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Hensel's Lemma:

Let $F(x)=\alpha_0 + \alpha_1 x+ \dots + \alpha_n x^n \in \mathbb{Z}_p[x]$.

We suppose that there is a $p$-adic $a_1 \in \mathbb{Z}_p$ such that

$$F(a_1) \equiv 0 \mod p \mathbb{Z}_p \\ \text{ and } F'(a_1) \not\equiv 0 \mod p \mathbb{Z}_p$$

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

The proof is the following:

We will prove the existence of a root $a$ n $\mathbb{Z}_p$, constructing a Cauchy sequence, $a_1, a_2, \dots $ that converges to $\alpha$.

For each $n \geq 1$, we will construct $a_n$ such that

  1. $F(a_n) \equiv 0 \mod p^n \mathbb{Z}_p$
  2. $a_n \equiv a_{n+1} \pmod {p^n}$

for $a_2$

$F(a_2) \equiv 0 \mod p^2 \mathbb{Z}_p$

and $a_1 \equiv a_2 \pmod p$

$a_1 \equiv a_2 \pmod p \Rightarrow a_2=a_1+p b_1 \ | \ b_1 \in \mathbb{Z}_p$

So, $F(a_2)=F(a_1+pb_1) \overset{ \text{ Taylor }}{ = }F(a_1)+F'(a_1) \cdot b_1 \cdot p+ (\text{ terms in } p^n \mathbb{Z}_p, \text{ for } n \geq 2 )$

We want $F(a_2) \equiv 0 \pmod {p^2 \mathbb{Z}_p}$.

So, it should be $F(a_1)+F'(a_1) \cdot b_1 \cdot p \equiv 0 \pmod {p^2}$

From the hypothesis we have that $F(a_1)=p \cdot x$, so $p\cdot x+F'(a_1)\cdot b_1 \cdot p \equiv 0 \pmod {p^2}$

$\Rightarrow x+F'(a_0=1) \cdot b_1 \equiv 0 \pmod p$

$\Rightarrow F'(a_1)\cdot b_1 \equiv (-x) \pmod p$

Since $F'(a_1) \not\equiv 0 \pmod {p \mathbb{Z}_p}$, we have that the congruence has an unique solution

$b_1=(-x)[F'(a_1)]^{-1} \pmod p$

So, we calculated $a_2$.

Similarily, $a_n$ and so on.

So, we constructed a sequence $$a_1, a_2, \dots , a_n, a_{n+1}, \dots $$ where $a_n \equiv a_{n+1} \pmod {p^n}, \forall n \in \mathbb{N}$.

So, $a_{n+1}-a_n=p^n \cdot k \Rightarrow |a_{n+1}-a_n|_p \leq \frac{1}{p^n}$

$\lim_{n \rightarrow \infty} |a_{n+1}-a_n|_p=0$

$\Rightarrow \{a_n\}$ Cauchy

So, it converges, i.e.

$$\lim_{n \rightarrow \infty} a_n=\alpha$$

(By continuity of the polynomial function $F(x)$ we have that $F(\alpha)=0$)

Also $a_1 \equiv \alpha \pmod {p \mathbb{Z}_p}$, from the construction of $\alpha$.


  1. Could you explain to me why at the beginning we construct $a_n$ for each $n \geq 1$ such that

    1. $F(a_n) \equiv 0\mod p^n \mathbb{Z}_p$
    2. $a_n \equiv a_{n+1} \pmod {p^n}$
  2. At $$a_1 \equiv a_2 \pmod p \Rightarrow a_2=a_1+p b_1 \ | \ b_1 \in \mathbb{Z}_p$$ why is $b_1 \in \mathbb{Z}_p$ and not $b_1 \in > \mathbb{Z}$?

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$1.1$ We need that the solutions continue to be congruent to $0\mod p^n\Bbb Z_p$ because we are using the fact that the values converge to $\Bbb 0$ in $\Bbb Z_p$. Recall that $|x|_p=0\iff x=0$, so that the only number which is $\equiv 0\mod p^n$ for every $n$ is $0$ itself.

$1.2$ The point of constructing the sequence so that the reductions modulo $p^n$ have that property is reflected in the structure of the $p$-adic integers.

Note that a $p$-adic integer is formally a sum

$$\sum_{n=0}^\infty a_np^n,\quad 0\le a_n\le p-1$$

So the integers $a_0, a_0+a_1p, a_0+a_1p+a_2p^2,\ldots$ uniquely determine the $p$-adic integer just as does the sequence $3, 3.1, 3.14, 3.141,\ldots$ determines $\pi$. This means that a solution in $\Bbb Z_p$ is equivalent to a solution modulo $p^n$ for every $n$, which reduces down properly modulo all the lower powers. If your prime is say $3$ and your first candidate solution modulo $3$ is say $a_0=1$, but your second solution is $5=2+1\cdot 3^1$, then you have shifted your $3$-adic answer, it's no longer possible that these represent the same $3$-adic number as the sequence goes on, because any difference in the digits, changes the number, just like, $3.12\ldots$ cannot be $\pi$, no matter what comes after the $2$, because once you differ in one digit you are forever different.


$2$. Because in $\Bbb Z_p$ you have more options than you did in $\Bbb Z$. It might help to go back to definitions:

$$a\equiv b\mod p\iff a-b=pk$$

what ring $k$ comes from depends on your context, it is only classically that the symbol $\text{mod}$ has been used for integers, if you remember some of your ring theory, quotient rings are often written as $a\equiv b\mod I$ for some ideal $I$ when the two elements $a,b$ have the same image in the quotient ring, $R/I$. Here in this proof they are using that $p\Bbb Z_p$ is an ideal in $\Bbb Z_p$ and mean the terminology $\mod p$ to mean modulo the ideal $p\Bbb Z_p$. The writers are taking a notational shortcut because the context is $\Bbb Z_p$, so they understand $a_1\equiv a_2\mod p$ as shorthand for

$$a_1\equiv a_2\mod p\Bbb Z_p\iff a_1=a_2pb_1,\quad b_1\in\Bbb Z_p.$$

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  • $\begingroup$ Thank you for your answer... I will read it in a little while...Before I have also an other question... At the proof at the part: "For each $n \geq 1$, we will construct $a_n$ such that $$ 1. F(a_n) \equiv 0 \mod p^n \mathbb{Z}_p$$ $$ 2. a_n \equiv a_{n+1} \pmod {p^n}$$ for $a_2$ $F(a_2) \equiv 0 \mod p^2 \mathbb{Z}_p$ and $a_1 \equiv a_2 \pmod p$ " at the general formula at the second identity for $n=2$ (since at the first identity we used $n=2$) we get $a_2 \equiv a_{3} \pmod {p^2}$ but we have $a_1 \equiv a_2 \pmod p$... So, is the general formula wrong?? $\endgroup$ – Mary Star Jan 29 '15 at 21:57
  • $\begingroup$ Ah, no, that's not a problem: a typical example here is $p=5$, $a_0=1, a_1 = 6, a_2 =56$. Note that $a_2\equiv a_1\mod p^2$ and also $a_1\equiv a_0\mod p$, there's no contradiction, they're just noting that $a\equiv b\mod p^n\implies a\equiv b\mod p^m$ for any $m\le n$. $\endgroup$ – Adam Hughes Jan 29 '15 at 21:59

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