TITLE: Show that $\sqrt[3]{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt[3]{2})$
QUESTION [8 upvotes]: Setting $\alpha = \sqrt[3]{2}$ and $a+b\alpha+c\alpha^2=\sqrt[3]{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ since if it were reducible that would imply the existence of $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ which is easily shown not to be true)  . This comes down to proving that the equations $$ a^3+9ab^2=1 $$ and $$ 3b^3+3a^2b=1$$ have no solutions $a,b$ in $\mathbb{Q}(\sqrt{3})$. This is obviously extremely difficult to prove so there should be a better approach.
This is too long for a comment.
Considering the the irreducible polynomials for $\sqrt[3]{2}$ and $\sqrt[3]{1+\sqrt{3}}$ over $\mathbb{Q}(\sqrt{3})$ we see they are $x^3-(1+\sqrt{3})$ and $x^3-2$ respectively. If the two extensions in the comments are equal that would imply that there is an automorphism of the field $\mathbb{Q}(\sqrt{3})$ which fixes $\mathbb{Q}$ and sends $1+\sqrt{3}$ to $2$ but there is only one nontrivial automrphism of this field that fixes $\mathbb{Q}$ and it sends $\sqrt{3}$ to $-\sqrt{3}$. So the two extensions can't be isomorphic, let alone equal. Is this correct?

REPLY [2 votes]: I want to use Dedekind's theorem on the cycle structure of elements of the Galois group when viewed as a group of permutations of roots
Everything happens inside the splitting field $F$ of the polynomial $$f(x)=(x^3-1)^2-3=x^6-2x^3-2$$ over $\Bbb{Q}$. The zeros of $f(x)$ are $(1\pm\sqrt3)^{1/3}\omega^j, j=0,1,2$, $\omega=(-1+i\sqrt3)/2$, so
$$
F=\Bbb{Q}(\omega,\root3\of{1+\sqrt3},\root3\of{1-\sqrt3}).
$$
As $(1+\sqrt3)(1-\sqrt3)=-2$ we see that $F$ contains the cube roots of $2$. For the same reason $F=\Bbb{Q}(\omega,\root3\of{1+\sqrt3},\root3\of2)$.
Let's view the Galois group $G=Gal(F/\Bbb{Q})$ as a subgroup of $S_6$ permuting those six roots.

Lubin's calculations in the extensions of $\Bbb{F}_5$ imply that $f(x)$ remains irreducible modulo $5$. Therefore Dedekind's theorem implies that $G$ contains a 6-cycle $\tau$, and consequently also a product of two disjoint 3-cycles, namely $\tau^2$.
Modulo $13$ the polynomial $f(x)$ has three linear factors and a single cubic factor. This is because in $\Bbb{F}_{13}$ we can choose $\sqrt{3}=4$. Consequently $1+\sqrt{3}=5=(-2)^3$ has a cube root in $\Bbb{F}_{13}$. As $3\mid(13-1)$ it actually has three distinct cube roots in there. But $1-\sqrt3=-3$ is not a cubic residue modulo $13$. Consequently $x^3+3$ is an irreducible cubic factor of $f(x)$ over $\Bbb{F}_{13}$. This means that $G$ contains a 3-cycle.
The two previous bullets imply that the Sylow $3$-subgroup of $G$ cannot have order three. We can thus conclude that $9\mid [F:\Bbb{Q}]$. The main claim follows from this (and the tower law) immediately.