2$. Then the theorem is \begin{thm} Under the above hypothesis, $\rho$ comes from a modular form $f$ of weight $k(\rho)$, level $N(\rho)$, and under a certain {\em extra assumption}, character $\varepsilon$ of order prime to $\ell$. \end{thm} The extra assumption alluded to in the theorem is the following. \par\noindent{\bfseries Extra Assumption.} \par Not all of the following are true. \begin{enumerate} \item $\ell=3$, \item $\rho|\Gal(\Qbar/\Q(\sqrt{-3}))$ is abelian, and \item $\det(\rho)$ is not a power of the mod 3 cyclotomic character $\chi$. \end{enumerate} \begin{example} If $\det(\rho)=\chi$ is the mod 3 cyclotomic character, for example when $\rho$ comes from the Galois representation on the 3-torsion of an elliptic curve, then the extra assumption holds so the theorem applies. [[as long as the additional assumption hold?]] \end{example} Let $\rho$ be a representation as above. Then $$\det \rho:G\into\F_{\ell^{\nu}}$$ is very likely ramified at $\ell$. Let $\chi$ be the mod $\ell$ cyclotomic character. As an exercise one can show that $\det\rho=\chi^i\theta$ for some $i$ and some $\theta$ where $\theta:G\into\F_{\ell^{\nu}}$ is unramified at $\ell$. Using the Kronecker-Weber theorem and class-field theory one can prove the following two facts. \begin{enumerate} \item One can write $\theta$ as a Dirichlet character $(\Z/M\Z)^*\into\F_{\ell^{\nu}}^*$. [[Does this mean that $\theta$ factors through the mod $M$ cyclotomic character?]] \item In fact $M|N(\rho)$, since the conductor of the determinent of a representation should divide the conductor of the representation. [[?? My notes have the opposite remark?]] \end{enumerate} Thus we can view $\theta$ as a map $(\Z/N(\rho)\Z)^*\into\F_{\ell^{\nu}}^*$. Let $H=\ker\theta\subset(\Z/N(\rho)\Z)^*$. Define a congruence subgroup $\Gamma_H(N)$ by $$\Gamma_H(N)=\{ \Bigl(\begin{matrix}a&b\\c&d\end{matrix}\Bigr) \in \Gamma_0(N) : a, d\in H \}.$$ We have the following theorem. \begin{thm} Suppose $\ell>2$ and $\rho$ satisfies the above assumptions including the extra assumption. Then $\rho$ arises from a form $f$ in $S_{k(\rho)}(\Gamma_H(N(\rho)))$.\end{thm} In particular, the theorem is true if $\rho$ comes from the $\ell$-torsion representation on an elliptic curve since $\det=\chi$ so the extra assumption is satisfied. One might ask if the extra assumption is really necessary. At first Serre\index{Serre} did not think that it was. But he was surprised to discover the following example which shows that the extra assumption can not be eliminated. \begin{example}(Serre) The space $S_2(\Gamma_1(13))$ is 2 dimensional. Let $\ell=3$. There is a newform $f\in S_2(\Gamma_1(13))$ of character $\varepsilon:(\Z/13\Z)^*\into\C^*$. This character must be trivial on $\{\pm 1\}$ and has order 6. Using [[cite Ribet's LNM 601 article]] one shows that $$\Q(\ldots,a_n,\ldots)\supset\Q(\ldots,\varepsilon(d),\ldots) =\Q(\sqrt{-3}).$$ Furthermore, since $S_2(\Gamma_1(13))$ has dimension 2 there can be at most 2 embeddings of $\Q(\ldots,a_n,\ldots)$ into $\C$ (each embedding gives an independent newform). Thus one actually has that $\Q(\sqrt{-3})=\Q(\ldots,a_n,\ldots)$. Let $\lambda=\sqrt{3}$ and let $\rholamfbar:G\into\GL(2,\F_3)$ be the associated Galois representation . Then $\det(\rholamfbar)=\chi\theta$ where $\chi$ is the mod 3 cyclotomic character and $\theta\equiv\varepsilon\pmod{3}$ [[I am not really sure what this congruence means]]. In particular, $\theta$ has order $2$. Thus $H=\ker(\theta)\subset (\Z/13\Z)^*$ is exactly the squares in $(\Z/13\Z)^*$. The conclusion of the theorem can not hold since $S_2(\Gamma_H(13))=0$. This is because any form would have to have a character whose order is at most two since it must be trivial on $H$, but $S_2(\Gamma_H(13))\subset S_2(\Gamma_1(13))$ and $S_2(\Gamma_1(13))$ consists of forms whose character is of order $6$. [[There are a lot of compatibilities to check here which I have not checked.]] In this example \begin{itemize} \item $\ell=3$, \item $\det\rho$ is not a power of $\chi$, and \item $\rholamfbar|\Gal(\overline{\Q}/\Q(\sqrt{-3}))$ is abelian. \end{itemize} A good way to see the last assertion is to look at the following formula: $$f\tensor\varepsilon^{-1}=\overline{f}$$ (up to an Euler factor at 13) in the sense that $$\rho_{f,\lambda}\tensor\varepsilon^{-1}=\rho_{\overline{f}}.