Discriminant of Elliptic Curves

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."

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.

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