# Multiplicative relations and roots in different splitting fields

Let $f \in \mathbb{Z}[x]$ be a separable monic polynomial, with $f(0) \neq 0$, and $p$ be a prime number. Also, let $L$ be the splitting field of $f$ over $\mathbb{Q}_p$ and let $a_1, \ldots, a_n \in L$ be all the roots of $f$. Finally, let $b_1, \ldots, b_n \in \mathbb{C}$ also be the roots of $f$, but this time taken in the complex numbers.

Is is true that the multiplicative group generated by $a_1, \ldots, a_n$ is isomorphic to the multiplicative group generated by $b_1, \ldots, b_n$?

I would say YES, since the splitting field is unique up to isomorphism, but $L$ is a splitting field over $\mathbb{Q}_p$, while the roots in $\mathbb{C}$ are considered as in a splitting field over $\mathbb{Q}$. So I'm not convinced...

Thank you.

Well, since $f\in\Bbb Q[x]$, all its roots $\{\rho_i\}$ are algebraic over $\Bbb Q$. Call the group generated by these $G$. It’s a finitely generated abelian group that is injected both into an algebraic closure of $\Bbb Q_p$ and into $\Bbb C$. What the splitting fields of $f$ are, as subfields of the nonarchimedean and the archimedean algebraically closed fields, is immaterial, I believe.
• Thanks Lubin. Would you kindly tell me more about the injection. It has something to do with the uniqueness of algebraic closure? Is it right that, being $\mathbb{C}$ and $\overline{\mathbb{Q}_p}}$ both algebraic closure of $\mathbb{Q}$, there is a $\mathbb{Q}$-isomorphism $f$ from one algebraic closure to the other such that $f(b_i) = a_i$ ? – sercej Oct 19 '16 at 6:18
• Let $K$ be the field generated by the roots of $f$ — its splitting field. $K$ injects into any algebraically closed field of characteristic zero — in many ways, of course, just choose one of them, and call if $\varphi$. We’re interested in the restriction of $\varphi$ to your group $G$: automatically an injection into the multiplicative group of your algebraically closed field. And by the way, $\Bbb C$ is not an algebraic closure of $K$, because it’s not algebraic over $K$. – Lubin Oct 19 '16 at 13:50