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Glorfindel
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The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this referencethis reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$ n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$

The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$ n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$

The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$ n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$

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The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$ n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$