Suppose that is an elliptic curve over . Note, a change of variable gives . So, we can always assume that (sometimes we require also that the discriminant has maximal absolute value , but here we don’t need it).

For each point , we can choose a representative such that yet not all of them lie in (just to make sure that after modulo , this point lies in ). And, we associate to the point . We see easily that this also satisfies the equation . Thus we have defined an application . We can verify that is a group homomorphism. Yet we are not sure that, is still an elliptic curve because we do not know, a priori, whether or not in $latex\mathbb{F}_p$. In any case, we define

.

We can verify that if , then . What is more, since is not singular in , which shows that is a subgroup of .

Moreover, we consider the preimage of in . In other words, we set

.

This shows that is a subgroup of . In a similar way, we can define,

.

So, we get a filtration:

Then there is an important theorem:

**(1) is a finite set;**

**(2)We have an isomorphism ;**

**(3)For each , is a subgroup of , and the map is an isomorphism;**

**(4) .**

An important corollary is

**If , then the multiplication map 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 , then the isomorphism between induces a multiplication map . So, the image of in is not trivial, and , which shows that the image of in is not trivial, either, thus this multiplication map is injective. To show the surjectivity, suppose again that , then there is an element such that . So, there is a such that , and thus . We can continue this process, and find a series of such that . Push this process to infinity, and using the fact that is a complet space, we find . so, we conclude the proof.

So, we see that the isomorphisms are rather important maps. We first prove (3). Suppose that is a subgroup of . Then for any , we can write with units in , yet since , we have that . We know that . It is easy to see that we must have . If , by taking on both sides, we have that , thus . But this contradicts the definition of (**here we use the essential hypothesis that we are working in , not ,nor , and that is why this proposition starts from instead of these latter two**). So, we must . Yet we also have , so we can set . So, if , we can assume that . Thus the equation becomes . We can divide both sides by and modulo , thus we see that satisfies the equation in .

A very important fact about this singular curve is that, it has an affine group structure! We can assume and the equation becomes . There is an important parametrization of this curve. That is . Using this parametrization, we can show that, the map is a group morphisme.

So, now we can conclude the proof for (2) by noting that the curve over is just isomorphic to .

The proof of the first point (1) is a bit tricky. Since is compact, so is . What is more, is an open subgroup of , then is compact, discret, and thus finite.

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