Let $E/\Bbb{Q}:y^2=x^3+x$ has CM over $ \Bbb{Q}( \sqrt{-1})$. How can I prove a reduction of isogeny $[1+2\sqrt{-1}]$ is inseparable?

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Let $E/\Bbb{Q}:y^2=x^3+x$ has CM by $ \Bbb{Q}( \sqrt{-1})$. How can I prove a reduction of isogeny $[1+2\sqrt{-1}] \pmod{(1+2\sqrt{-1}}$) is inseparable ?

I know one way to judge this. Examing invariant differential with the theory of complex multiplication gives this is purely inseparable and reduction at $\pmod{5}$ to $5$-th Frobenius.

But maybe there are easy way to prove this. Thank you in advance.

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Let $f$ be a non constant map of curves. Then $f$ is separable if and only if the map of pull back of differential form $ω_E$, $ \ f*$ is nonzero.

Let $\tilde E$ be a reduction mod $(1+2\sqrt{-1})$,$E$ has good reduction at this prime ideal.

$ \tilde{[1+2\sqrt{-1}]}^*\tilde ω_E= \tilde {(1+2\sqrt{-1})} \tilde {ω_E}=(1+2・2) \tilde ω_E=5\tilde ω_E=0$ implies $\tilde{[1+2\sqrt{-1}]}$is inseparable(Especially, degree of morphism is $5$, the map is purely inseparable).