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I need to show that this is an algebraic number by showing that it is a solution to: $x^{6} - 9x^{4} - 4x^{3} + 27x^{2} + 36x -23 = 0.$

I'm really struggling with this one, I tried squaring and cubing $\sqrt{2} + \sqrt[3]{2}$, but I only end up with more squares and cubes.

Seji
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    How are you at linear algebra? You can show a polynomial dependence by systematically working with the "squaring and cubing" as vectors over the rational numbers. – hardmath Oct 18 '18 at 19:06

3 Answers3

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$$(x-\sqrt2)^3=2$$

$$\iff x^3-2+3x\cdot2=(3x^2+2)\sqrt2$$

Square both sides

lab bhattacharjee
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The minimal polynomial of $\alpha=\sqrt{2}$ is $a^2-2$ and the minimal polynomial of $\beta=\sqrt[3]{2}$ is $b^3-2$.
$\{1,a,b,b^2,ab,ab^2\}$ is a base of $\mathbb{A}=\mathbb{Q}[a,b]/(a^2-2,b^3-2)$, hence $\alpha+\beta$ is a root of a polynomial belonging to $\mathbb{Q}[x]$ with degree $\leq 6$. Indeed in $\mathbb{A}$ the terms $(a+b)^k$ decompose as follows: $$\begin{array}{|c|c|c|c|c|c|c|}\hline & 1 & a & b & b^2 & ab & ab^2\\ \hline 1 & 1 &&&&&\\ \hline (a+b) & 1 & 1 &&&&\\ \hline (a+b)^2 & 2 & & & 1 & 2 & \\ \hline (a+b)^3 & 2 & 2 & 6 &&& 3\\ \hline (a+b)^4 & 4 & 8 & 2 & 12 & 8 & \\ \hline (a+b)^5 & 40 & 4 & 20 & 2 & 10 & 20\\ \hline (a+b)^6 & 12 & 80 & 60 & 60 & 24 & 12\\ \hline\end{array}$$ hence by Gaussian elimination (we have seven rows and six columns) $$ (a+b)^6 = 4+24(a+b)-12(a+b)^2+4(a+b)^3+6(a+b)^4 $$ and $\alpha+\beta$ is a root of $x^6-6x^4-4x^3+12x^2-24x-4$. This is the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$ since the involved matrix has rank $6$: the determinant of the matrix formed by the first six rows and columns is

$$ 924 = 3\cdot 2^2\cdot\det\left(\begin{smallmatrix}0&1&2&0\\2&0&0&1\\1&6&4&0\\10&1&5&10\end{smallmatrix}\right). $$

Your polynomial cannot vanish at $\sqrt{2}+\sqrt[3]{2}$ since the conjugates of this algebraic number are $\pm\sqrt{2}+\omega^k\sqrt[3]{2}$, hence the norm of $\sqrt{2}+\sqrt[3]{2}$ is $(-4)\color{red}{\neq(-23)}$.

Jack D'Aurizio
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  • I continue to be mystified by all the problems we get where some student is trying to prove something that is just wrong. Trying to find out the source usually fails, often we are told some friend gave them the problem; the "friend gave me" aspect is by no means restricted to wrong problems, of course, neither is it universal. – Will Jagy Oct 18 '18 at 19:54
  • Could you recommend some free sources about this topic? As far as I understand it is something called field extention, isn't it? – Mikalai Parshutsich Oct 19 '18 at 11:45
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Set

$\alpha = \sqrt 2 + \sqrt[3] 2; \tag 1$

then,

$\alpha - \sqrt 2 = \sqrt [3] 2; \tag 2$

$(\alpha - \sqrt 2)^3 = 2; \tag 3$

$\alpha^3 - 3 \sqrt 2 \alpha^2 + 6 \alpha - 2\sqrt 2 = 2; \tag 4$

$-\sqrt 2(2 + 3\alpha^2) = 2 - 6\alpha - \alpha^3; \tag 5$

$2(2 + 3\alpha^2)^2 = (2 - 6\alpha - \alpha^3)^2. \tag 6$

If all we want to do is show $\alpha$ algebraic over $\Bbb Q$, we can stop here; (6) is evidently a $6$ degree polynomial over $\Bbb Q$; also, a finer inspection of (6) shows $\alpha$ to be an algebraic integer, since the leading coefficient, of $\alpha^6$, is $1$; to see exactly what polynomial we get, however, we must expand out each side of (6):

$2(4 + 12\alpha^2 + 9\alpha^4) = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 7$

$8 + 24\alpha^2 + 18\alpha^4 = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 8$

finally,

$\alpha^6 - 6\alpha^4 - 4\alpha^3 + 12 \alpha^2 -24\alpha - 4 = 0, \tag 9$

a sixth-degree polynomial satisfied by $\alpha$, in agreement with that obtained by Jack D'urizio.

Robert Lewis
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