Isomorphism of projective varieties and coordinate rings

Question

Let $K$ be a field.

Let $f: \mathbb{P}^1(K) \rightarrow \mathbb{P}^2(K)$ be a morphism given by $$f([x_1:x_2])=[x_1^2:x_1x_2:x_2^2].$$

(a) Are $\mathbb{P}^1(K)$ and $f(\mathbb{P}^1(K))$ isomorphic as projective varieties?

(b) Are the rings $\dfrac{K[X,Y]}{I_P(\mathbb{P}^1(K))}$ and $\dfrac{K[X,Y,Z]}{I_P(f(\mathbb{P}^1(K)))}$ isomorphic?

My attempt:

So for the first question, we want to show that $f$ is a morphism that has an inverse.

$f$ is a morphism since for $P=[x:y] \in \mathbb{P}^1(K)$, we can write $f(P)=(f_0(P), f_1(P), f_2(P))$ where $f_0=X^2, f_1=XY, f_2=Y^2 \in K[\mathbb{P}^1(K)]$ are homogeneous. Moreover, $\mathbb{P}^1(K) \subset \mbox{dom } f$.

Now, I want to find an inverse. I'm not really sure how to compute it, but my guess is for any $[x:y:z] \in \mathbb{P}^2(K)$,with $x \neq 0$, $f^{-1}([x:y:z])=[x:y]$. It will follow that if $x_1 \neq 0$, $f^{-1}([x_1^2:x_1x_2:x_2^2])=[x_1^2:x_1x_2]=[x_1:x_2]$. If $x=0$, it will follow that $y=0$. We have the point $[0:0:1]$ so $f^{-1}([0:0:1])=[0:1]$. I am not sure for the other cases.

For the second question, I observed that $f(\mathbb{P}^1(K)) =V(Y^2-XZ)$ where $V(Y^2-XZ)$ is the zero locus of $g=Y^2-XZ$. We can show that $g$ is irreducible in $K[X,Y,Z]$ so

$$\dfrac{K[X,Y,Z]}{I_P(f(\mathbb{P}^1(K)))}=\dfrac{K[X,Y,Z]}{I_P(V(Y^2-XZ))}=\dfrac{K[X,Y,Z]}{(Y^2-XZ)}$$. Meanwhile,

$\dfrac{K[X,Y]}{I_P(\mathbb{P}^1(K))}= \dfrac{K[X,Y]}{\{0\}} \cong K[X,Y].$

Thus, this problem is reduced to showing if $$K[X,Y] \cong \dfrac{K[X,Y,Z]}{(Y^2-XZ)}.$$ But these two rings are not isomorphic since the latter is not a UFD.

Are my thoughts correct? Any help would be appreciated. Thank you!

Answer 1

Your work in the post is correct. One easier way to write down an inverse morphism (which has the benefit of being more obviously a morphism) is as follows: on the affine patch in $\Bbb P^2$ given by $X\neq 0$, the inverse morphism can be defined by $[1:y:y^2]\mapsto [1:y]$, and on the affine patch given by $Z\neq 0$, the inverse morphism can be given as $[x^2:x:1]\mapsto [x:1]$. These maps are clearly regular and agree everywhere they're both defined, and their domain of definition is the whole of the image of $\Bbb P^1$.