Let $F(x_0, x_1, x_2)$ be a homogeneous polynomial with $deg( f) = 2$. Let $V:= Z(F) \in \mathbb{P}^2$ be the zero-set of $F$.
The task is to show that $V \simeq \mathbb{P}^1$.
Firstly, we can change co-ordinates such that $F = X_0^2 - X_1X_2$. After some discussion with a friend, I realised that, by employing the morphism,
$\phi: \mathbb{P}^1 \to V \\ \phi: (u;t) \mapsto (u^2, t^2, ut)$
and then considering the restriction of $\phi$ to the each open set in the standard affine open cover of $\mathbb{P}^1$, we can show that the map is well-defined, and indeed an isomorphism of varieties.
My question is this: is it possible to establish the same result, but instead, by establishing a $k$-algebra isomorphism of the underlying co-ordinate rings? Indeed, what really are the co-ordinate rings of projective varieties?
Moreover, does the following argument hold?
Consider the k-algebra homomorphism induced by,
$\psi: k[x_0,x_1,x_2] \to k[u,t] \\ x_0 \mapsto u^2 \\ x_1 \mapsto t^2 \\ x_2 \mapsto ut$
Then, we have $ker (\psi) = (F)$ and so, $Im(\psi) \simeq \frac{k[x_0,x_1,x_2]}{F} = A(V)$ by the isomorphism theorem and the definition of the co-ordinate ring of V.
But then, $Im(\psi) = k[u^2, t^2, ut]$, which is the co-ordinate ring of $\mathbb{P}^1$.
Therefore, as $A(V) \simeq A(\mathbb{P}^1)$, we have $V \simeq \mathbb{P}^1$.