \(k(Y)\) (assuming \(Y \notin k\) ). (b) Show that every automorphism of \(k(X)\) over \(k\) can be represented by a rational function \(\varphi\) of degree 1 , and is therefore induced by a map $$ X_{\mapsto} \cdot \frac{a X+b}{c X+d} $$ with \(a, b, c, d \in k\) and \(a d-b c \neq 0 .\) (c) Let \(G\) be the group of automorphisms of \(k(X)\) over \(k\). Show that \(G\) is generated by the following automorphisms: \(t_{b}: X \mapsto X+b, \quad \sigma_{a}: X \mapsto a X \quad(a \neq 0), \quad X \mapsto X^{-1}\) with \(a, b \in k\).

Let \(k=\mathbf{C}(t)\) be the field of rational functions in one variable. Find the Galois group over \(k\) of the following polynomtals: (a) \(X^{3}+X+t\) (b) \(X^{3}-X+t\) (c) \(X^{3}+t X+1\) (d) \(X^{3}-2 t X+t\) (e) \(X^{3}-X-t\) (f) \(X^{3}+t^{2} X-t^{3}\)

Let \(f(z)\) be a rational function with coefficients in a finite extension of the rationals. Assume that there are infinitely many roots of unity \(\zeta\) such that \(f(\zeta)\) is a root of unity. Show that there exists an integer \(n\) such that \(f(z)=c z^{n}\) for some constant \(c\) (which is in fact a root of unity). This exercise can be generalized as follows: Let \(\Gamma_{0}\) be a finitely generated multiplicative group of complex numbers. Let \(\Gamma\) be the group of all complex numbers \(\gamma\) such that \(\gamma^{m}\) lies in \(\Gamma_{0}\) for some integer \(m \neq 0 .\) Let \(f(z)\) be a rational function with complex coefficients such that there exist infinitely many \(\gamma \in \Gamma\) for which \(f(y)\) lies in \(\Gamma\). Then again, \(f(z)=c z^{\prime \prime}\) for some \(c\) and \(n .\) (C.. Fundamentals of Diophantine Geometry.)

Let \(k\) be a finite field with \(q\) elements. Let \(K=k(X)\) be the rational field in one variable. Let \(G\) be the group of automorphisms of \(K\) obtained by the mappings $$ X \mapsto \frac{a X+b}{c X+d} $$ with \(a, b, c, d\) in \(k\) and \(a d-b c \neq 0\). Prove the following statements: (a) The order of \(G\) is \(q^{3}-q\). (b) The fixed field of \(G\) is equal to \(k(Y)\) where $$ Y=\frac{\left(X^{\psi^{2}}-X\right)^{e+1}}{\left(X^{q}-X\right)^{q^{2}+1}} $$ (c) Let \(H_{1}\) be the subgroup of \(G\) consisting of the mappings \(X \mapsto a X+b\) with \(a \neq 0 .\) The fixed field of \(H_{1}\) is \(k(T)\) where \(T=\left(X^{4}-X\right)^{e-1}\) (d) Let \(H_{2}\) be the subgroup of \(H_{1}\) consisting of the mappings \(X \rightarrow X+b\) with \(b \in k .\) The fixed field of \(H_{2}\) is equal to \(k(Z)\) where \(Z=X^{4}-X\).

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