Is it true that every singular quartic curve is hyper-elliptic?

Question

Every hyper-elliptic curves are singular. Is the converse holds in case of quartic curve? I can't find any counterexamples..

Answer 1

The definition of a hyperelliptic curve you may be familiar with is "a curve which is given by $y^2 = f(x)$, where $f(x)$ is a degree $n$ polynomial with $n$ distinct roots".

What you are misunderstanding is the meaning of "given". In this case, the definition means that any such equation defines an affine curve, and the corresponding hyperelliptic curve is then the smooth completion of the affine curve. For $n\ge 4$, the projective plane curve obtained from the given equation is singular, and thus is not the hyperelliptic curve corresponding to that equation. To obtain the actual hyperelliptic curve, one must desingularize that singular projective plane curve, which for curves is equivalent to taking the normalization.

Another way to think about it is this: Given the affine plane curve, its function field is the fraction field of $k[x,y]/(y^2-f(x))$. This function field is then a field of transcendence degree 1 over $k$, inside which $k$ is algebraically closed. By the curves-fields correspondence, this function field corresponds to a smooth projective curve, which by definition is the hyperelliptic curve associated to $y^2 = f(x)$.