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) $\iff$ $4a^3 + 27b^2 = 0$


1 Answer 1


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 $b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.