The chosen answer here claims that for rational $u$ and $v$, $u^{\frac{1}n} + v^{\frac{1}n}$ is rational iff $u^{\frac{1}n}$ and $v^{\frac{1}n}$ are both rational. However, the link to a proof seems to be broken so I was wondering if my version of a proof is correct. (This is a proof adapted from Robert Israel's proof that I saw somewhere.)
Suppose that $u = \frac{a}b, \gcd(a,b) = 1,$ both $u^{\frac{1}n} ,v^{\frac{1}n}$ are irrational, and $$u^{\frac{1}n} + v^{\frac{1}n} = r$$ for some rational $r$.
This implies $$(u^{\frac{1}n}-r)^n = v.$$ Now by Eisenstein's Criterion, $f(x) = ax^n-b$ is irreducible so by Gauss, so is $F(x) = x^n-u.$ Let $\alpha$ be the positive root of $f$ so $\alpha = u^{\frac{1}n}$. Then if $r$ is rational, $$g(x) = (z-r)^n-v-(z^n-u)$$ is a polynomial of lower degree with $\alpha$ as its root which is a contradiction.
Is this a valid proof? If not, can someone point me towards a correct proof?
