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I am looking at Hensel's Lemma:

Let $F(x)=a_0+a_1x+ \dots + a_n x^n \in \mathbb{Z}_p[x]$.

We suppose that there is a $p$-adic number ($p>2$) $\alpha_1 \in \mathbb{Z}_p$, such that:

$$F(\alpha_1) \equiv 0 \pmod{p\mathbb{Z}_p}$$

$$F'(\alpha_1) \not\equiv 0 \pmod{p\mathbb{Z}_p}$$

Then, $\exists $ a $p$-adic integer $\alpha \in \mathbb{Z}_p$ such that $F(\alpha)=0$.

Could you explain what $p \mathbb{Z}_p$ means? What property does an element, that belongs in $p \mathbb{Z}_p$, satisfy?

EDIT: Suppose that we have the equation $ax^2+by^2+cz^2=0$ where $a,b,c$ are odd and $2 \nmid abc$ and also $a+b \equiv 0 \mod 4$.

According to my lecture notes, we cannot apply this form of Hensel's Lemma in this case because of the following:

$$F(x)=ax^2+by_0^2+cz_0^2$$

$$F'(x)=2ax$$

It should hold that: $$F'(\alpha_1) \not\equiv 0 \mod 2$$

but this can't happen.

So we have to use this form of Hensel's Lemma:

Let $F(X)=a_nX^n+ \dots+ a_1X+a_0 \in \mathbb{Z}_p[X]$ and $\exists \alpha_1 \in \mathbb{Z}_p$ such that $F(a_1) \equiv 0 \mod {p\mathbb{Z}_p}$ and $|F(\alpha_1)|_p<|F'(\alpha_1)|_p^2$. Then $\exists a \in \mathbb{Z}_p$ such that $a \equiv \alpha_1 \mod{p \mathbb{Z}_p}$ and $F(a)=0$.

And then we work $\mod 8$.

But, couldn't we also use the first version of Hensel's Lemma with $2 \mathbb{Z}_2=8$ ?

Adam Hughes
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evinda
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    A $p$-adic number $z\in\Bbb{Q}_p$ is in $p\Bbb{Z}_p$ iff it can be written in the form $z=pa$ where $a\in\Bbb{Z}_p$. This happens if and only if $|z|_p<1$. A useful thing to know is that $p\Bbb{Z}_p\cap \Bbb{Z}=p\Bbb{Z}$. – Jyrki Lahtonen Nov 01 '14 at 18:59
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    Do you know $\Bbb{Z}_p$ means? As a subset of $\Bbb{Q}_p$ that is? In the context of $p$-adic numbers it no longer means the quotient ring $\Bbb{Z}/p\Bbb{Z}$. – Jyrki Lahtonen Nov 01 '14 at 19:06
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    Another way of looking at $p\Bbb{Z}p$ is to observe that it is the closure of $p\Bbb{Z}$. So $z\in p\Bbb{Z}_p$ if and only there exists a sequence of integers $(a_n){n>0}$, all divisible by $p$, such that $z=\lim_{n\to\infty}a_n$. Yet another reformulation is that for all $n\in\Bbb{N}$ there exists an integer $a_n\in p\Bbb{Z}$ such that $|z-a_n|_p<p^{-n}$. – Jyrki Lahtonen Nov 01 '14 at 19:14
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    Or would examples of when the conditions of Hensel's lemma are/aren't satisfied be more helpful to you? – Jyrki Lahtonen Nov 01 '14 at 19:16
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    $p\mathbb{Z}_p$ means the ideal $(p)$ in $\mathbb{Z}_p$. In the same way, $p\mathbb{Z}=(p)$ is the ideal generated by $p$ in $\mathbb{Z}$. – Dietrich Burde Nov 01 '14 at 19:31
  • @JyrkiLahtonen Why does it stand that $p \mathbb{Z}_p \cap \mathbb{Z}=p \mathbb{Z}$ ? Also, why is $p \mathbb{Z}_p$ a closure of $p \mathbb{Z}$ ? Could you also explain me why an other reformulation is that $\forall n \in \mathbb{N}, \exists$ an integer $a_n \in p \mathbb{Z}$ such that $|z-a_n|_p<p^{-n}$ ? I would like to see also examples of when the conditions of Hensel's lemma are/aren't satisfied.. – evinda Nov 01 '14 at 19:48
  • Representative on-site examples of Hensel's lemma here, here and here – Jyrki Lahtonen Nov 01 '14 at 20:22
  • The reformulation is just the usual closure in a metric space. If $(X,d)$ is a metric space, and $A\subseteq X$, then a point $z\in X$ is in the closure $\overline{A}$, iff for all $\varepsilon>0$ there is a point $a_{\varepsilon}\in A$ such that $d(z,a_{\varepsilon})<\varepsilon$. – Jyrki Lahtonen Nov 01 '14 at 20:26
  • To prove that $p\Bbb{Z}p=\overline{p\Bbb{Z}}$ use the following facts about the $p$-adic power series representations. An element is in $\Bbb{Z}_p$, iff its power series contains no terms with negative powers of $p$. Therefore an element $z$ is in $p\Bbb{Z}_p$, iff its power series only contains terms with positive powers of $p$. You get the integers $a{\varepsilon}$ by truncating the series of $z$ at selected points. Because the $p^0$ -term has coefficient zero, all those integers are multiples of $p$. – Jyrki Lahtonen Nov 01 '14 at 20:30
  • @JyrkiLahtonen So if we would want to apply the Hensel's Lemma could we also find an $\alpha_1 \in \mathbb{Z}_p$ such that $F(\alpha_1) \equiv 0 \mod{2 \cdot (7 \cdot 2 +1)}$ ? – evinda Jan 29 '15 at 21:07
  • @evinda $p\Bbb Z_p\cap \Bbb Z=p\Bbb Z$ because if $z\in p\Bbb Z_p$ then $|z|_p<1$, if $z\in\Bbb Z$ is an integer, then $|z|_p<1\implies z=pk$ for some $k\in\Bbb Z$. By definition $z\in p\Bbb Z$, hence one inclusion is proved. The other direction is trivial. – Adam Hughes Jan 30 '15 at 02:11

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