Elliptic curves, reduction map, $E_n$ Let $E$ be the elliptic curve and set $\phi: E(\mathbb{Q}_p) \rightarrow E(\mathbb{F}_p)$ to be the reduction morphism. Define $E_n := \{(x:y:z) \in \ker \phi | x/y \in p^n\mathbb{Z}_p\}$. I'm busy studying elliptic curves and found this exercise online: Show that $\forall n \geq 1$ and $(x:y:z) \in E_n$ it holds that $z/y \in p^{3n}\mathbb{Z}_p$. Now, I've been busy on this for a while now and this is what I've come up with:
In order to show that, I started with dividing the projective equation by $y^3$, which is allowed because of the fact that $(x:y:z) \in \ker \phi$: 
$$z/y= x^3/y^3 + axz^2/y^3 + bz^3/y^3.$$
Because $(x:y:z) \in E_n$, it follows that $x/y \in p^n \mathbb{Z}_p$ and because $(x:y:z) \in \ker \phi$, we have that $z/y \in p \mathbb{Z}_p$. Any ideas on how to continue? 
As always, any help would be greatly appreciated. 
 A: Consider the equation you wrote down:
$$\frac{z}{y} = \left(\frac{x}{y}\right)^3 + a\frac{x}{y}\left(\frac{z}{y}\right)^2+b\left(\frac{z}{y}\right)^3,$$
and for simplicity give names $X=z/y$ and $Z=z/y$, so you have:
$$Z = X^3+aXZ^2+bZ^3.$$
You are trying to find the $p$-adic valuation of $Z$, i.e., $\nu_p(Z)$, knowing that $\nu_p(X)\geq n$. You need to use two properties of $\nu_p$:


*

*$\nu_p(s+t)\geq \min \{\nu_p(s),\nu_p(t)\}$, with equality if $\nu_p(s)\neq \nu_p(t)$. 

*$\nu_p(st)=\nu_p(s)+\nu_p(t)$.


We shall use these two properties to show that $\nu_p(Z)\geq 3\nu_p(X)$. We assume $a,b\in\mathbb{Z}_p$ (which can always be achieved by an appropriate change of variables in the Weierstrass equation), so that $\nu_p(a),\nu_p(b)\geq 0$. We distinguish two cases:


*

*Suppose first that $\nu_p(Z)<\nu_p(X)$. Then,
$$\nu_p(Z)=\nu_p(X^3+aXZ^2+bZ^3)\geq \min\{\nu_p(X^3),\nu_p(aXZ^2),\nu_p(bZ^3)\}\geq \min\{3\nu_p(X),\nu_p(a)+\nu_p(X)+2\nu_p(Z),\nu_p(b)+3\nu_p(Z)\}\geq 3\nu_p(Z).$$
Thus, $\nu_p(Z)\geq 3\nu_p(Z)$ and so $\nu_p(Z)\leq 0$. However, $Z=z/y$ and $[x,y,z]$ is in the kernel of reduction, which means $\nu_p(z/y)>0$, a contradiction. 

*Hence, we must have $\nu_p(Z)\geq \nu_p(X)$. Then,
$$\nu_p(Z)=\nu_p(X^3+aXZ^2+bZ^3)\geq \min\{\nu_p(X^3),\nu_p(aXZ^2),\nu_p(bZ^3)\}\geq \min\{3\nu_p(X),\nu_p(a)+\nu_p(X)+2\nu_p(Z),\nu_p(b)+3\nu_p(Z)\}\geq 3\nu_p(X)\geq 3n,$$
as desired.
