Degree of an algebraic number (${\mathrm{deg}}_{\mathbb{Q}}\alpha =[\mathbb{Q}(\alpha ):\mathbb{Q}]$)

The degree of an algebraic number $\alpha$ is the degree of a polynomial of its minimal polynomial of an algebraic element over $\mathbb{Q}$. Equivalently, it is the field extension degree $[\mathbb{Q}(\alpha ):\mathbb{Q}]$. If that degree is $d$, then $1,\alpha ,\dots ,{\alpha }^{d-1}$ form a basis over the rational numbers: higher powers reduce using the minimal polynomial of an algebraic element, while a dependence among these first $d$ powers would contradict minimality.^[ref]

For $\alpha =\sqrt{2}$, the irreducible polynomial $X^2-2$ gives degree two and expressions $a+b\sqrt{2}$. For $\alpha =\sqrt[3]{2}$, the polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ because a reducible cubic would have a rational root of a polynomial, which the rational root theorem excludes. Its degree is three. Merely finding a polynomial equation gives an upper bound: $\sqrt{2}$ also satisfies $X^4-4=0$, but its degree is still two.