Is the difference of two irrationals which are each contained under a single square root irrational?

Question

Is $ x^\frac{1}{3} - y^\frac{1}{3}$ irrational, given that both $x$ and $y$ are not perfect cubes, are distinct and are integers (i.e. the two cube roots are yield irrational answers)?

I understand that the sum/difference of two irrationals can be rational (see this thread: Is the sum and difference of two irrationals always irrational?). However, if my irrationals are contained under one root (so for example $3^\frac{1}{3}$ and not $2^\frac{1}{2} + 1$), can one generalise to show that $ x^\frac{1}{p} - y^\frac{1}{q} $ is irrational, where of course $x$ and $y$ are not powers of $p$ and $q$ respectively?

Answer 1

If $ x $ and $ y $ are not perfect cubes, then the polynomials $ T^3 - x $ and $ T^3 - y $ are irreducible in $ \mathbf Q[T] $. Consider the splitting field $ L $ of this family over $ \mathbf Q $. Let $ G = \textrm{Gal}(L/\mathbf Q) $, and consider the stabilizer subgroups $ G_x, G_y $ of $ x^{1/3}, y^{1/3} $ respectively. By isomorphism extension, the different conjugates of $ x $ are in correspondence with the different cosets of $ G_x $ in $ G $, and likewise for $ y $. Thus, we obtain

$$ \textrm{Tr}_{L/\mathbf Q}(x^{1/3}) = \sum_{\sigma \in G} \sigma(x^{1/3}) = |G_x|( x^{1/3} + \zeta x^{1/3} + \zeta^2 x^{1/3} ) = 0 $$

where $ \zeta $ is a primitive third root of unity, and likewise for $ y^{1/3} $. Therefore, $ x^{1/3} - y^{1/3} $ lies in the kernel of the field trace of $ L/\mathbf Q $. But the trace of a rational number is a nonzero integer multiple of it, therefore the only rational number in the kernel is zero, hence if this number is rational it must be zero, and we must have $ x = y $.

The argument can be adapted to the case with prime(!) $ p, q $ using the criterion that $ T^p - x $ is irreducible in $ K[T] $ for a field $ K $ if and only if $ x $ is not a perfect $ p $th power. In fact, a stronger result is true: $ x^{1/n} $ always lies in the kernel of the field trace as long as $ x $ is not a perfect $ n $th power (no primality required.) For this, see Theorem 3 in this article.

Answer 2

Denote the cube roots by $X,Y$, so that $X^3=x$ and $Y^3=y$ with $x,y\in \mathbb Z$.

Suppose, with slightly greater generality, that we have $X-Y-R=0$ where $X^3,Y^3,R\in \mathbb Q$.

We first want to argue that $XY\in \mathbb Q$. To do so, observe the identity:

$$X^3-Y^3-R^3-3XYR=(X-Y-R)(X^2+Y^2+R^2+XY+XR-YR)$$

This immediately implies that $$3XYR=X^3-Y^3-R^3$$ which, as desired, implies that $XY\in \mathbb Q$.

But then we have two real numbers, $X,Y$ such that both $X-Y$ and $XY$ are rational. It follows that $X,-Y$ both satisfy a quadratic equation with rational coefficients, namely \begin{align} X^2 - (X-Y)X -XY &= 0,& Y^2 +(X-Y)Y -XY&=0. \end{align} But this is not possible, as $Z^3-x=0$ is the minimal polynomial for the cube root of $x$ over $\mathbb Q$.