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Suppose that $f \in \mathbb{Z}[x]$ is irreducible. Then $f$ factors as $\prod\limits_{i=1}^n {f_i}^{e_i}$ over $\mathbb{Z}/p \mathbb{Z}$ and by Hensel's Lemma, each term ${f_i}^{e_i}$ lifts to some factor $F_i$ of $f$ over $\mathbb{Z}_p$.

Must $F_i$ be irreducible over $\mathbb{Z}_p$?

Alexander
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  • You're neglecting some minor details, I think, which are important to your question. e.g. (this one is not directly relevant to your question) do you realize that for $f = x^2 + 2$ and $p=2$, $f$ factors as $x \cdot x$ over $\mathbb{F}_2$, but $x^2+2$ is the only (monic) factor of $f$ over $\mathbb{Z}_2$? –  Aug 02 '13 at 20:54
  • I am aware of the fact you state but that doesn't affect my question. What sort of details do you mean? – Alexander Aug 02 '13 at 20:57

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No. For example consider the case $f(x)=x^2+7$, $p=2$. Modulo $p$ we have $$ f(x)\equiv x^2+1\equiv(x+1)^2. $$ The lifted factor in $\mathbb{Z}_2[x]$ has to be $f(x)$ itself. But there is a $w=\sqrt{-7}$ in the ring $\mathbb{Z}_2$, so the lifted polynomial factors as $(x-w)(x+w)$ over the 2-adics.

Jyrki Lahtonen
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  • Here the only modulo $2$ factor $x+1$ has two distinct lifts $x\pm w$ in $\mathbb{Z}_2[x]$ that are both factors of $f(x)$. – Jyrki Lahtonen Aug 02 '13 at 20:45
  • If an $f_i$ has a lifting, can this lifting be also reducible? – OR. Aug 02 '13 at 20:49
  • No. If $f_i$ has a lifting $\tilde f_i\in\mathbb{Z}_p[x]$, and $\tilde f_i$ factors, then you can reduce that factorization modulo $p$ to show that $f_i$ will factor further. – Jyrki Lahtonen Aug 02 '13 at 20:51
  • But all the eventual factors of $F_i$ in $\mathbb{Z}_p[x]$ must become $f_i^d$ for some $d\le e_i$, when they are reduced modulo $p$ – Jyrki Lahtonen Aug 02 '13 at 20:53
  • Also, is there an easy way to know that $\sqrt{-7}$ is in $\mathbb{Z}_2$? – Alexander Aug 02 '13 at 21:02
  • A generalization of Hensel lifting works. $1$ is a "good enough" approximation to $\sqrt{-7}$, because $1^2-(-7)$ is divisible by eight. If an approximation is good enough the recursive process of Hensel lifting works out fine. With $f(x)=x^2+7$ we have here $f'(1)=2$ that is not a unit. This prohibits the usual Hensel's lemma from working, but there is a generalization that does work! If the approximation of a root is " much better than just mod $p$", then this allows us to overcome the fact that $f'$ is not a unit at that point. I don't remember the exact statement of the general version. – Jyrki Lahtonen Aug 02 '13 at 21:09
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    @Alexander: See this excellent answer by Cantlog for an argument proving the existence of $\sqrt{-7}\in\Bbb{Q}_2$. – Jyrki Lahtonen Dec 18 '13 at 12:50