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

I havent covered galois theory, this is an exercise from the chapter algebraic extensions in gallian.

I can see that $\sqrt[3]{2} + \sqrt[3]{4}$ $\in \mathbb{Q}(2^{1/3})$, after this I am clueless.

Need hints will finish the proof.

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  • $\begingroup$ See math.stackexchange.com/questions/2252741/… $\endgroup$ – lab bhattacharjee Apr 26 '17 at 8:03
  • $\begingroup$ @labbhattacharjee How is this related? $\endgroup$ – Wojowu Apr 26 '17 at 8:05
  • $\begingroup$ @Wojowu, You can use the same pattern, right? Here $\alpha=\sqrt[3]2$ $\endgroup$ – lab bhattacharjee Apr 26 '17 at 8:06
  • $\begingroup$ @labbhattacharjee Unless I'm missing something, that'd give you a quadratic with roots $\sqrt[3]{2},\sqrt[3]{4}$, coefficients of which are not in $\mathbb Q$. I still fail to see the connection with this problem. $\endgroup$ – Wojowu Apr 26 '17 at 8:15
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    $\begingroup$ @Wojowu : I don't see any relation either and I'd be curious to see a complete solution obtained by using lab bhattacharjee's link. $\endgroup$ – Georges Elencwajg Apr 26 '17 at 20:37
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Considering what you've said about $\mathbb Q(2^{1/3})$, cube should suffice. Calculate the powers up to 3rd, and do some linear combinations. $$\begin{array}{c|c|c|c|} & 1& \sqrt[3]2& \sqrt[3]4 \\ \hline \alpha^0& 1& 0& 0 \\ \alpha^1& 0& 1& 1 \\ \alpha^2& ?& ?& ? \\ \alpha^3& ?& ?& ? \\ \hline \end{array}$$

$x^3-6x-6=0$

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let $x=\sqrt[3]{2},\alpha=x+x^2$ $$ \alpha^3 = (x+x^2)^3 =x^3(1+x)^3=2(1+3x+3x^2+2)=6+6\alpha $$

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  • $\begingroup$ Funny, we came up with the same solution, only with mirror-image notation! $\endgroup$ – Barry Cipra Apr 26 '17 at 8:30
  • $\begingroup$ yes, though your exposition is more elegant and complete than mine $\endgroup$ – David Holden Apr 26 '17 at 8:36
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Let $x=a(1+a)$, where $a^3=2$. Then

$$x^3=a^3(1+3a+3a^2+a^3)=2(1+3a(1+a)+2)=6x+6$$

Thus $x=\sqrt[3]2+\sqrt[3]4$ is a root of the (irreducible) cubic $x^3-6x-6=0$.

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Let $\beta=\sqrt[3]{2}$ and $\alpha=\beta+\beta^2$.

A systematic way to find the minimal polynomial of $\alpha$ is to consider the map $\mu: x \mapsto \alpha x$ on $\mathbb Q(\beta)$, write its matrix with respect to the basis $1, \beta, \beta^2$, and then find the minimal polynomial of this matrix. This works because $p(\mu)(x)=p(\alpha)x$ for every polynomial $p \in \mathbb Q[T]$.

The matrix is $\pmatrix{ 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 }$ and its characteristic polynomial is $x^3 - 6 x - 6$. Since this polynomial is irreducible over $\mathbb Q$ by the rational root theorem, it is the minimal polynomial.

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