Problem 1 - Abstract Algebra: Theory and Applications

Question

Compute each of the following Galois groups. Which of these field extensions are normal field extensions? If the extension is not normal, find a normal extension of \(Q\) in which the extension field is contained. (a) \(G(\mathbb{Q}(\sqrt{30}) / \mathbb{Q})\) (d) \(G(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q})\) (b) \(G(\mathbb{Q}(\sqrt[4]{5}) / \mathbb{Q})\) (c) \(G(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q})\) (e) \(G(\mathbb{Q}(\sqrt{6}, i) / \mathbb{Q})\)

Step-by-Step Solution

Answer

Question: Determine the Galois groups for the following field extensions, and indicate if the extensions are normal or not. (a) \(\mathbb{Q}(\sqrt{30}) / \mathbb{Q}\) (b) \(\mathbb{Q}(\sqrt[4]{5}) / \mathbb{Q}\) (c) \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q}\) (d) \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q}\) (e) \(\mathbb{Q}(\sqrt{6}, i) / \mathbb{Q}\) Answer: (a) The Galois group for the extension \(\mathbb{Q}(\sqrt{30}) / \mathbb{Q}\) has order 2, and the extension is normal. (b) The Galois group for the extension \(\mathbb{Q}(\sqrt[4]{5}) / \mathbb{Q}\) has order 4, and the extension is not normal. (c) The Galois group for the extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q}\) has order 6, and the extension is normal. (d) The Galois group for the extension \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q}\) has order 12, and the extension is normal. (e) The Galois group for the extension \(\mathbb{Q}(\sqrt{6}, i) / \mathbb{Q}\) has order 4, and the extension is normal.

Step 1: Find the minimal polynomial

The minimal polynomial for \(\sqrt{30}\) over \(\mathbb{Q}\) is \(x^2 - 30\), as this is the smallest irreducible polynomial in \(\mathbb{Q}[x]\) that has \(\sqrt{30}\) as a root.

Step 2: Determine the order of the Galois group

The order of the Galois group is equal to the degree of the minimal polynomial, so we have \(|G(\mathbb{Q}(\sqrt{30})/ \mathbb{Q})| = 2\).

Step 3: Determine if the extension is normal

Since the minimal polynomial \(x^2 - 30\) splits completely in \(\mathbb{Q}(\sqrt{30})\), the extension is normal. (b) \(G(\mathbb{Q}(\sqrt[4]{5}) / \mathbb{Q})\)

Step 1: Find the minimal polynomial

The minimal polynomial for \(\sqrt[4]{5}\) over \(\mathbb{Q}\) is \(x^4 - 5\), as this is the smallest irreducible polynomial in \(\mathbb{Q}[x]\) that has \(\sqrt[4]{5}\) as a root.

Step 2: Determine the order of the Galois group

The order of the Galois group is equal to the degree of the minimal polynomial, so we have \(|G(\mathbb{Q}(\sqrt[4]{5})/ \mathbb{Q})| = 4\).

Step 3: Determine if the extension is normal

Since the minimal polynomial \(x^4 - 5\) does not split completely in \(\mathbb{Q}(\sqrt[4]{5})\), the extension is not normal. (c) \(G(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q})\)

Step 1: Find the minimal polynomials

The minimal polynomials for \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{5}\) over \(\mathbb{Q}\) are \(x^2 - 2\), \(x^2 - 3\), and \(x^2 - 5\), respectively.

Step 2: Determine the order of the Galois group

We know that the extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\) is a splitting field for the polynomial \((x^2 - 2)(x^2 - 3)(x^2 - 5)\). The degree of this polynomial is 6, so the order of the Galois group is 6.

Step 3: Determine if the extension is normal

Since the above polynomial splits completely in \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})\), the extension is normal. (d) \(G(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q})\)

Step 1: Find the minimal polynomials

The minimal polynomials for \(\sqrt{2}\), \(\sqrt[3]{2}\), and \(i\) over \(\mathbb{Q}\) are \(x^2 - 2\), \(x^3 - 2\), and \(x^2 + 1\), respectively.

