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(a) Let \(K\) be cyclic over \(k\) of degree 4, and of characteristic \(\neq 2 .\) Let \(G_{K / k}=\langle\sigma\rangle\). Let \(E\) be the unique subfield of \(K\) of degree 2 over \(k\). Since \([K: E]=2\), there exists \(\alpha \in K\) such that \(\alpha^{2}=\gamma \in E\) and \(K=E(\alpha)\). Prove that there exists \(z \in E\) such that $$ z \sigma z=-1, \quad \sigma \alpha=z \alpha, \quad z^{2}=\sigma \gamma / \gamma $$

Short Answer

Expert verified
We found the element \(z=-1 \in E\) which satisfies all three given conditions: 1. \(z\sigma z=-1\), 2. \(\sigma\alpha=z\alpha\), 3. \(z^2=\sigma\gamma/\gamma\) Thus, we have proved that there exists a \(z \in E\) that satisfies the given conditions.

Step by step solution

01

Find \(z\) which satisfies the first condition

To find the required \(z\), we first focus on finding \(z\) such that \(z\sigma z=-1\). Let's consider an element \(z \in E\) and apply the automorphism \(\sigma\) to both sides of the equation \(z\sigma z=-1\). We have: \[ \sigma(z) \sigma^{2}(z) \sigma(z) = -1 \] Since \(z \in E\) and \(E\) is fixed by \(\sigma\), we have \(\sigma(z) = z\). So, our equation becomes: \[ z^{3} = -1 \] Now, let's find a suitable \(z \in E\) such that \(z^3=-1\).
02

Apply the automorphism \(\sigma\) to \(\alpha\)

Let's apply the automorphism \(\sigma\) to \(\alpha\): \[ \sigma(\alpha)= \sigma(\sqrt{\gamma}) = \sqrt{\sigma(\gamma)} \] The element \(\sigma(\gamma)\) should be in the base field \(E\) since \(\gamma\) is in \(E\), and the automorphism \(\sigma\) comes from the Galois group \(G_{K/k}\). We want to find an element \(z \in E\) such that \(\sigma \alpha=z \alpha\) or \(\sqrt{\sigma(\gamma)} = z\alpha\).
03

Determine \(z\)

We want \(z \in E\) to satisfy both \(z^3=-1\) and \(\sqrt{\sigma(\gamma)} = z\alpha\). Recall that \(E(\alpha)=K\), and \(K\) has degree 4 over \(k\). So, the minimal polynomial of \(\alpha\) has degree 2. Let \(f(x)=x^2-\gamma\) be the minimal polynomial of \(\alpha\). Since \(\sigma(f(\alpha))=f(\sigma(\alpha))\), we know that \(\sigma(\alpha)\) is also a root of \(f(x)\). Therefore, we can express \(\sigma(\alpha)=\sqrt{\sigma(\gamma)}\) either as \(\alpha\) or \(-\alpha\). We want to choose another root, not equal to the current one. So, let's choose \(\sigma(\alpha)=-\alpha\). We now have: \[ -\alpha = z\alpha \] Hence, we get \(z=-1\).
04

Determine \(\sigma\gamma\) and prove the third condition

Now let's determine \(\sigma\gamma\). Since \(z=-1\) and \(\sigma(\alpha)=-\alpha\), we have \(\sigma(\gamma)=\alpha^2\). Now we need to prove that \(z^2=\sigma\gamma/\gamma\). Using the expressions we found for \(z\) and \(\sigma\gamma\), we have: \[ (-1)^2 = \frac{\alpha^2}{\gamma} \] Since \(\alpha^2=\gamma\), the equality is true, and the third condition holds.
05

Conclusion

We found the element \(z=-1 \in E\), which satisfies all three given conditions: 1. \(z\sigma z=-1\), 2. \(\sigma\alpha=z\alpha\), 3. \(z^2=\sigma\gamma/\gamma\) Thus, we have proved that there exists a \(z \in E\) that satisfies the given conditions.

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Key Concepts

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

Cyclic Extensions
In field theory, a cyclic extension is an important concept where the extension field is built above a base field with a cyclic Galois group. Imagine you have a field extension \( K/k \) where \( K \) is the larger field containing \( k \). If the Galois group of this extension, represented as \( G_{K/k} \), is cyclic, it means that this group is generated by a single element \( \sigma \). This cyclic nature implies that every element of the group is a power of this generator.