$$ Now reduce mod $3$ to obtain $$\rholamfbar\tensor\varepsilon^{-1}\isom\rholamfbar$$ since $f\equiv\overline{f}\pmod{\sqrt{-3}}$ (since $3$ is ramified). Thus $\rholamfbar$ is isomorphic to a twist of itself by a complex character thus by elementary facts $\rholamfbar$ is reducible and abelian over the field corresponding to its kernel. [[So why is the Galois group of $\Q(\sqrt{-3})$ in the kernel of $\rholamfbar$?]] [[Ribet suddenly gets very excited and notes that $\rholamfbar$ is also abelian on the Galois groups of $\Q(\sqrt{13})$ and $\Q(\sqrt{39})$ by the same argument.]] \end{example} There are two main ingredients in this theorem. The first is companion forms. \subsection{Companion Forms} The key papers dealing with companion forms are Gross and Coleman-Voloch. The context is the following. Suppose we are interested in the recipe for $k(\rho)$. There are two cases to consider, the level 1 case and the level 2 case. We have already discussed the level 2 case. In the level 1 case what happens? If we semisimplify and restrict to inertia we obtain a direct sum of two representations. There are two possibilities for $\rho|I$: $$\rho|I=\Bigl(\begin{matrix}\chi^{\alpha}&*\\0&\chi^{\beta}\end{matrix}\Bigr) \text{ or } \rho|I=\Bigl(\begin{matrix}\chi^{\alpha}&0\\0&\chi^{\beta}\end{matrix}\Bigr).$$ In the second case it is easy to guess $k(\rho)$. The thing to do is to look at the exponents and normalize as best you can. Since $\alpha$ and $\beta$ are only defined mod $p-1$ we may, after relabeling if necessary, assume that $0\leq\alpha\leq\beta\leq p-2$. Factor out $\chi^{\alpha}$ to obtain $$\chi^{\alpha}\tensor \Bigl(\begin{matrix}1&0\\0&\chi^{\beta-\alpha}\end{matrix}\Bigr).$$ Next (secretly) recall that if $f$ is ordinary of weight $k$ then $f$ gives rise to the representation $$\Bigl(\begin{matrix}\chi^{k-1}&*\\0&\chi\end{matrix}\Bigr)$$ (here * can be trivial). [[Something must be wrong here because nothing makes sense in the following argument unless the lower right entry of this matrix is 1 instead of $\chi$.]] If $f$ is of weight $\beta-\alpha+1$ twisting $\alpha$ times by $q\frac{d}{dq}$ gives the desired representation. Thus the weight $k(\rho)$ is $$(p+1)\alpha+\beta-\alpha+1=\beta+p\alpha+1.$$ There is one caveat. Since Serre\index{Serre} was terrified of weight 1 forms, if $\alpha=\beta=0$ then Serre defines $k(\rho)=p$ instead of $k(\rho)=1$. There is no distinction between Serre's formula and the above formula when the weight is $>1$ since, as Katz proved, any form of weight $>1$ lifts to a characteristic $0$ form. But this is not true for weight $1$. To understand the weight $1$ forms we must use sophisticated tools of algebraic geometry. Suppose $f$ is of weight $k$, $2\leq k\leq p+1$. Let $\rholamfbar$ be the associated representation where $\lambda$ is some prime dividing $\ell=p$. Then $$\rholamfbar|I_p=\Bigl(\begin{matrix}\chi^{k-1}&*\\0&1\end{matrix}\Bigr).$$ [[I really wonder why this should be so in such brave generality? It seems like things were not like this just a minute ago.]] To introduce the idea of a companion form suppose that somehow by chance $*=0$ so we have $\Bigl(\begin{smallmatrix}\chi^{k-1}&0\\0&1\end{smallmatrix}\Bigr)$. Take $\rho$ and twist by $\chi^{1-k}$ then $$\rho\tensor\chi^{1-k}|I_p=\rho\tensor\chi^{p-k}|I_p= \Bigl(\begin{matrix}1&0\\0&\chi^{p-k}\end{matrix}\Bigr).$$ What is the minimum weight giving rise to this representation. The level is 1 [[why??]] the representation is semisimple and reducible and $\alpha=0$, $\beta=p-k$, so the natural weight is $p-k+1$. Thus the conjecture predicts that there should exist another form $g$ of weight $p+1-k$ such that $\rho_g=\rho_f\tensor\chi^{p-k}$. Here $g$ is a {\em companion form} of $f$. Thus $g=\theta^{p-k}f$ where $\theta=q\frac{d}{dq}$ is the $\theta$-operator we defined when studying $\theta$-series. This conjecture was proved by Gross when $k\neq 2,p$. It was proved by Coleman and Voloch when $k=p$. This is how we know about the weight. Next time we will talk about the character. [[What is the other main ingredient in the proof.]] