Pullback of a translation map of a divisor

Question

Let X be an elliptic curve, D be a divisor on X with degree 0. Then $\tau^{*}_{Q} D = D $ for all $Q \in X$. Is this true?

$\tau_Q$ is a translation map defined by $X \rightarrow X, P \mapsto P+Q$ and $\tau^{*}_Q$ is the pullback.

Answer 1

I always find these sorts of questions tricky because there are two group operations:

• the addition of divisors modulo linear equivalence, within the group $Pic(X)$;

• the elliptic curve group law.

These two group operations are closely related, and before we can answer your question, we need to spell out exactly how they are related:

Fix a "zero point" $p_0$ on the elliptic curve $X$. Let $Pic^0(X)$ denote the group of degree-zero divisors on $X$, up to linear equivalence. As explained in Hartshorne IV.4, every degree-zero divisor is linearly equivalent to $p - p_0$ for some point $p \in X$, and this point $p$ is unique. Thus we have a bijection between points in $X$ and divisor classes in $Pic^0(X)$, given by: $$ p \ \ \ \ \longleftrightarrow \ \ \ \ \ p -_{\rm divisor \\ minus} p_0 $$

The elliptic curve group law is inherited from the group structure of $Pic^0(X)$ via this correspondence. In other words, $$ p +_{\rm group \\ \ law} q \ =\ r $$ holds precisely when $$ \left( p -_{\rm divisor \\ \ minus} p_0 \right) +_{\rm divisor \\ \ \ add} \left( q -_{\rm divisor \\ \ minus} p_0 \right) \sim_{\rm \ divisor \\ equivalence} \left( r -_{\rm divisor \\ \ minus} p_0\right), $$ and this is true precisely when $$r \sim_{\rm \ divisor \\ equivalence} \left( p +_{\rm divisor \\ \ add} q -_{\rm divisor \\ \ minus} p_0 \right)$$

[Here, "divisor equivalence" means "linear equivalence".]

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Returning to the question you asked, we are presented with a divisor $D$ of degree zero. As mentioned above, we have $$D \sim_{\rm \ divisor \\ equivalence} \left( p -_{\rm divisor \\ \ minus} p_0 \right)$$ for some unique $p \in X$.

$\tau_q$ is the transformation $p \mapsto p +_{\rm group \\ law} q$, which implies that $$\tau_q^\star D \sim_{\rm \ \ \ divisor \\ equivalence} \left(p -_{\rm group \\ \ \ law} q \right) -_{\rm divisor \\ \ \ minus} \left(p_0 -_{\rm group \\ \ law} q\right)$$

Your question is whether we can prove that $$ \tau_q^\star D \sim_{\rm \ divisor \\ equivalence} D.$$

Well, let's try and prove this. In view of the fact that $$\left( p -_{\rm group \\ \ \ law} q \right) +_{\rm group \\ \ \ law} q = p,$$ and in view of our above characterisation of the group law on the elliptic curve, we have $$\left( p -_{\rm group \\ \ \ law} q \right) \sim_{\rm \ divisor \\ equivalence} \left( p +_{\rm divisor \\ \ \ add} p_0 -_{\rm divisor \\ \ minus} q \right)$$

Similarly, replacing $p$ with $p_0$, we have $$\left( p_0 -_{\rm group \\ \ \ law} q \right) \sim_{\rm \ divisor \\ equivalence} \left( p_0 +_{\rm divisor \\ \ \ add} p_0 -_{\rm divisor \\ \ minus} q \right).$$

Taking the difference of the above two equations, we have $$ \left(p -_{\rm group \\ \ \ law} q \right) -_{\rm divisor \\ \ minus} \left(p_0 -_{\rm group \\ \ law} q\right) \sim_{\rm \ \ \ divisor\\ equivalence} \left( p -_{\rm divisor \\ \ minus} p_0 \right)$$ which confirms that $\tau_q^\star D \sim_{\rm \ \ \ divisor \\ equivalence} D$, as required.