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Chapter 7: Problem 7

Let \(F\) be a field containing \(i=\sqrt{-1}\). Let \(K\) be a splitting field of the polynomial \(t^{4}-a\) with \(a \in F\). Show that the Galois group of \(K\) over \(F\) is a subgroup of a cyclic group of order 4 . If \(t^{4}-a\) is irreducible over \(F\), show that this Galois group is cyclic of order 4 . If \(\alpha\) is a root of \(t^{4}-a\), express all the other roots in terms of \(\alpha\) and \(i\)

Short Answer

Expert verified
The Galois group of the splitting field \(K\) over field \(F\) that contains \(i=\sqrt{-1}\) is a subgroup of a cyclic group of order 4. If the polynomial \(t^4 - a\) is irreducible, the Galois group is cyclic of order 4. The roots of the polynomial are given by \(\alpha, i\alpha, -\alpha,\) and \(-i\alpha\).

Step by step solution

01

Identify the roots of the polynomial

To find the roots of the polynomial \(t^4 - a\), let \(r\) be a root of \(t^4 - a\). Then by definition, we have \(r^4 - a = 0\), so \(r^4 = a\). Since \(F\) contains \(i = \sqrt{-1}\), we can rewrite \(r^4=a\) as \(r^4 = (a^{\frac{1}{4}})^4\), meaning \(r\) should be one of the 4 distinct roots of a, denoted by \(r_k = a^{\frac{1}{4}}e^{\frac{2k\pi i}{4}}\) where \(k=0,1,2,3\). Note that if \(\alpha\) is one root, the other roots can be expressed in terms of \(\alpha\) and \(i\).
02

Show that the Galois group is a subgroup of a cyclic group of order 4

Let's denote the Galois group of \(K\) over \(F\) by \(G(K/F)\). Additionally, let's denote our four roots by \(\alpha, \beta, \gamma, \delta\). We need to show that \(G(K/F)\) is a subgroup of a cyclic group of order 4. Consider the automorphisms of \(G(K/F)\) that fix \(F\): 1) The identity automorphism, 2) The automorphism that maps \(\alpha\) to another root, say \(\beta\). Note that when \(\sigma\) is an automorphism that fixes \(F\), then \(\sigma\) maps \(i\) to either \(i\) or \(-i\). However, due to the properties of roots, the automorphism cannot map \(i\) to \(-i\), so \(\sigma\) is determined by its action on the roots of the polynomial. Thus, there are at most 4 automorphisms in \(G(K/F)\), and since the order of \(G(K/F)\) divides 4, it is a subgroup of a cyclic group of order 4.
03

Show that if the polynomial is irreducible, the Galois group is cyclic of order 4

Assume the polynomial \(t^4 - a\) is irreducible in \(F\). First, we want to show that the extension degree \([K:F]\) is 4. Since \(a \in F\), we know the minimal polynomial of \(\alpha\) over \(F\) has degree at most 4. Since the polynomial is irreducible, the degree of the minimal polynomial is 4. Hence, \([F(\alpha):F]=4\). Also, we know that the splitting field of the polynomial has degree 4 over \(F\). Thus \(K=F(\alpha)\) and \([K:F]=[F(\alpha):F]=4\). Let \(H = G(K/F)\), and recall that \(|H|\) divides 4. If \(|H| = 2\), then \(H\) is a subgroup of a cyclic group of order 4. However, this contradicts our assumption that the polynomial is irreducible, as when \(|H| = 2\), the polynomial would not be irreducible in \(F\). Thus, \(|H|=4\), and \(G(K/F)\) is cyclic of order 4.
04

Express all other roots in terms of \(\alpha\) and \(i\)

Let \(\alpha\) be a root of \(t^4 - a\). We can write the other roots, \(\beta, \gamma, \delta\), as complex numbers with magnitude equal to the square root of the magnitude of \(\alpha^4=a\), and arguments differing by a multiple of \(\frac{\pi}{2}\) radians. Specifically, the other roots can be expressed as follows: 1. \(\beta = i\alpha = (\sqrt{-1})\alpha\) 2. \(\gamma = -\alpha = -1\alpha\) 3. \(\delta = -i\alpha = -(\sqrt{-1})\alpha\) Thus, the other roots can be expressed in terms of \(\alpha\) and \(i\).

