[Math] An Isomorphism of Projective Varieties

algebraic-geometry

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 $V \simeq \mathbb{P}^1$.

Best Answer

1) No, it is not possible to
"establish the same result...by establishing a $k$-algebra isomorphism of the underlying co-ordinate rings"
because these coordinate rings are not isomorphic!
Indeed the ring $k[x_0,x_1,x_2]:=\frac{k[X_0,X_1,X_2]}{(X_0^2-X_1X_2)}$ is not isomorphic to $k[T_0,T_1]$ since the first is not a UFD ($x_0^2=x_1x_2$ are different factorizations of the same element!), whereas the second is a UFD, like all polynomial rings over a field.

2) The notion of "coordinate ring" for a projective variety $X$ depends on the embedding of $X$ into projective space and is by far not as satisfying as in the affine case.
In particular that coordinate ring $S(X)$ is not at all equal to the ring of regular functions $\mathcal O(X)=k$.

3) It is not true that "we can change co-ordinates such that $F = X_0^2 - X_1X_2$".
This is valid only for irreducible polynomials $F$.
For example no change of coordinates will force the conic $X_0X_1=0$ to have as equation $X_0^2 - X_1X_2$ .

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