fields containing roots Let $f(x) = x^3 - 2 \in \mathbb{Q}[x]$. Then $f(x)$ has roots $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$, and $\sqrt[3]{2}\omega^2$, where $\omega$ is a complex cube root of $1$. In Birkhoff and Maclane's Algebra, 2nd edition, they say (page 448) that if we take $K = \mathbb{Q}[x]/(f(x))$ then $K \cong \mathbb{Q}(\sqrt[3]{2})$. Why do we have this instead of $K \cong \mathbb{Q}(\omega)$?
 A: $\omega$ is a root of $x^3 - 1$, not $f(x)$, but it plays a tangential role. Your mistake seems to be that you think $\omega$ is a root of $f(x)$. Now you're right that $\omega$ does play a part here: The roots of $f(x)$ are $\sqrt[3]{2}$, $\sqrt[3]{2}\omega$, and $\sqrt[3]{2}\omega^2$.
This is a good question, because the splitting field of $f(x)$ has degree greater than the degree of $f(x)$! This is what you were getting at, since the splitting field is $\mathbb{Q}(\omega, \sqrt[3]{2})$. (Since $x^3 - 2$ is irreducible in $\mathbb{Q}$, a field of characteristic zero, it has three distinct roots, by the way).
A: As a matter of fact, if $\omega$ is a root of $f(x)$, it's the case that $K$ is isomorphic to both. The roots of an irreducible polynomial generate isomorphic field extensions over the base field. Where $\omega = \zeta_{3}$, see AlexM's excellent post. 
${\bf Remark:}$ It is imperative, however, that $f$ be irreducible over your base field $F$. For example, if you instead had $f(x) = x^{3} - 1$, the roots of $f(x)$ are $1, \zeta_{3}, \zeta_{3}^{2}$, but clearly $\mathbb{Q}(1)$ is not isomorphic to $\mathbb{Q}(\zeta_{3})$. 
