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I am preparing for an exam in (mostly classical) algebraic geometry, and I have some preparatory questions, among which:

Can you write the equations of any nonsingular curve in any projective space which is not rational?

A problem with this question is that we never really defined what a "rational curve" is in class, but from what I can understand looking around, it should be a curve which is birationally equivalent to $\mathbb{CP}^1$.

I have found this beautiful answer on MO, saying that cubic curves are an example since they have genus $1$ and $\mathbb{CP}^1\cong S^2$ has genus $0$. However, if I'm not mistaken, this relies on the fact that two smooth curves are birational iff they are isomorphic, which we didn't see in class.

Is there some simple (and simple to prove) example for this question?

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Daniel Robert-Nicoud
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    The twisted cubic is in fact rational: your parametrisation has a rational inverse, defined away from the singularities $[0 : 0 : 0 : 1]$ and $[1 : 0 : 0 : 0]$ by $[x : y : z : w] \mapsto [y : z]$. – Zhen Lin Aug 16 '14 at 22:21
  • @ZhenLin I see. I'll have to edit my answer, then. Thanks you. – Daniel Robert-Nicoud Aug 16 '14 at 22:42
  • The intersection of two quadrics in $\mathbb P^3$ is an elliptic curve. – Fredrik Meyer Aug 16 '14 at 23:01
  • Does $V(x^2+y^2+z^2)\subseteq\mathbb{P}^2_\mathbb{R}$ work for you? It's smooth, but it's not birationally equivalent (over $\mathbb{R}$) to $\mathbb{P}^1_\mathbb{R}$, since any open of $\mathbb{P}^1_\mathbb{R}$ contains an $\mathbb{R}$-point, but no open subset of $V(x^2+y^2+z^2)$ contains no $\mathbb{R}$-points. – Alex Youcis Aug 17 '14 at 05:45
  • @AlexYoucis Thanks, but I think I have to do it over $\mathbb{C}$... – Daniel Robert-Nicoud Aug 17 '14 at 09:27
  • @DanielRobert-Nicoud What are you allowed to use? Suppose that $E$ is an elliptic curve over $\mathbb{C}$, and it was birational to $\mathbb{P}^1$. Let $U\subseteq E$ and $V\subseteq\mathbb{P}^1$ be isomorphic over $\mathbb{C}$. You have SESs $\text{Cl}(E-U)\to \text{Cl}(E)\to\text{Cl}(U)\to 0$ and $\text{Cl}(\mathbb{P}^1-V)\to\text{Cl}(\mathbb{P}^1)\to\text{Cl}(V)\to 0$. But, $\text{Cl}(E)$ is uncountable, and since $E-U$ is a finite union of points, this implies that $\text{Cl}(E-U)$ is countable, so $\text{Cl}(U)$ is uncountable. But, since $\text{Cl}(\mathbb{P}^1)=\mathbb{Z}$ you see – Alex Youcis Aug 17 '14 at 09:33
  • @DanielRobert-Nicoud that $\text{Cl}(V)$ is countable. So $U\not\cong V$. Here $\text{Cl}$ denotes the class group. – Alex Youcis Aug 17 '14 at 09:34
  • Ingenious idea that, @Alex! – Georges Elencwajg Aug 17 '14 at 09:53
  • @GeorgesElencwajg Thanks :) – Alex Youcis Aug 17 '14 at 09:56
  • @AlexYoucis This looks like a nice way to do it, however we didn't do class groups either. – Daniel Robert-Nicoud Aug 17 '14 at 10:13
  • @DanielRobert-Nicoud Can you say what topics you did cover? I'm afraid Alex is going to write a third answer that doesn't help you :) –  Aug 17 '14 at 10:16
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    @DanielRobert-Nicoud The real back-to-basics, low-browed way you could do that is take an elliptic curve, say $y^2=x^3-1$. Obviously every affine open in $\mathbb{P}^1$ is a UFD, but with a bit of work you can show that no affine open of the ellpitic curve is a UFD. This is a low-brow way of doing what I did in my last comment. – Alex Youcis Aug 17 '14 at 10:19
  • @DanielRobert-Nicoud I think Roy Smith's answer on MO might be the easiest one, if it's true that an equivalent characterization of rationality of a curve is that you can parametrize it by rational functions (in one variable)..I have not yet thought about it too much but it looks promising. – Paul Aug 17 '14 at 18:24
  • Since you deleted your answer, I can't comment on what you said. Both elliptic curves and the projective line have all local rings UFDs--this is because they are both regular. It's that all affine opens of projective space are UFDs, but no affine open of an elliptic curve is a UFD> – Alex Youcis Aug 17 '14 at 22:03
  • NB: I made a late night mistake above. In the argument using class groups, I was thinking about Chow groups. The argument still works, but you should replace $\text{Cl}(E-U)$ with $\mathbb{Z}^m$, where $m=#(E-U)$, and similarly for $\text{Cl}(\mathbb{P}^1-V)$. – Alex Youcis Aug 18 '14 at 00:42
  • @AlexYoucis Ok, thanks for your help. I will have to think about it for a while. – Daniel Robert-Nicoud Aug 18 '14 at 09:44

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The Fermat curve $X:=V(f) \subset \mathbb P^2$ with equation $$f(x,y,z) = x^d + y^d + z^d, d \in \mathbb N,$$ is smooth with geometric genus $g(X) = \frac {(d-1)*(d-2)}{2}$. The curve X is not rational iff $g(X) > 0$, i.e. iff degree $d > 2$.

A curve is rational by definition iff it is birational equivalent to projective space $\mathbb P^1$. Note. A smooth projective curve is rational iff it is biholomorphic to $\mathbb P^1$.

Jo Wehler
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