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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.

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    $\begingroup$ Might it be easier to consider your bihomogeneous equation $q^2n=p^2z$ as defining a (non-singular) curve of type $(2,1)$ in $\mathbb P^1\times\mathbb P^1$, where the first $\mathbb P^1$ has coordinates $[p,q]$ and the second has coordinates $[n,z]$. $\endgroup$
    – Joe Silverman
    Commented Nov 26, 2020 at 23:23
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    $\begingroup$ I think just thinking about rationality of the cubic surface (or in @JoeSilverman's formulation the curve) isn't particularly useful: they are rational, for instance in your case via the parametrization $n=\frac{p^2}{q^2}$. The more relevant fact is that there is no nontrivial section of the map to $\mathbb{P}^1$, given by forgetting $p$ and $q$. $\endgroup$
    – dhy
    Commented Nov 26, 2020 at 23:52
  • $\begingroup$ @dhy yes thank you! I've started reasoning like this: by the useful suggestion of JoeSilverman I have that the curve is a twisted cubic and so, up to a linear transformation (which does not affect the problem) we have that the curve is locally of the form (t,t^2,t^3) and so the second coordinate is forced to be a perfect square. Sorry to bother you but can you please explain why there is no nontrivial section of the projection into the second factor $\mathbb{P}^1$? $\endgroup$
    – gigi
    Commented Nov 27, 2020 at 0:08

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