Going through Hensel's Lemma, I feel I read somewhere that the limit of sequence of integers $a_0,a_1,a_2,...$=$ a$ is root of the $f(X)\in\mathbb{Z}_p[X]$, where, $$F(a_0)\equiv0\pmod{p}\;\;\;,\;\;\;F'(a_0)\not\equiv 0\pmod{p}$$ and $a_n$s satisfy $$f(a_n)\equiv 0\pmod {p^{n+1}}\;\;\;,a_{n+1}\equiv a_{n}\pmod{p^{n+1}}$$.

Does $f(a_n)\equiv0\pmod{p^\infty} $ mean $f(a_n)=0$?

Clearly, $f(a_n)\equiv 0 \pmod {x} \implies \exists y:xy=f(a_n)$. How does this imply $f(a_n)=0$? Are we supposed to deal with p-adic norm and hence $\|p^{\infty}\|_p= 0$

Edit I am also curious how Hensel came up with this idea, that finding root of $\mathbb{Z}_p[X]$ might have something to do with modular arithmetic.

P.S. I am utterly new to this topic so excuse silly ideas.

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    $\begingroup$ Grab some book which deals with p-adics -- Gouvea's "p-adic Numbers" is great -- and learn there how we can define $\,\Bbb Z_p\,$ as an inverse limit of the finite groups $\,\Bbb Z/p^n\Bbb Z\,$ , which would help a lot to understand why $\,a_{n+1}=a_n\pmod {p^n}\,$ and etc. $\endgroup$ – DonAntonio Apr 7 '13 at 12:08
  • $\begingroup$ Thanks. Great. Already grabbed in my hand. $\endgroup$ – user59756 Apr 7 '13 at 12:13

There is no such thing as $\mod p^{\infty}$. However, by the congruences with respect to growing powers of $p$, the sequence $a_n$ of elements of $\mathbb Z$ determines an element of $\mathbb Z_p$, $a_n\to a$. You have $a\equiv a_n\pmod{p^n}$ for all $n$. You also have another sequence $b_n$ given by $b_n=f(a_n)$ and n eed to argue with the properties of that specific $f$ that $b_n$ also converges to some $b$. After that, you can conclude $b=0$ because $b\equiv b_n\equiv 0\pmod {p^n}$ for all $n$. If you had $b\ne 0$, then there would be at least some $n$ with $b\not\equiv 0\pmod {p^n}$.

As to the "idea", $\mathbb Z_p$ is all about arithmetic modulo powers of $p$. Indeed, the argument may become easier to grasp if you reformulate everything with $p$-adic absolute value. Then you will notice that this is "just" a reinvention of Newton's iteration formula for finding a root.

  • $\begingroup$ you say the properties of that specific $f$. Does being a polynomial not guarantee that $a_n\rightarrow a \implies f(a_n)\rightarrow f(a)$? I cannot think counter examples in $\mathbb{R}[X]$. Is it different in $\mathbb{Z}_p[X]$? Thank you for the nice answer. $\endgroup$ – user59756 Apr 7 '13 at 11:58
  • $\begingroup$ @700resu Yes, I forgot that you mentioned it being a polynomial explicitly - that is enough "speific property" indeed as $x\equiv b\pmod {p^n}\Rightarrow f(x)\equiv f(y)\pmod {p^n}$. $\endgroup$ – Hagen von Eitzen Apr 7 '13 at 13:05

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