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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    $\begingroup$ 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 $[x : y : z : w] \mapsto [y : z]$. $\endgroup$
    – Zhen Lin
    Commented Aug 16, 2014 at 22:21
  • $\begingroup$ @ZhenLin I see. I'll have to edit my answer, then. Thanks you. $\endgroup$ Commented Aug 16, 2014 at 22:42
  • $\begingroup$ The intersection of two quadrics in $\mathbb P^3$ is an elliptic curve. $\endgroup$ Commented Aug 16, 2014 at 23:01
  • $\begingroup$ 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. $\endgroup$ Commented Aug 17, 2014 at 5:45
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    $\begingroup$ @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. $\endgroup$ Commented Aug 17, 2014 at 10:19

1 Answer 1


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 $\mathbb P^1$.


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