Let $\beta = 1 + \frac{\sqrt[3]{5}}{3}$. Is this an algebraic integer?
Now, let $F = \mathbb{Q}(\sqrt[3]{5}) = \mathbb{Q}(\beta)$ be the number field generated by $\sqrt[3]{5}$.
I know that $\beta$ is an algebraic integer if and only if its minimal polynomial $f_{\beta}$ is in $\mathbb{Z}[x]$.
Hence, I've tried to find the minimal polynomial of $\beta$, but this gets extremely messy and cumbersome. Is there a better way to do this?
It's not that cumbersome.
We have $\beta - 1 = \frac{\sqrt[3]5}3$, which gives us $$ (\beta - 1)^3 = \frac 5{27}\\ 27(\beta - 1)^3 - 5 = 0\\ 27\beta^3 - 81\beta^2 + 81\beta - 32 = 0 $$ and we see that the minimal integer polynomial is not monic.