 Pole orders on a smooth cubic in Weierstrass form

For $y^2={\prod }_{j=1}^3(x-{\lambda }_j)$ with distinct roots in characteristic different from two, the unique point at infinity has local coordinates $u=X/Y$, $v=Z/Y$ satisfying $v=u^3+O(u^5)$. Thus $x=u/v$ and $y=1/v$ have poles two and three. Regular functions off infinity have the unique form $A(x)+yB(x)$; the even and odd leading pole orders cannot cancel. A basis of $L(n{P}_{\infty })$ consists of $x^j$ with $2j\le n$ and $y{x}^j$ with $3+2j\le n$. This gives dimension one at $n=0$ and dimension $n$ for $n\ge 1$ without Riemann-Roch theorem.