Suppose the usual modular curve $E=X\_0(N)$ over $\mathbb{Q}$ has genus 1 (e.g. $N=15$). Define the conductor of $E/\mathbb{Q}$ as the ideal/integer:


$$M=\prod\_{p}p^{f(E/\mathbb{Q}\_p)},$$


where


$$f(E/\mathbb{Q}\_p)=\begin{cases}0 & E\text{ has good reduction mod }p\\1 & E\text{ has multiplicative reduction mod }p\\ 2 & E\text{ has additive reduction mod }p\end{cases}$$


is the "exponent of the conductor of $E/\mathbb{Q}\_p$" (see for example Silverman's book Advanced Topics in the Arithmtic of Elliptic Curves, chapter IV, section 10). In the case of $p=2,3$ the exponent $f(E/\mathbb{Q}\_p)$ might have extra terms depending on its wild ramification.


How do you prove that $N=M$?


If $E$ had a Weierstrass equation, the proof would be straight forward since $f(E,\mathbb{Q}\_p)$ is easy to calculate. You could also use the Ogg-Saito formula if you can calculate the Neron model of $E/\mathbb{Q}\_p$, but this also requires a Weierstrass equation.


I know that the function field of $E/\mathbb{C}$ is equal to $\mathbb{C}(j,j\_N)$ and that an algebraic relation between $j$ and $j\_N$ gives an equation for $E$, but this polynomial is extremely inconvenient for calculations.


Is there a way to calculate a simple equation for $E/\mathbb{Q}$?



When is the conductor of an elliptic modular curve equal to its level?
Already in the beginning of the 20th century, Fricke had determined explicit equations of the modular curves $X\_0(N)$ which are of genus 1. To do this he constructed, in each case, two explicit functions $\sigma, \tau$ on $X\_0(N)$ such that $j$ and $j\_N$ are rational functions of $\sigma,\tau$ with coefficients in $\mathbb{Q}$. In this way he gets an equation of the form $\sigma^2 = P(\tau)$ where $P$ is a certain polynomial of degree 3 or 4 with integer coefficients.


From this, it is not difficult to compute a minimal Weierstrass equation and the conductor (which were not defined at the time of Fricke), using Tate's algorithm and Ogg's formula. As you have correctly guessed, one finds that the conductor is equal to $N$.


The very question of determining the conductor of $X\_0(N)$ in the case of genus 1 has in fact been solved a long time ago by Ligozat in his PhD thesis (1974), where he proved BSD for such curves (assuming the Tate-Shafarevich group is trivial). His article can be found here: [Courbes modulaires de genre 1](https://eudml.org/doc/94716). The book by Fricke is Die elliptischen Funktionen und ihre Anwendungen (II), which I believe is available on archive.org (Ligozat gives all the precise references).


Since this proof is essentially case-by-case, you may ask for a more abstract-oriented proof. This is indeed possible but uses deep results. It is a theorem of Carayol that the conductor of a modular elliptic curve is equal to the level of the associated modular form. One way to explain this is to consider the associated $L$-functions. The conductor $M$ of an elliptic curve $E$ is determined by the $L$-function $L(E,s)$, in the sense that it is the only integer such that $M^{s/2} (2\pi)^{-s} \Gamma(s) L(E,s)$ has a functional equation of the standard type. On the modular side, it is easy to show that the $L$-function of a modular form of level $N$ has conductor $N$ (in the previous sense). So if you accept to use Carayol's theorem, you get $N=M$ (Wiles's theorem is not needed in your case since $X\_0(N)=E$).


Using this more advanced method, you can in fact determine the conductor of the abelian variety $J\_0(N)$, the Jacobian variety of $X\_0(N)$, for any integer $N$. The precise formula depends on the dimensions of the space of cusp forms of weight 2 and level $N'$ dividing $N$ (see Ligozat's article, at that time it was only conjectured, and proved only in the case $N$ is squarefree using Deligne's results).


