The sum of square and cube roots is an algebraic function How can we prove that the complex valued function $\sqrt z + \sqrt[3] z$ is an algebraic function?
Context
According to my understanding, an algebraic function can be defined as a root of a polynomial equation, I just cannot find a polynomial equation which has this as a root.
 A: In general, the sum of two algebraic function is algebraic. To find an explicit algebraic equation with root $\sqrt z + \sqrt[3] z$ expand the product 
$$\prod_{k \in \{0,1\}, l \in \{0,1,2\}}( u - ( (-1)^k \sqrt{z} + \omega^l \sqrt[3]{z} ))$$
$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ is a third root of $1$. All the roots of $v^3 = z$ are $\omega^l \sqrt[3]{z}$, $l \in \{0,1,2\}$. You combine them all together in a product which will contain only integral powers of $z$.  You get 
$$u^6   - 3 z\,  u^4  -2z\, u^3 + 3 z^2 u^2 - 6 z^2 u + (z^2 - z^3) =0$$
satisfied by $\sqrt z + \sqrt[3] z$ . 
As an illustration, take a particular value $z = 2^6 = 64$. Then $$u^6   - 3 z\,  u^4  -2z\, u^3 + 3 z^2 u^2 - 6 z^2 u + (z^2 - z^3) = \\ =u^6-192 u^4-128 u^3+12288 u^2-24576 u-258048$$ with roots 
$$12 = \sqrt{64} + \sqrt[3]{64}\\
- 4 =-\sqrt{64}+ \sqrt[3]{64} \\
- 10 + 2 \sqrt{3} i = -\sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \sqrt[3]{64}\\
- 10 - 2 \sqrt{3} i = -\sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) ^2\sqrt[3]{64}\\
6 + 2 \sqrt{3} i = \sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \sqrt[3]{64}\\
6 - 2 \sqrt{3} i = \sqrt{64} + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) ^2\sqrt[3]{64}
$$
A: $\sqrt{z}$ and $\sqrt[3]{z}$ are both algebraic over $\mathbb{C}(z)$, and hence by a basic theorem of field extensions you get that $\mathbb{C}(z)(\sqrt{z},\sqrt[3]{z})/\mathbb{C}(z)$ is algebraic. Are you familiar with the theorem about the compositum of two algebraic (equivalently finite) extensions (within a common field) being algebraic?
