# Looking for a smooth curve that is not rational

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?

#### Solutions Collecting From Web of "Looking for a smooth curve that is not rational"

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 $\mathbb P^1$.