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 4/17/96 %%%%%%%%% \section{The exceptional level 1 case} Let $$\rho:G=\galq\into\GL(2,\fln)$$ be a Galois representation with $\ell>2$. Assume that $\rho$ is irreducible and modular. Then $\rho$ comes from a modular form in $S_{k(\rho)}(\Gamma_1(Np))$. Here $p=\ell$ [I think] and $2\leq k(\rho)\leq p^2-1$. We still must talk about the exceptional case in level 1. In this situation $$\rho|I_{\ell}=\Bigl(\begin{matrix}\chi^{\alpha}&*\\0&\chi^{\beta}\end{matrix}\Bigr)=\chi^{\beta}\tensor\Bigl(\begin{matrix}\chi&*\\0&1\end{matrix}\Bigr)$$ since $\alpha=\beta+1$. [[I do not remember why.]] Define a representation $\sigma$ of $\galq$ by $$\sigma:=\rho\tensor\chi^{-\beta}\isom\Bigl( \begin{matrix}\chi&*\\0&1\end{matrix}\Bigr).$$ Fortunately we know enough to say something about the weight $k(\sigma)$. In fact, $$k(\sigma)=\begin{cases}2&\sigma\text{ if is finite at $\ell$}\\ \ell+1& \text{ otherwise}\end{cases}$$ This is enough to determine $k(\rho)$ since $$k(\rho)=k(\chi^{\beta}\tensor\sigma)=(p+1)\beta+k(\sigma).$$ We have not yet said what we mean for $\sigma$ to be finite at $\ell$. \begin{defn} $\sigma|{D_{\ell}}$ is finite if $\sigma|D_{\ell}$ is equivalent to a representation defined by $\mathcal{G}(\Qlbar)$, where $\mathcal{G}$ is a finite flat\index{finite flat schemes} $\fln$ vector space scheme over $\Zl$. \end{defn} This definition is not terribly enlightening. To help fathom what is going on we consider the following special case. Suppose $E/\Q$ is an elliptic curve with semistable\index{semistable} (=good or multiplicative) reduction at $\ell$. Then $E[\ell]$ defines a representation $$\sigma:G\into\Aut E[\ell]=\GL(2,\F_{\ell}).$$ Let $\Delta_E$ be the discriminant of $E$, then $\sigma$ is finite at $\ell$ iff $$\ord_{\ell}\Delta_E\equiv 0\pmod{\ell}.$$ Thus when $\Delta_E$ satisfies this requirement, $k(\sigma)=2$. Next suppose $p\neq\ell$ and assume $E$ has semistable\index{semistable} reduction at $p$. Then the representation $\sigma$ attached to $E[\ell]$ is unramified at $p$ iff $$\ord_p\Delta_E\equiv 0\pmod{\ell}.$$ The reference for everything we have been discussing is \cite{serre5}. The bridge is Kummer theory. Let $E$ be an elliptic curve with multiplicative reduction at $p$. Suppose $V=E[\ell]=\fl\oplus\fl$. We have a map $$\tau:D=\galql\into \Aut V.$$ We treat the case $p=\ell$ and $p\neq\ell$ at the same time. By the theory of Tate curves\index{Tate curve} there is an exact sequence of $D$-modules $$0\into X\into V\xrightarrow{\alpha} Y\into 0.$$ [[I do not remember him explicitly saying that these maps should be $D$-modules maps but they must be for the twist $\presup{d}s$ to be a splitting.]] Each of these terms is a $D$ module and $X$ and $Y$ are 1 dimensional. The action of $D$ on $Y$ is given by an unramified character $\varepsilon$ of order dividing 2 (i.e., the order is 1 or 2). The action of $D$ on $X$ is given by $\chi\varepsilon$. Next we define an element of $H^1(D,\Hom(Y,X))$. Choose a splitting $s:Y\into V$. Thus $s$ is a linear map (not necessarily a map of $D$-modules) [[otherwise all twists of a splittings would be trivial]] such that $\alpha s=1$. For each $d\in D$ consider the twisting $\presup{d}s:Y\into V$ defined by $\presup{d}s(y)=ds(d^{-1}y)$. Since $$\alpha (d s (d^{-1}y)) = d (\alpha (s (d^{-1} y))) = d(d^{-1}y)=y$$ if follows that $\presup{d}s$ is again a splitting. Thus $\presup{d}s-s:Y\into V$ followed by $\alpha:V\into Y$ is $0$. We therefore have that $\presup{d}s-s\in \Hom(Y,X)$. The map $d\mapsto \presup{d}s$ defines an element of $H^1(D,\Hom(Y,X))$. We have isomorphisms $$H^1(D,\Hom(Y,X))\isom H^1(D,\mu_{\ell})\isom\Qp^*/\Qp^{*\ell}.