# Short Weierstrass equation is non-singular for not 2 or 3 characteristic

Consider a field $K$ of characteristic $p \neq 2,3$. Consider a curve $E$ over $K$ defined by the equation $y^2 = x^3 + ax + b$.

How can I show that:

$E$ is not an elliptic curve (it is not singular) $\iff$ $4a^3 + 27b^2 = 0$

By definition, a point on the affine curve $$E\colon\quad \underbrace{y^2-x^3-ax-b}_{=:f}=0$$ is singular if and only if the Jacobi matrix $$J_f =\Big(\frac{\partial}{\partial x}f,\frac{\partial}{\partial y}f\Big) =(-3x^2-a,2y)$$ does not have maximal rank at that point, that is (here), vanishes. Hence precisely the points $(x,0)$, where $x$ annihilates both $x^3+ax+b$ and its derivative $3x^2+a$, are singular points of $E$. Therefore, such a point exists if and only if $x^3+ax+b$ has a multiple root, and this is the case if and only if its discriminant $$\Delta=-4a^3-27b^2$$ is zero.
To be precise, it still remains to prove that the point at infinity $[0:1:0]$ is always non-singular: On the affine patch $\{y=1\}$, the (projectivized) curve equation $y^2z-x^3-axz^2-bz^3$ dehomogenizes to $z-x^3-axz^2-bz^3$. Its Jacobi matrix $(-3x^2-az^2,1-2axz-3bz^2)$ becomes $(0,1)$ at the point at infinity $(0,0)$, hence is non-degenerate for any $a$ and $b$.