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 ($\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}$

  • $\begingroup$ 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... $\endgroup$ – Jyrki Lahtonen Jun 1 '16 at 20:01
  • $\begingroup$ 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 $6$? $\endgroup$ – amiz9 Jun 1 '16 at 20:09
  • 2
    $\begingroup$ Again your index equation is false here. Left hand side should have biggest and smallest field. $\endgroup$ – Dietrich Burde Jun 1 '16 at 20:10
  • $\begingroup$ OK thanks, so in general if we have fields $C \subset B \subset A$ then we would have : $[A : C]=[A : B] \times [B : C]$? $\endgroup$ – amiz9 Jun 1 '16 at 20:19
  • $\begingroup$ Yes, exactly. And just a typo in your last line: $\mathbb{Q}(\sqrt{2})\subset \mathbb{Q}$ is not true. $\endgroup$ – 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 $[\mathbb{Q}(\alpha, \sqrt{2}) : \mathbb{Q}(\sqrt{2})]=3$.

  • $\begingroup$ 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 $3$? $\endgroup$ – amiz9 Jun 1 '16 at 20:03
  • $\begingroup$ 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. $\endgroup$ – 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 $$t^3 - (0+0+0) t^2 + 3 \cdot \mathrm{det} \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} t - \mathrm{det} \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= t^3 - 6t - 6.$$

  • $\begingroup$ How does this change over $\mathbb{Q}(\sqrt{2})$? $\endgroup$ – amiz9 Jun 1 '16 at 19:50
  • $\begingroup$ also did the coefficients require first calculating $\alpha, \alpha^2...$ etc? $\endgroup$ – amiz9 Jun 1 '16 at 19:52
  • 1
    $\begingroup$ @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 $\alpha \cdot \sqrt[3]{2} = 2 + \sqrt[3]{4}$. $\endgroup$ – user343900 Jun 1 '16 at 19:53
  • 1
    $\begingroup$ @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 $\mathbb{Q}(\sqrt{2}).$ $\endgroup$ – user343900 Jun 1 '16 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.