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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!

Alphonse
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  • What is a "projective model" of an affine variety, resp. an "affine model" of a projective variety? – Alphonse Oct 21 '18 at 15:40
  • I found : https://math.stackexchange.com/questions/2777751/ – Alphonse Oct 21 '18 at 16:41
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    I don't know if there's a "definition" of model anywhere, but to me, a projective model of an affine variety $X$ would be a projective variety $Y$ such that $X$ is a subscheme of $Y$ (and similarly for affine model of a projective variety). In each case, there is a kind of natural choice of model - you can take the projective closure of an affine variety under one of the standard open embeddings $\Bbb A^n\to\Bbb P^n,$ and you can take the affine cone over a projective variety (basically, take the equations defining the variety, and interpret them in $\Bbb A^{n+1}$ instead of in $\Bbb P^n$). – Stahl Oct 21 '18 at 22:33
  • See also https://mathoverflow.net/questions/99128 – Watson Jul 05 '20 at 17:47

1 Answers1

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  1. 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."
  2. 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 $C_1.$

Stahl
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  • Thank you very much! I have two little questions: 1) in our case, the curve $w^2 = v^{2g+2} f(1/v) = v^4( v^{-4} + 1) = 1+v^4$ is the same as $y^2=f(x)$. I don't understand what it means to glue these two identical curves along the maps you've written. What do we get concretely? $$ $$ 2) You write "not all curves can be embedded in $P^2$", but what is an example? Typically, the twisted cubic is isomorphic to $P^1$, which can of course be embedded in $P^2$... – Alphonse Oct 22 '18 at 06:46
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    @Alphonse 1) The equation might be the same, but the points on the original curve correspond to different points on the other via the gluing map. For example, the point $(x,y) = (2,\sqrt{17})$ is on the curve, but under the transformation, this is sent to $(1/2,\sqrt{17}/4).$ You're gluing two copies of the same curve, but not along the identity map. If this seems strange, think about how $\Bbb P^1$ is constructed by gluing two copies of $\Bbb A^1$ together via $x\mapsto\frac{1}{x}.$ – Stahl Oct 22 '18 at 07:23
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  • The genus of a plane curve of degree $d$ is given by $g= (d - 1)(d - 2)/2,$ and this can never be equal to $2.$ So any curve of genus $2$ provides a counterexample!
    • – Stahl Oct 22 '18 at 07:23
  • Thank you for your comments. I think you might be interested in my question https://math.stackexchange.com/questions/2965725, which should be the last of this series. – Alphonse Oct 22 '18 at 08:28
  • @Stahl. Could you give me a comment or answer to this question regarding this question ?https://math.stackexchange.com/questions/4679093/definition-of-gluing-two-affine-curves?noredirect=1#comment9894938_4679093 – Poitou-Tate Apr 16 '23 at 05:39