# Isogenous elliptic curve with integral j-invariant

Let $E$ be an elliptic curve over a local field $K$.

If $E$ has non-integral $j$-invariant, under what conditions will there exist an isogenous curve with integral $j$-invariant?

Here, saying an element $a \in K$ is integral means that $a$ belongs to the ring of integers of $K$.

I feel that the following result may be relevant.

Let $E$ be an elliptic curve over a local field $K$. We say $E$ has potential good reduction over $K$ if there is a finite extension $K^{\prime}/K$ such that $E$ has good reduction over $K^{\prime}$. The $j$-invariant of $E$ is integral if and only if $E$ has potential good reduction over $K$.

We can assume that the local field $K$ is algebraically closed and complete for the valuation $v$. Then $E$ is $v$-adically analytically isomorphic to a Tate curve, which means that there is a $v$-adic analytic isomorphism $E(K)\to K^*/q^{\mathbb{Z}}$, where $q\in K^*$ satisfies $|q|_v<1$. Further, $|j(E)|_v = |q|_v^{-1}$. Let $E'$ be isogenous to $E$, say $\phi:E\to E'$. In the analytic model, the kernel of $\phi$ is generated by two elements $\zeta$ and $Q$, where $\zeta$ is an $n$'th root of unity and $Q$ is some root of $q$, say $Q^m=q$. Then$$E'(K) \cong K^*/\zeta^{\mathbb{Z}}Q^{\mathbb{Z}} \cong K^*/Q^{n\mathbb{Z}},$$ where the second isomorphism is raising to the $n$'th power. Thus $E'$ has a Tate parametrization, and the analytic formula gives $$|j(E')|_v = |Q^n|_v^{-1} = |q|^{-n/m}_v > 1.$$ Hence $E'$ has non-integral $j$-invariant.
By the criterion of Neron-Ogg-Shafarevich, an elliptic curve has good reduction if and only if its $\ell$-adic Tate module is unramified for $\ell\neq p$, where $p$ is the residue characteristic of $K$. This is clearly an isogeny invariant. Using the result you mentioned, it follows that integrality of the $j$-invariant is an isogeny invariant.