Chapter II Example 3.3 -p.28 Silverman I have a question about Chapter II Example 3.3 -p.28 in Silverman "Arithmetic of Elliptic Curves". I feel like I'm misreading it and would like clarification.
Let $K$ be a field such that $\mathrm{char}(K)\neq 2.$ Let $e_1,e_2,e_3\in \bar{K}$ be distinct, and consider the curve :
$$C:y^2=(x-e_1)(x-e_2)(x-e_3).$$
One can check that $C$ is smooth and that it has a single point at infinity, which  we denote by $P_{\infty}.$ For $i=1,2,3,$ let $P_i=(e_i,0)\in C.$ He claims that
$$\mathrm{div}(x-e_i)=2(P_i)-2(P_{\infty}).$$
What I don't see is why
$$\mathrm{ord}_P(x-e_i)=0\;\;\text{for all } P\in C\;\; P\neq P_i,P_\infty$$
and
$$\mathrm{ord}_{P{_i}}(x-e_i)=2\;\;\mathrm{ord}_{P{_{\infty}}}(x-e_i)=-2.$$
Thanks.
 A: Silverman's book is advancing at a relatively quick pace early on if you have zero experience with local parameters of curves and projective spaces in general. The material prior to this is to some extent only a quick review. Spend more time with chapter I and its exercises.
A general useful fact about all algebraic plane curves is:
Fact. Assume that the point $P=(x_0,y_0)$ is on the curve $F(x,y)=0$. If $\dfrac{\partial F}{\partial x}(P)\neq0$ then $y-y_0$ is a local parameter at $P$. Similarly,
If $\dfrac{\partial F}{\partial y}(P)\neq0$ then $X-x_0$ is a local parameter at $P$.
Applied to your example curve this implies that $y$ is a local parameter at all the three points $P_i,i=1,2,3.$ We can also see this directly as follows. The maximal ideal $M=M_{P_1}$ is generated by $y-0=y$ and $x-e_1$. But the equation of the curve implies that
$$
x-e_1=\frac{y^2}{(x-e_2)(x-e_3)}.
$$
Here $y^2\in M^2$ and $(x-e_2)(x-e_3)$ is non-zero at $P_1$, so we see that $x-e_1\in M^2$. Therefore $M/M^2$ is spanned by the coset of $y$, and (in accordance with The Fact) $y$ is a local parameter at $P_1$. A similar argument works for $P_2$ and $P_3$.
Anyway, $\operatorname{ord}_{P_1}((x-e_2)(x-e_3))=0$ and we arrive at 
$$\operatorname{ord}_{P_1}(x-e_1)=\operatorname{ord}_{P_1}(y^2)=2.$$
Because $P_1$ is the only affine point with $x$-coordinate equal to $e_1$ we automatically get $\operatorname{ord}_{P}(x-e_1)=0$ for all $P\neq P_1,P_\infty$.
To see what happens at $P_\infty$ we need to move to a different affine chart. In homogeneous coordinates $[X,Y,Z]$ the equation of the curve is (the usual homogenization business) $$Y^2Z=(X-e_1Z)(X-e_2Z)(X-e_3Z).\qquad(*)$$ The point $P_\infty=[0,1,0]$
is only on the affine chart $Y\neq0$, so we dehomogenize by introducing the variables
$u=X/Y=x/y$ and $v=Z/Y=1/y$ and rewrite $(*)$ as
$$
v=(u-e_1v)(u-e_2v)(u-e_3v).\qquad(**)
$$
At $P_\infty$ we have $u=v=0$. Because the partial derivative w.r.t. $v$ of
$v-(u-e_1v)(u-e_2v)(u-e_3v)$ does not vanish at $(u,v)=(0,0)$ The Fact implies that $u$
is a local parameter at $P_\infty$. Because also $v\in M_{P_\infty}$ the equation $(**)$ then implies that $\operatorname{ord}_{P_\infty}(v)=3$. But $u=x/y$ and $v=1/y$, so we can solve that $\operatorname{ord}_{P_\infty}(x)=-2$ and
$\operatorname{ord}_{P_\infty}(y)=-3$. Therefore also
$$
\operatorname{ord}_{P_\infty}(x-e)=-2,
$$
because the constant $e_1$ does not affect the pole order.
