How to calculate the degree of a map

Question

I was finishing up on some material related to this question Weierstrass Equation of an Elliptic Curve and to finalize it, I am trying to prove that the map \begin{align*} \phi: &E \rightarrow \mathbb{P}^2\\ & P \mapsto [x(P),y(P),1] \end{align*} such that $\phi(O)=[0,1,0]$, where $O$ is the point at infinity and $P$ is a point on our elliptic curve, is in fact an isomorphism. Edit: E is an elliptic curve defined over a finite field $K$.

Throughout literature I have seen that the most common way to do so is by proving that the map $\phi$ has degree one, which, to me, is the same as proving that $K(E)=K(x,y)$.

In order to do so I started by considering the following two projections:

\begin{align*} \psi_1: &E \rightarrow \mathbb{P}^1\\ & P \mapsto [x(P),1]\\ & O \mapsto [1,0] \end{align*}

and

\begin{align*} \psi_2: &E \rightarrow \mathbb{P}^1\\ & P \mapsto [y(P),1]\\ & O \mapsto [1,0] \end{align*}

My goal is to prove that $deg(\psi_1)=2$ and $deg(\psi_2)=3$. But I have no idea how: everywhere I looked it was used a formula to calculate the degree of a map in terms of ramifications, which is something that was never approached in my class. This formula was presented to me like: \begin{align*} deg(\psi_1)=\sum_{P \in \psi_1^{-1}} e_\phi(P) \hspace{3mm} (1) \end{align*}

In this sense, I have two questions:

1. Is there an easier way to calculate the degree of these two maps that don't involve ramifications? If so, could you please give a hint?

2. If not, can someone give me a brief explanation on the expression (1)? On how to work with it?

I am very thankful for all the help!

Answer 1

Let me first note a couple of things.

1. You are trying to prove that there is an isomorphism between the points of an elliptic curve $E$, and a curve $C$, given by the Weirestrass equation you found on your mentioned post.
   So if you define your map: \begin{align*} \phi: &E \rightarrow \mathbb{P}^2\\ & P \mapsto [x(P),y(P),1] \end{align*} and set $C=\phi(Ε)$, what you want to prove is that $\phi$ defines an isomorphism $E\rightarrow \phi(E)=C$.

2. Both projections $ψ_1$ and $ψ_2$ are maps $E\rightarrow\mathbb{P}^1$, rather than $\mathbb{P}^2$ like you've written

3. Finally, the correct formula states that if $\psi:C_1\rightarrow C_2$ is a non-constant morphism of smooth curves, then for any $Q\in C_2$ we have: $$\deg(\psi)=\sum_{P \in \psi^{-1}(Q)} e_\psi(P)$$ (Notice that we can say for any $Q\in C_2$, rather than $Q\in\psi(C_1)$ since every non-constant morphism between smooth curves is surjective, so $\psi(C_1)=C_2$.)

Now let's look at the case of $ψ_1$ (The case for $ψ_2$ follows analogously).

By construction, $x$ has a double pole only at $\mathcal{O}$ and nowhere else.
Hence, if we choose for our $Q$ the point at infinity in $\mathbb{P}^1$, namely $Q=[1,0]$, then we must have that the only point in the inverse image of $Q$ is $\mathcal{O}$, so $ψ_1^{-1}([1,0])=\left\{\mathcal{O}\right\}$.

Thus by using the formula above, we conclude that: $$\deg(\psi_1)=\sum_{P \in \psi_1^{-1}([1,0])} e_{\psi_1}(P)=e_{ψ_1}(\mathcal{O})$$

All that's left to do is calculate $e_{ψ_1}(\mathcal{O})$.

By definition, we have: $$e_{ψ_1}(\mathcal{O})=\operatorname{ord}_{\mathcal{O}}(ψ_1^*t) $$

where $t$ is a uniformizer at $ψ_1(\mathcal{O})=[1,0]=$ point at infinity.
If we denote by $X$ (capital X) the first coordinate function for $\mathbb{P}^1$, then a uniformizer at infinity is: $t=\frac1X$.

Then $ψ_1^*t$ is the same as taking the reciprocal of the first coordinate of $ψ_1$, that is: $$ψ_1^*t=\frac1x$$ (Note that here is a lowercase $x$, the function that you got from the construction on your other post)

Now putting all these together, we get:

$$e_{ψ_1}(\mathcal{O})=\operatorname{ord}_{\mathcal{O}}(ψ_1^*t) =\operatorname{ord}_{\mathcal{O}}\left(\frac1x\right)=-\operatorname{ord}_{\mathcal{O}}(x)=-(-2)=2$$

where we got $\operatorname{ord}_{\mathcal{O}}(x)=-2$ from the fact that $x$ has pole of order $2$ at $\mathcal{O}$.

And so we proved that $\deg(ψ_1)=2$ as required.