# Minimal polynomial of $\sqrt{2}+\sqrt[3]{3}$.

Is there a way to show that the minimal polynomial of this number over $\mathbb Q$ has degree $6$ without too much annoying computations? I have "reduced" it to showing that both $\sqrt{2}$ and $\sqrt[3]{3}$ are in $\mathbb Q(\sqrt{2}+\sqrt[3]{3})$...

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Reference: Primitive Root Theorem

Just follow the proof, and you can show that the number $f$ in the proof, can be chosen to be $1$. Then $\mathbb{Q}(\sqrt 2, \sqrt[3]{3}) = \mathbb{Q}(\sqrt 2 + \sqrt[3]{3})$.

As $\alpha_1 = \sqrt 2$, and $\alpha_2 = -\sqrt 2$,

$\beta_1=\sqrt[3]{3}$, $\beta_2=\sqrt[3]{3} \omega$, $\beta_3= \sqrt[3]{3}\omega^2$, where $\omega=\exp(2\pi/3)$.

You just need the following equations are not true (which is obvious because one is real while the other is not),

$$-2\sqrt 2 = \sqrt[3]{3}-\sqrt[3]{3}\omega,$$ and $$-2\sqrt 2 = \sqrt[3]{3}-\sqrt[3]{3}\omega^2.$$

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There is a standard procedure for computing the minimal polynomial for the sum of two numbers from their minimal polynomials.

Let $f(X)$ and $g(Y)$ be the minimal polynomials of the algebraic numbers $\alpha$ and $\beta$. Set $Z=X+Y$. Now consider the polynomial $f(Z-Y)$ and eliminate from this the variable $Y$ using the relation $g(Y)=0$. Then thhe resulting polynomial involving just $Z$ is the minimal polynomial for $\alpha+\beta$.

Here it boils down to eliminating $Y$ from $(Z-Y)^2-2$ using $Y^3-3=0$.

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