How to determine the type of a divisor $D$ on a product of elliptic curves?

Question

Say $E_1, \dotsc, E_n$ are elliptic curves (over $\mathbb C$), and $$D \subset E_1 \times \dotsc \times E_n$$ is an (effective) divisor. Then the line bundle $\mathcal O(D)$ has a type $(d_1, \dotsc, d_n)$, which we also call the type of $D$. Is it true that $$d_i = \deg( D \cdot E_i),$$ where I embed $E_i$ as a curve in $E_1 \times \dotsc \times E_n$ by $z \mapsto (0, \dotsc, z, \dotsc, 0)$?

This is clear if $D = \sum_{i=1}^n \operatorname{pr}_i^* D_i$, where $D_i \subset E_i$ are effective divisors of degree $d_i$. But I think that $$\operatorname{Pic}(E_1 \times \dotsc \times E_n) \neq\operatorname{Pic}(E_1) \times \dotsc \times \operatorname{Pic}(E_n),$$ so there might be other effective divisors?

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Specifically, I would like to understand the two examples:

1. Let $E = \mathbb C / \Gamma$ and elliptic curve, with $\Gamma = \mathbb Z + \tau \mathbb Z$. Consider the divisor $D \subset E \times E$ given by $$D = \{(z,z):z \in E\} \cup \{(z,-z):z \in E\},$$ so the union of the diagonal and the "anti-diagonal". What is it's type?
2. Take the quotient $\pi: E \times E \to A := E \times E / \langle(\frac 1 2, \frac 1 2)\rangle$. Here we consider $(\frac 1 2, \frac 1 2)$ as a $2$-torsion point in $E \times E$. What is the type of $\overline D = \pi(D)$? Using the automorphism $\varphi:E \times E \to E \times E$, $(z_1, z_2) \mapsto (z_1, z_1 - z_2)$ (which satisfies $\varphi^2 = \operatorname{id}$), one sees that $A \cong F \times E$, where $F = E / \langle \frac 1 2 \rangle$. In that description, $\overline D$ becomes $$\overline D = \{(z,0) : z \in F\} \cup \{(z,2z):z \in \mathbb C\} \subset F \times E.$$

Answer 1

I managed to calculate my examples. In the first one, $D_0$ has indeed polarization type $(2,2)$. To see this, let $E = \mathbb C / (\mathbb Z + \tau \mathbb Z)$ be an elliptic curve, and consider the isogeny $$\varphi: E \times E \to E \times E, (z_1, z_2) \mapsto (z_1 + z_2, z_1 - z_2).$$ Then $\varphi^2(z_1, z_2) = (2z_1, -2z_2)$, so $\deg \varphi^2 = 16$ and $\deg \varphi = 4$.

Note that $D_0 = \varphi^* \tilde D$, where $\tilde D$ is the divisor $$\tilde D = E \times \{0\} \cup \{0\} \times E.$$ Clearly $\tilde D$ has type $(1,1)$, and the corresponding hermitian form on $\mathbb C^2$ is $$H\left(\left(\begin{smallmatrix}u_1 \\ u_2 \end{smallmatrix}\right),\left(\begin{smallmatrix}v_1 \\ v_2 \end{smallmatrix}\right)\right) = \frac 1 {\operatorname{Im}\tau} (u_1 \bar v_1 + u_2 \bar v_2).$$ If $\phi: \mathbb C^2 \to \mathbb C^2$ is the analytic representation of $\varphi$, an easy calculation shows $\phi^* H = 2H$. Hence $D$ has type $(2,2)$.

I think my second example actually serves as a counterexample to the claim that the $d_i$ can be computed as $E_i \cdot D$. The projection $E \times E \to A$ has degree $2$, and the above isogeny $\varphi$ factors over $A$, so $\tilde D$ has to pull-back to a polarization of type $(1,2)$ on $A$ (for combinatorial reasons). But under the identification $A \cong F \times E$, with $F = E / \langle \frac 1 2 \rangle$, $D_1$ is given by $$D_1 = (F \times 0) \cup \{(z,2z)\} \subset F \times E.$$ Then $0 \times E$ intersects both components in the point $(0,0)$; $F \times 0$ doesn't intersect itself, but intersects the second component in the points $(0,0)$ and $(\frac \tau 2, 0)$. All intersections are transversal, so this would give a polarization type $(2,2)$.↯