On the modularity of elliptic curves over \(\mathbb Q\): wild 3-adic exercises (Q2723516)

In this paper the authors complete the work begun by \textit{A. Wiles} [Ann. Math. (2) 141, 443--551 (1995; Zbl 0823.11029)] and \textit{R. Taylor} and \textit{A. Wiles} [Ann. Math. (2) 141, 553--572 (1995; Zbl 0823.11030)] to show that all elliptic curves over \(\mathbb Q\) are modular. By earlier work of \textit{B. Conrad, F. Diamond} and \textit{R. Taylor} [J. Am. Math. Soc. 12, 521--567 (1999; Zbl 0923.11085)], it is sufficient to prove the following result. NEWLINENEWLINENEWLINETheorem B. If \(\bar{\rho} : \text{Gal}(\bar{\mathbb Q}/{\mathbb Q}) \to \text{GL}_2({\mathbb F}_5)\) is an irreducible continuous representation with cyclotomic discriminant, then \(\bar{\rho}\) is modular.NEWLINENEWLINENEWLINEFor the proof of Theorem B, the authors divide the representations into six classes according to their 3-adic behaviour. In each case, they find an elliptic curve \(E\) over \(\mathbb Q\) with \(\bar{\rho} = \bar{\rho}_{E,5}\) and a very specific form of the mod-3 representation \(\bar{\rho}_{E,3}\). By the Langlands-Tunnell theorem, \(\bar{\rho}_{E,3}\) is modular. Then by techniques à la Wiles and Taylor-Wiles, the authors show that \(\rho_{E,3}\) is modular as well. Whence \(\bar{\rho} = \bar{\rho}_{E,5}\) is also modular. NEWLINENEWLINENEWLINEThree of the six cases (3-adic conductor at most \(3^2\)) had been dealt with earlier by \textit{F. Diamond} [Ann.\ Math. (2) 144, 137--166 (1996; Zbl 0867.11032)] and Conrad, Diamond and Taylor [loc. cit.]. Quoting from the eminently readable introduction: ``This leaves the cases \(f = 27\), \(81\), and \(243\), which are complicated by the fact that \(E\) now only obtains good reduction over a wild extension of \({\mathbb Q}_3\). In these cases our argument relies essentially on the particular form we have obtained for \(\bar{\rho}_{E,3}\) (. . .). We do not believe that our methods for deducing the modularity of \(\rho_{E,3}\) from that of \(\bar{\rho}_{E,3}\) would work without this key simplification. It seems to be a piece of undeserved good fortune that for each possibility for \(\bar{\rho}|_{I_3}\) we can find a choice for \(\bar{\rho}_{E,3}\) for which our methods work''.