📂Abstract Algebra

Algebraic Extension

Definition ¹

Let $E$ be an extension of $F$ and call it $n \in \mathbb{N}$.

1. If every element of $E$ is an algebraic number over $F$, then $E$ is called an algebraic extension of $F$.
2. If $E$ is a $n$-dimensional vector space over $F$, then $E$ is called a $n$-dimensional finite extension over $F$.
3. The degree of a finite extension $E$ over $F$ is denoted as $[ E : F ]$.

Theorem

Let $E$ be a finite extension of $F$, and let $K$ be a finite extension of $E$.

• [1]: A finite extension is an algebraic extension.
• [2]: $$[E : F] = 1 \iff E = F$$
• [3]: $$[K : F] = [K : E ] [E : F]$$

Description

To get a feel for the degree, let’s take a look at the most familiar cases of the real numbers and the rational numbers.

$\mathbb{R}$ is obviously $[ \mathbb{R} : \mathbb{R} ] = 1$, so it is fine to call it a $1$rd degree real number field. Meanwhile, since a complex number can be expressed with two independent real numbers, it can also be seen as a Euclidean space $\mathbb{R} \times \mathbb{R} = \mathbb{C}$ or a simple extension $\mathbb{R} ( i ) = \mathbb{C}$ with the addition of the imaginary unit $i$. In this case, since $[ \mathbb{C} : \mathbb{R} ] = 2$, the complex number field can be called a $2$-degree real number field.

Adding the irrational number $\sqrt{2}$ to the finite extension $\mathbb{Q} ( \sqrt{2} )$ of $\mathbb{Q}$, and taking $\left\{ 1 , \sqrt{2} \right\}$ as a basis over $\mathbb{Q}$, the degree becomes $$\left[ \mathbb{Q} ( \sqrt{2} ) : \mathbb{Q} \right] = 2$$ When adding one more $\sqrt{3}$, since $\sqrt{2} \cdot \sqrt{3} = \sqrt{6}$, the basis of $\left( \mathbb{Q} ( \sqrt{2} ) \right) ( \sqrt{3} ) = \mathbb{Q} ( \sqrt{2} , \sqrt{3} )$ becomes $\left\{ 1 , \sqrt{2} , \sqrt{3} ,\sqrt{6} \right\}$. Therefore, calculating the degree over $\mathbb{Q}$ yields $$\left[ \mathbb{Q} ( \sqrt{2} , \sqrt{3} ) : \mathbb{Q} \right] = 4$$

If we add $2^{1/3}$ instead of $\sqrt{3}$, the basis of $\mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right)$ becomes complex as shown in $\left\{ 1 , 2^{1/2} , 2^{1/3} , 2^{2/3} , 2^{5/6} , 2^{7/6} \right\}$ considering all combinations. Calculating the degree yields $$\left[ \mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right) : \mathbb{Q} \right] = 6$$ However, according to theorem [3], it can be calculated as follows: $$\left[ \mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right) : \mathbb{Q} \left( 2^{1/6} \right) \right] \left[ \mathbb{Q} \left( 2^{1/6} \right) : \mathbb{Q} \right]$$ On the right-hand side, the basis of $\left[ \mathbb{Q} \left( 2^{1/6} \right) : \mathbb{Q} \right]$ is trivially $\left\{ 1 , 2^{1/6} , 2^{2/6} , 2^{3/6} , 2^{4/6} , 2^{5/6} \right\}$, so $$\left[ \mathbb{Q} \left( 2^{1/6} \right) : \mathbb{Q} \right] = 6$$ Summarizing, $$\left[ \mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right) : \mathbb{Q} \left( 2^{1/6} \right) \right] = 1$$ Therefore, by theorem [2], $$\mathbb{Q} \left( 2^{1/2} , 2^{1/3} \right) = \mathbb{Q} \left( 2^{1/6} \right)$$ is concluded.