Let $\alpha$ and $\beta$ be real algebraic numbers satisfying $\alpha+\beta\notin\mathbb{Q}$ and $\alpha^{3}+\beta^{3}\in\mathbb{Q}$.
Let $K\subseteq\mathbb{C}$ be the splitting field of the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
Suppose that the Galois group $G:=\textrm{Gal}(K/\mathbb{Q})$ is abelian.
Show that:
- $x := \alpha\omega+\beta\omega^{2}$
- $y := \alpha\omega^{2}+\beta\omega$
are elements of $K$
where $\omega\in\mathbb{C}$ with $\omega^{3}=1,\omega\neq1$.
We find that
- $x + y = -(\alpha + \beta)$
- $x \cdot y = \alpha^2 - \alpha \beta + \beta^2 = \frac{\alpha^3 + \beta^3}{\alpha + \beta}$
- $\alpha\beta=\frac{(\alpha+\beta)^{3}-(\alpha^{3}+\beta^{3})}{3(\alpha+\beta)}\in K$.
- The polynomial $P(X)=X^{3}-3\alpha\beta\cdot X-(\alpha^{3}+\beta^{3})$ has $\alpha+\beta$ as a root.
- The terms $x$ and $y$ look like the cubic Lagrange Resolvent.
but how can we finish?
I think that the conditions are contradictory.
As $K/\mathbb{Q}$ is abelian, every subfield of $K$ is Galois, thus $K=\mathbb{Q}(\alpha+\beta)$. In particular $K \subset \mathbb{R}$.
But $(x-y)^2=(\alpha-\beta)^2(\omega-\omega^2)^2$ can be written as $(x+y)^2-4xy \in K$. Now, $(\alpha-\beta)^2=(\alpha^2-\alpha\beta+\beta^2)-\alpha\beta$ is already in $K$, so $(\omega-\omega^2)^2=2\omega-2 \in K$ so that $\omega \in K$. We get a contradiction, to which the only escape is when $\alpha=\beta$ is the cubic root of a rational number (which stll isn't abelian).