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I have just begun delving into p-adic number theory. I was wondering, given a poynomial $f(x)$ with integer coefficients, what does it mean when we say, $f(x)$ has a root in $\mathbb{Z}_2$, for instance.

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Exactly what it is supposed to mean: since $\mathbb{Z}$ is contained as a subring in the $p$-adics, you can view $f(x)\in\mathbb{Z}[x]$ as a polynomial in $\mathbb{Z}_p[x]$, and asking for a root there is asking for $a\in\mathbb{Z}_p[x]$ such that $f(a)=0$, with the operations involved in evaluating $f$ at $a$ being done in $\mathbb{Z}_p$. –  Arturo Magidin Nov 27 '10 at 3:51
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It means that for every positive integer k there is an integer x (depending on k) such that f(x)=0 (mod p^k). The statements "f(x)=0 has an integer solution modulo arbitrarily high powers of p" and "f(x)=0 has a p-adic integer solution" are equivalent. The second statement is stronger a priori but follows from the first by a compactness argument or an algorithm for locating the roots. –  T.. Nov 27 '10 at 4:06
    
@SamB I feel I have to ask. Why do keep editing all these old (number-theory) questions?! Not that there's something wrong with that, but the frontpage is now full of old questions, rather the new questions! –  user2468 Mar 5 '12 at 2:06
    
@J.D.: Well, mostly because I keep spotting titles that could use formatting... –  SamB Mar 5 '12 at 2:10
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1 Answer

If $f(x) = f_0 + f_1 x + \ldots + f_d x^d \in \mathbb{Z}[x]$, and $f(r) = 0$ over $\mathbb{Z}_2$ then $$f_0 + f_1 r + \ldots + f_d r^d \equiv 0 \bmod 2.$$ Equivalently, $f(x) \equiv (x-r) g \bmod 2$ for some polynomial $g(x) \in \mathbb{Z}_2[x].$

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