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Let $f,g$ be distinct irreducible factors of $x^n-1$ over $\mathbb{Z}_p[x]$ (polynomials over $p$-adic integers). Suppose $\overline{f},\overline{g}$ are coprime in $\mathbb{F}_p[x]$ - thus, the ideal generated by them $(\overline{f},\overline{g}) = 1$ in $\mathbb{F}_p[x]$. Must $(f,g) = 1$ in $\mathbb{Z}_p[x]$?

Note that $f,g$ are certainly coprime, but $\mathbb{Z}_p[x]$, coprime doesn't mean comaximal (e.g. $p,x$ are coprime but not comaximal).

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Suppose $(f,g)\ne 1$, then they are contained in some maximal ideal $m\supset (f,g)$, but the maximal ideals of $\mathbb{Z}_p[x]$ are precisely the ideals of the form $(p,h(x))$, where $h(x)$ is irreducible and remains irreducible mod $p$. Thus, $\mathbb{Z}_p[x]/m\cong \mathbb{F}_p[x]/(\overline{h})$. This implies that $(\overline{h})\supset(\overline{f},\overline{g})$, but since $\overline{f},\overline{g}$ are comaximal, they generate the unit ideal, and so $\overline{h}$ must be a unit, contradicting the fact that $h$ is irreducible mod $p$.

This implies that $(f,g) = 1$.

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    To prove $m$ is of that form, do you do something like this?: If not, we'd have $1\in (p)+m$, so $\exists a\in\mathbb Z_p : 1+ap\in m$, but $1+ap\in\mathbb Z_p^\times$, contradiction. Now $\mathbb Z_p[x]/m\cong\mathbb F_p[x]/\overline m$, so $\overline m=(\overline h)$ with $\overline h$ irreducible mod $p$, so $m=(p,h)$ for any lift $h$. (In general, with the same argument: if $a\in A$ satisfies $1+aA\subset A^\times$, $a$ is contained in every maximal ideal. As it seems, this is a characterization of the Jacobson radical, commutative case) – 2017-02-07
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    @barto Hmm, your proof seems to work (I didn't have one in mind actually - the statement seemed obvious to me). – 2017-02-07