$$ The last isomorphism follows from Kummer theory. [[I do not know where the first one comes from.]] Thus $\tau$ defines an element of $\Qp^*/\Qp^{*\ell}$. Serre proved that $\tau$ is finite iff the corresponding element of $\Qp^*/\Qp^{*\ell}$ defined by $\tau$ is in the image of $\Zp^*$. If $E/\Qp$ is an elliptic curve with multiplicative reduction then there exists a Tate parameter $q\in\Qp^*$ with $\Val_p(q)>0$ such that $$E\isom E_q:=\G_m/q^{\Z}$$ over the unique quadratic unramified extension of $\Qp$. The kernel of multiplication by $p$ gives rise to an exact sequence as above. [[Work out the details sometime!]] Furthermore, the element of $\Qp^*/\Qp^{*^\ell}$ defined by the representation comming from $E[\ell]$ is just the image of $q$. In this context there is a formula for $\Delta_E$ which is $$\Delta_E=q\prod_{n=1}^{\infty}(1-q^n)^{24}.$$ Note that the product factor is a unit in $\Qp$. Thus $\Val_p\Delta_E=\Val_p q$. [[At this moment Ogus\index{Ogus} asks if Serre uses Raynaud's result. Ribet first says no, but then changes his mind and says ``Yes, he must be able to explicitly describe finite flat\index{finite flat schemes} vector space schemes.'' Then Ogus asks what is wrong with weight 1. Ribet replies that Serre didn't know how to define forms in weight 1. But L. Merel says something about Serre being frustrated because he couldn't compute in weight 1.]] \chapter{Fermat's Last Theorem} \section{The application to Fermat} \begin{quote} ``As part of this parcel, I can sketch the application to Fermat.'' \end{quote} Suppose that semistable\index{semistable} elliptic curves over $\Q$ are modular. Then FLT is true. Why? ``As I have explained {\em so many times}\ldots'' Suppose $\ell>5$ and $$a^{\ell}+b^{\ell}+c^{\ell}=0$$ with $abc\neq 0$, all relatively prime, and such that $A=a^{\ell}\equiv -1\pmod{4}$, $B=b^{\ell}$ is even and $C=c^{\ell}\equiv 1\pmod{4}$. Then we consider the elliptic curve $$E:y^2=x(x-A)(x-B).$$ The {\em minimal discriminant} is $$\Delta_E=\frac{(ABC)^2}{2^8}$$ as discussed in Serre's \cite{serre5} and Diamond-Kramer [[cite Math Research Letters, 1995]]. The conductor $N_E$ is equal to the product of the primes dividing $ABC$ (so in particular $N_E$ is square-free). Furthermore, $E$ is semistable\index{semistable} -- the only hard place to check is at $2$. Diamond-Kramer checks this explicitly. Here is how to get Fermat's theorem. View $E[\ell]$ as a $G=\galq$-module. The idea is to show that this representation must come from a modular form of weight 2 and level 2. This will be a contradiction since there are no modular forms of weight 2 and level 2. But to apply the level and weight theorem we need to know that $E[\ell]$ is irreducible. The proof of this is due to Mazur\index{Mazur}. Let $\rho:G\into\Aut E[\ell]$ be the Galois representation on the $\ell$ torsion of $E$. Since $E$ is semistable\index{semistable} for $p\neq\ell$, $$\rho|I_p\isom \Bigl(\begin{matrix}1&*\\0&1\end{matrix}\Bigr).$$ Assume $\rho$ is reducible. Then $\rho$ has an invariant subspace so $$\rho\isom\Bigl(\begin{matrix}\alpha&*\\0&\beta\end{matrix}\Bigr).$$ Then from the form of $\rho|I_p$ we see that the characters $\alpha$ and $\beta$ could be ramified only at $\ell$. Thus $\alpha=\chi^i$ and $\beta=\chi^{1-i}$ where $\chi$ is the mod $\ell$ cyclotomic character. The exponents are $i$ and $1-i$ since $\alpha\beta$ is the determinent which is $\chi$. [[Why is $\chi$ supposed to be the only possible unramified character? probably since whatever the character is, it is a product of $\chi$ times something else, and the other factor is ramified.]] What happens to $\rho$ at $\ell$, i.e., what is $\rho|I_{\ell}$? There are only two possibilities. Either $\rho|I_{\ell}$ is the direct sum of the two fundamental characters or it is the sum of the trivial character with $\chi$. The second possibility must be the one which occurs. [[I do not understand why... something about ``characters globally are determined by local information.'']] So either $i=0$ or $i=1$. If $i=0$, $$\rho=\Bigl(\begin{matrix}1&*\\0&\chi\end{matrix}\Bigr).$$ This means that there is an element of $E[\ell]$ whose subspace is left invariant under the action of Galois. Thus $E$ has a point of order $\ell$ rational over $\Q$. If $i=1$ then $$\rho=\Bigl(\begin{matrix}\chi&*\\0&1\end{matrix}\Bigr).