Suppose that an Abelian surface $A$ is isogenous to the product of two elliptic curves $E \times E'$. When can we say that the Néron–Severi group is generated by the classes of these two elliptic curves, $NS(A) \cong \mathbb{Z}E \oplus \mathbb{Z}E'?$
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2$\begingroup$ $\ldots\oplus \mathrm{Hom}(E, E')$ $\endgroup$– Piotr AchingerCommented Aug 28, 2023 at 16:47
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2$\begingroup$ See Lemma 3.1 in the following reference for a more general fact: Remy van Dobben de Bruyn "A variety that cannot be dominated by one that lifts" Duke Math. J. 170.7, p. 1251–1289 (2021). $\endgroup$– Piotr AchingerCommented Aug 28, 2023 at 16:51
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2$\begingroup$ (Since you assumed isogenous, not isomorphic, to $E\times E'$, the formula maybe only works after tensoring with $\mathbb{Q}$.) $\endgroup$– Piotr AchingerCommented Aug 28, 2023 at 16:53
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2$\begingroup$ ... but surely this was known well before my paper! Here is a simple argument over $\mathbf C$: then $\operatorname{NS}(X)$ is the image of $c_1 \colon \operatorname{Pic}(X)\to H^1(X,\mathbf Z)$ (up to a finite kernel in $\operatorname{NS}$), which by the Lefschetz (1,1)-theorem is the intersection $H^{1,1}(X)\cap H^2(X,\mathbf Z)$. Then use that $$H^{1,1}(X\times Y) \cong \bigoplus_{i,j=0}^1 H^{i,j}(X)\otimes H^{1-i,1-j}(Y),$$ so besides $H^{1,1}(X)\cap H^2(X,\mathbf Z)$ and $H^{1,1}(Y)\cap H^2(Y,\mathbf Z)$, you get something like $\operatorname{Hom}_{\mathbf Z\text{-HS}}(H^1(X),H^1(Y))$. $\endgroup$– R. van Dobben de BruynCommented Aug 28, 2023 at 20:50
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$\begingroup$ I’m interested in understanding if the formula holds over $\mathbb{Z}$. What about in the case when $E$ and $E’$ are not isogenous so the Picard rank should be 2. $\endgroup$– StormblessedCommented Sep 21, 2023 at 12:08
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