A question about cubic roots of rational numbers I'm trying to understand if, given $K$ a cubic cyclic extension of $\mathbb{Q}(\zeta_3)$, where $\zeta_3$ is a third primitive root of unity, it always exists $b \in \mathbb{Q}$ such that $\sqrt[3]{b} \in K$. I believe it exists, but I can't find a proof of it.
We know from Hilbert 90 that it exists a $b \in \mathbb{Q}(\zeta_3)$ with this property, but is it possible to show that we can take it rational? If it's not true, is there a counterexample to this?
 A: If $k$ contains $\zeta_3$, then two cubic extensions $k(\sqrt[3]a)$ and $k(\sqrt[3]b)$ are equal if and only if $a/b$ or $ab$ is a cube in $k$.
(if $\sigma$ generates the Galois group of $K$ over $k$, the Galois action decomposes $K$ as $k.1 \oplus k.\alpha \oplus k.\alpha^{-1}$ where $\sigma(\alpha) = \zeta_3\alpha$ and $\sigma(\alpha^{-1}) = \zeta_3^2 \alpha^{-1}$ ; if $x \in K^*$ is a cube root of an element of $k$ then $x$ has to belong to either $k.\alpha$ or $k.\alpha^{-1}$)
This leads to considering the quotient of the $\Bbb F_3$-vector space $(k^*/(k^*)^3)$ by $\Bbb F_3^*$. Cubic extensions of $k$ are actually points of the projective space  $X = ((k^*/(k^*)^3) \setminus \{0\})/\Bbb F_3^*$.
Let $k = \Bbb Q(\zeta_3)$.
Since here $\mathfrak O_k$ is principal, $(k^*/(k^*)^3)$ is the direct sum of one $\Bbb F_3$ from the unit group and one $\Bbb F_3$ from every prime ideal of $\mathfrak O_k$.
You ask if there are cubic extensions that don't come from rational $a$. Since the Galois group of $k$ over $\Bbb Q$ still acts on $X$, it is enough to show that there are some points in $X$ that are not fixed by the Galois group of $k$. 
There are many such points : for example, if $a \in k$ is an algebraic integer of norm $p$ with $p \equiv 1 \pmod 3$, then the cubic extensions $ k(\sqrt[3]a)$ and $k(\sqrt[3]{\bar a})$ are distinct, so they cannot com from some $k(\sqrt[3]b)$ with rational $b$.
There also are cubic extensions that are not of the form $k(\sqrt[3]b)$ with rational $b$, but are "rational" (invariant by the Galois group of $k$), such as $k(\sqrt[3]{\zeta_3}) = k(\zeta_9)$
