# Finding all morphism from a variety to itself

Let $$C:=X^2+Y^2-Z^2$$ be a projective variety in $\Bbb P^2$. What are all the morphisms $C\to C$ ?
More generally, how does one find all morphisms from a given variety to itself?

Related: I would also like to know if any software can compute this.
Alternatively, I would also accept an answer that gives a complete reference that I can follow.

Definitions:
Let the field be $K=\Bbb Q$. The coordinate ring of $C$ is $$K[C]=K[X,Y,Z]/(X^2+Y^2-Z^2)$$ with function field $K(C)=\text{Quot}(K[C])$. A rational map $\phi\in K(C)$ is regular at $P\in C$ if $\phi(P)$ is defined. $\phi$ is a morphism if it is regular at every $P\in C$. i.e. $\phi(C)\subseteq C$. In particular, $\phi=F/G$ where $F,G\in K[C]$ are homogeneous polynomials of the same degree.

Some workings:
$C$ is just the group of rational points in the circle. They form an abelian group with the group law $$(x_1,y_1,z_1)\oplus(x_2,y_2,z_2)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1,z_1z_2)$$ so that $N=(1,0,1)$ is the identity. So similar to elliptic curves, we can define $\phi=[m]$ to be the multiplication by $m$ map and $\phi$ is a morphism by the definitions above. For example, let $m=3$: $$\require{enclose} \enclose{horizontalstrike}{[3](x,y,z)=(x^2-y^2,2xy,z^2)\oplus(x,y,z)=(x^4-6 x^2 y^2+y^4,4 x y (x^2-y^2),z^4)}$$ Edit_1:
$$[3](x,y,z)=(x^2-y^2,2xy,z^2)\oplus(x,y,z)=(x^3-3 x y^2,3 x^2 y-y^3,z^3)$$ and it is easy to check that $\phi$ is defined for all $P\in C$.

Clearly we can also compose any two morphisms, so it suffices to find the "different types".

Another morphism can be a translation. Fix a $Q\in C$, then define \begin{align*} \phi: C &\to C\\ P &\mapsto P\oplus Q \end{align*} It is clear that $\phi(P)\in C$ and is defined, so this seems to work.

The main question is: are there other kind of morphisms and how do I find all of them? The only fact known to me would be that any morphism $\phi:C\to C$ is either constant or surjective.

Some thoughts:
In the case of an elliptic curve, if we want to set $\phi(\mathcal O)=\mathcal O$, then $\phi$ is an isogeny. It is known that this forces $\phi$ to be endomorphisms and the endomorphism ring contains $\Bbb Z$ due to the scalar multiplications $[m]$. But there can also be complex multiplication when the endomorphism ring is larger than $\Bbb Z$. Can something similar happen here? I do note that if I consider $\Bbb Q(i)$, then \begin{align*} \phi: C &\to C\\ (x,y,z) &\mapsto (x,iz,y) \end{align*} gives me something similar. Note that $$x^2+(iz)^2+y^2=x^2-z^2+y^2=0$$ Thanks for reading!

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Something's wrong with your $[3]$ I guess (should be cubic, yours seems to be $[4]$), but it doesn't really matter. – Hagen von Eitzen Jan 25 '14 at 21:46
@HagenvonEitzen Thanks, I have corrected it. – Yong Hao Ng Jan 26 '14 at 7:50

If $\mathrm{char}(k) \neq 2$, there is an isomorphism $\mathbb{P}^1 \cong C$ ("classification of Pythagorean triples") given by mapping $[a:b] \mapsto [a^2-b^2 : 2ab : a^2+b^2]$. The non-constant morphisms $\mathbb{P}^1 \to \mathbb{P}^1$ correspond to $k$-homomorphisms $k(t) \to k(t)$, i.e. to elements in $k(t) \setminus k$.
My understand is that non-constant morphisms $\Bbb P^1\to \Bbb P^1$ are of the form $[f(x,y):g(x,y)]$ where $f(x,y),g(x,y)$ are homogenous polynomials with no common factors. The no common factor condition ensures that $[f(x,y):g(x,y)]$ cannot be simultaneously both zero so it is well defined at all $[x:y]\in\Bbb P^1$. Is this correct? In addition, do you mind explaining how this translates to $k$-homomorphisms $k(t)\to k(t)$? (say $[x^2:x^2-y^2]$) I assume $$k(t) =\{f(t)/g(t): f(t),g(t)\in k[t]\}$$ – Yong Hao Ng Jan 26 '14 at 7:22
If $f=p/q \in k(t) \setminus k$ with coprime $p,q \in k[t]$, then the corresponding morphism $\mathbb{P}^1 \to \mathbb{P}^1$ maps $[a:1] \mapsto [p(a):q(a)]$, suitably extended to the point at infinity $[1:0]$. At the moment I'm not sure how to write this down explicitly, but basically this extension from the fact that $\mathbb{P}^1$ is proper. In general, these morphisms don't have a "global polynomial expression". – Martin Brandenburg Jan 26 '14 at 11:41