Proving Non-Normality: A Deep Dive Into Field Extensions

Hey there, math enthusiasts! Today, we're diving into the fascinating world of abstract algebra, specifically field theory. Our mission? To prove that the field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is not normal. Sounds intriguing, right? Let's break it down step by step and make this journey as clear and engaging as possible. We'll explore the core concepts, unravel the reasoning, and hopefully gain a deeper appreciation for the beauty and intricacies of mathematical structures.

Understanding the Basics: Fields, Extensions, and Normality

Before we jump into the proof, let's refresh our understanding of the fundamental concepts. A field is essentially a set equipped with two operations: addition and multiplication, satisfying certain properties (like commutativity, associativity, and the existence of identities and inverses). The set of rational numbers, denoted by $\mathbb{Q}$, is a prime example of a field.

Now, what about a field extension? It's simply a situation where we have a field $F$ and another field $K$ containing $F$. Think of it as $F$ being a subset of $K$. We write this as $F \subset K$. The field $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is an extension of $\mathbb{Q}$. It includes all elements that can be expressed using rational numbers, $\sqrt{3 + \sqrt{7}}$, and the usual field operations.

The concept of normality is crucial here. A field extension $K/F$ is normal if every irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K$. In simpler terms, if a polynomial with coefficients in $F$ has a root in $K$, all its roots must also be in $K$. This property is key to understanding the structure of field extensions.

To demonstrate that the extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is not normal, we need to show that there exists an irreducible polynomial in $\mathbb{Q}[x]$ which has a root in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$ but does not split completely within this field. This is the heart of our proof, and we'll meticulously explore it. We'll use the minimal polynomial of $\alpha = \sqrt{3 + \sqrt{7}}$ to achieve this.

Unveiling the Minimal Polynomial

The minimal polynomial of an algebraic number $\alpha$ over a field $F$ is the monic polynomial of smallest degree in $F[x]$ that has $\alpha$ as a root. It's irreducible over $F$ and plays a vital role in understanding the field extension $F(\alpha)$.

Let's find the minimal polynomial of $\alpha = \sqrt{3 + \sqrt{7}}$ over $\mathbb{Q}$. Here's how we can proceed:

1. Square the expression: We start by squaring $\alpha$: $\alpha^2 = 3 + \sqrt{7}$.
2. Isolate the square root: Rearranging, we get $\alpha^2 - 3 = \sqrt{7}$.
3. Square again: Squaring both sides yields $(\alpha^2 - 3)^2 = 7$.
4. Simplify and Rearrange: Expanding and simplifying, we get $\alpha^4 - 6\alpha^2 + 9 = 7$, or $\alpha^4 - 6\alpha^2 + 2 = 0$.

Thus, we have found a polynomial $f(x) = x^4 - 6x^2 + 2$ that has $\alpha$ as a root. Now, we need to show that this polynomial is irreducible over $\mathbb{Q}$. Proving irreducibility can be done using various methods, such as Eisenstein's criterion (though it doesn't apply directly here), or by checking for rational roots and then factoring it into quadratic factors if no rational roots are found.

In this case, since the polynomial has no rational roots (by the Rational Root Theorem), we can test whether it can be factored into quadratics with rational coefficients. If it can be factored, the product of the roots will be rational, and the sum of roots will be rational. After checking this condition, we discover it cannot be factored into quadratics. Therefore, $f(x) = x^4 - 6x^2 + 2$ is irreducible over $\mathbb{Q}$. This is the minimal polynomial of $\alpha$. This minimal polynomial is important for further exploration.

The Roots of the Minimal Polynomial

Let's find the roots of the polynomial $f(x) = x^4 - 6x^2 + 2 = 0$. We can use the quadratic formula to solve for $x^2$:

$x^2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 8}}{2} = \frac{6 \pm \sqrt{28}}{2} = 3 \pm \sqrt{7}$.

Therefore, we have:

$x^2 = 3 + \sqrt{7} \implies x = \pm \sqrt{3 + \sqrt{7}}$ $x^2 = 3 - \sqrt{7} \implies x = \pm \sqrt{3 - \sqrt{7}}$

So, the four roots of the polynomial $f(x)$ are $\alpha_1 = \sqrt{3 + \sqrt{7}}$, $\alpha_2 = -\sqrt{3 + \sqrt{7}}$, $\alpha_3 = \sqrt{3 - \sqrt{7}}$, and $\alpha_4 = -\sqrt{3 - \sqrt{7}}$.

Now, let's consider the field $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$. Clearly, $\alpha_1$ is in this field, as it's the generator. Also, $\alpha_2 = -\alpha_1$ is in this field. However, are $\alpha_3$ and $\alpha_4$ also in the field? Let's investigate further. In the context of a field extension, the key is to determine whether the other roots are members of the base field and how those roots can be constructed using elements of the base field and the generator.

Showing Non-Normality

Our aim is to demonstrate that the field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is not normal. To do this, we need to show that the minimal polynomial $f(x) = x^4 - 6x^2 + 2$ does not split completely in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$. In other words, not all roots of $f(x)$ are contained in the field $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$.

We know that $\alpha_1 = \sqrt{3 + \sqrt{7}}$ and $\alpha_2 = -\sqrt{3 + \sqrt{7}}$ are in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$. However, let's examine $\alpha_3 = \sqrt{3 - \sqrt{7}}$.

Notice that $\alpha_3$ is not in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$. To see why, observe that any element in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$ can be written in the form $a + b\sqrt{3 + \sqrt{7}}$, where $a, b \in \mathbb{Q}$. However, it is not possible to express $\sqrt{3 - \sqrt{7}}$ in this form. The root $\alpha_3$ involves a different square root and cannot be constructed using the generator $\sqrt{3 + \sqrt{7}}$ and rational numbers alone. Since $\alpha_3$ is not in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$, and similarly, $\alpha_4$ is also not in $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$, the minimal polynomial $f(x)$ does not split completely in the field. Therefore, the field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is not normal.

Conclusion: The Extension is Not Normal

We have successfully proven that the extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3 + \sqrt{7}})$ is not normal. By finding the minimal polynomial of $\sqrt{3 + \sqrt{7}}$, determining its roots, and showing that not all roots lie within the field $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$, we've provided a clear demonstration of non-normality.

This exercise highlights the crucial interplay between polynomials, roots, and field extensions. It shows that the structure of a field extension, particularly whether it's normal or not, significantly depends on the properties of the minimal polynomials and their roots.

I hope this step-by-step breakdown has been helpful and insightful. Keep exploring the wonders of abstract algebra, and happy math-ing!

Further Exploration: If you're eager to delve deeper, consider exploring other examples of non-normal field extensions and investigating the concept of the Galois group, which provides a powerful tool for understanding field extensions. You could also explore how to construct the smallest normal extension containing $\mathbb{Q}(\sqrt{3 + \sqrt{7}})$.

For more in-depth information, you can also check out resources like the following:

• MathWorld: (https://mathworld.wolfram.com/) - A comprehensive online resource for mathematical definitions, concepts, and formulas. This website provides detailed explanations and examples of field theory and related topics. MathWorld is an excellent resource for anyone looking to deepen their understanding of abstract algebra concepts and delve into the nuances of field extensions.

I hope this article has helped you understand the concepts of non-normality and field extensions! Keep exploring and enjoy the beauty of math! If you have any further questions, please do not hesitate to ask. Happy learning!