$$ Therefore $\mu_{\ell}\hookrightarrow E[\ell]$. Divide to obtain $E'=E/\mu_{\ell}$. The representation on $E[\ell]/\mu_{\ell}$ is constant (this is basic linear algebra) so the resulting representation on $E'[\ell]$ has the form $\Bigl(\begin{smallmatrix}1&*\\0&\chi\end{smallmatrix}\Bigr)$ so $E'$ has a rational point of order $\ell$. Now $E[2]$ is a trivial Galois module since it contains $3$ obvious rational points, namely $(0,0), (A,0),$ and $(B,0)$. Thus the group structure on the curve $E$ (or $E'$) (which we constructed from a counterexample to Fermat) contains $$\Z/2\Z\oplus\Z/2\Z\oplus\Z/\ell\Z.$$ In Mazur's\index{Mazur} paper [[``rational isogenies of prime degree'']] he proves that $\ell\leq 3$. Since we assumed that $\ell>5$, this is a contradiction. Notice that we have {\em not} just proved FLT. We have demonstrated the {\em irreducibility} of the Galois representation on the $\ell$ torsion of the elliptic curve $E$ arising from a hypothetical counterexample to FLT. We now have a representation $$\rho:\galq\into\GL(2,\Fl)=\Aut E[\ell]$$ which is irreducible and modular of weight $2$ and level $N=N_E$ (the conductor of $E$). Because $\rho$ is irreducible we conclude that $\rho$ is modular of weight $k(\rho)$ and level $N(\rho)$. Furthermore $$\ord_{\ell}\Delta_E=\ord_{\ell}(ABC)^2 =2\ell\ord_{\ell}abc\equiv 0\pmod{\ell}$$ so $k(\rho)=2$. We can also prove that $N(\rho)=2$. Clearly $N(\rho)|N_E$. This is because $N(\rho)$ computed locally at $p\neq\ell$ divides the power of $p$ in the conductor of the $\ell$-adic representation for $E$ at $p$. [[I do not understand this.]] Since $\rho$ is only ramified at $2$ or maybe $\ell$, $N(\rho)$ must be a power of $2$. For $p\neq \ell$, $\rho$ is ramified at $p$ iff $\ord_p\Delta_E\not\equiv 0\pmod{\ell}$ which does happen when $p=2$. Since $N_E$ is square free this implies that $N(\rho)=2$. But $S_2(\Gamma_1(2))=0,$ which is the ultimate contradiction! But how do we know semistable\index{semistable} elliptic curves over $\Q$ are modular? %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 4/19/96 \section{Modular Elliptic Curves} \begin{thm}[Theorem A] Every semistable\index{semistable} elliptic curve over $\Q$ is modular. \end{thm} There are several ways to define a modular elliptic curve. \begin{defn} Let $E$ be an elliptic curve. Then $E$ is {\em modular} if there is a prime $\ell>2$ such that the associated $\ell$-adic Galois representation $$\rho_{E,\ell^{\infty}}:G=\galq\into\GL(2,\Zl)$$ defined by the $\ell$-power division points on $E$ is modular (i.e., it is a $\rholamf$). \end{defn} \begin{defn} Let $E$ be an elliptic curve of conductor $N_E$. For each prime $p$ not dividing $N_E$ one associates a number $a_p$ related [[in a simple way!]] to the number of points on the reduction of $E$ modulo $p$. Then $E$ is {\em modular} if there exists a cusp form $f=\sum b_n q^n$ which is an eigenform for the Hecke operators such that $a_p=b_p$ for almost all $p$. In the end one deduces that $f$ can be chosen to have weight $2$, trivial character, and level $N_E$. \end{defn} \begin{defn}[Shimura]\index{Shimura} An elliptic curve $E$ is modular if there is nonconstant map $X_0(N)\into E$ for some $N$. \end{defn} \begin{thm}[Theorem B. Wiles, Taylor-Wiles] Suppose $E$ is a semistable\index{semistable} elliptic curve over $\Q$ and suppose $\ell$ is an odd prime such that $E[\ell]$ is irreducible and modular, then $\rho_{E,\ell^{\infty}}$ is modular and hence $E$ is modular. \end{thm} Now we sketch a proof that theorem B implies theorem A. First take $\ell=3$. If $E[3]$ is irreducible then by work of Langlands-Tunnel we win. The idea is to take $$\rho:G\into\GL(2,\F_3)\hookrightarrow\GL(2,\Z[\sqrt{-2}])\subset\GL(2,\C).