How to find a minimal polynomial

Question

I need to find minimal polynomial of $\alpha = \sqrt 2 + \sqrt [3] 3 $ over $\mathbb Q$ and prove that my result is minimal polynomial. How do I do that?

Answer 1

The way to find the minimal polynomial is to start with the supposition that $$ x - \left( \sqrt{2} + \sqrt[3]{3} \right) = 0 $$ then move the $\sqrt[3]{3}$ to the right side of the equality and cube, getting $$ x^2 + 6x - \sqrt{2} \left( 3x^2+2 \right) = 3 $$ Now isolate the terms involving $\sqrt{2}$ on the right and square, getting (after some grouping of terms) $$ x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1 = 0 $$ from which you read off the minimal polynomial.

It is easy to verify that this is a polynomial of $\alpha$ over $\Bbb{Q}$. To prove minimality you need to invoke the fact that $\gcd(2,3) = 1$, which implies that a minimal polynomial of a square root and a cube root of two numbers sharing no common factors must be of degree at least $6$.

Answer 2

It can be done in two steps:

$\left(\alpha - \sqrt{2}\right)^3 = \left(\sqrt[3]{3}\right)^3 = 3$, then you get:

$p(\alpha)\sqrt{2} = q(\alpha)$, and square both sides again:

$2p(\alpha)^2 - q(\alpha)^2=0$, where $p(\alpha), q(\alpha)$ are integer-coefficient polynomials in $\alpha$, and this is the minimal polynomial you seek.