Nonsingular projective curve model corresponding to $y^2 = x^4+1$ Consider the affine curve $C_1 = V(y^2 - (x^4+1)) \subset \Bbb A^2_k$.
In the answers to this question, they claim that there is a (unique?) nonsingular projective curve $C_2$ corresponding to $C_1$ (using "using weighted projective space, or by gluing affine models", or "blow-ups").
Could someone explain to me : 1) what does it mean that a nonsingular projective curve $C_2$ "corresponds" to $C_1$ ? and 2) what is an explicit equation for $C_2$ (they might be several $C_2$...) ?
The naive idea of a projective curve associated to $C_1$ is the projective closure under the inclusion map $\Bbb A^2 \to \Bbb P^2, (x,y) \mapsto [x:y:1]$, which gives $S_2 = V(y^2z^2 - (x^4 + z^4)) \subset \Bbb P^2_k$. But this is a singular curve.
My question 1) is to understand what we define as being a non-singular projective curve associated to $C_1$. And then my question 2) is to know what this definition gives very explicitly in our specific case.
For question 1), a possible definition would be "there exists an open immersion $j : \Bbb A^2 \to \Bbb P^2$ such that the closure of $j(C_1)$ is a nonsingular curve $C_2$". Or maybe "the (unique) projective smooth curve $C_2$ with function field equal to $Frac(k[x,y]/(y^2-x^4-1))$" as here? Would that answer correctly my question 1)?
Thank you!
 A: *

*The answer to your first question is what you've written: there is an equivalence of categories $$\{\textrm{f.g. field extensions }K/k\textrm{ with }\operatorname{trdeg}_k K = 1\}\simeq\{\textrm{nonsingular projective curves }C/k\}$$ (see theorem 50.2.6 here). When people use the description of an affine curve to refer to a nonsingular projective curve, they typically mean "the nonsingular projective curve whose function field is the same as the function field of the affine curve."

*As for how to find the equation of such a $\tilde{C}$ given the original affine curve $C,$ there are multiple procedures which will compute this for you, as you can see here (start with the projective closure of the original affine curve, and then resolve the singularities).


In general, I don't know of a way to write down the normalization given the equation of the original curve, but for a hyperelliptic curve you can be very explicit. Suppose for simplicity you are given an equation $$y^2 = f(x)$$ corresponding to a hyperelliptic curve over an algebraically closed field of characteristic 0. Then one can show that the curve in $\Bbb A^2$ given by the equation $$w^2 = v^{2g + 2}f(1/v)$$ ($g$ here being the genus of the curve, which you can write down in terms of the degree of $f$) glues together with the original curve via
\begin{align*}
(x,y)&\mapsto (v,w) = \left(\frac{1}{x},\frac{y}{x^{g+1}}\right)\\
(v,w)&\mapsto (x,y) = \left(\frac{1}{v},\frac{w}{v^{g+1}}\right)
\end{align*}
and that the glued curve is smooth. It is a good exercise to work all this out for yourself - try to show that a hyperelliptic curve over an algebraically closed field of characteristic $0$ is always given by an equation of this form! :)
As a last remark, you could not hope that the answer to question 1 would be "there exists an open immersion $j:\Bbb A^2\to\Bbb P^2$ such that the closure of $j(C_1)$ is a nonsingular curve $C_2,$" because not all curves can be embedded in $\Bbb P^2$! If you replace "$\Bbb P^2$" by "$\Bbb P^n$ for some $n$" (even only using $\Bbb P^3$ will suffice, in fact), then you would have a less canonical but equivalent description of the nonsingular projective curve corresponding to $C_1.$
