elliptic curves over $latex \mathbb{Q}_p$ and modulo $latex p$

Suppose that $E:y^2=x^3+ax+b$ is an elliptic curve over $\mathbb{Q}_p$. Note, a change of variable $x=p^{2n}x',y=p^{3n}y'$ gives $E:y'^2=x'^3+ap^{-4n}+bp^{-6n}$. So, we can always assume that $a,b\in\mathbb{Z}_p$(sometimes we require also that the discriminant $\Delta=4a^3+27b^2$ has maximal absolute value $|\Delta|_p$, but here we don’t need it).

For each point $P\in E(\mathbb{Q}_p)$, we can choose a representative $P=[x:y:z]$ such that $a,y,z\in\mathbb{Z}_p$ yet not all of them lie in $p\mathbb{Z}_p$(just to make sure that after modulo $p$, this point lies in $\mathbb{P}^2_{\mathbb{F}_p}$). And, we associate $[x:y:z]$ to the point $[x':y':z']\in\mathbb{P}^2_{\mathbb{F}_p}$. We see easily that this $P'=[x':y':z']$ also satisfies the equation $y^2=x^3+a'x+b'$. Thus we have defined an application $f:E(\mathbb{Q}_p)\rightarrow E(\mathbb{F}_p)=E',P\mapsto P'$. We can verify that $f$ is a group homomorphism. Yet we are not sure that, $E'$ is still an elliptic curve because we do not know, a priori, whether $\Delta'=0$ or not in $latex\mathbb{F}_p$. In any case, we define

$E^0(\mathbb{Q}_p)=\{P\in E(\mathbb{Q}_p)|P'\text{ is not singular in }E'\}$.

We can verify that if $P,Q\in E^0(\mathbb{Q}_p)$, then $P+Q\in E^0(\mathbb{Q}_p)$. What is more, since $e_E=[0:1:0]$ is not singular in $E'$, which shows that $E^0(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$.

Moreover, we consider the preimage of $[0:1:0]$ in $E(\mathbb{Q}_p)$. In other words, we set

$E^1(\mathbb{Q}_p)=f^{-1}([0:1:0])$.

This shows that $E^1(\mathbb{Q}_p)$ is a subgroup of $E^0(\mathbb{Q}_p)$. In a similar way, we can define,

$E^n(\mathbb{Q}_p)=\{P=[x:y:z]\in E^1(\mathbb{Q}_p)|x/y\in p^n\mathbb{Z}_p\}$.

So, we get a filtration:

$E(\mathbb{Q}_p)\supset E^0(\mathbb{Q}_p)\supset E^1(\mathbb{Q}_p)\supset...$

Then there is an important theorem:

(1)$E(\mathbb{Q}_p)/E^0(\mathbb{Q}_p)$ is a finite set;

(2)We have an isomorphism $E^0(\mathbb{Q}_p)/E^1(\mathbb{Q}_p)\rightarrow E'^{ns}(\mathbb{F}_p)$;

(3)For each $n>0$, $E^n(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$, and the map $f_n:E^n(\mathbb{Q}_p)\rightarrow\mathbb{F}_p,P=[x:y:z]\mapsto p^{-n}x/y(mod p)$ is an isomorphism;

(4) $\bigcap_nE^n(\mathbb{Q}_p)=\{0\}$.

An important corollary is

If $gcd(m,p)=1$, then the multiplication map $E^1(\mathbb{Q}_p)\rightarrow E^1(\mathbb{Q}_p),P\mapsto mP$ is a bijection(hence an isomorphism).

The corollary is a direct consequence of the third proposition in the theorem above and the Hensel’s lemma. If $P\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_P)$, then the isomorphism between $E^n(\mathbb{Q}_p)/E^{n+1}(\mathbb{Q}_p)=\mathbb{F}_p$ induces a multiplication map $\mathbb{F}_p\rightarrow\mathbb{F}_p,x\mapsto mx$. So, the image of $P$ in $\mathbb{F}_p$ is not trivial, and $gcd(m,p)=1$, which shows that the image of $mP$ in $\mathbb{F}_p$ is not trivial, either, thus this multiplication map is injective. To show the surjectivity, suppose again that $Q\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_p)$, then there is an element $x\in\mathbb{F}_p$ such that $mx=f_n(Q)$. So, there is a $P_n\in E^n(\mathbb{Q}_p)$ such that $x=f_n(P_n)$, and thus $Q-mP_n\in E^{n+1}(\mathbb{Q}_p)$. We can continue this process, and find a series of $P_m\in E^{m}(\mathbb{Q}_p)$ such that $Q-m(P_n+P_{n+1}+...+P_k)\in E^{k+1}(\mathbb{Q}_p)$. Push this process to infinity, and using the fact that $\mathbb{Q}_p$ is a complet space, we find $Q=m\sum_nP_n$. so, we conclude the proof.

So, we see that the isomorphisms $f_n:E^n(\mathbb{Q}_p)/E^{n+1}(\mathbb{Q}_p)\rightarrow\mathbb{F}_p$ are rather important maps. We first prove (3). Suppose that $E^n(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$. Then for any $P=[x:y:z]\in E^n(\mathbb{Q}_p)$, we can write $x=p^Ax',y=p^By',z=p^Cz'$ with $x',y',z'$ units in $\mathbb{Z}_p$, yet since $y^2z=x^3+axz^2+bz^3$, we have that $p^{2B+C}y'^2z'=p^{3A}x'^3+ap^{A+2C}+bz'^3p^{3C}$. We know that $A-B\geq n$. It is easy to see that we must have $B\leq C$. If $C\leq A$, by taking $ord_p$ on both sides, we have that $2B+C=3C$, thus $B=C$. But  this contradicts the definition of $E^1(\mathbb{Q}_p)$(here we use the essential hypothesis that we are working in $E^1(\mathbb{Q}_p)$, not $E^0$,nor $E$, and that is why this proposition starts from $E^1$ instead of these latter two). So, we must $B\leq C\leq A$. Yet we also have $2B+C=A$, so we can set $A-B=r,C-A=2r$. So, if $P\in E^n(\mathbb{Q}_p)-E^{n+1}(\mathbb{Q}_p)$, we can assume that $P=[p^nx':y':p^{3n}z']$. Thus the equation becomes $p^{3n}y'^2z'=p^{3n}x'^3+ap^{7n}x'z'^2+bp^{9n}z'^3$. We can divide both sides by $p^{3n}$ and modulo $p$, thus we see that $P'=[x',y',z']$ satisfies the equation $Y^2Z=X^3$ in $\mathbb{P}^2_{\mathbb{F}_p}$.

A very important fact about this singular curve $Y^2Z=X^3$ is that, it has an affine group structure! We can assume $Y=1$ and the equation becomes $Z=X^3$. There is an important parametrization of this curve. That is $[x:1:x^3](x\in K)$. Using this parametrization, we can show that, the map $E(K)-\{[0:0:1]\}\rightarrow K,[x:1:x^3]\mapsto x$ is a group morphisme.

So, now we can conclude the proof for (2) by noting that the curve $Y^2Z=X^3$ over $\mathbb{F}_p$ is just isomorphic to $\mathbb{F}_p$.

The proof of the first point (1) is a bit tricky. Since $\mathbb{P}^2_{\mathbb{F}_p}$ is compact, so is $E(\mathbb{Q}_p)$. What is more, $E^0(\mathbb{Q}_p)$ is an open subgroup of $E(\mathbb{Q}_p)$, then $E(\mathbb{Q}_p)/E^0(\mathbb{Q}_p)$ is compact, discret, and thus finite.

The other two propositions (2) and (4) are not hard to prove.