# Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$

Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$ as a root).

I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals.

• Perhaps this answer about finding the minimal polynomial using resultants will help. – Noble Mushtak May 3 '16 at 19:50
• Hint: Look at $(a-\sqrt5)^3$. Also note that you will, at the end, have a sixth degree polynomial. – Arthur May 3 '16 at 19:51
• A better method is to note that $\sqrt[3]{2}$ and $\sqrt{5}$ are both clearly algebraic over $\mathbb{Q}$, so their sum is as well. – anomaly May 3 '16 at 21:28
• You seem to have quite an odd way of using MathJaX. Please have a look at the edits I made. Also, you might find this MathJaX tutorial to be usefull. – gebruiker May 4 '16 at 14:39

## 4 Answers

$x = \sqrt[3]{2} + \sqrt{5}$

$x - \sqrt{5} = \sqrt[3]{2}$

$x^3 - 3\sqrt{5}x^2 + 15x - 5\sqrt{5} = 2$

$x^3 - 15x - 2 = \sqrt{5}(3x^2 + 5)$

$(x^3 - 15x - 2)^2 = 5(3x^2 + 5)^2$

$x^6 -75x^4 - 4x^3 - 400x^2 + 60x - 121 = 0$

Now, ignoring accuracy and assuming arithmetical errors are both inevitable and irrelevant, there is another correctly figured out 6th degree polynomial with $\sqrt[3]{2} + \sqrt{5}$ as a solution.

Unless, I didn't make an arithmetical error calculating this. I might not have. There's a first time for everything.

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So obviously I made an arithmetical error. I knew I would. But it doesn't matter as the solving for the polynomial doesn't affect that there will be such a polynomial.

$x = \sqrt[3]{2} + \sqrt{5} \iff$

$x + k_1\sqrt{5} = \sqrt[3]{2} \iff$

$x^3 + k_2\sqrt{5}x^2 + k_3*5x + k_4*5*\sqrt{5} = 2 \iff$

$x^3 + k_5x + k_6 = \sqrt{5}(k_7x^2 + k_8) \iff$

$(x^3 + k_5x + k_6)^2= 5(k_7x^2 + k_8)^2 \iff$

$x^6 + k_9x^5 + k_{10}x^4 + k_{11}x^3 + k_{12}x^2 + k_{13}x + k_{14} = 0$

Where $k_i$ are integers and easily calculatable by someone who can count to 20 with his shoes on.

• You switched the sign on the $15x$ and then expanded wrongly. Should be $x^6-15x^4-4x^3+75x^2-60x-121=0$ – almagest May 3 '16 at 20:40
• Like I said, arithmetical errors are both inevitable and irrelevant. – fleablood May 3 '16 at 20:59
• Unless you are an accountant.:)....+1 – DanielWainfleet May 3 '16 at 22:00

$$r-\sqrt5=\sqrt[3]2\implies r^3-3\sqrt5\,r^2+15\,r-5\sqrt5=2\implies$$

$$r^3+15r-2=(3r^2+5)\sqrt5\implies r^6+30r^4+225r^2-4r^3-60r+4=45r^4+150r^2+125$$

Now group up elements and put $\;x=r\;$ and there you have a rational polynomial for which $\;\sqrt[3]2+\sqrt5\;$ is a root.

• @Winther Thank you, right. Yet I also think like fleablood above: arithmetical mistakes are almost inevitable and very, very non-important. Nevertheless, I already corrected. – DonAntonio May 3 '16 at 21:20

As a vector space over $\mathbb Q$, $\mathbb Q(\sqrt[3]{2},\sqrt 5)$ has a basis $\{1,\sqrt 5,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{2}\cdot\sqrt 5,\sqrt[3]{4}\cdot\sqrt 5\}$. Now consider the linear map $\phi:\mathbb Q(\sqrt[3]{2},\sqrt 5)\to\mathbb Q(\sqrt[3]{2},\sqrt 5)$ given by $\phi(y)=(\sqrt[3]{2}+\sqrt 5)y$. With respect to this basis, its matrix is given by $$\begin{bmatrix} \phi(1)&\phi(\sqrt 5)&\phi(\sqrt[3]{2})&\phi(\sqrt[3]{4})&\phi(\sqrt[3]{2}\cdot\sqrt 5)&\phi(\sqrt[3]{4}\cdot\sqrt 5) \end{bmatrix}$$ which comes out to $$A=\begin{bmatrix} 0&5&0&2&0&0\\ 1&0&0&0&0&2\\ 1&0&0&0&5&0\\ 0&0&1&0&0&5\\ 0&1&1&0&0&0\\ 0&0&0&1&1&0 \end{bmatrix}.$$ The characteristic polynomial $f(x)=\det(xI-A)$ has $\sqrt[3]{2}+\sqrt 5$ as a root, since $\sqrt[3]{2}+\sqrt 5$ is an eigenvalue of $\phi$. This turns out to be $$\det\begin{bmatrix} x&-5&0&-2&0&0\\ -1&x&0&0&0&-2\\ -1&0&x&0&-5&0\\ 0&0&-1&x&0&-5\\ 0&-1&-1&0&x&0\\ 0&0&0&-1&-1&x \end{bmatrix}=x^6-15x^4-4x^3+75x^2-60x-121.$$

• Thank seem a little over complicated but very well thought out and a nice way to look at it from a different view point. – Ernest Michael Nelson May 4 '16 at 15:26

Hint. Let $a,b$ be algebraic over field $F$ and $m_a(x), m_b(x)\in F[x]$ be minimal polynomials of $a$ and $b$, respectively. Then $a+b, ab$ are algebraic over $F$ and $$\operatorname{deg}m_{a+b}(x)\leq \operatorname{deg}m_{a}(x)\operatorname{deg}m_{b}(x); \ \operatorname{deg}m_{a\cdot b}(x)\leq \operatorname{deg}m_{a}(x)\operatorname{deg}m_{b}(x)$$