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Let $f \in \Bbb Z[X]$ be monic and assume that $f$ has a root $a_n$ modulo $p^n$ for every $n \geq 1$ (where $p$ is a fixed prime). Does it follow that $f$ has a root in $\Bbb Z_p$?

The problem is that we might have $f'(a_n) \equiv 0 \pmod p$ so that Hensel's lemma doesn't apply directly. If this is true, what reference/book gives a proof of this fact (stated as above)?

Alphonse
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  • If $a_n \equiv a_{n-1} \pmod{p^{n-1}}$ for all $n$ then the answer is yes, so it feels to me this is more of a "definition of $\mathbb{Z}_p$" problem rather than something to do with Hensel's Lemma. But I am not familiar with the set theoretic issues to say whether or not we can WLOG assume this in your question. – dyf Nov 04 '18 at 21:09
  • @dalbouvet : thank you. But I precisely did not assume that condition (under which the statement is indeed trivial). – Alphonse Nov 04 '18 at 21:10
  • yes indeed, I would also be interested if someone else can shed some light on this issue (and thus answers the set theoretic issues I'm concerned about). – dyf Nov 04 '18 at 21:12
  • Such an $f$ has a root in $\mathbb{Z}_p$ if and only if there exists a compatible sequence as in @dalbouvet's answer. As they said, this is really a question about understanding the definition of $\mathbb{Z}_p$, not Hensel's lemma. – RghtHndSd Nov 04 '18 at 21:20
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    @dalbouvet: Don't worry about set-theoretic issues. Essentially all practical applications of mathematics does require any set theory at all, and can be proven in a weak system called ACA. And almost all modern mathematics (outside of set theory and higher proof/recursion/model theory) only requires finitely many applications of the powerset. If you cannot convince yourself that something is true (such as the identity you suggested in your comment), it is practically impossible that it is due to a set-theoretic issue. – user21820 Nov 05 '18 at 08:42
  • More general setting: https://math.stackexchange.com/questions/3032633 – Watson Dec 09 '18 at 17:24

2 Answers2

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Yes, and this is not a generalization of Hensel's lemma.

We will repeatedly pass to subsequences as follows. First, the set $R_1 = \{ a_n \bmod p \}$ has finitely many elements, so there must be some residue $r_1 \in R_1$ such that there are infinitely many $n$ with $a_n \equiv r_1 \bmod p$. Pass to this subsequence; that is, assume WLOG that every $a_n$ has this property (mostly in order to avoid having to come up with some annoying piece of notation for the new subsequence).

Next construct $R_2 = \{ a_n \bmod p^2 : n \ge 2 \}$, which again is finite, so again there is some residue $r_2 \in R_2$ such that there are infinitely many $n$ with $a_n \equiv r_2 \bmod p^2$. Again pass to this subsequence. Etc.

In this way we construct a sequence of residues $r_k \in \mathbb{Z}/p^k\mathbb{Z}$ such that $r_k \equiv r_{k-1} \bmod p^{k-1}$ and such that there exist infinitely many $a_n$ such that $a_n \equiv r_k \bmod p^k$. The $r_k$ define an element $r \in \mathbb{Z}_p$ which is a root of $f$. Note that we don't need to assume that $f$ is monic.

Qiaochu Yuan
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Consider $f$ as a map $\mathbb{Z}_p\to\mathbb{Z}_p$. Since $f$ has a root mod $p^n$ for all $n$, the image of $f$ contains points arbitrarily close to $0$ in the $p$-adic metric. But $f$ is continuous and $\mathbb{Z}_p$ is a compact Hausdorff space, so the image of $f$ must be closed. Thus $0$ is in the image of $f$, i.e. $f$ has a root in $\mathbb{Z}_p$.

(If you unwind this proof it's essentially the same as Qiaochu's, but the topological language really makes it very simple!)

Eric Wofsey
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    Ah, thank you for this and the reference to Konig’s lemma! It felt very much like a compactness argument but I couldn’t quite see exactly how. – Qiaochu Yuan Nov 05 '18 at 02:06
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    Or: since $\mathbf Z_p$ is compact, the original sequence $a_n$ has a convergent subsequence, say $a_{n_i}$ with limit $a$. Then by continuity of $f(x)$ we have $f(a_{n_i}) \rightarrow f(a)$ as $i \rightarrow \infty$, and at the same time $|f(a_{n_i})|_p \leq 1/p^{n_i} \rightarrow 0$, so $|f(a)|_p = 0$ and thus $f(a) = 0$. – KCd Nov 14 '18 at 06:36