Pullback of a translation map of a divisor 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.
 A: 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".]

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.
