 Solution

Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence