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 $M_p$.
I want to understand the proof of this, that is, the uniformizer is a generator. I firstly tried to show $t$ is a generator of $M_p/M_p^2$ and use Nakayama's lemma. Maybe this is overthinking.I believe this statement has quite easy proof.
Thank you in advance.
Given $g \in M_p$, then $g(p) = 0$ so $\DeclareMathOperator{\ord}{ord} \ord_p(g) \geq 1$. Since $\ord_p(t) = 1$, then $$ \ord_p(g/t) = \ord_P(g) - \ord_p(t) = \ord_P(g) - 1 \geq 0 $$ so $g/t \in \overline{K}[C]_p$ since $\overline{K}[C]_p$ is the valuation ring of $\ord_p$. Then there exists a function $h \in \overline{K}[C]_p$ such that $g = ht$, so $g \in (t)$. Thus $M_p \subseteq (t)$ and the other inclusion is immediate.
(You certainly could use Nakayama's lemma to prove this. Since $M_p/M_p^2$ is $1$-dimensional and $t \notin M_p^2$, then $\{t\}$ is a basis for $M_p/M_p^2$.)