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Given any elliptic curve over $\mathbb{Q}$ of conductor $N$, by modularity of elliptic curves, there exists a surjective morphism from $X_0(N)$ $\rightarrow$ $E$.There may be several such 'N' and morphisms from $X_0(N)$ $\rightarrow$ $E$. One such $N$ is the conductor of $E$.

If there exists an elliptic curve $E$ and a morphism $f$: $X_0(p)$ $\rightarrow$ $E$ for a fixed prime $p$, then "will it imply that the conductor of $E$ is $p$".

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Yes. If there is a surjection $X_0(N) \to E$ then the conductor of $E$ must divide $N$. Since there is no elliptic curve of conductor 1, an elliptic curve uniformized by $X_0(p)$ for $p$ prime must have conductor $p$.

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  • $\begingroup$ Dear David, Thanks for your comment. $\endgroup$ – Srilakshmi May 9 '12 at 11:29

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