 Solution

For a divisor $D$ on a smooth projective curve, let $L(D)=\{f\in k(X):\mathrm{div}(f)+D\ge 0\}\cup \{0\}$, and write $\ell (D)={\dim }_{k}L(D)$. The Riemann-Roch theorem stateswhere $g$ is the genus, $K_X$ is a canonical divisor given by a nonzero rational differential, and the degree of a divisor is the sum of its coefficients since $k$ is algebraically closed.

For genus one, a nonzero regular differential has a nonnegative divisor of degree $2g-2=0$, hence no zeros; therefore $K_X\sim 0$. Negative-degree divisors have no nonzero sections. Thus $\ell (n{P}_{\infty })=n$ for $n>0$, $\ell (0)=1$, and $L(n{P}_{\infty })=0$ for $n<0$.

Choose $x\in L(2P_{\infty })\setminus L(P_{\infty })$ and $y\in L(3P_{\infty })\setminus L(2P_{\infty })$. Their unique poles have orders two and three respectively. Distinct pole orders prove linear independence in the following complete list:Since $y^2$ also lies in the last six-dimensional space, it satisfiesThe coefficient $a$ cannot vanish because only $x^3$ among the displayed basis has a sixth-order pole available to cancel the leading pole of $y^2$. Homogenizing gives a plane cubic relation, valid also in characteristics two and three without completing squares or dividing by three.

It remains to prove that the map actually embeds the curve. For $D=3P_{\infty }$, Riemann-Roch gives $\ell (D)=3$, $\ell (D-Q)=2$ for every point $Q$, and $\ell (D-Q-R)=1$, including $R=Q$. The first equality makes the linear system base-point-free; the second separates distinct points, and the double-point case separates tangent directions. Therefore $D$ is very ample and ${\varphi }_D$ is a closed immersion into ${\mathbb{P}}^2$. Its pullback of a line has degree three, so its image has degree three. The displayed cubic equation consequently defines the image, which is smooth because it is isomorphic to $X$. Thus $\boxed{{\varphi }_{3P_{\infty }}:X\cong \text{a smooth plane cubic}}$. The affine expression $(1:x:y)$ extends across $P_{\infty }$ to $(0:0:1)$ by the pole orders.