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Key Concepts

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

Splitting Field
Understanding the concept of a splitting field is crucial when studying field theory and the Galois group. A splitting field for a polynomial is the smallest field extension in which the polynomial splits into linear factors, or in simpler terms, where the polynomial can be expressed as a product of factors of the form \text{\((t - root)\text{\)}, with each root being a solution of the polynomial within this extended field. For the polynomial in our exercise, \text{\((t^4 - a)\text{\)}, the splitting field includes all four complex roots of the polynomial, and it is constructed over the field \text{\((F)\text{\)}, which already contains the complex unit \text{\((i)\text{\)}. When discussing the connection between splitting fields and the Galois group, it's important to note that the Galois group consists of all field automorphisms that leave the original field invariant but can permute the roots of the polynomial within the splitting field.
Polynomial Irreducibility
Polynomial irreducibility refers to a polynomial that cannot be factored into the product of two or more non-constant polynomials with coefficients in the given field. In other words, it cannot be broken down into simpler polynomial pieces within that field. In the context of our exercise, if the polynomial \text{\((t^4 - a)\text{\)} is irreducible over field \text{\((F)\text{\)}, this signifies that there are no factors of \text{\((t^4 - a)\text{\)} with coefficients in \text{\((F)\text{\)} other than itself and one. Irreducibility is a pivotal concept in field theory because it can determine the properties of the Galois group. For instance, irreducibility implies a certain size and structure for the Galois group, which in our case would be cyclic of order 4.
Field Theory
Field theory is a branch of algebra that involves the study of fields, which are sets equipped with two operations, addition and multiplication, and follow certain properties similar to those of rational numbers. Within field theory, the concept of field extensions is fundamental, as it allows the construction of larger fields containing the roots of a polynomial that are not present in a smaller field. Our exercise is a practical problem in field theory, as it requires understanding field extensions like the splitting field and how polynomials behave within these fields, specifically regarding their roots and factorization.
Cyclic Group
A cyclic group is a type of group in mathematics that can be generated by a single element, meaning that every element of the group can be obtained by repeatedly combining this generator with itself. Cyclic groups are classified based on the number of elements (order) they contain. The Galois group of a polynomial can sometimes be a cyclic group. In our exercise, the problem guides us through the reasoning that the Galois group is a subgroup of a cyclic group of order 4. This means the Galois group can have 1, 2, or 4 elements, and if the polynomial is irreducible, the Galois group will precisely have 4 elements (i.e., it will be cyclic of order 4) and be generated by a single automorphism acting on the roots of the polynomial.
Root of a Polynomial
A root of a polynomial is a solution to the equation formed by setting the polynomial equal to zero. Basically, it's a number such that when you plug it into the polynomial, the output is zero. For the polynomial \text{\((t^4 - a)\text{\)}, a root is a number which, when raised to the power of four, gives \text{\((a)\text{\)}. In our exercise, complex numbers play a vital role, as the roots include complex solutions. This is why the field \text{\((F)\text{\)} must contain the complex unit \text{\((i)\text{\)}, as it allows for the expression of all roots in terms of complex numbers.
Complex Numbers
Complex numbers are numbers of the form \text{\((a + bi)\text{\)}, where \text{\((a)\text{\)} and \text{\((b)\text{\)} are real numbers, and \text{\((i)\text{\)} is the imaginary unit satisfying \text{\((i^2 = -1)\text{\)}. They extend the real numbers and provide a way to express solutions to equations that have no real solutions, like the square root of a negative number. In the context of the exercise, the roots of the polynomial \text{\((t^4 - a)\text{\)} may be complex, and being familiar with complex numbers allows us to understand the structure of these roots and how they relate to each other. Additionally, this complex number framework is essential to express all other roots in terms of one root and the complex unit \text{\((i)\text{\)}, as required by the exercise.

