It is known that the smooth projective curve $C$ over $\mathbb C$ defined by $$C=\operatorname{Proj}(\mathbb C[X,Y,Z]/(X^2+Y^2+Z^2))$$ has genus $0$ (by Riemann-Roch theorem or by Hurwitz formula for instance) and as a consequence, it is isomorphic to $\mathbb P^1$. I was asked to make such an isomorphism explicit, but I am facing difficulties in doing so.
My starting point is very down-to-earth, I am looking for a morphism of graded $\mathbb C$-algebras $$f^\#:\mathbb C[X,Y,Z]/(X^2+Y^2+Z^2)\rightarrow \mathbb C[S,T]$$ such that there is no homogeneous prime ideal of $C[S,T]$ containing all of $f^\#(X), f^\#(Y), f^\#(Z)$ aside from the irrelevant one. Such a morphism would indeed give rise to a morphism of schemes over $\mathbb C$ $$f:\mathbb P^1\rightarrow C$$ and in order to check that $f$ is an isomorphism, it would be enough to check it for the restrictions to principal open subschemes $f^{-1}(D_+(A))\rightarrow D_+(A)$ for $A=X,Y,Z$. At the level of algebra, it means that $f^\#$ must induce an isomorphism after taking the homogeneous localization at $A=X,Y,Z$.
But now, I fail to see any good way to define such a morphism $f^\#$. In all generality, writing $f^\#(A) = \alpha_AS+\beta_AT$ for $\alpha_A,\beta_A\in \mathbb C$ and $A=X,Y,Z$, given the condition that the image of $X^2+Y^2+Z^2$ must be $0$, I arrived at the conclusion that the complex coefficients must satisfy $$\begin{align} \alpha_X^2+\alpha_Y^2+\alpha_Z^2 &= 0 \\ \beta_X^2+\beta_Y^2+\beta_Z^2 &= 0 \\ \alpha_X\beta_X+\alpha_Y\beta_Y+\alpha_Z\beta_Z&=0 \end{align} $$
I have considered taking these coefficients as $6$-roots of unity in order to exploit the fact that $1+j+j^2=0$, but I wonder if this would lead to anything satisfying. I lack intuition/conviction about how to proceed further.