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!
Best Answer
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 $\Bbb P^1$.