Why 16 in the discriminant of short weierstrass elliptic curve

Question

Let $E:y^2=x^3+ax+b$ be a elliptic curve over finite field $\mathbb{F}_p$ with $char\neq2,3$. Some books say that discriminant is $-16(4a^3+27b^2)$ and some say $-(4a^3+27b^2)$. My calculation using the discriminant formula of polynomial says that the later one is correct. I could not find any reason of 16. I would like to know which one is correct and why.

Answer 1

Question: "Let $E:y^2=x^3+ax+b$ be a elliptic curve over finite field $k:=F_p$ with char≠2,3. Some books say that discriminant is $−16(4a^3+27b^2)$ and some say $−(4a^3+27b^2)$."

Answer: Given an equation $F(x,y):=y^2-f(x)$ where $f(x) \in k[x]$ is a polynomial of degree $d\geq 3$, it follows the jacobian ideal $J(F):=(y,f(x),f'(x)) \subseteq k[x,y]$ gives rise to the singular subscheme $C_{sing} \subseteq C:=Spec(k[x,y]/(F)$. And $C_{sing}=\emptyset$ iff $J(F)=(1)$ is the unit ideal. The ideal $J(F)$ is the unit ideal iff $f(x)$ and $f'(x)$ have no common root in $k$ and this is iff $Discr(f(x),f'(x))\neq 0$. In this case $-16$ is a unit in $k$ hence $−16(4a^3+27b^2)\neq 0$ iff $−(4a^3+27b^2) \neq 0$ in $k$. There is the "Sylvester formula" calculating the discriminant $Discr(f(x),f'(x))$ as a determinant.

Note: In general if $f(x) \in k[x]$ is a polynomial, the discriminant $Discr(f(x),f'(x))$ gives a criterion to test if $f(x)$ has mutiple roots without calculating the roots: There is no general formula for the solutions to the equation $f(x)=0$ when $n:=deg(f(x))\geq 5$. The discriminant test is valid for any $n$.

Note: If you begin with a polynomial $F(x,y):=y^2-f(x)$ with $f(x):=x^3+ax+b)\in \mathbb{Z}[x,y]$ and reduce the polynomial modulo primes $p$, it follows the discriminant $D(f)=u(a,b)$ which is a polynomial in $a,b$, will tell you for which primes $p$ the curve $V(F_p(x,y)) \subseteq \mathbb{A}^2_{\mathbb{F}_p}$ is regular.

Relationship between discriminants and smoothness of curves