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Most popular questions from this chapter

\begin{array}{l} \text { Let } K \text { be Galois over } F \text { and let } E \text { be finite over } F \text { . Assume that } K \cap E=F \text { . }\\\ \text { Prove that }[K E: K]=[E: F] \end{array}

(a) Let \(K\) be an abelian extension of \(F\) and let \(E\) be an arbitrary extension of \(F\). Prove that \(K E\) is abelian over \(E\). (b) Let \(K_{1}, K_{2}\) be abelian extensions of \(F\). Prove that \(K_{1} K_{2}\) is abelian over \(F_{.}\) (c) Let \(K\) be a Galois extension of \(F\). Prove that there cxists a maximal abelian subextension of \(K\). In other words, there exists a subfield \(K^{\prime}\) of \(K\) containing \(F\) such that \(K^{*}\) is abelian, and if \(E\) is a subfield of \(K\) abelian over \(F\) then \(E \subset K^{\prime}\). (d) Prove that \(G_{K \mathbb{K}}\). is the commutator group of \(G_{K, \mathcal{F}}\), in other words, \(G_{K / K}\) is the group generated by all elements

By a radical tower over a field \(F\) we mean a sequence of finite extensions $$ F=F_{0} \subset F_{1} \in \cdots \in F_{r} $$ having the property that there exist positive integers \(d_{i}\), elements \(a_{1} \in F_{i}\) and \(x_{\text {i }}\) with \(x_{i}^{2}=a_{i}\) such that $$ F_{i+1}=F_{A}\left(x_{j}\right) . $$ We say that \(E\) is contained in a radical tower if there exists a radical tower as above such that \(\bar{E} \subset F\), Let \(E\) be a finite extension of \(F\). Prove: (a) If \(E\) is contained in a radical tower and \(E\) is a conjugate of \(E\) over \(F\), then \(E\) is contained in a radical tower. (b) Suppose \(E\) is contained in a radical tower. Let \(L\) be an extension of \(F\). Then \(E L\) is contained in a radical tower of \(L\). (c) If \(E_{1}, E_{2}\) are finite extensions of \(F\), each one contained in a radical tower. then the composite \(E_{1} E_{2}\) is contained in a radical tower. (d) If \(E\) is contained in a radical tower, then the smallest normal extension of F containing \(E\) is contained in a radical tower. (e) If \(F_{0} \subset \cdots \subset F_{,}\) is a radical tower, let \(K\) be the smallest normal extension of \(F_{0}\) containing \(F_{r}\). Then \(K\) has a radical tower over \(F\). (f) Let \(E\) be a finite extension of \(F\) and suppose \(E\) is contained in a radical tower. Show that there exists a radical tower $$ F \subset E_{0} \subset E_{1} \subset \cdots \subset E_{m} $$ such that: \(E_{m}\) is Galois over \(F\) and \(E \subset E_{m}\) \(E_{0}=F(\zeta)\) where \(\zeta\) is a primitive \(n\) -th root of unity; For each \(i, E_{i+1}=E_{i}\left(x_{i}\right)\) where \(x_{i}^{t}=a_{i} \in E_{i}\) and \(d_{i} \mid n\). Thus if \(E\) is contained in a radical tower, then \(E\) is solvable by radicals in the sense given in the text. The property of being contained in a radical tower is closer to the naive notion of an algebraic element being expressible in terms of radicals, and so we gave the development of this exercise to show that this naive notion is equivalent to the notion given in the text.

Let \(E=F(x)\) where \(x\) is algebraic over \(F\), of odd degree. Show that \(E=F\left(x^{2}\right) .\)

Determine the Galois groups of the following polynomials over the rational numbers. Find the discriminants. (a) \(t^{3}-3 t+1\) (b) \(t^{3}+3\) (c) \(t^{2}-5\) (d) \(t^{3}-a\) where \(a\) is rational, \(\neq 0\), and is not a cube of a rational number (c) \(t^{3}-5 t+7\) (f) \(t^{3}+2 t+2\) (g) \(t^{3}-t-1\)
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