Question about uniformizers of Elliptic Curves Let $k$ be a field with $\operatorname{char}(k)\neq 2,3$ and $E: y^2=x^3+Ax+B$ an elliptic curve over $k$ , where $4A^3+27B^2\neq 0$ and let $P=(\alpha,\beta)$ be a point defined over $k$.
Show that if $\beta\neq 0$, then $x-\alpha$ is a uniformizer of $P$
So basically we want to show that the maximal ideal $(y-\beta,x-\alpha)$ is equal to $(x-\alpha)$ by somehow utilizing the curve $E$. Our teacher showed the following example during a lecture so I'd like to apply a similar strategy, but don't know how.
Example: Let $k$ be as above, $E: x^2+y^2=1$ and $P=(1,0)$, show that $(y)$ is a uniformizer. Then, using $E$:
$y^2=1-x^2=(1-x)(1+x)$. 
Now (1+x) is nonzero at p, so $\frac{1}{1+x}\in\mathcal{O}_{E,p}$, thus $(1-x)=\frac{1}{1+x}\cdot y^2\in(y)$ and thus $(x-1,y)=(y)$ which proves our statement. 
However, I have no idea how to apply this to the complicated example since it won't factor in such a nice way.
 A: Let $f(x) = x^3 + Ax + B$.  Expand $f(x)$ as a Taylor series in $x-\alpha$: $$f(x) = f(\alpha) + A_1 (x-\alpha) + A_2 (x-\alpha)^2 + A_3 (x-\alpha)^3.$$
We know that $f(\alpha) = \beta^2$.  We don't care what $A_1, A_2, A_3$ are.  Thus we can rewrite the equation of the curve $E$ as $$y^2 - \beta^2 = A_1 (x-\alpha) + A_2 (x-\alpha)^2 + A_3 (x-\alpha)^3.$$  Now the left side factors and we can use the same technique as in your example with $x^2+y^2=1$.
This is a useful technique to remember - when you are focusing on a point $(\alpha, \beta)$ of a curve, you should rewrite the equation of the curve using a Taylor expansion around $(x-\alpha)$ (or maybe $y-\beta$, depending on the problem).
A: This question is closely related to your other question, which I recommend you solve first. As this is homework, I'll be brief.
Once you have verified that every $(\alpha,\beta)\in C(k)$ satisfies
$$(y-\beta)(y+\beta)=(x-\alpha)(x^2+\alpha x+\alpha^2+A),$$
all that remains is to show that $y+\beta$ is a unit in $k[C]_P$ if (and only if) $2\beta\neq0$.
