In the study of elliptic curves, specifically in Weierstrass form, you have the equation

$E : y^2 = x^3 +ax +b$.

However I have found the discriminant comes in two different forms:

$\Delta = -16(4a^3 + 27b^2) $ or $\Delta = 4a^3 + 27b^2$

I understand how to get the second equation, but where does the $-16$ come from?

From the Wiki page: "Although the factor −16 is irrelevant to whether or not the curve is non-singular, this definition of the discriminant is useful in a more advanced study of elliptic curves."


1 Answer 1


A cubic over $k$ in Weierstrass form (affine form) is given by $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ The discriminant is defined by $$\Delta = -b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6,$$ where $ b_2=a_1^2+4a_2$, $ b_4=2a_4+a_1a_3 $, $b_6=a_3^2+4a_6$ and $b_8 = a_{1}^{2} a_{6}+4 a_{2} a_{6}-a_{1} a_{3} a_{4}+a_{2} a_{3}^{2}-a_{4}^{2}$.

Finally, a elliptic curve over $k$ is a cubic in Weiertrass form, where $\Delta \neq 0$ (i.e., is a non-singular cubic in Weiertrass form).

We can make some substitutions to simplify the equation of cubic in Weiertrass form, the first assumes $char(k)$ is not $2$. Replacing $y$ by $ \frac{1}{2} \left(y-a_1x-a_3\right)$, the result is $$y^2=4x^3+b_2x^2+2b_4x+b_6.$$

The second assumes in addition that $char(k) \neq 3$. Replace $(x,y)$ by $\left( \frac{x-3b_2}{36}, \frac{y}{108} \right) $, and the result is $$y^2=x^3-27c_4-54c_6,$$ where $c_4=b_2^2-24b_ 4$ and $c_6=-b_2^3+36b_2b_4-216b_6.$

Moreover, when $char(k)$ not is $2$ or $3$, we have $$ 1728\Delta=c_4^3 - c_6^2. $$

Now, consider the cubic $y^2=x^3+ax+b$ over $k$. If $char(k)$ not is $2$ or $3$, we have $c_4=-48a $ and $c_6=-864b$, so $$\Delta = \frac{(-48a)^3-(-864)^2}{1728} = -16(4a^3+27b^2).$$

Thus, assuming that $char(k)\neq2$ and $char(k) \neq 3$, an elliptic curve over $k$ is given by $$y^2=x^3+ax+b,$$ where $\Delta=-16(4a^3+27b^2) \neq 0$.

Note that, $\Delta=-16(4a^3+27b^2) = 0$ if, and only if, $4a^3+27b^2 = 0$, because $16=2^4 \neq 0$ in $k$ with $char(k) \neq 2$. Thus, the factor $−16$ is irrelevant in this case.

See Chapter III of the book 'Elliptic Curves' of Anthony Knapp for more information.

  • 1
    $\begingroup$ I think there might be a typo in the answer, because mathworld gives $b_2 = a_1^2 + 4a_2$. link: mathworld.wolfram.com/…. $\endgroup$
    – eatfood
    Commented Sep 12, 2020 at 7:47

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