$\mathbb{Q}(\sqrt{2})\cong \mathbb{Q}[x]\mathrm{/}(x^2-2)$ with basis $1,\sqrt{2}$

Given: The element $a=\sqrt{2}$ over $\mathbb{Q}$.

Eisenstein's criterion makes a primitive integer polynomial irreducible over $\mathbb{Q}$ when some prime divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient (Eisenstein criterion over the integers).

A simple algebraic extension is its minimal-polynomial quotient and has the associated power basis (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1,a,\dots ,a^{n-1}$ and degree $n$).

The real numbers are a complete ordered field, so $\sqrt{2}$ exists and satisfies $(\sqrt{2}{)}^2=2$ (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$).

An element is algebraic when a nonzero polynomial vanishes at it (Algebraic and transcendental elements and algebraic extensions); for an algebraic element with minimal polynomial $m_a$, $f(a)=0$ exactly when $m_a$ divides $f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).