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Let $E$ be an elliptic curve of conductor $N\cdot p^2$ over $\mathbb{Q}$, defined by the equation $$y^2=x^3+p^2b\cdot x + p^3\cdot c$$ and parametrized by a map $$X_{0}(N\cdot {p}^{2})\rightarrow E$$ and let $E_{\sqrt(p)}$ be the curve $$y^2=x^3+b\cdot x+c$$ parametrized by a map $$X_{0}(N)\rightarrow E_{\sqrt(p)}$$ (Here, $p$ is a prime greater than or equal to $5$, and we assume that $E_{\sqrt(p)}$ reduces well at $p$.)

Both become isomorphic over $\mathbb{Q}[\sqrt{p}]$. Is there a lift of that isomorphism to a morphism: $$X_{0}(N\cdot {p}^{2})\rightarrow X_{0}(N)$$ or $$X_{0}(N)\rightarrow X_{0}(N\cdot {p}^{2})$$ and if so, can that morphism be described explicitly?

If it is not possible in general, can we classify the cases where such a lift exists?

(EDIT: If this question is stupid, feel free to tell me and to tell me why.)

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    $\begingroup$ Your second curve $E_{\sqrt{p}}$, am I right assuming it is the curve defined over $\mathbb{Q}$ obtained by twisting $E$ by $p$ ? If so the conductor of it is $p^2N$ in most cases, unless $p$ divided $N$ and you are lucky. $\endgroup$
    – Chris Wuthrich
    Commented Jan 11, 2016 at 14:32
  • $\begingroup$ No. $E$ is the quadratic twist of $E_{\sqrt{p}}$ by $p$. $\endgroup$
    – The Thin Whistler
    Commented Jan 13, 2016 at 18:47
  • $\begingroup$ Well there's always a forgetful morphism $X_0(Np^2)\to X_0(N)$. $\endgroup$
    – stankewicz
    Commented Jan 15, 2016 at 16:23

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