$$ The point is that there are two maps $\Z[\sqrt{-2}]\into\F_3$ given by reduction modulo each of the primes lying over $3$. Choosing one gives a map $\GL(2,\Z[\sqrt{-2}])\into\GL(2,\F_3)$ which, for some amazing reason [[which is?]], has a section. Then $\rho:G\into\GL(2,\C)$ is a continuous representation with odd determinent which must still be irreducible. By Langlands-Tunnel we know that $\rho$ is modular and in fact $\rho$ comes from a weight $1$ cusp form $f$ of level a power of 3 times powers of most primes dividing $\cond(E)$. Reducing $f$ modulo some prime of $\Z[\sqrt{-2}]$ lieing over $3$ we obtain a mod $3$ modular form which corresponds to $\rho:G\into E[3]$. The proof of all this uses the immense base-change business in Langlands' book. [[Ribet next says: ``have to get 3's out of the level! This jacks up the weight, and the level is still not square free. Then have to adjust the weight again.'' I do not know what the point of this is.]] Kevin Buzzard\index{Buzzard} asked a question relating to how one knows the hypothesis needed for the theorem on weights and levels applies in our situation. To answer this suppose $\ell=3$ or $5$. Form the associated representation $\rho:G\into\GL(2,\Fl)$ coming from $E[\ell]$ and assume it is irreducible, modular and semistable\index{semistable}. \begin{defn} A mod $\ell$ Galois representation is {\em semistable} if for all $p\neq\ell$, the inertia group $I_p$ acts unipotently and the conjectured weight is $2$ or $\ell+1$. \end{defn} Note that $\det\rho=\chi$. \begin{lem} Under the above assumptions $\ell$ divides the order of the image of $\rho$. \end{lem} \begin{proof} If not, then $\rho$ is finite at all primes $p$, since for primes $p\neq\ell$ inertia acts trivially [[some other argument for $\ell$]]. Inertia acts trivially because if the order of the image of $\rho$ is prime to $\ell$ then $\rho$ acts diagonally. For if not then since $\rho|I_p$ is unipotent (hence all eigenvalues are 1), in a suitable basis something like $\Bigl(\begin{smallmatrix}1&\psi\\0&1\end{smallmatrix}\Bigr)$ is in the image of $\rho$ and has order $\ell$, a contradiction. Because of finiteness $k(\rho)=2$ and $N(\rho)=1$ which is a contradiction since there are no forms of weight 2 and level 1. \end{proof} Next we consider what happens if $E[3]$ is reducible. There are two cases to consider. First suppose $E[5]$ is also reducible. Then $E$ contains a rational subgroup of order $15$. We can check by hand that all such curves are modular. The key result that is that $X_0(15)(\Q)$ is finite. The second possibility is that $E[5]$ is irreducible. In this case we first find a curve $E'$ which is semistable\index{semistable} over $\Q$ such that \begin{itemize} \item $E'[5]\isom E[5]$ (this is easy to do because of the lucky coincidence that $X_0(5)$ has genus $0$) \item $E'[3]$ is irreducible \end{itemize} Next we discover that $E'$ is modular since $E'[3]$ is irreducible. This implies $E'[5]$ is modular hence $E[5]$ is modular. Theorem B then implies $E$ is modular. \chapter{Deformations} \index{deformations} \section{Introduction} For the rest of the semester let $\ell$ be an odd prime. Let $\rho:G\into\GL(2,\Fln)$ be such that \begin{itemize} \item $\rho$ is modular \item $\rho$ is irreducible \item $\rho$ is semistable\index{semistable} \end{itemize} \begin{defn}The representation $\rho$ is {\em semistable} if \begin{itemize} \item for all $p\neq\ell$, $$\rho|I_p\isom\Bigl(\begin{matrix}1&*\\0&1\end{matrix}\Bigr),$$ ($*$ is typically trivial since most primes are unramified.) \item $k(\rho)=2\text{ or }\ell+1$. \end{itemize} \end{defn} This means that there are 2 possibilities. \begin{enumerate} \item $\rho$ is finite at $D_{\ell}$. \item $$\rho|D_{\ell}\isom \Bigl(\begin{matrix}\alpha&*\\0&\beta\end{matrix}\Bigr)$$ where $\beta$ is unramified. (Since $\det(\rho)=\chi$ we can add that $\alpha|I_{\ell}=\chi$.) \end{enumerate} If $k(\rho)=2$ then possibility 1 occurs. If $k(\rho)=\ell+1$ then we are in case 2, but being in case 2 does not imply that $k(\rho)=\ell+1$. If $\rho$ comes from an elliptic curve $E/\Q$, then case 1 occurs if $E$ has good reduction at $\ell$ whereas case 2 occurs if $E$ has ordinary multiplicative reduction [[I am not sure about this last assertion because I missed it in class.]]