# Elliptic Curve Discriminant

Here is my attempted approach to prove that the discriminant $$\triangle = 4a^3+27b^2$$ of an elliptic curve in the form of $$y^2 = x^3 + ax +b$$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the process. I try to approach this from the definition of non-singularity (part of the definition of an elliptic curve), which is equivalent to the statement: the equation $$y^2 = x^3 + ax +b$$ is differentiable everywhere on the graph. From that definition, I try to derive the derivative of $$y$$ in respect to $$x$$ through implicit differentiation: \begin{align} y^2 &= x^3+ax+b\\ \label{ref1} 2yy' &= 3x^2+a\\ y' &= \frac{3x^2+a}{2y} \end{align} If the graph is singular, then $$y'$$ does not exist, in other words: \begin{align} 2y &= 0\\ 3x^2+a &\neq 0 \end{align} I replace $$y = 0$$ in the equation of elliptic curve which yields $$$$0 \ = \ x^3 + ax + b$$$$ I apply Cardano's method, who realize that the form could be represented as $$$$(\alpha-\beta)^3 + 3\alpha\beta(\alpha-\beta) = \alpha^3 - \beta^3$$$$ in which \begin{align} \alpha\beta &= \frac{a}{3} \\ \alpha^3 - \beta^3 &= -b \end{align} By substituting $$\alpha = \frac{a}{3\beta}$$ in the second equation from above, I obtain $$$$(\frac{a}{3\beta})^3 - \beta^3 = -b$$$$ I further simplify this by considering $$\beta^3$$ as a whole, i.e. \begin{align} \frac{a^3}{27}-\beta^6 &= -b\beta^3\\ (\beta^3)^2 - b\beta^3 -\frac{a^3}{27} &= 0 \end{align} By quadratic formula, $$$$\beta^3 = \frac{b\pm \sqrt{b^2+\frac{4a^3}{27}}}{2} = \frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}$$$$ From $$\alpha^3 - \beta^3 = -b$$, I get $$$$\alpha^3 = \beta^3 - b = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}$$$$ From $$3x^2+a \neq 0$$: $$$$3\left(\left(-\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3} - \left(\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3}\right)^2 + a \neq0$$$$ The equation above is from $$3(\alpha-\beta)^2+a \neq 0$$. By De Moivre's Formula there are two other equations, $$3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$$ and $$3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})^2-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$$. These result in several representations of b in terms of a and one integer solution, $$a=0,b=0$$ shouldn't exist at the same time, $$b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$$. These two meet my expectation, $$\triangle = 0$$. However, there are other four representations that do not meet my expectation $$b \neq \pm \sqrt{\frac{a^3}{54}\pm\frac{5ia^3}{6\sqrt{3}}}$$ and $$b \neq \pm \sqrt{\frac{a^3}{54}\mp\frac{5ia^3}{6\sqrt{3}}}$$. I also try to see what I could get from $$$$3\left(\left(-\frac{b}{2}\right)^\frac{1}{3} - \left(\frac{b}{2}\right)^\frac{1}{3}\right)^2 + a \neq0$$$$ which yields that $$a=0,b=0$$ shouldn't exist at the same time, or $$b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$$ these satisfy $$\triangle = 0$$ and also a weird pair of $$b \neq \pm \frac{ia^{3/2}}{12\sqrt{3}}$$, which doesn't even satisfy $$\triangle = 0$$. Could someone explain this?

When $$3x^2 + a \ne0$$ and $$2y = 0$$, the curve is still nonsingular, its just that the slope of the tangent line is infinite. The singularity occurs when $$3x^2 + a = 0 \text{ and } 2y =0$$ simultaneously i.e. $$a = - 3x^2$$ so $$0 = x^3 + (-3x^2)x + b$$ giving $$b = 2x^3$$ this is what leads to the equation $$27b^2 + 4a^3 = 0$$