# Calculating minimum polynomial

I'm trying to calculate the minimal polynomial $f^\alpha_\mathbb{Q}$ for $$\alpha = \sqrt{2} + \sqrt{4}.$$ But I don't know how to do this. Does anyone have a hint? I tried to square $\alpha$, then I get $\alpha^2 = \sqrt{4} + 4 + \sqrt{16}$. But I'm not sure where that's going.

Does anyone know a structured method for solving these kind of problems, instead of just trying some things manually?

• $$\alpha^3=2+4+6\alpha$$ – lab bhattacharjee Feb 10 '18 at 11:55

## 3 Answers

Let $t=\sqrt  2$ so that $t^3=2$ and $a=t+t^2$

Every polynomial expression in $t$ can be reduced to a quadratic in $t$ using $t^3=2$. Three such expressions will enable $t$ to be eliminated.

$a^2=t^4+2t^3+t^2=2t+4+t^2$

$a^3=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2=\text{ (we notice immediately) } 6a+6$

Had that easy observation not been available, standard elimination of $t$ and $t^2$ from $a, a^2, a^3$ using your favourite (linear) method would leave at most a cubic.

Using the Binomial Theorem $$\alpha^3=(\sqrt{2} + \sqrt{4})^3=2+3\sqrt{16}+3\sqrt{32}+4=6+3\sqrt{8}(\sqrt{2}+\sqrt{4})=6(1+\alpha)$$ so the required polynomial is $$\alpha^3-6\alpha-6$$ Your best bet when dealing with expressions containing $n$th roots is to raise the expression to the power $n$ and hope for a nice simplification.

• Great thanks! I hadn't thought at all about Binomial theorem but that seems to simplify it a big deal! Lastly, to show that this is irreducible we can say that it is Eisenstein for $p = 3$ right? – Sigurd Feb 10 '18 at 12:03
• Yes! This example works perfectly due to $6=2\cdot3$. – TheSimpliFire Feb 10 '18 at 12:04

I suppose that you know the formula $(a+b)^3=a^3+b^3+3ab(a+b)$

Hence by cubing both sides we have $$\alpha^3= 2+4+3*2*(\sqrt {2}+\sqrt {4})$$ but we have $\sqrt {2}+\sqrt {4}=\alpha$ hence we get $$\alpha^3= 2+4+6\alpha$$ $$\Rightarrow \alpha^3-6\alpha-6=0$$