uniformizer is a generator for the maximal ideal from Silverman's AEC I cannot understand why uniformizer is a generator for the maximal ideal in "Joseph H. Silverman, The Arithmetic of Elliptic Curves", which is as follows:
$$\text{ord}_p: \bar K[C]_p \to \{ 0,1,\dots \} \cup \{ \infty \}$$
where $C$ is a curve and $$\text{ord}_p(f) = \sup \{ d \in \mathbb N : f \in M^{d} _p \} , \\ M_p = \{ g \in \bar K [C] : g(p) = 0 \} .$$
A uniformizer for $C$ at $P$ is any function  $ t \in \bar K [C]$ with $\text{ord}_p(t)＝1$
Silverman says uniformizer is a generator for the ideal $M_p$.
But I cannot understand this. For example,
$C$:$Y^2＝X^3＋X$
Then, $\text{ord}_p(Y)＝1$
But $Y$ does not generate  $M_p =(X,Y)$.
I think I'm making a mistake, so I'd like you to correct the mistake and give me proof of the fact that the uniformizer is a generator. Thank you in advance.
 A: I think you're confusing $p$ with $P$. Assuming you meant $P$, your question follows from standard commutative algebra. As stated in the other answer, $\bar{K}[C]_P$ is the local ring of $\bar{K}[x,y]$ at $P$:
$$
\bar{K}[C]_P = \{f/g \in \bar{K}(x,y) : g(P) \neq 0\}
$$
$\bar{K}[C]_P$ has exactly one maximal ideal, namely $M_P \bar{K}[C]_P$. To prove this, simply observe that $M_P$ is a maximal ideal (essentially proved in Silverman, on p. 5 -- the quotient is $\bar{K}[C]_P / M_P \bar{K}[C]_P = \bar{K}$ which is a field), and every element of $\bar{K}[C]_P \setminus M_P$ is a unit, so there can be no other maximal ideals. By Proposition II.1.1 of Silverman (equivalently, Exercise 2.1 of Silverman), $\bar{K}[C]_P$ is a discrete valuation ring. (A discrete valuation ring is, by definition, a principal ideal domain with exactly one nonzero maximal ideal.)
From here, the theorem that you want is proved as follows:
Theorem: Let $R$ be a discrete valuation ring with maximal ideal $M$. Let $\pi \in M \setminus M^2$. Then $\pi$ is a generator of $M$ (i.e. $M = \pi R$).
Proof: By hypothesis, $R$ is a principal ideal domain, so $M$ is a principal ideal. Let $t$ be a generator of $M$. By a standard result in algebra, every principal ideal domain is a unique factorization domain. Hence $\pi = u\cdot t^k$ for some unit $u \in M$ and some non-negative integer $k$. If $k=0$, then $\pi = u$ is a unit, contradicting $\pi \in M$. If $k \geq 2$, then $\pi \in M^2$, contradicting $\pi \notin M^2$. Hence $k=1$ and $\pi = ut$, so $t = \pi u^{-1}$. Since $t$ is a multiple of $\pi$, every element of $M$ is a multiple of $\pi$. Furthermore, $\pi \in M$. Hence $M = \pi R$.
A: $\overline{K}[C]_P$ is the local ring at $P$. Taking $P = (0,0)$, we have
$$\overline{K}[C]_P = \frac{\overline{K}[x,y]_{(x,y)}}{(y^2 - x^3 - x)}.$$
We have $y^2 = x^3 + x = x(x^2 + 1)$, but $x^2 + 1$ is a unit in the local ring and so $y^2 = x \cdot (unit)$ in $\overline{K}[C]_P$. Hence $M_P = (x,y)$ is generated only by $y$ and $y$ is a uniformizer at $P$.
