Determine the minimal polynomial of $\frac{1}{\sqrt[5]{2}}+\frac{1}{11}$ over $\mathbb{Q}$.
Put $x=\frac{1}{\sqrt[5]{2}}+\frac{1}{11}$. Put $x=\frac{1}{\sqrt[5]{2}}+\frac{1}{11}$. We need to find a nonzero polynomial with rational coefficients that has $x$ as a root. This is achieved by rewriting our expression as $x-\frac{1}{11}=\frac{1}{\sqrt[5]{2}}$. Raising each side by $5$ gives $\left(x-\frac{1}{11} \right)^5=\frac{1}{2}$, which can be expanded into $x^5-\frac{5}{11}x^4+\frac{10}{121}x^3-\frac{10}{1331}x^2+\frac{5}{14641}x-\frac{161053}{322102}=0$.
I now claim this is the minimal polynomial. However, I am not entirely sure as to how I should show this. After all, if we rewrite the equation so that coefficients are integers, namely by multiplying both sides by $2\times11^5 =322102$,then we still cannot use Eisenstein's criterion to show that it is irreducible over $\mathbb{Q}$. Do you have any suggestions?
Added comment:
It is easy to show that $\mathbb{Q}(\sqrt[5]{2})$ has dimension $5$ as a vector space over $\mathbb{Q}$. Is there some result that I can apply to make use of this fact, namely some lemma that enables me to argue that the minimal polynomial must be of an order that is the lowest common multiple of the terms in question?