# Smoothness of elliptic curves and separable polynomials

Let $$E: y^2 = f(x)$$ be plane curve defined over a field of characteristic $$0$$ where $$f \in K[x]$$ is a cubic polynomial. In order for $$E$$ to be an elliptic curve, it must be smooth which means $$(\frac{\partial g}{\partial x}(P), \frac{\partial g}{\partial y}(P)) \neq (0,0)$$ for all points $$P \in E$$ where $$g = y^2 - f(x)$$.

Question Is it true that $$E$$ is smooth if $$f$$ is a separable polynomial?

I think that we may can use the fact that $$f$$ is separable if and only if $$\gcd(f,f')=1$$. But I don't know if it is possible to establish a connection to the definition of smoothness.

• $\frac{\partial g}{\partial x}(a,b)=0$ means that $f'(a)=0$ and $\frac{\partial g}{\partial y}(a,b)=0$ means that $b=0$ so $f(a)=0$. Commented Oct 13, 2020 at 18:30
The discriminant of a polynomial of positive degree is zero if and only if the polynomial has a multiple root. If $$f$$ is separable over a field of characteristic zero, this is not the case. Hence the discriminant is nonzero and $$E$$ is smooth.