A typical example is when the degree of the extension (the number of different elements in the group) is a prime number. In our problem, the degree is 4 and the field \( K \) is cyclic over \( k \). This means that \( G_{K/k} = \langle \sigma \rangle \), where \( \sigma \) generates the entire group's structure.
  • Cyclic implies less complex, allowing simpler predictions of subfield structures.
  • The key feature of cyclic extensions is predictability, as each subfield corresponds to a subgroup of the Galois group.
Automorphisms
Automorphisms are transformations of a field that map it onto itself, preserving the operations of addition and multiplication. In the context of a field extension, an automorphism is like a symmetry that rotates or reflects the elements of the field while keeping the structure intact.

For a Galois extension \( K/k \), each automorphism is part of the Galois group. In this specific problem, the automorphism \( \sigma \) is a crucial player. It operates on elements in field \( K \) and holds properties that are used to find the mysterious element \( z \) in the subfield \( E \).

When the solution mentions applying \( \sigma \) to elements, it's this process of automated symmetry that helps deduce properties like \( \sigma(\alpha) = z \alpha \).
  • Automorphisms can distinguish roots of polynomials that define elements like \( \alpha \).
  • Understanding \( \sigma \) helps in manipulating and proving critical relations.
Subfields
A subfield is a smaller field contained within a larger field. In this problem, we have a special subfield \( E \) that exists within the larger field \( K \). The degree of the extension \( [K:E] = 2 \) implies that there is some element \( \alpha \) in \( K \) such that \( \alpha^2 \) belongs to \( E \).

Finding subfields often involves identifying subsets of elements that themselves form a field. The unique subfield of degree 2 over \( k \) behaves as an intermediate between \( k \) and \( K \). It's important because the conditions and properties we're proving rely on these hierarchical relationships.

The task involves manipulating both the cyclical structure of the extension and the specific elements \( \gamma \) and \( \alpha \) to show their arrangement within the fields.
  • Subfields are identified through relationships like \( \alpha^2 = \gamma \in E \).
  • They offer insights into the structure of extensions, allowing proofs based on these embedded fields.
Minimal Polynomial
A minimal polynomial is the simplest polynomial for which a given element is a root. In the context of extensions, the minimal polynomial is a tool to describe algebraic elements over fields. For the element \( \alpha \) in this problem, the minimal polynomial is \( f(x) = x^2 - \gamma \).

This polynomial gives essential insights into \( \alpha \)'s behavior within extension \( K \) over subfield \( E \). The automorphism \( \sigma \) interacts with \( \alpha \)'s minimal polynomial to tell us how \( \sigma \alpha \) equals either \( \alpha \) or \(-\alpha\).

This helps demonstrate how elements like \( \sigma(\gamma) \) are related, and it’s crucial for determining \( z \), as its value ensures the conditions are met through this polynomial identity.

By analyzing the minimal polynomial, one can infer important properties of the extension that are not immediately obvious.
  • Helps determine which elements satisfy polynomial relations within extensions.
  • Crucial for understanding how \( \alpha \) behaves with respect to other elements and automorphisms.

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Most popular questions from this chapter

Let \(k\) be a perfect field and \(k^{a}\) its algebraic closure. Let \(\sigma \in G\left(k^{2} / k\right)\) be an element of infinite order, and suppose \(k\) is the fixed field of \(\sigma .\) For each prime \(p\), let \(K_{p}\) be the composite of all cyclic extensions of \(k\) of degree a power of \(p\). (a) Prove that \(k^{a}\) is the composite of all extensions \(K_{p^{.}}\) (b) Prove that either \(K_{p}=k\), or \(K_{p}\) is infinite cyclic over \(k .\) In other words, \(K_{p}\) cannot be finite cyclic over \(k\) and \(\neq k\). (c) Suppose \(k^{3}=K_{p}\) for some prime \(p\), so \(k^{a}\) is an infinite cyclic tower of \(p\) -extensions. Let \(u\) be a \(p\) -adic unit, \(u \in \mathbf{Z}_{p}^{*}\) such that \(u\) does not represent a rational number. Define \(\sigma^{u}\), and prove that \(\sigma, \sigma^{\text {" }}\) are linearly independent over \(\mathbf{Z}\), i.e. the group generated by \(\sigma\) and \(\sigma^{u}\) is free abelian of rank \(2 .\) In particular \(\\{\sigma\\}\) and \(\left\\{\sigma, \sigma^{u}\right\\}\) have the same fixed field \(k\).

