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Consider an affine variety $C$ in $\mathbb{P}^2$ determined by $x^2+y^2-z^2=0$, then how to prove the image of the following morphism is dense in $\mathbb{P}^1$?

$$\begin{align} \varphi:C&\to \mathbb{P}^1\\ [x:y:z]&\mapsto [x+z:y] \end{align}$$

Frank
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    This morphism does not seem to be well-defined. The point $[-1 : 0 : 1]$ lies in $C$, but its image is supposed to be $[0 : 0]$. Also, I think the variety $C$ you described is not affine. – diracdeltafunk Apr 27 '23 at 03:14
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    @diracdeltafunk When defining maps of nonsingular projective curves, people often just give expressions that are defined on a dense open: $D(x+z) \cup D(y)$ in this case. By the curve-to-projective extension theorem or the valuative criterion for properness, such a map must extend to all of $C$. In this case we have $[x+z : y] = [y : z-x]$ on $D(x+z) \cap D(z-x)$, and the second expression can be used near the point $[-1:0:1]$. This particular example is discussed here. – Viktor Vaughn Apr 27 '23 at 06:48

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Facts from general topology is that, image of an irreducible set is irreducible. By properness of $C$, the image is closed. In Zariski topology, only irreducible closed subsets of $\mathbb{P}^1$ are one point sets or the whole curve. So, you check that the image contains at least two points (or, just the morphism is nontrivial). For example, $$\varphi (-1:0:1)=(0:1),\ \varphi(1:0:1)=(1:0).$$ (I assumed characteristic $\neq 2$.)

(Note: Since both are smooth projective curves, $\varphi$ is a morphism, seen as $[x+z:y]=[y:z-x]$ on $C$.)

Ayaka
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