Proving that a rational map is an endomorphism of an elliptic curve

Question

I recently posted

Morphisms between projective varieties

but haven't yet had a response, so I've tried to address the problem from a different perspective (namely, following the definitions given the 1st chapter of Silverman's Arithmetic of Elliptic Curves).

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Consider the curve (defined over an algebraically closed field) in $\mathbb{P}^2,$ given by $E:\;y^2z=x^3-xz^2.$

Problem. I want to show that the rational map $\phi=(xz:yz:-x^2)$ is actually a morphism $E\to E.$

Attempt. It's clear that $\phi$ is regular everywhere except at $P=(0:0:1)$ and $Q=(0:1:0).$

We show it's regular at these points too as follows:

Observe that $$(xz:yz:-x^2)=(x^2z:xyz:-x^3)=(x^2z:xyz:-y^2z-xz^2)=(x^2:xy:-y^2-xz)$$ and this last term is clearly regular at $Q;$ in particular, we have $\phi(Q)=P.$

Similarly, observe that $$(xz:yz:-x^2)=(xyz:y^2z:-x^2y)=(xyz:x^3-xz^2:-x^2y)=(yz:x^2-z^2:-xy)$$ and this last term is clearly regular at $P$; in particular, we have $\phi(P)=Q.$

This shows that $\phi$ is indeed a morphism $E\to E.$

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I'd really like to know whether what I've done is right!

Moreover, I'd be very interested to know how this matches up with the seemingly more abstract proof given by Mumford in his red book (see link at top of question).

Answer 1

This seems like a straight forward application of lemma 3.6 in Hartshorne

Lemma 3.6 Let $X$ be any variety and let $Y \subset \mathbb{A}^n$ be an affine variety. A map of sets $\psi : X \longrightarrow Y$ is a morphism if and only if $x_i \circ \psi$ is a regular function on $X$ for each $i$, where $x_1,...,x_n$ are the coordinate functions on $\mathbb{A}^n$.

This is proven quite early in the book (p. 20 to be precise) so it's not some arcane deus ex machina.

Therefore proving that the function $\phi$ you have is a morphism is equivalent to proving that

1. $x_1 \circ \phi = (x,y,z) \mapsto xz$
2. $x_2 \circ \phi = (x,y,z) \mapsto yz$
3. $x_3 \circ \phi = (x,y,z) \mapsto -x^2$

are all regular. But they are by definition polynomial functions and therefore regular. QED