. What is a deformation\index{deformations} of $\rho$ and when can we prove that it is modular? Let $A$ be a complete local Noetherian ring with maximal ideal $\m$ and residue field $\Fln$ (so $A$ is furnished with a map $A/\m\iso\Fln$). Let $$\tilde{\rho}_A:G\into\GL(2,A)$$ be a representation which is ramified at only finitely many primes. Assume $\tilde{\rho}=\tilde{\rho}_A$ lifts $\rho$, i.e., the reduction of $\tilde{\rho}$ mod $\m$ gives $\rho$. \begin{thm} Let the notation be as above, then $\tilde{\rho}$ is modular iff it satisfies ($*$).\end{thm} Neither of the terms in this theorem have been defined yet. %%%%%% I do not have enough in my notes to define these terms yet. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%4/23/96, Ribetnotes.tex \section{Condition $(*)$} Let $$\rho:G=\galq\into \GL(2,\Fln)$$ be a Galois representation. The statement we wish to understand is \begin{quote} ``All $\tilde{\rho}$ which satisfy $(*)$ are modular.'' \end{quote} Let $A$ be a complete local Noetherian ring with residue field $\Fln$. This means that we are given a map $A/\m\isom\Fln$. Suppose $\tilde{\rho}:G\into\GL(2,A)$ satisfies the following conditions: \begin{itemize} \item $\tilde{\rho}$ lifts $\rho$, \item $\det\tilde{\rho}=\tilde{\chi}$, \item $\tilde{\rho}$ is ramified at only finitely many primes, and \item condition $(*)$. \end{itemize} What is condition $(*)$? It the requirement that $\tilde{\rho}$ have the same qualitative properties as $\rho$ locally at $\ell$. There are two cases to consider. {\bfseries Case 1.} (arising from supersingular reduction at $\ell$) Suppose $\rho$ is finite and flat at $\ell$. Then $\rho|I_{\ell}$ is given by the 2 fundamental characters $$I_{\ell}\into\F^*_{\ell^2}$$ of level 2 (instead of from powers of these characters because of the semistability assumption). Condition $(*)$ is that the lift of $\rho$ is also constrained to be finite and flat. This means that for every $n\geq 1$, $$\tilde{\rho}|D_{\ell} \mod \m^n : D_{\ell}\into \GL(2,A/\m^n)$$ is finite and flat, i.e., it comes from a finite flat\index{finite flat schemes} group scheme over $\Zl$ which is provided with an action of $A/\m^n$ as endomorphisms. {\bfseries Case 2.} (bad multiplicative reduction at $\ell$ or good ordinary reduction at $\ell$) In case 2 $$\rho|D_{\ell}\isom \Bigl(\begin{matrix}\alpha&*\\0&\beta\end{matrix}\Bigr)$$ where $\beta$ is unramified (equivalently $\alpha|I_{\ell}=\chi$). [[In the case of good ordinary reduction $\beta$ is given by the action of Galois on $E[\ell]$ in characteristic $\ell$.]] Condition $(*)$ is the requirement that $$\tilde{\rho}|D_{\ell}\isom \Bigl(\begin{matrix}\tilde{\alpha}&*\\0&\tilde{\beta}\end{matrix}\Bigr),$$ where $\tilde{\beta}$ is unramified, $\tilde{\alpha}|I_{\ell}=\tilde{\chi}$. It follows automatically that $\tilde{\beta}$ lifts $\beta$. \subsection{Finite flat representations}\index{finite flat representations} In general what does it mean for $\rho$ to be finite and flat. It means that there exists a finite flat\index{finite flat schemes} group scheme $\sG$ over $\Zl$ such that $\sG(\Qlbar)$ is the representation space of $\rho$. This definition is {\em subtle}. Coleman asked if there is a way to reformulate the definition without mentioning group schemes. Ribet mentioned Hopf algebras but then stopped. Buzzard\index{Buzzard} suggested some of the subtlety of the definition by claiming that in some situation $\chi$ is finite flat\index{finite flat representations} whereas $\chi^2$ is not. Ogus\index{Ogus} vaguely conjectured that Fontaine's language is the way to understand this. It is possible in case 2 above for $\rho|D_{\ell}$ to be finite flat without $\tilde{\rho}$ finite flat. The quintessential example is an elliptic curve $E$ with supersingular reduction at $\ell$ such that $\ell|\ord_{\ell}\Delta_E$. \section{Classes of Liftings} Let $\Sigma$ be a finite set of prime numbers. We characterize a class of liftings $\tilde{\rho}$ which depends on $\Sigma$. What does it mean for $\tilde{\rho}$ to be in the class of deformations \index{deformations} corresponding to $\Sigma$? \subsection{The case $p\neq\ell$} First we talk about the case when $p\neq\ell$. If $p\in\Sigma$ then there is no special condition on $\tilde{\rho}|I_p$. If $p\not\in\Sigma$ one requires that $\tilde{\rho}$ is qualitatively the same as $\rho$. This means \begin{enumerate} \item If $\rho$ is unramified at $p$ (which is the usual case), then we just require that $\tilde{\rho}$ is unramified at $p$. \item If $\rho$ is ramified but unipotent at $p$ so $$\rho|I_p\isom \Bigl(\begin{matrix}1&*\\0&1\end{matrix}\Bigr)$$ we require that $$\tilde{\rho}|I_p\isom \Bigl(\begin{matrix}1&*\\0&1\end{matrix}\Bigr).