Isogenous elliptic curves with the same j-invariant are isomorphic

Question

Let $K$ be a field. Consider two elliptic curves $E,E^{\prime}$ over $K$ isomorphic over $K^{\textrm{sep}}$ or equivalently having the same $j$-invariant. Suppose there exists a non-zero $K$-isogeny $E^{\prime}\to E$. Is it always true that $E$ is isomorphic to $E^{\prime}$ over $K$? I am particularly interested in the case when $K$ is a number field.

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The statement turned out to be false for curves with complex multiplication though it is true for non CM-curves.

Answer 1

• If $E'$ doesn't have CM then yes. Let $f:E'\to E$ be the isogeny, since $j(E)=j(E')$ there is an isomorphism $g:E\to E'$ defined over $K^{sep}$ so that $g\circ f$ is an endomorphism of $E'$, which must be $[n]$. The $Gal(\overline{K}/K)$ Galois action commutes with $f$ and $[n]$ so it must commute with $g$ which is defined over $K$.

• If $E'$ has CM then it can fail. Let $$E'/\Bbb{Q}:y^2=x^3+x$$ then $$[1+i](x,y)=(x,y)+(-x,iy)$$ is an endomorphism defined over $\Bbb{Q}(i)$ but whose kernel $$\{O,(0,0)\}$$ belongs to $E'(\Bbb{Q})$, and all the endomorphisms $[\pm 1\pm i]$ with the same kernel are defined over $\Bbb{Q}(i)$.

  So $E=E'/\ker(1+i)$ is an elliptic curve defined over $\Bbb{Q}$ and we have an isogeny $E'\to E$ defined over $\Bbb{Q}$,

  $j(E')=j(E)$ since $[1+i]$ is an endomorphism,

  but $E$ can't be isomorphic to $E'$ over $\Bbb{Q}$,

  as composing that isomorphism $E\to E'$ with the isogeny $E'\to E$ would give an endomorphism of $E'$ defined over $\Bbb{Q}$ and with kernel $\ker(1+i)$.

  Edit: $E$ has the equation $Y^2=X^3-4X$, this is because $(x,y)+(0,0)=(1/x,-y/x^2)$

  so $\Bbb{Q}(E)=Tr_{\Bbb{Q}(E')/\Bbb{Q}(E)}(\Bbb{Q}(E'))=\Bbb{Q}( x+1/x, y-y/x^2)$

  and the isogeny $E'\to E$ is $(x,y)\to (x+1/x,y-y/x^2)$

Answer 2

Expanding on @reuns answer, there is a whole family of counterexamples for CM curves:

First, let's deal with CM elliptic curves over $\mathbb{Q}$, where the theory is the simplest. Suppose that $E/\mathbb{Q}$ has CM by an order $\mathcal{O}$ in an imaginary quadratic field $K$.

Pick a prime $p$ of $\mathbb{Z}$ that ramifies in $K$, i.e., $p\mathcal{O}_K = \mathfrak{p}^2$ for some prime $\mathfrak{p}$ of $\mathcal{O}_K$. Let $\mathfrak{q} = \mathcal{O} \cap \mathfrak{p}$ and define $$E[\mathfrak{q}] =\{P \in E(\bar{\mathbb{Q}})\; |\; [\alpha]_E(P) = O, \mbox{ for all } \alpha \in \mathfrak{q} \}$$ Now, picking the so-called normalized isomorphism $[\cdot]_E : \mathcal{O}\to \mathrm{End}(E)$, means we have $$\sigma(E[\mathfrak{q}]) = E[\sigma(\mathfrak{q})] $$ for every $\sigma \in \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. Since $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ acts transitively on the primes above $p$, it follows that $\sigma(\mathfrak{q})=\mathfrak{q}$ and so $$\sigma(E[\mathfrak{q}]) = E[\mathfrak{q}] .$$ Therefore, we have a quotient isogeny $$E \to E/E[\mathfrak{q}]$$ defined over $\mathbb{Q}$. Since $\mathfrak{q}$ has norm $q$ in $\mathcal{O}$, we have that $E[\mathfrak{q}]$ has cardinality $q$ and so the degree of the isogeny $E \to E/E[\mathfrak{q}]$.

Even though $E$ has CM, the only endomorphisms defined over $\mathbb{Q}$ are of the form $[n]_E$, which has degree $n^2$. So, if $E/E[\mathfrak{q}] \cong_{\mathbb{Q}} E$, then this would imply that $E$ has an endomorphism of degree $q$ defined over $\mathbb{Q}$, which cannot happen.

The same works for any CM elliptic curve $E$ defined over a number field $F$ that doesn't contain the CM field.