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Francesco Polizzi
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Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback invia $A^{\ast}$$f^*$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{\ast}$ would produce elliptic curves in $A^{\ast}$. This is a contradiction: indeed $A^{\ast}$ is simple, since by assumption $A$ is simple.

Poincaré Reducibility Theorem [Birkenhake-Lange, Complex Abelian Varieties, Chapter 5] says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback in $A^{\ast}$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{\ast}$ would produce elliptic curves in $A^{\ast}$. This is a contradiction: indeed $A^{\ast}$ is simple, since by assumption $A$ is simple.

Poincaré Reducibility Theorem says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback via $f^*$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{\ast}$ would produce elliptic curves in $A^{\ast}$. This is a contradiction: indeed $A^{\ast}$ is simple, since by assumption $A$ is simple.

Poincaré Reducibility Theorem [Birkenhake-Lange, Complex Abelian Varieties, Chapter 5] says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.

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Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \ldots \times E_n \to A$$$$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\vee} \colon A^{\vee} \to E_1^{\vee} \times \ldots \times E_n^{\vee},$$$$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback in $A^{\vee}$$A^{\ast}$ of any elliptic curve in $E_1^{\vee} \times \ldots \times E_n^{\vee}$$E_1^{\ast} \times \cdots \times E_n^{\ast}$ would giveproduce elliptic curves in $A^{\vee}$$A^{\ast}$. This is a contradiction: indeed $A^{\vee}$$A^{\ast}$ is simple, since by assumption $A$ is simple.

Poincaré Reducibility Theorem says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \ldots \times E_n \to A$$ would give a dual isogeny $$f^{\vee} \colon A^{\vee} \to E_1^{\vee} \times \ldots \times E_n^{\vee},$$ and the pullback in $A^{\vee}$ of any elliptic curve in $E_1^{\vee} \times \ldots \times E_n^{\vee}$ would give elliptic curves in $A^{\vee}$. This is a contradiction: indeed $A^{\vee}$ is simple, since by assumption $A$ is simple.

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback in $A^{\ast}$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{\ast}$ would produce elliptic curves in $A^{\ast}$. This is a contradiction: indeed $A^{\ast}$ is simple, since by assumption $A$ is simple.

Poincaré Reducibility Theorem says that any abelian variety $A$ is isogenous to a product $A_1^{n_1} \times \cdots \times A_k^{n_k}$, where the $A_j$ are simple abelian variety, pairwise non-isogenous. The integers $n_j$ are uniquely determined, whereas the factors $A_j$ are determined up to isogeny.

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Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

Not in general.

For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \ldots \times E_n \to A$$ would give a dual isogeny $$f^{\vee} \colon A^{\vee} \to E_1^{\vee} \times \ldots \times E_n^{\vee},$$ and the pullback in $A^{\vee}$ of any elliptic curve in $E_1^{\vee} \times \ldots \times E_n^{\vee}$ would give elliptic curves in $A^{\vee}$. This is a contradiction: indeed $A^{\vee}$ is simple, since by assumption $A$ is simple.