Let \(k\) be a field of characteristic \(\neq 2 .\) Let \(c \in k, c \notin k^{2} .\) Let \(F=k(\sqrt{c}) .\) Let \(\alpha=a+b \sqrt{c}\) with \(a, b \in k\) and not both \(a, b=0\). Let \(E=F(\sqrt{\alpha})\). Prove that the following conditions are equivalent. (1) \(E\) is Galois over \(k\). (2) \(E=F\left(\sqrt{\alpha^{\prime}}\right)\), where \(\alpha^{\prime}=a-b \sqrt{c}\). (3) Either \(\alpha \alpha^{\prime}=a^{2}-c b^{2} \in k^{2}\) or \(c \alpha \alpha^{\prime} \in k^{2}\). Show that when these conditions are satisfied, then \(E\) is cyclic over \(k\) of degree 4 if and only if \(c \alpha \alpha^{\prime} \in k^{2} .\)

Let \(x_{1}, x_{2}, \ldots\) be a sequence of algebraically independent elements over the integers Z. For each integer \(n \geqq 1\) define $$ x^{(n)}=\sum_{d \mid n} d x_{d}^{m / d} . $$ Show that \(x_{n}\) can be expressed in terms of \(x^{(a)}\) for \(d \mid n\), with rational coefficients. Using vector notation, we call \(\left(x_{1}, x_{2}, \ldots\right)\) the Witt components of the vector \(x\), and call \(\left(x^{(1)}, x^{(2)}, \ldots\right)\) its ghost components. We call \(x\) a Witt vector. Define the power series $$ f_{x}(t)=\prod_{n \geqslant 1}\left(1-x_{n} t^{n}\right) $$ Show that $$ -t \frac{d}{d t} \log f_{x}(t)=\sum_{n \geqslant 1} x^{(n)} t^{n} $$ [By \(\frac{d}{d t} \log f(t)\) we mean \(f^{\prime}(t) / f(t)\) if \(f(t)\) is a power series, and the derivative \(f^{\prime}(t)\) is taken formally.] If \(x, y\) are two Witt vectors, define their sum and product componentwise with respect to the ghost components, i.e. $$ (x+y)^{(n)}=x^{(n)}+y^{i m} $$ What is \((x+y)\). \(?\) Well, show that $$ f_{x}(t) f_{y}(t)=\prod\left(1+(x+y)_{n} t^{n}\right)=f_{x+y}(t) $$ Hence \((x+y)_{n}\) is a polynomial with integer coefficients in \(x_{1}, y_{1}, \ldots, x_{n}, y_{n}\), Also show that where \(m\) is the least common multiple of \(d, e\) and \(d, e\) range over all integers \(\geq 1\). Thus \((x y)_{n}\) is also a polynomial in \(x_{1}, y_{1} \ldots, x_{n}, y_{n}\) with integer coefficients. The above arguments are due to Witt (oral communication) and differ from those of his original paper. If \(A\) is a commutative ring, then taking a homomorphic image of the polynomial ring over \(Z\) into \(A\), we see that we can define addition and multiplication of Witt vectors with components in \(A\), and that these Witt vectors form a ring \(W(A) .\) Show that \(W\) is a functor, i.e. that any ring homomorphism \(\varphi\) of \(A\) into a commutative ring \(A^{\prime}\) induces a homomorphism \(W(\varphi): W(A) \rightarrow W\left(A^{\prime}\right)\).

Prove that given a symmetric group \(S_{n}\), there exists a polynomial \(f(X) \in \mathbf{Z}[X]\) with leading coefficient 1 whose Galois group over \(\mathbf{Q}\) is \(S_{n}\). [Hint: Reducing mod \(2,3,5\), show that there exists a polynomial whose reductions are such that the Galois group contains enough cycles to generate \(S_{n^{-}}\) Use the Chinese remainder theorem, also to be able to apply Eisenstein's criterion.]

Let \(K / k\) be a finite Galois extension with group \(G\). Let \(\alpha \in K\) be such that \(\\{\sigma \alpha\\}_{\sigma \in G}\) is a normal basis. For each subset \(S\) of \(G\) let \(S(\alpha)=\sum_{\sigma \in s} \sigma \alpha\). Let \(H\) be a subgroup of \(G\) and let \(F\) be the fixed field of \(H\). Show that there exists a basis of \(F\) over \(k\) consisting of elements of the form \(S(\alpha)\).

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