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


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.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s