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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?

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2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Lubin
    Jul 8, 2019 at 1:37
  • $\begingroup$ 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 $[\mathbb{Q}(\alpha):\mathbb{Q}]$? $\endgroup$ Jul 8, 2019 at 6:01
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    $\begingroup$ The minimal poly for $\alpha$ divides $f$ over $\mathbb Z$, so divides $f$ over $\mathbb F_3$. $\endgroup$ Jul 8, 2019 at 15:51
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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.

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  • $\begingroup$ 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? $\endgroup$ Jul 8, 2019 at 5:39
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    $\begingroup$ @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 $a^3+3b^3+9c^3-9abc$. $\endgroup$
    – JWL
    Jul 8, 2019 at 11:05

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