Problem 3
Question
Let \(L: \mathbb{Q}\) be a splitting field extension for \(x^{4}-5\) over \(\mathbb{Q}\). What is its Galois group? List the fields intermediate between \(L\) and \(\mathbb{Q}\), and determine which of them are normal over \(Q\).
Step-by-Step Solution
Verified Answer
The Galois group is \( \text{Gal}(L/\mathbb{Q}) \) of order 8, isomorphic to a subgroup of \( S_4 \), with \( \mathbb{Q}(i) \) as a normal intermediate field.
1Step 1: Identify the Roots
Identify the solutions to the polynomial \( x^4 - 5 = 0 \). The roots are \( \sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, \) and \(-i\sqrt[4]{5} \).
2Step 2: Determine the Splitting Field
Since \( x^4 - 5 \) has the roots \( \sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, \) and \(-i\sqrt[4]{5} \), the splitting field is \( L = \mathbb{Q}(\sqrt[4]{5}, i) \).
3Step 3: Find the Degree of the Extension
To find \( [L: \mathbb{Q}] \), notice that \( [L: \mathbb{Q}] = [\mathbb{Q}(\sqrt[4]{5}, i): \mathbb{Q}(\sqrt[4]{5})] \cdot [\mathbb{Q}(\sqrt[4]{5}): \mathbb{Q}] \). This results in a degree of 8.
4Step 4: Determine the Galois Group
The Galois group \( \text{Gal}(L/\mathbb{Q}) \) must be isomorphic to a subgroup of \( S_4 \) as the degree of the polynomial is 4, and it has order 8 as this degree indicates simple group symmetry.
5Step 5: Identify Intermediate Fields and Normal Extensions
The intermediate fields are \( \mathbb{Q}(\sqrt[4]{5}), \mathbb{Q}(i), \mathbb{Q}(\sqrt[4]{5}i) \). Out of these, \( \mathbb{Q}(i) \) is normal as it contains all roots of the minimal polynomial \( x^2 + 1 \).
Key Concepts
Splitting FieldGalois GroupNormal Extension
Splitting Field
When dealing with polynomials and field extensions, a **splitting field** plays a crucial role. A splitting field of a polynomial is a field extension over which the polynomial splits into linear factors. For example, consider the polynomial \( x^4 - 5 \). This polynomial doesn't have rational roots. Therefore, we find its roots in a larger field.
The roots of \( x^4 - 5 = 0 \) are \( \sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, \) and \(-i\sqrt[4]{5} \). These roots require us to extend \( \mathbb{Q} \) to include \( \sqrt[4]{5} \) and \( i \), since \( i \) is the imaginary unit. Thus, the splitting field \( L \) is \( \mathbb{Q}(\sqrt[4]{5}, i) \), the smallest field containing all the roots of the polynomial. This field ensures the polynomial splits entirely into linear factors within it.
The roots of \( x^4 - 5 = 0 \) are \( \sqrt[4]{5}, -\sqrt[4]{5}, i\sqrt[4]{5}, \) and \(-i\sqrt[4]{5} \). These roots require us to extend \( \mathbb{Q} \) to include \( \sqrt[4]{5} \) and \( i \), since \( i \) is the imaginary unit. Thus, the splitting field \( L \) is \( \mathbb{Q}(\sqrt[4]{5}, i) \), the smallest field containing all the roots of the polynomial. This field ensures the polynomial splits entirely into linear factors within it.
Galois Group
A fascinating area in Galois Theory is the **Galois group**. This group consists of field automorphisms of a field extension. For our extension \( L = \mathbb{Q}(\sqrt[4]{5}, i) \), the Galois group \( \text{Gal}(L/\mathbb{Q}) \) captures the symmetries of this field extension.
The order of the Galois group is equal to the degree of the field extension. Our extension \( [L: \mathbb{Q}] \) is of degree 8. Thus, the Galois group here must have order 8. This group can be isomorphic to well-known groups such as the dihedral group of order 8, which corresponds to the symmetries of a square. Each element of the Galois group can be thought of as permuting the roots of the polynomial, maintaining the structure induced by the polynomial's coefficients.
The order of the Galois group is equal to the degree of the field extension. Our extension \( [L: \mathbb{Q}] \) is of degree 8. Thus, the Galois group here must have order 8. This group can be isomorphic to well-known groups such as the dihedral group of order 8, which corresponds to the symmetries of a square. Each element of the Galois group can be thought of as permuting the roots of the polynomial, maintaining the structure induced by the polynomial's coefficients.
Normal Extension
A key feature of field extensions in Galois Theory is whether they are **normal extensions**. An extension is normal if it is the splitting field of some polynomial. In our scenario, \( L = \mathbb{Q}(\sqrt[4]{5}, i) \) is obviously a normal extension because we constructed it as the splitting field of \( x^4 - 5 \).
Intermediate fields can also be considered to determine their normality. In this case, the intermediate fields are \( \mathbb{Q}(\sqrt[4]{5}) \), \( \mathbb{Q}(i) \), and \( \mathbb{Q}(\sqrt[4]{5}i) \). Among these, \( \mathbb{Q}(i) \) is normal as it contains both roots of its minimal polynomial \( x^2 + 1 \). A field is normal if it includes all roots of the minimal polynomial of any of its elements.
Intermediate fields can also be considered to determine their normality. In this case, the intermediate fields are \( \mathbb{Q}(\sqrt[4]{5}) \), \( \mathbb{Q}(i) \), and \( \mathbb{Q}(\sqrt[4]{5}i) \). Among these, \( \mathbb{Q}(i) \) is normal as it contains both roots of its minimal polynomial \( x^2 + 1 \). A field is normal if it includes all roots of the minimal polynomial of any of its elements.
Other exercises in this chapter
Problem 1
Show that \(x^{6}+3\) is irreducible over \(Q\), but is not irreducible over \(Q(\omega)\), where \(\omega\) is a primitive sixth root of unity
View solution Problem 4
Suppose that \(q\) is a prime, that char \(K \neq q\) and that \(x^{4}-\theta\) is irreducible in \(K[x]\). Let \(\omega\) be a primitive \(q\) th root of unity
View solution Problem 5
Suppose that \(q\) is a prime, that char \(K=q\) and that \(\theta \in K\). Describe the splitting field for \(x^{q}-\theta\) over \(K\).
View solution Problem 6
Suppose that \([L: K]\) is a prime \(p\), that \(p \neq\) char \(K\) and that \(L\) is algebraically closed. Suppose (if possible) that \(p>2\). (i) Show that t
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