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!

  • $\begingroup$ You mean are $\Bbb{P}^1(K)$ and $f(\Bbb{P}^1(K))$ isomorphic. It is a birational map, so find the inverse and check if it is a morphism (given locally by a non-vanishing regular map). Also I'm a bit inconfortable to call $K[x_1,x_2]$ the coordinate ring of $\Bbb{P}^1(K)$, it is clearly not the same as the coordinate ring of an affine variety. $\endgroup$
    – reuns
    Jun 11, 2020 at 2:49
  • $\begingroup$ Thanks for the insights. That was the definition given to us in our class. Would you mind to explain the coordinate ring for projective varieties? $\endgroup$ Jun 11, 2020 at 2:52
  • 1
    $\begingroup$ The idea is that there is something analogous to the coordinate ring of affine varieties : it is the sheaf of regular functions, for $\Bbb{P^1(C)}$ it means gluing $\Bbb{C}[z]$ and $\Bbb{C}[1/z]$ in a space where we can extract all the information. $\endgroup$
    – reuns
    Jun 11, 2020 at 2:55
  • $\begingroup$ Given that the projective varieties are isomorphic in this case, does it follow that the coordinate rings will now be isomorphic? $\endgroup$ Jun 11, 2020 at 3:02
  • 1
    $\begingroup$ There's a reason the author of this exercise is asking you this question. I'll be happy to help you once you've edited your post with your own efforts in determining these coordinate algebras. $\endgroup$
    – KReiser
    Jun 11, 2020 at 3:14

1 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 $\Bbb P^1$.

  • $\begingroup$ Can you explain the part where $Z \neq 0$? Why is the form of the point $[x^2:x:1]?$ $\endgroup$ Jun 14, 2020 at 4:35
  • $\begingroup$ If a point of the form $[x_1^2:x_1x_2:x_2^2]$ has $Z\neq 0$, then we can divide all three entries by $Z=x_2^2$ to get a point of the form $[\frac{x_1^2}{x_2^2}:\frac{x_1}{x_2}:1]$, which is exactly of the form $[x^2:x:1]$ for $x=\frac{x_1}{x_2}$. $\endgroup$
    – KReiser
    Jun 14, 2020 at 4:51
  • $\begingroup$ Makes sense. Thank you so much! $\endgroup$ Jun 14, 2020 at 5:44
  • $\begingroup$ @MashedPotato If this resolves your question, please consider accepting the answer by clicking the checkmark to the left (you should also consider doing this to some of your other posts - I checked, and you have never accepted an answer!). $\endgroup$
    – KReiser
    Jun 14, 2020 at 5:50
  • $\begingroup$ Done. I didn't know that was a thing. Forgive my noobness. $\endgroup$ Jun 14, 2020 at 6:09

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