I've read somewhere an MSE that we can understand the normalized valuation of a polynomial in the function field of a curve at a smooth point as the first non-vanishing coefficient (or exponent) of a corresponding laurent series, meaning that (using Silvermans notation) for the function $f \in \overline{K}(C)$ it's sufficient to get find a local parametrization g of C in a neighborhood of P with $g(0) = P, g'(0) \neq 0$. Then if $$f(g(t)) = \sum_{i \geq n} a_i t^i$$ with $a_n \neq 0$ then $\text{ord}_P(f) = a_n$ (or just the exponent n?).
The poster then argued that $t \mapsto (x(t),t^2)$ is a parametrization of $y^2 = x^3 + x$ with $x(t) = t - t^3+3t^5-12t^7+O(t^9)$ being the inverse series of $x^3+x$ and with this $\text{ord}_P(y) = 1$.
But I don't understand this argument at all. Why use the inverse series? Why is this indeed a parametrization? How exactly can I now get the order? Can I also conclude now a statement about the order of x?
I just can't find the post for the life of me and almost forgot about it until I read in the MIT course notes on elliptic curves:
This is clear when $k = \mathbb{C}$ and in general one can formally define the Laurent series expansion g ∈ k((t)) of f at P so that $\text{ord}_P(f)$ is the exponent of its leading term.
So there is indeed a connection but it's quite difficult for me to grasp it and use it to find seeked orders.
Edit: Found the link (What does the power of an ideal *mean*?), but I think there has to be a mistake since he actually argues that $$\text{ord}_P(x) = 1$$, which cant be the case since $$\text{ord}_P(x) = \text{ord}_P(y^2-x^3) = \min\{\text{ord}_P(y^2),\text{ord}_P(x^3)\} = 2$$