Computing the order of $dx$ at the infinity point of an elliptic curve I am having trouble to figure out the order of the differential form $dx$ is on the infinity point $P=[0:1:0]$ of the elliptic curve $C$:
$$y^2 = (x-e_1)(x-e_2)(x-e_3)$$
I want to compute this straight from the definitions instead of using any theorem. My first step was to find a uniformizer. For that, I computed
$$\mathfrak m_{[0:1:0]}/\mathfrak m^2_{[0:1:0]} = (x,z)/(x^2,xz,z^2)$$
From homogenizing the equation of the elliptic curve and expanding the left hand side, one sees that $zy^2= x^3 +... \in \mathfrak m^2_{[0:1:0]}$. Since $y^2$ doesn't vanish at $P=[0:1:0]$, it is a unit in the local ring $C_P$. Hence $z\in \mathfrak m^2_{[0:1:0]}$ and $x$ is a uniformizer, which we name '$t$'.
Now, by definition $\operatorname{ord}_P(dx) = \operatorname{ord}_P(dx/dt) = 0$. I know that I made a mistake, for according to my reference this should be $=-3$. But where was it?
 A: Let $X,Y,Z$ be the homogeneous coordinates of $\mathbb P^2$ so that  $x=X/Z,y=Y/Z$.
At $P$ we choose the coordinates $t=X/Y, u=Z/Y$ and then $P$ is given by $t=u=0$.
As you write, $t$ is a uniformizing parameter for the curve $C$ at its point $P$ since the equation of $C$ in the affine plane $Y\neq0$  (with coordinates $t,u$) is $$u=t^3-\cdots-e_1e_2e_3u^3 =\text {a homogeneous polynomial of degree 3 in t,u}\quad (\ast)$$ Let us come back to your question.
 We have $x=X/Z=t/u$. Now in the local ring $\mathcal O_{C,P}$, a discrete valuation ring,  equation $(\ast)$ implies that $u$ has valuation $3$.
Hence $x=X/Z=t/u$  has valuation $-2$ so that indeed $\operatorname{ord}_P(dx)=-3$, just as your text claims.   
Edit: a complement
It is easy to see that the order of $dx$ is $1$ at $P_i=(e_i:1:0)$, so that the divisor of $dx$ is $$\text {div}\:dx=1.P_1+1.P_2+1.P_3-3.P$$ The degree of that divisor is $0$ : this is in line with the fact that the canonical  bundle $K_C$, of which $dx$ is a rational section, has degree $0$. 
