Maybe it is a stupid question but I’m not able to find the answer anywhere else.

My goal is to prove in an “algebraic geometry fashion” that $sqrt{n}$ is not a rational number for $n$ not a perfect square.

Thus suppose instead that $sqrt{n}=frac{p}{q}$ where $p,q in mathbb{Z}$. After some simple calculation we have $q^2n=p^2$. I can complete this equation in $mathbb{P}^3$ with homogeneous coordinates $(p,q,n,z)$ yielding $q^2n=p^2z$.

Now it is easy to check that the projective cubic surface $X=(q^2n-p^2z=0)$ is singular along the line $p=q=0$.

If this surface would be rational then it is possible to find a parametrization $varphi:mathbb{C}^2 rightarrow U$ where $U subset X$ is open such that $$varphi(x,y)=(varphi_1(x,y),varphi_2(x,y),varphi_3(x,y),1)$$ where each $varphi_i(x,y)$ is a rational function (I can eventually shrink $U$ such that does not contain the hyperplane at infinity).

If this holds then I can choose $x=frac{p}{q}$ and $y=frac{r}{s}$ and find infinite solutions to the problem.

This in some sense makes me thinking that $X$ is not rational, but I’m not able to prove it.

What is wrong with this idea? I’m not able to figure it out.

Thanks in advance.