# Minimal polynomial of $\sqrt{2+\sqrt[3]{3}}$ over $\mathbb{Q}$

About 2 weeks ago, I tried to solve the following problem.

Find the minimal polynomial of $$\alpha=\sqrt{2+\sqrt[3]{3}}$$ over $$\mathbb{Q}$$.

My attempt

First, I tried to find the polynomial with rational(integer) coefficients having $$\alpha$$ as a root, and $$f(x)=x^6-6x^4+12x^2-11$$ is a polynomial such that $$f(\alpha)=0$$. Unfortunately, $$6$$ and $$12$$ is not divided by $$11$$, so I could not use the Eisenstein's criterion.

Instead of directly showing that $$f(x)$$ is irreducible over $$\mathbb{Q}$$, I tried to show that $$[\mathbb{Q}(\alpha):\mathbb{Q}]=6$$. Since $$[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2)][\mathbb{Q}(\alpha^2):\mathbb{Q}]$$ and $$\alpha^2=2+\sqrt[3]{3}$$, we know that $$[\mathbb{Q}(\alpha^2):\mathbb{Q}]=3$$. Thus if we succeed to show that $$[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2)]=2$$, then the proof is over. However, I could not do it.

I supposed that $$\alpha\in\mathbb{Q}(\alpha^2)=\mathbb{Q}(\sqrt[3]{3})$$, then there are $$a,b,c\in \mathbb{Q}$$ such that $$\alpha=a+b\sqrt[3]{3}+c\sqrt[3]{9}.$$ By squaring both sides of the equation, we obtain $$(a^2-6bc-2)+(3c^2-2ab-1)\sqrt[3]{3}+(b^2+2ca)\sqrt[3]{9}=0$$ and $$a^2-6bc-2=3c^2-2ab-1=b^2+2ca=0$$. However, I don't know how to show that this system of equations does not have a rational root and I'm stuck here.

Question: Is there a (or an alternative) way to solve the problem?

Here is an alternative solution. Working in $$\mathbb F_3$$, the minimal polynomial for $$\alpha$$ factorises as $$(x^2+1)^3$$, and $$x^2+1$$ is irreducible. Thus 2 divides $$[\mathbb Q(\alpha):\mathbb Q]$$, and you showed 3 divides this as well.

• I was going to give my overblown and overadvanced argument, and saw that yours encapsulated all my ideas in a short and comparatively elementary manner. Plus one. Jul 8, 2019 at 1:37
• I understand that $f(x)=(x^2+1)^3$ in $\mathbb{F}_3[x]$, but how can I know that the working in $\mathbb{F}_3$ implies that 2 divides $[\mathbb{Q}(\alpha):\mathbb{Q}]$? Jul 8, 2019 at 6:01
• The minimal poly for $\alpha$ divides $f$ over $\mathbb Z$, so divides $f$ over $\mathbb F_3$. Jul 8, 2019 at 15:51

This is not an alternative solution, but I think I can continue from where you have left.

So you were trying to show that there is no rational solution to $$2 + \sqrt[3]{3} = (a + b \sqrt[3]{3} + c\sqrt[3]{9})^2$$

Taking the field norm (in $$\mathbb{Q}(\sqrt[3]{3})$$ over $$\mathbb{Q}$$) of both sides, you get $$11 = r^2$$ for some $$r\in\mathbb{Q}$$ which is absurd.

• I learned the "norm" in $\mathbb{Z} [\sqrt{d}]$, but I have not heard the generalized concept. Could you tell me where I can learn that? Jul 8, 2019 at 5:39
• @choco_addicted The wikipedia page has the definition. The norm of $x\in\mathbb{Q}(\sqrt[3]{3})$ is the determinant of $\mathbb{Q}$-linear map $a\mapsto ax$ from $\mathbb{Q}(\sqrt[3]{3})$ to itself. From the definition, it is quite clear that the norm function is multiplicative. In $\mathbb{Q}(\sqrt[3]{3})$, the norm of $a+b\sqrt[3]{3}+c\sqrt[3]{9}$ is $a^3+3b^3+9c^3-9abc$.
– JWL
Jul 8, 2019 at 11:05