Let $f$ be the minimal polynomial of $\alpha = \sqrt{-1} + \sqrt{17} + \sqrt{-17}$ over $\mathbb{Q}$ (with degree 4). Prove $f$ is reducible over $\mathbb{Q}_p$ (p-adic rationals) for all primes p and over $\mathbb{R}$.
I understand and can show that it must be reducible in $\mathbb{R}$ because the only irreducibles in $\mathbb{R}$ are of degree $\leq 2$. I'm thrown off by $\mathbb{Q}_p$. I first tried to calculate $f(x)$ explicitly, but couldn't quite figure out how to do that, and I'm not sure that it would even help.
Where do I go from here? Any hints or guidance would be greatly appreciated. We only know basic field theory and p-adic theory.
Note that the degree of the minimal polynomial of $\alpha$ over $F$ is $[F(\alpha):F]$. For any $p(x)\in F[x]$, if it has a root $\alpha\in K$ (i.e. $p(\alpha)=0$ in $K$), then ${\rm minpoly}_{\alpha,K}(x)\mid p(x)$ over $K$. Thus if we have an inequality $\deg{\rm minpoly}_{\alpha,F}>\deg{\rm minpoly}_{\alpha,K}$, we know ${\rm minpoly}_{\alpha,F}$ is reducible over $K$.
Show: for every prime $p$ at least one of $\{-1,17,-17\}$ is a quadratic residue modulo $p$, hence the field extension $K={\bf Q}_p(\sqrt{-1}+\sqrt{17}+\sqrt{-17})$ must be a subfield of ${\bf Q}_p(\sqrt{-1})$ or ${\bf Q}_p(\sqrt{17})$, both of which are at most quadratic extensions of ${\bf Q}_p$, hence $[K:{\bf Q}_p]\le 2$.
You will need to consider only three cases: (i) $p=2$, (ii) $p=17$, (iii) $p\ne 2,17$.