Let $f(X)$ be a cubic with 3 real roots, integer coefficients irreducible over $\mathbb{Q}$. Let $\alpha$ be one of these roots, and consider the number field $\mathbb{Q}(\alpha)$.
Dirichlet's unit theorem says we have two fundamental units $u_1,u_2$.
Apparently, the assertion that the integers $m$ and $n$ satisfy $\pm u_1^m u_2^n=x+y\alpha$ for some $x,y \in \mathbb{Z}$ is equivalent to the assertion that $Trace((\beta-\gamma) u_1^m u_2^n)=0$ where $\beta$ and $\gamma$ are the other roots of $f$.
This confuses me since it is not obvious that $\beta$ and $\gamma$ actually lie in $\mathbb{Q}(\alpha)$ - this certainly woudn't be true if we dropped the assumption that all the roots are real. Can anyone help with this?
[EDIT: The Diophantine eqns tag comes from the fact that this statement is from a paper by Ljunggren which solves the equation x^3-3xy^2-y^3=1. It was written in German, so it's possible that I have misunderstood it and the assertion in my question isn't what he actually said...]
[Second edit: For this particular case the discriminant is 81, a perfect square, so it Galois group is A3 and so the splitting field has degree 3 over $\mathbb{Q}$. So $\beta$ and $\gamma$ really are in $\mathbb{Q}(\alpha)$. However I don't see how to show the given result (other than by computing the other roots in terms of $\alpha$ and verifying it, which would be hard to do in general.]
Indeed, you are right. It can be proved that the splitting field of a cubic polynomial $f$ over $\Bbb{Q}$ is $$ \Bbb{Q}(\alpha, \sqrt{D}) $$ where $\alpha$ and $D$ are a root and the discriminant of $f$, respectively. In particular, the splitting field of $f$ has degree $3$ over $\Bbb{Q}$ if and only if $\sqrt{D}$ is rational, which isn't always the case even if $f$ has only real roots.