Relationship between discriminants and smoothness of curves

Question

My understanding of the use of the discriminant in elliptic curve theory is to test whether an elliptic curve in Weierstrass normal form over a field not of characteristic either 2 or 3, $y^{2} = x^{3}+Ax+B$, is smooth, i.e., non-singular. What I am confused about is the relationship between the discriminant and smooth curves. For instance, $y=x^{2}$ over $\mathbb{R}$ has a zero discriminant, yet is a smooth curve.

• Why does a zero discriminant of an elliptic curve tell us that the curve is not smooth, and why does this not seem to apply likewise to other types of curves, such as, quadratics?

• Is it perhaps because smoothness implies non-singularity for elliptic curves, but in general this is not the case?

Answer 1

Given an elliptic curve, we mean a smooth projective plan curve of the form $Y^2Z = X^3+ AXZ^2 +BZ^3 \subset\mathbb{P}^2.$

You can check that the point $(0: 1: 0)$ is always smooth, by choosing the chart $y\neq 0$.

Then you work on the chart $z\neq 0$, then the points satisfy $ 0= F(x,y) = -y^2 + x^3 +ax+ b$. Smoothness means the differential doesn't vanish in this case. If $\frac{\partial F}{\partial y} =0, $ then $y=0. $ Then $0=x^3 + ax + b$, whose partial w.r.t. x zero iff have double roots iff discriminant is zero. So singular iff discriminant = 0.

In conclusion, discriminant, in this case, is related to the part of degree three equations, which can be linked to smoothness. Degree 2 is another case, and it has a different definition of discriminant.

Answer 2

Question: "Why does a zero discriminant of an elliptic curve tell us that the curve is not smooth, and why does this not seem to apply likewise to other types of curves, such as, quadratics? Is it perhaps because smoothness implies non-singularity for elliptic curves, but in general this is not the case?"

Answer: If $f(x,y):=y-x^2$ it follows the jacobian ideal $J(f)$ equals

$$J(f)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(-2x, 1)=(1)$$

Hence the singular scheme $C_{sing}$ of the curve $C$ (which is the zero scheme of the jacobian ideal) is empty.

If $k$ is any field and $F(x,y):=y^n-f(x)$ with $n\geq 2$ an integer and $deg(f)=d$, you may consider the algebraic curve $C:=Spec(A)$ where $A:=k[x,y]/(F(x,y))$. The jacobian ideal $J$ given by

$$J:=(ny^{n-1},-f'(x))$$

defines the singular subscheme $C_{sing}:=V(J) \subset C$. Assume for simplicity $char(k) \neq n$. Let $(a,b)\in C_{sing}(K)$ be a $K$ rational point of $C_{sing}$, where $k\subseteq K$ is a field. It follows $a,b\in K$ and $f(a)=f'(a)=0$. Hence the curve $C$ has a $K$-rational singular point $(a,0)\in K^2$ iff

$$f(a)=f'(a)=0.$$

If $f(x):=a_0+a_1x+\cdots +a_dx^d$ we define the discriminant $\Delta(f)$ to be the resultant $\Delta(f):=Res(f(x),f'(x))$ of the polynomial $f(x)$ and it's derivative $f'(x)$. It follows $\Delta(f)=0$ iff $f(x)$ has a multiple root in $K$, where $K$ is a finite extension of $k$. Hence the discriminant $\Delta(f)=0$ iff the curve $C$ has a $K$-rational singular point, where $k\subseteq K$ is a finite extension.

Note: The discriminant $\Delta(f):=Res(f(x),f'(x))$ is a polynomial in the coefficients of $f(x)$, hence you can calculate $\Delta(f)$ without knowing anything on the roots of $f(x)$ or if the roots of $f(x)$ live in a finite extension $k \subseteq K$. There is a well known determinantal formula for the resultant $Res(f(x),g(x))$ of any pair of polynomials $f(x),g(x)$ - the "Sylvester determinant".

As remarked in the comments: If $k$ is not algebraically closed you may get singular points defined over non-trivial extensions $k \subseteq K$. The curve $C$ is finite type over $k$ and closed points $x\in C$ have residue field $k \subseteq \kappa(x)$ which is a finite extension in general. Hence points of $C_{sing}$ are not in general defined over the base field $k$.

Example: If $k=\mathbb{Q}$ is the field of rational numbers and $f(t):=g(t)^2$ where $g(t):=t^2-2$. Let $F(x,y):=y^n-f(t)$. It follows

$$f'(t)=2g(t)g'(t)=4t(t^2-2) \in k[t].$$

Hence $f(t)=0$ and $f'(t)=0$ does not have a solution in $k$. It has a solution in $k(\sqrt{2})$. This phenomenon also happens in characteristic $p>0$.

Note: If the polynomial $f(x)\in k[x^p]$ is a polynomial in $x^p$ it follows $f'(x)=0$ hence $\Delta(f)=Res(f(x), f'(x))$ is identically zero in characteristic $p>0$. Over the algebraic closure $\overline{k}$ it follows $f(x^p)=g(x)^p$, hence $f(x)$ has multiple roots and the curve is singular.