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I'm learning some basic content about Riemann Surface with Farkas&Kra's Riemann Surface. In Section III.10.3 I saw The Corollary 2(Outline of the proof can be found in How to prove that a Compact Riemann Surface is a projective algebraic curve? too):

a. Every non-hyperelliptic surface of genus 3 is a submanifold of $\Bbb{C}P^2$

b. Every other surface of genus $g \geq 2$ is a submanifold of $\Bbb{C}P^3$.

And from Is there a complex surface into which every Riemann surface embeds? I know that no Riemann surface of genus 2 can be embedded in $\Bbb{C}P^2$.

Hence it means every Riemann surface of genus 2 cannot be wrote as $F(z, w) = 0$ for some polynomial in $\Bbb{C}[z, w]$, or equivalently it cannot be wrote as $F(Z, W, X) = 0$ for some homogenous polynomial in $\Bbb{C}[Z, W, X].$ It has to be some algebraic curve of the form $\{(z, w, u)| f(z,w, u)=0, g(z, w,u)=0\}$, or equivalently $\{[Z, W, U, X]| F(Z, W, U, X)=0, G(Z, W, U, X)=0\}$.

But I saw a construction for a genus 2 Riemann Surface in https://mathoverflow.net/questions/401141/:

$$ y^3=\frac{z^2-1}{z^2+1} $$

Hence it means $\{[Z, Y, X]|(Z^2+X^2)Y^3=X^3(Z^2-X^2)\}$ is a genus 2 Riemann Surface in $\Bbb{C}P^2$. Since it's the zero set of a homogeneous polynomial of three complex variables, it can be holomophicly embeded into $\Bbb{C}P^2$(is this argument correct?), which seems to be a confliction with the fact that genus 2 Riemann surface cannot be embeded in $\Bbb{C}P^2$.

Did I understand something wrong? Originally I thought all compact Riemann surface can be wrote as $f(z, w)=0$ but then the corollary tells me it's no the case. And then I found this example, it seems to be a contradiction and makes me confused.

Edit: Setting $X=0$ in $(Z^2+X^2)Y^3=X^3(Z^2-X^2)$, I got $Z^2Y^3 = 0$, which seems to contain some singularities in the infinity line. Is that for this reason the curve does not contradict with the fact that no Riemann surface of genus 2 can be embbed in $\Bbb{C}P^2$?

And in Farkas&Kra's Book section 0.3, I saw

Examples of compact Riemann surfaces of genus $g$ are the surfaces of the algebraic functions: $$ w^2 = \prod_{j=1}^{2g+2}(z-e_j), e_j \neq e_k \text{ for } j\neq k $$

For $g > 1$ it has some singularities in the infinity line so it doesn't contradict the fact genus 2 Riemann surface cannot be embeded in $\Bbb{C}P^2$ too? Is this right?

Thanks!

onRiv
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    Nowhere did the OP in the linked question say that it can be embedded in CP^2. – Cranium Clamp Jun 20 '22 at 15:58
  • @CraniumClamp Does writting a Riemann Surface as a zero set of a polynomial of two variables mean that it can be embedded in $\Bbb{C}P^2$? I thought it’s the way – onRiv Jun 20 '22 at 16:08
  • @CraniumClamp Thanks. You are right. The algebraic equations I posted doesn't mean it‘s an embedding. In Griffiths‘s Introduction to algebraic curve， I found a basic fact(Theorem 2.2) that solves my confusion too. – onRiv Jun 21 '22 at 10:32
  • It says that any plane algebraic curve can be normalized to a compact Riemann surface and any compact Riemann surface $\tilde{C}$ can be through the normalization of a certain plane algebraic curve $C$ at most ordinary double points. – onRiv Jun 21 '22 at 10:41

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