Rational parametrization of a 'singular elliptic curve' I consider a curve $C$ on a algebraically closed field  $\mathbb{K}$ of equation $y^2=x^3+ax+b$.
I suppose that the curve is singular (it's equivalent to $4a^3+27b^2= 0$) and i want proof that there is a rational parametrization of $C$. The book where i have read this exercise book from give this hint: let P a singular point of $C$ and observe that a line for $P$ meet $C$ in only one another point of $C$.
It's easy observe that $P=(x_0,0)$ with $x_0$ a multiple root of $x^3+ax+b$.
Then the system between $C$ and a generic line for $P$ is $\begin{cases} y^2=(x-x_0)^2(x-x_1)\\ y=t(x-x_0) \end{cases}$
Then $t^2(x-x_0)^2=(x-x_0)^2(x-x_1)$ and (suppose $x\neq x_0$)  $t^2=x-x_1$ and $x=t^2+x_1$.
Then $y=t(t^2+x_1-x_0)$
My quest is if is necessary suppose $\mathbb{K}$ algebraically closed?
 A: No, there's no need to suppose that $k$ is algebraically closed. In fact, if $\operatorname{char} k \neq 2,3$ we don't need any assumptions at all to reach the conclusions of this problem: in this situation, any elliptic curve can be put in short Weierstrass form with an (affine) equation of the form $y^2=x^3+ax+b$. This has a singularity iff the equations $y^2=x^3+ax+b$, $2y$, and $3x^2+a$ have a common solution. By the characteristic assumptions, $2\neq 0$, so $y=0$ and we need to find a common solution to $x^3+ax+b$ and $3x^2+a$. This happens iff $x^3+ax+b$ and $3x^2+a$ have a common factor.
The common factor must divide $(x^3+ax+b)-\frac{x}{3}(3x^2+a)=\frac23ax+b$, so if $a\neq 0$ then it's $\frac23ax+b$, giving a singular point with coordinates in our base field. If $a=0$, then $b=0$ as well, and the singular point is at $(0,0)$. Either way, we find that $x^3+ax+b=(x-x_0)^2(x-x_1)$ for $x_0,x_1\in k$.
From there, we can construct the map of rings $k[x,y]/(y^2=(x-x_0)^2(x-x_1))\to k[t]$ by $x\mapsto x_1+t^2$ and $y\mapsto t(t^2+x_1-x_0)$, which is well-defined because $t^2(t^2+x_1-x_0)^2=(t^2+x_1-x_0)^2(x_1+t^2-x_1)$. This map of rings gives a  non-constant map from $\Bbb A^1$ to our elliptic curve $X$, which extends to a surjective map $\Bbb P^1\to X$ by the curve-to-projective extension theorem.
None of the above required any assumptions on $k$ besides $\operatorname{char} k\neq 2,3$. As usual in the theory of elliptic curves, the case of characteristic 2 or 3 is a little more involved, but if one assumes that the curve has equation $y^2=x^3+ax+b$ and the base field is perfect, the same conclusion that the result is independent of whether the base field is algebraically closed may be reached. Please leave me a comment if you're interested in the details.
