How to find the degree of the extension $[\mathbb{Q}(\sqrt[4]{3+2\sqrt{5}}):\mathbb{Q}]$?

Question

How to find the degree of extension for $[\mathbb{Q}(\sqrt[4]{3+2\sqrt{5}}):\mathbb{Q}]$? I believe that the minimal polynomial of $\sqrt[4]{3+2\sqrt{5}}$ is $x^8-6x^4-11$, but I don't know how to show that it is irreducible over $\Bbb{Q}$. I also tried to show that $x^4-(3+2\sqrt{5})$ is irreducible over $\Bbb{Q}(\sqrt{5})$, but it is still too complicated for me.

Answer 1

Here's a nice trick how to show that $f(x)=x^8-6x^4-11$ is irreducible over $\mathbb{Q}$.

By Gauss' lemma it is irreducible over $\mathbb{Q}$ iff it is irreducible over $\mathbb{Z}$. Let $f=gh$ be a product of two polynomials with integer coefficients. Looking at the constant term of $f$ which is $-11$, you see that the constant terms of $g$ and $h$ have to be $\pm 1$ and $\pm 11$. Since the constant term is the product of all roots of the polynomial up to a sign, $g$ or $h$, and as a result $f$, has a root $\alpha$ with $\lvert \alpha \rvert \leq 1$. But then $\lvert \alpha^8 - 6 \alpha^4 -11 \rvert \geq 11-6-1 > 0$ and so $\alpha$ is not a zero of $f$, contradiction.

Hence $f$ is irreducible.

Answer 2

Let $\omega=\dfrac{-1+\sqrt{5}}{2}$. The element $3+2\sqrt{5}\in\mathbb{Z}[\omega]$ has norm $\vert 3^2-2^2\cdot 5\vert=11$, which is prime. Hence $3+2\sqrt{5}$ is irreducible.

Since $\mathbb{Z}[\omega]$ is a PID with quotient ring $\mathbb{Q}(\sqrt{5})$, Eisenstein criterion+Gauss lemma allows us to conclude that $X^4-(3+2\sqrt{5})$ is irreducible over $\mathbb{Q}(\sqrt{5})$.