Understanding the Congruence in Hensel's Lemma

[evinda]

Hi! (Wave)I am looking at the proof of Hensel's Lemma.

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 $\alpha_1 \in \mathbb{Z}$ such that:
$$F(\alpha_1) \equiv 0 \mod p\mathbb{Z}_p$$
and
$$F'(\alpha_1) \not\equiv 0 \mod p\mathbb{Z}_p$$

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

Proof:

We will prove the existence of a root $\alpha$ in $\mathbb{Z}_p$, constructing a sequence Cauchy, $\alpha_1, \alpha_2, \dots$, that converges to $\alpha$.

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

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

For $a_2$:

$$F(\alpha_2) \equiv 0 \mod p^2 \mathbb{Z}_p$$
$$\alpha_1 \equiv \alpha_2 \pmod p$$

$$\alpha_1 \equiv \alpha_2 \pmod p \Rightarrow \alpha_2=\alpha_1+pb_1$$
$$F(\alpha_1)+F'(\alpha_1)b_1p$$

Therefore, $F(\alpha_2)=F(\alpha_1+pb_1)=F(\alpha_1)+F'(\alpha_1)b_1p+(\text{ terms in }p^n \mathbb{Z}_p, \forall n \geq 2)$

We want $F(\alpha_2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$

So, it must be: $F(\alpha_1)+F'(\alpha_2)b_1 p \equiv 0 \pmod{p^2}$

From $F(\alpha_2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$, we conclude that $$F(\alpha_1)+F'(\alpha_1)b_1p+(\text{ terms in }p^n \mathbb{Z}_p, \forall n \geq 2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$$
but how do we conclude that it must stand: $F(\alpha_1)+F'(\alpha_2)b_1 p \equiv 0 \pmod{p^2}$ ?

 

[jvicens]

Hi there! It's great to see that you are studying Hensel's Lemma. To answer your question, we can conclude that $F(\alpha_1)+F'(\alpha_1)b_1p+(\text{ terms in }p^n \mathbb{Z}_p, \forall n \geq 2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$ because we have already shown that $F(\alpha_2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$ and we know that $F(\alpha_2)$ can be rewritten as $F(\alpha_1)+F'(\alpha_1)b_1p+(\text{ terms in }p^n \mathbb{Z}_p, \forall n \geq 2)$. Therefore, by substitution, we have $F(\alpha_1)+F'(\alpha_1)b_1p+(\text{ terms in }p^n \mathbb{Z}_p, \forall n \geq 2) \equiv 0 \pmod{p^2 \mathbb{Z}_p}$, which is what we wanted to show. I hope this helps clarify things for you. Keep up the good work!