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

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.

• Can you show that $x^2-(3+2\sqrt5)$ is irreducible over $\Bbb Q(\sqrt5)$? – Angina Seng May 31 at 10:17
• @AnginaSeng Yes, $(a+b\sqrt{5})^2=3+2\sqrt{5}$ can lead to a contradiction, but I think that I need to show the irreducibility for $x^4-(3+2\sqrt{5})$ over $\Bbb{Q}(\sqrt{5})$, instead of $x^2-(3+2\sqrt{5})$. I do not know how to reach a contradiction when $(a+b\sqrt{5})^4=3+2\sqrt{5}$, and I need to consider the quadratic cases. – MathEric May 31 at 10:22
• I saw two post explaining the irreducibility via Eisenstein criterion with prime $2$, which is not valid. I want to prevent the third wrong answer so I am writing this comment. – dust05 May 31 at 10:45

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.
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 $$\mathbb{Q}(\sqrt{5})$$.