# Using the power series expansion for $w(t)$ to construct a subgroup of the rational points on an elliptic curve.

I'm doing a course on elliptic curves, and I'm stuck on a line in a proof which is supposedly using the uniqueness in Hensel's lemma. Starting with an elliptic curve

$$E:Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ^2+a_6$$

we work in the affine piece $Y \not=0$.

So we let $t=\frac{-X}{Y}$, $w=\frac{-Z}{Y}$, getting

$$w=t^3+a_1tw+a_2t^2w+a_3w^2+a_4tw^2+a_6w^3=:f(t,w).$$

We've found a power series $w(t)$ such that $w(t)=f(t,w(t))$.

Now we're trying to prove

Lemma 7.2

Let $R$ be an integral domain complete w.r.t. an ideal $I$, let $a_1,...,a_6 \in R$ and let $K=\mbox{Frac}(R)$. Then $$\hat{E}(I)=\{(t,w(t))\in E(K)\;|\; t \in I\}$$ is a subgroup of $E(K)$.

In trying to prove closure, we take two points $P_1,P_2\in \hat{E}(I)$ and try to show that the third point on the line $P_1P_2$, $P_3=-P_1-P_2=(t_3,w_3)$, is in $\hat{E}(I)$.

The lecturer claims that once we show $t_3, w_3 \in I$ then we are done, by the uniqueness of the power series in Hensel's lemma. I don't see why this is true. Our statement of Hensel's lemma is:

Let $R$ be an integral domain complete with respect to $I$, and let $F \in R[X]$, $s \geq 1$. Suppose we are given $a \in R$ such that

$\begin{array}{lll}F(a) &\equiv &0 \hspace{5mm} \mbox{(mod }I^s) \\ F'(a) &\in &R^\times.\end{array}$

Then there exists a unique $b \in R$ such that

$\begin{array}{lll}F(b) &= &0 \\ b &\equiv &a \hspace{5mm} \mbox{(mod }I^s).\end{array}$

If we had $w_3 \equiv 0 \;(\mbox{mod }t_3^3)$, then I think by the uniqueness in Hensel's lemma we should have $w_3=w(t_3)$, as required. But I don't see how this is true or how it has anything to do with $w_3$ being in $I$.

I think we are using Hensel's lemma with $F(X) = X - f(t_3,X)$, $a = 0$ and $s = 1$. We have $F(0) = -t_3^3 \equiv 0 \bmod I$ and $F'(0) = 1 - a_1 t_3 - a_2 t_3^2 \in 1+I \subseteq R^\times$, so Hensel's lemma tells us that there is a unique $b \in R$ such that $F(b) = 0$ and $b \equiv 0 \bmod I$. Since $w_3$ and $w(t_3)$ both satisfy this, we get $w_3 = w(t_3)$.