What is the minimal degree of a polynomial with integer coefficients with $a^{1/b}$ as a root where a and b are integers?
Assume it can't be simplified, as in, there are no integer $x$ and $y<b$ such that $a^{1/b}=x^{1/y}$.
It feels like the answer really should be $b$. In some cases it can be shown using Eisenstein, and for $b=3$ we can show it with the general second degree solution. I can't see a proof that it should be $b$ generally, though.
By Karpilovsky's theorem (see Bill Dubuque's answer here), the minimal polynomial is $X^b-a$ whenever $b$ is not a multiple of $4$, or $a$ is not $-4$ times a fourth power.
If, on the other hand, we are in the "nasty" case when $b$ is a multiple of $4$ (say $b=4\beta$) and $a=-4c^4$, the theorem only tells us that the minimal polynomial will have degree strictly less than $b$. We have
$$ X^b-a=X^{4\beta}+4c^4=(X^{2\beta}-2cX^{\beta}+2c^2)(X^{2\beta}+2cX^{\beta}+2c^2) $$ So we have two "ex aequo" candidates of degree $2\beta=\frac{b}{2}$. It is not clear if they are irreducible however.