# Minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$

I have been working on these two minimal polynomial questions and am particularly concerned about (b)

Find the minimal polynomial for $\sqrt[3]{4}+\sqrt[3]{2}$

(a) over $\mathbb{Q}$

By setting $\alpha=\sqrt[3]{4}+\sqrt[3]{2}$, I found the minimal polynomial to be $\alpha^3-6\alpha-6=0$ (edited). This method involved just squaring out terms and was fairly lengthy - is this the standard procedure?

(b) over $\mathbb{Q}(\sqrt{2})$

What does it mean to be the minimal polynomial over $\mathbb{Q}(\sqrt{2})$? Over $\mathbb{Q}$, I see the minimal polynomial as the polynomial of lowest degree such that $\alpha$ is a root but I cannot see what is going on here.

From (a), we know that $[\mathbb{Q}(\alpha) : \mathbb{Q}]=3$. So:

$$[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}]=[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha) : \mathbb{Q}]=3[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\alpha)]$$

So since the degree is a multiple of $3$, the degree of the minimal polynomial of $\alpha$ over $\sqrt{2}$ is also a multiple of $3$. Can we automatically conclude it is $3$? I do not believe so since we are considering a larger field ($\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}$

• Yes. In the last paragraph you can conclude that the degree is $3$. You may need the following the convince your teacher and classmates. Because $2=[\Bbb{Q}(\sqrt2):\Bbb{Q}]$ you can conclude that $[\Bbb{Q}(\sqrt2,\alpha):\Bbb{Q}]$ is also a multiple of $2$. Therefore the degree of the other minimal polynomial is also at least $3$. But clearly it cannot be higher, so... – Jyrki Lahtonen Jun 1 '16 at 20:01
• I am getting confused.. $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\sqrt{2})]=[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}] \times [\mathbb{Q} : \mathbb{Q}(\sqrt{2})]$ which are multiples of $3$ and $2$ respectively.... so shouldn't the degree be a multiple of $6$? – amiz9 Jun 1 '16 at 20:09
• Again your index equation is false here. Left hand side should have biggest and smallest field. – Dietrich Burde Jun 1 '16 at 20:10
• OK thanks, so in general if we have fields $C \subset B \subset A$ then we would have : $[A : C]=[A : B] \times [B : C]$? – amiz9 Jun 1 '16 at 20:19
• Yes, exactly. And just a typo in your last line: $\mathbb{Q}(\sqrt{2})\subset \mathbb{Q}$ is not true. – Dietrich Burde Jun 1 '16 at 20:35

For $(a)$ compute $$x^3=(2^{1/3}+4^{1/3})^3=2+3\cdot 2^{4/3}+3\cdot 2^{5/3}+4=6+6(2^{1/3}+4^{1/3})=6+6x.$$ Hence the minimal polynomial is given by $x^3-6x-6$.

For $(b)$, the degree of the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{2})$ is given by $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\sqrt{2})]=3$.

• Thanks. Just so I understand $(b)$, we get $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\sqrt{2})]=[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}] \times [\mathbb{Q} : \mathbb{Q}(\sqrt{2})]$.. We know that $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}]$ is a multiple of $3$, and isn't $[\mathbb{Q} : \mathbb{Q}(\sqrt{2})]=2$? So how does this product equal $3$? – amiz9 Jun 1 '16 at 20:03
• Your index equation is false. You need to start with $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}]$ on the left hand side: biggest and smallest field. In the product the middle field twice. Then you get $6=3\cdot 2$, so nothing exiting. – Dietrich Burde Jun 1 '16 at 20:07

Another way to do this is to interpret $\alpha$ as a $\mathbb{Q}$-linear map from $\mathbb{Q}(\sqrt[3]{2})$ to itself. With respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, this is represented by the matrix $$\begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}.$$ The characteristic and minimal polynomial is $$t^3 - (0+0+0) t^2 + 3 \cdot \mathrm{det} \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} t - \mathrm{det} \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= t^3 - 6t - 6.$$

• How does this change over $\mathbb{Q}(\sqrt{2})$? – amiz9 Jun 1 '16 at 19:50
• also did the coefficients require first calculating $\alpha, \alpha^2...$ etc? – amiz9 Jun 1 '16 at 19:52
• @amiz9 No. $\alpha$ acts as the map $x \mapsto \alpha x$ so you need to pick a basis of $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ and multiply those by $\alpha$. For example, the second column looks like it does because $\alpha \cdot \sqrt[3]{2} = 2 + \sqrt[3]{4}$. – user343900 Jun 1 '16 at 19:53
• @amiz9 It doesn't change over $\mathbb{Q}(\sqrt{2})$ since $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is also a basis of $\mathbb{Q}(\sqrt[3]{2},\sqrt{2})$ over $\mathbb{Q}(\sqrt{2}).$ – user343900 Jun 1 '16 at 19:53