I'm confused about a relatively simple algebraic geometry computation. Let's take the affine plane curve $x^3-y^2=1$; this is an open subset of an elliptic curve. Indeed, if we embed affine space into projective space in the usual way, we get the genus one projective curve $x^3-y^2z=z^3$; so far so good.
Now consider the weighted projective space $\mathbb{P}(2,3,1)$ with coordinates $x_2,y_3,z_1$. The affine chart away with $z_1=1$ is just the $(x_2,y_3)$-plane, so we can embed our above curve into $\mathbb{P}(2,3,1)$ via $x\mapsto x_2$, $y\mapsto y_3$; the completion inside the weighted projective space is then given by the homogenous equation $x^3_2-y^2_3 = z_1^6$.
Consider now the affine chart $x_2=1$; this is the quotient by $\mu_2$ acting with weights $(3,1)=(1,1)$ on $y_3,z_1$, so its ring of functions is $\mathbb{C}[y_3^2,z_1^2,y_3z_1]$. Write $a=y_3^2,b=z_1^2,t=y_3z_1$, so that this ring can be written $\mathbb{C}[a,b,t]/(ab-t^2)$.
Then the restriction of the complete curve $x^3_2-y^2_3 = z_1^6$ to the chart is given by $\mathbb{C}[a,b,t]/(ab-t^2,1-a-b^3)= \mathbb{C}[b,t]/((1-b^3)b-t^2)$. But now we see that this affine piece of the curve is cut out in affine $(b,t)$-space by the degree-$4$ equation $-b^4 +b-t^2=0$; it's easy to verify this is non-singular, so it's a Zariski open inside a smooth genus-$3$ curve.
So the conclusion seems to be that the curve defined by $x^3_2-y^2_3 = z_1^6$ has a piece which is a Zariski open inside a genus-$1$ curve, and also a piece which is a Zariski open inside a genus-$3$ curve, which is impossible. What's going on?
