Problem I.3.18 in Hartshorne

Question

Problem I.3.18b-c in Hartshorne is concerned with the surface $Y$ of $\mathbb{P}^3$ given parametrically by $(x,y,z,w) = (t^4,t^3u,tu^3,u^4)$. In particular, part c asks to prove that $Y$ is isomorphic to $\mathbb{P}^1$. I did check two different solutions available online, and they both claim that $Y$ is the image of the $4$-tuple embedding of $\mathbb{P}^1$. However, the $4$-tuple embedding takes $\mathbb{P}^1$ into $\mathbb{P}^4$ and not $\mathbb{P}^3$. This is not the case with $Y$, since there is one monomial term missing, i.e. $t^2u^2$. So i wonder what is a rigorous argument for showing that $Y$ is isomorphic to $\mathbb{P}^1$.

Answer 1

The 4-tuple embedding with $x^2 y^2$ missing is still surjective, as seen in the paramaterization of Y, and it is still well-defined without that term, i.e. it's image is not everywhere 0.

We only need to check injectivity, which we can do by finding a well-defined inverse morphism. But given the morphism sending $[x:y]$ to $[x^4:x^3y:xy^3:y^4]$ we can see how we want to define it. Namely, on the affine open set where $w$ is nonzero, we'll send our point to $[z:1]$, and on the affine open set where $x$ is nonzero, we'll send our point to $[1:y]$. Note that on the intersection of the open sets where $x$ and $y$ are nonzero, the image of $[x:y]$ was also equal to $[t^4:t^3:t:1]$ and $[1:u:u^3:u^4]$, where $t = \frac{x}{y}$ and $u=\frac{y}{x}$, hence why we see this is the inverse.

One checks that this is well-defined on the intersection of these two open sets (which cover Y as we always have $t$ or $u$ not zero), and so our morphism was injective and thus an isomorphism.