Under the assumption that $E[n]\subseteq E(\mathbb{F}_q)$, to compute $E[n]$ I'd avoid using division polynomials, as they rapidly become cumbersome. Rather I would generate random elements of $E[n]$ until I have a generating set.

Assume that we know the order of $E(\mathbb{F}_q)$, by Schoof's algorithm or by one of its improvements. Also assume that we can factor this order completely.

This is how I'd generate random elements of $E[n]$. Pick a random point $P$ on $E(\mathbb{F}_q)$. One can do this by picking an $x$-coordinate randomly and solving, if possible, a quadratic equation to get the $y$-coordinate. As we know the prime factors of the order of $E(\mathbb{F}_q)$ we can find the order of $P$ in this group, and write this order as $mn'$ where $n'$ is the highest common factor of the order of $P$ and $n$. Then computing $[m]P$ gives an element of $E[n]$.

After generating enough elements of $E[n]$, we should be able to find a two-element "basis" of the group. (I'll omit details here; this can certainly be done by reducing to the prime power case, but perhaps it can be done more generally). One ends up with two points $P$ and $Q$. These points have order $n$ and their Weil pairing $e(P,Q)$ is a primitive $n$-th root of unity $\zeta$.

Note that the Weil pairing can be computed using Miller's algorithm or more recent alternatives. Given a point $R\in E[n]$ we can find $a$ and $b$ with $R=aP+bQ$ by using the Weil pairing: $$e(R,Q)=e(aP+bQ,Q)=e(P,Q)^ae(Q,Q)^b=\zeta^a$$ etc.

Under the assumption that $E[n]\subseteq E(\mathbb{F}_q)$, to compute $E[n]$ I'd avoid using division polynomials, as they rapidly become cumbersome. Rather I would generate random elements of $E[n]$ until I have a generating set.

Assume that we know the order of $E(\mathbb{F}_q)$, by Schoof's algorithm or by one of its improvements. Also assume that we can factor this order completely.

This is how I'd generate random elements of $E[n]$. Pick a random point $P$ on $E(\mathbb{F}_q)$. One can do this by picking an $x$-coordinate randomly and solving, if possible, a quadratic equation to get the $y$-coordinate. As we know the prime factors of the order of $E(\mathbb{F}_q)$ we can find the order of $P$ in this group, and write this order as $mn'$ where $n'$ is the highest common factor of the order of $P$ and $n$. Then computing $[m]P$ gives an element of $E[n]$.

After generating enough elements of $E[n]$, we should be able to find a two-element "basis" of the group. (I'll omit details here; this can certainly be done by reducing to the prime power case, but perhaps it can be done more generally). One ends up with two points $P$ and $Q$. These points have order $n$ and their Weil pairing $e(P,Q)$ is a primitive $n$-th root of unity $\zeta$.

Note that the Weil pairing can be computed using Miller's algorithm or more recent alternatives. Given a point $R\in E[n]$ we can find $a$ and $b$ with $R=aP+bQ$ by using the Weil pairing: $$e(R,Q)=e(aP+bQ,Q)=e(P,Q)^ae(Q,Q)^b=\zeta^a$$ etc.

Under the assumption that $E[n]\subseteq E(\mathbb{F}_q)$, to compute $E[n]$ I'd avoid using division polynomials, as they rapidly become cumbersome. Rather I would generate random elements of $E[n]$ until I have a generating set.

Assume that we know the order of $E(\mathbb{F}_q)$, by Schoof's algorithm or by one of its improvements. Also assume that we can factor this order completely.

This is how I'd generate random elements of $E[n]$. Pick a random point $P$ on $E(\mathbb{F}_q)$. One can do this by picking an $x$-coordinate randomly and solving, if possible, a quadratic equation to get the $y$-coordinate. As we know the prime factors of the order of $E(\mathbb{F}_q)$ we can find the order of $P$ in this group, and write this order as $mn'$ where $n'$ is the highest common factor of the order of $P$ and $n$. Then computing $[m]P$ gives an element of $E[n]$.

After generating enough elements of $E[n]$, we should be able to find a two-element "basis" of the group. (I'll omit details here; this can certainly be done by reducing to the prime power case, but perhaps it can be done more generally). One ends up with two points $P$ and $Q$. These points have order $n$ and their Weil pairing $e(P,Q)$ is a primitive $n$-th root of unity $\zeta$.

Note that the Weil pairing can be computed using Miller's algorithm or more recent alternatives. Given a point $R\in E[n]$ we can find $a$ and $b$ with $R=aP+bQ$ by using the Weil pairing: $$e(R,Q)=e(aP+bQ,Q)=e(P,Q)^ae(Q,Q)^b=\zeta^b$$ etc.

Under the assumption that $E[n]\subseteq E(\mathbb{F}_q)$, to compute $E[n]$ I'd avoid using division polynomials, as they rapidly become cumbersome. Rather I would generate random elements of $E[n]$ until I have a generating set.

Assume that we know the order of $E(\mathbb{F}_q)$, by Schoof's algorithm or by one of its improvements. Also assume that we can factor this order completely.

This is how I'd generate random elements of $E[n]$. Pick a random point $P$ on $E(\mathbb{F}_q)$. One can do this by picking an $x$-coordinate randomly and solving, if possible, a quadratic equation to get the $y$-coordinate. As we know the prime factors of the order of $E(\mathbb{F}_q)$ we can find the order of $P$ in this group, and write this order as $mn'$ where $n'$ is the highest common factor of the order of $P$ and $n$. Then computing $[m]P$ gives an element of $E[n]$.

After generating enough elements of $E[n]$, we should be able to find a two-element "basis" of the group. (I'll omit details here; this can certainly be done by reducing to the prime power case, but perhaps it can be done more generally). One ends up with two points $P$ and $Q$. These points have order $n$ and their Weil pairing $e(P,Q)$ is a primitive $n$-th root of unity $\zeta$.

Note that the Weil pairing can be computed using Miller's algorithm or more recent alternatives. Given a point $R\in E[n]$ we can find $a$ and $b$ with $R=aP+bQ$ by using the Weil pairing: $$e(R,Q)=e(aP+bQ,Q)=e(P,Q)^ae(Q,Q)^b=\zeta^a$$ etc.