The Minimal Polynomial of √2 + √3 over Q

# The Minimal Polynomial of √2 + √3 over Q

Recall from The Minimal Polynomial of an Algebraic Element in a Field Extension page that if $(F, +, *)$ is a field and $(K, +, *)$ is a field extension and if $u \in K$ is algebraic over $F$ then the minimal polynomial of $u$ over $F$ is defined to be the unique irreducible monic polynomial $p \in F[x]$ such that:

(1)\begin{align} \quad p(u) = 0 \end{align}

We will now look at finding the minimal polynomial of $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$. Let:

(2)\begin{align} \quad x = \sqrt{2} + \sqrt{3} \end{align}

Then:

(3)\begin{align} \quad x - \sqrt{2} &= \sqrt{3} \\ \quad (x - \sqrt{2})^2 &= (\sqrt{3})^2 \\ \quad x^2 - 2\sqrt{2}x + 2 &= 3 \\ \quad x^2 - 1 &= 2\sqrt{2}x \\ \quad (x^2 - 1)^2 &= (2\sqrt{2}x)^2 \\ \quad x^4 - 2x^2 + 1 &= 8x^2 \\ \quad x^4 - 10x^2 + 1 &= 0 \end{align}

Therefore, the polynomial $f(x) = x^4 - 10x^2 + 1$ is such that:

(4)\begin{align} \quad f(\sqrt{2} + \sqrt{3}) = 0 \end{align}

This polynomial is monic. It is a bit technical to show that $f$ is actually the minimal polynomial of $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$. We will omit the details for now.