Step 2: Determine the order of the Galois group

We know that the extension \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i)\) is a splitting field for the polynomial \((x^2 - 2)(x^3 - 2)(x^2 + 1)\). The degree of this polynomial is 12, so the order of the Galois group is 12.

Step 3: Determine if the extension is normal

Since the above polynomial splits completely in \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i)\), the extension is normal. (e) \(G(\mathbb{Q}(\sqrt{6}, i) / \mathbb{Q})\)

Step 1: Find the minimal polynomials

The minimal polynomial for \(\sqrt{6}\) over \(\mathbb{Q}\) is \(x^2 - 6\), and for \(i\) is \(x^2 + 1\).

Step 2: Determine the order of the Galois group

We know that the extension \(\mathbb{Q}(\sqrt{6}, i)\) is a splitting field for the polynomial \((x^2 - 6)(x^2 + 1)\). The degree of this polynomial is 4, so the order of the Galois group is 4.

Step 3: Determine if the extension is normal

Since the above polynomial splits completely in \(\mathbb{Q}(\sqrt{6}, i)\), the extension is normal.

Key Concepts

Normal Field Extension

Understanding the nature of a normal field extension is crucial when studying Galois theory. A field extension, specifically a normal field extension, has a key property that all polynomials with coefficients in the base field, which have a root in the extension field, must split into linear factors within that extension. This means that every root of the polynomial can be found within the extension field. To exemplify this, let’s consider the extension \(\mathbb{Q}(\sqrt{30}) / \mathbb{Q}\), where every root of the minimal polynomial \(x^2 - 30\) can be expressed within this field, thereby making it a normal field extension.

A clearer indication of a normal extension involves the existence of a splitting field. For instance, the field \(\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \mathbb{Q}\) is normal, as it is the splitting field for the polynomial \(x^2 - 2)(x^2 - 3)(x^2 - 5)\), which completely splits into linear factors over the field extension.

Minimal Polynomial

The concept of a minimal polynomial is central to understanding field extensions and their Galois groups. It is the irreducible polynomial of the lowest degree with coefficients in the base field that has the element from the extension field as a root. For instance, the minimal polynomial for \(\sqrt[4]{5}\) over \(\mathbb{Q}\) is \(x^4 - 5\), which is irreducible in \(\mathbb{Q}[x]\).

The importance of the minimal polynomial lies in its role in determining the order of the Galois group, which is the size of the group. As seen in the solution for \(\sqrt[4]{5}\), the degree of the minimal polynomial, which is four in this case, is directly related to the order of the Galois group. This relation helps in comprehending the Galois group's structure and its implications on the field extension.

Splitting Field

When we talk about a splitting field, we are referring to a field extension over which a given polynomial splits into linear factors. It is, in essence, the smallest field extension in which the given polynomial can be completely factored. Considering \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, i) / \mathbb{Q}\), this field is a splitting field for the polynomial \(x^2 - 2)(x^3 - 2)(x^2 + 1)\).

Furthermore, the Galois group of a splitting field is tightly linked with the corresponding polynomial, as the order of the group gives us information about the number of permutations or symmetries of the roots of the polynomial. This connection reveals the deep interplay between field theory, algebraic equations, and symmetry, which is the heart of Galois theory.

Field Extension

In abstract algebra, a field extension is a set \(E\) containing a given field \(F\), where \(E\) is itself a field under the same operations of addition and multiplication as \(F\). Every element of \(F\) is also an element of \(E\), but \(E\) has additional elements and hence more complex structure. The relationship between Galois groups and field extensions is significant because the Galois group is comprised of automorphisms of the larger field \(E\) that fix the smaller field \(F\) elementwise.

For instance, in the extension \(\mathbb{Q}(\sqrt{6}, i) / \mathbb{Q}\), \(\mathbb{Q}\) is the base field, and \(\mathbb{Q}(\sqrt{6}, i)\) is the extended field that includes \(\sqrt{6}\) and \(i\), non-elements of \(\mathbb{Q}\). Understanding the size and structure of the extension aids in unraveling its algebraic properties and the behavior of its Galois group.