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Dear MO Community,

Let $X = X_0(N)_{/\mathbb{Q}}$, and $J$ its jacobian. Mazur defines the Eisenstein quotient of $J$, denoted $\widetilde{J}$, as

\[ 0 \rightarrow \gamma_IJ \rightarrow J \rightarrow \widetilde{J} \rightarrow 0. \]

Here, $I$ is the Eisenstein ideal of the Hecke algebra $\mathbb{T}$, $\gamma_I$ is the kernel of the map $\mathbb{T} \rightarrow \lim_{\leftarrow_m} \mathbb{T}/I^m$, and $\gamma_IJ$ is the sub-abelian variety generated by the images $\alpha J$ for $\alpha \in \gamma_I$.

My question is: why is it actually a quotient of $J^{new}_{/\mathbb{Q}}$, the new part of the Jacobian?

Mazur does prove that $\widetilde{J}$ is actually a quotient of $J^- = J/(1 + w)J$; is this the key?

References:

Mazur, B. "Modular curves and the Eisenstein Ideal", Publications Mathématiques de l'IHES, 1977

Mazur, B. "Rational isogenies of prime degree", Inventiones mathematicae, 1978

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    $\begingroup$ Wasn't $N$ prime in Mazur's work? So everything is new, right? Can you give an explicit reference to a claim of Mazur that the Eisenstein quotient is a quotient of the new part when $N$ isn't prime? That would help me to orient myself. $\endgroup$
    – Kevin Buzzard
    Commented Sep 30, 2010 at 12:33
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    $\begingroup$ Kevin is correct that in Mazur's context $N$ is always prime. Perhaps you mean to ask something else? (For example, the fact that $\tilde{J}$ is a quotient of $J^-$ is important, and while it is easy away from $2$, is one of the more subtle points of his paper in the case of an Eisenstein prime of char. 2.) $\endgroup$
    – Emerton
    Commented Sep 30, 2010 at 12:35
  • $\begingroup$ Sorry guys, I didn't realise that N was prime in the proposition. I should be more careful next time. $\endgroup$
    – Barinder Banwait
    Commented Sep 30, 2010 at 17:05

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