Degree of a simple extension is the degree of the minimal polynomial

Statement

Let $L/K$ be a field extension and let $a \in L$ be algebraic over $K$ with minimal polynomial $m_a$ of degree $n$. Then $[K(a) : K] = n$, and $1, a, a^2, \dots, a^{n-1}$ is a basis of $K(a)$ over $K$.

Remarks

The proof: by the structure theorem, $K(a) \cong K[x]/(m_a)$. By polynomial division, the classes of $1, x, \dots, x^{n-1}$ are a basis of $K[x]/(m_a)$. For example, $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3$, with the basis $1, \sqrt[3]{2}, \sqrt[3]{4}$. With the tower law, the degree of every element of a finite extension $L/K$ divides $[L : K]$.

Explanations

L1 human

The degree of the field you get by adding one algebraic element is the degree of its minimal polynomial. Each element of $K(a)$ is a polynomial in $a$ of degree less than $n$, in exactly one way.