Bounds for the number of points on projective hyperelliptic curves over finite fields

Question

Let $C$ be projective hyperelliptic curve over finite field $K$.

What are bounds for the number of points $\#C(K)$?

The Hasse-Weil bound requires smooth curves, and hyperelliptic curves are not smooth if the degree is greater than three, having only one singular point.

We have explicit example violating the Hasse-Weil bound with only one singular point.

abx claims "A projective hyperelliptic curve is smooth by definition."

For counterexample to abx take $x^6+z^6=y^2 z^4$ and observe that it has singular point over the rationals.

Take $p=29,K=GF(p), C: -4 x^4+z^4-z^2 y^2=0$.

According to sage, this hyperelliptic curve violate the Hasse-Weil bound, what is wrong with this argument?

sage session:

p=29
K.<x,y,z>=GF(p)[]
f= -4*x^4+z^4-y^2*z^2
C=Curve(f)
print(abs(p+1-len(C.rational_points())),2*sqrt(1.0*p))
#11 10.7703296142690

Added Related question Why this genus one curve over F_5 appear to violate Hasse-Weil bound?

Answer 1

For simplicity, let $q$ be odd. The question is to bound $\mathbb{F}_q$ points on $y^2 = f(x)$ when there are singularities. Write $f(x) = u(x) v(x)^2$ where $u$ is square free. Let $X$ be the affine curve $y^2 = f(x)$ and let $\tilde{X}$ be the normalization $z^2 = u(x)$, so these are birational by $(x,z) \mapsto (x, v(x) z)$.

Let the degree of $u$ be $2g+r$ where $r=1$ or $2$. Then $\tilde{X}$ is a genus $g$ curve with $r$ punctures, so the Weil bounds for $\tilde{X}$ are $$-2 g \sqrt{q}-r \leq \#\tilde{X}(\mathbb{F}_q) - q -1 \leq 2 g \sqrt{q}.$$

Let $m$ be the number of distinct roots of $v$ in $\mathbb{F}_q$. At each root of $v$, the map $(x,z) \mapsto (x, v(z) x)$ might create a new $\mathbb{F}_q$-point (if $x$ is in $\mathbb{F}_q$ and $z$ is not) or might merge two $\mathbb{F}_q$-points into one. (It also might send one $\mathbb{F}_q$-point to one such point, if $z=0$.) So the bounds for the original affine curve are $$-2 g \sqrt{q}-r-m \leq \#X(\mathbb{F}_q) - q -1 \leq 2 g \sqrt{q}+m.$$

Finally, if you want to count points on the closure of $X$ in $\mathbb{P}^2$, add in one more point at $\infty$, giving $$-2 g \sqrt{q}-r-m+1 \leq \#\overline{X}(\mathbb{F}_q) - q -1 \leq 2 g \sqrt{q}+m+1.$$