= Degree of an algebraic number
{title2=$\deg_{\mathbb Q}\alpha=[\mathbb Q(\alpha):\mathbb Q]$}
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 $1,\alpha,\ldots,\alpha^{d-1}$ 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.<A Computational Introduction to Number Theory and Algebra by Victor Shoup>[section 16.6, algebraic elements and simple extensions]{ref}
For $\alpha=\sqrt2$, the <irreducible polynomial> $X^2-2$ gives degree two and expressions $a+b\sqrt2$. 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: $\sqrt2$ also satisfies $X^4-4=0$, but its degree is still two.
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