$$ This situation can occur with an elliptic curve which has multiplicative reduction at $p$ and for which $\ell\nd\ord_p\Delta_E$. \end{enumerate} At this point I mention the prime example. \begin{example} Suppose $\tilde{\rho}=\rho_{f,\lambda}$ is the $\lambda$-adic representation attached to $f$. What can we say about $\ord_p N(f)$? We know that $$\ord_p N(f)=\ord_p N(\rho) + \dim (\rho)^{I_p} - \dim (\tilde{\rho})^{I_p}$$ where $(\rho)^{I_p}$ means the inertia invariants in the representation space of $\rho$. If $\rho$ is semistable then $$\ord_p N(\rho)+\dim (\rho)^{I_p} = 2.$$ Since we are assuming $\rho$ is semistable\index{semistable}, $\ord_p N(f)\leq 2$. Furthermore, the condition $p\not\in\Sigma$ is a way of saying $$\ord_p N(f) = \ord_p N(\rho). $$ Thus the requirement that $p\not\in\Sigma$ is that the error term $\dim(\rho)^{I_p}-\dim(\tilde{\rho})^{I_p}$ vanish. Note that $\ord_p N(\tilde{\rho})=\ord_p N(f)$ by Carayol's theorem. Thus $\ord_p N(f)$ is just a different way to write $\ord_p N(\tilde{\rho})$. \end{example} \begin{example} Imagine $\rho$ is ramified at $p$ and $\ord_p N(\rho)=1$. Then $\ord_p N(f)$ is either 1 or 2. The requirement that $p\not\in\Sigma$ is that $\ord_p N(f)=1$. \end{example} \subsection{The case $p=\ell$} Next we talk about the case $p=\ell$. There are two possibilities: either $\ell\in\Sigma$ or $\ell\not\in\Sigma$. If $\ell\in\Sigma$ then we impose no {\em further} condition on $\tilde{\rho}$ (besides the already imposed condition $(*)$, semistability at $\ell$, etc.). If $\ell\not\in\Sigma$ and $\rho$ if finite and flat (which is not always the case) then we require $\tilde{\rho}$ to be finite and flat. If $\ell\not\in\Sigma$ and $\rho$ is not finite flat\index{finite flat representations} then no further restriction (this is the Tate curve\index{Tate curve} situation). %%%%%% Suppose $\tilde{\rho}=\rho_{f,\lambda}$, and $\tilde{\rho}$ belongs to the class defined by $\Sigma$. We want to {\em guess} (since there is no Carayol theorem) an integer $N_{\Sigma}$ such that $N(f)|N_{\Sigma}$. What is $N_{\Sigma}$? It will be $$N_{\Sigma}=\prod_{\substack{p\neq\ell\\p\in\Sigma}} p^2 \cdot \prod_{\substack{p\neq\ell\\p\not\in\Sigma}} p^{\ord_p N(\rho)} \cdot \ell^{\delta}.$$ Here $\ord_p N(\rho)$ is 1 iff $\rho$ is ramified at $p$. The exponent $\delta$ is either $0$ or $1$. It is $1$ iff $k(\rho)=\ell+1$ or $\ell\in\Sigma$ and $\rho$ is ordinary at $\ell$, i.e., $$\rho|D_{\ell}\isom \Bigl(\begin{matrix}\alpha&*\\0&\beta\end{matrix}\Bigr).$$ [[This definition could be completely wrong. It was definitely not presented clearly in class.]] There is an exercise associated with this. It is to justify {\em a priori} the definition of $\delta$. Suppose, for example, that $\rho_{f,\lambda}$ satisfies $(*)$, then we want to show that $\ell^2\nd N(f)$. \begin{thm} Every $\tilde{\rho}$ of class $\Sigma$ comes from $S_2(\Gamma_0(N_{\Sigma}))$. \end{thm} Define an approximation $\T$ to the Hecke algebra by letting $$\T=\Z[\ldots,T_n,\ldots]\subset\End(S_2(\Gamma_0(N_{\Sigma})))$$ where we adjoin only those $T_n$ for which $(n,\ell N_{\Sigma})=1$. For some reason there exists a map $$\T\into\Fln: \quad T_r\mapsto \tr \rho(\Frob_r).$$ Why should such a map exist? The point is that we know by the theorem that $\rho$ comes by reduction from a $\rho_{f,\lambda}$ with $f\in S_2(\Gamma_0(N_{\Sigma}))$. But there is a wrinkle. [[I do not understand: He says: ``Clearly $N(\rho)|N_{\Sigma}$. $k(\rho)=2$ or $\ell+1$. If $k(\rho)=\ell+1$ then $\delta=1$ so $\ell|N_{\Sigma}$. Have to trickily slip over to weight 2 in order to get $f$.'' This does not make any sense to me.]] \section{Wiles' Hecke algebra} Composing the map $\T\into\sO_{E_f}$ with reduction mod $\lambda$ from $\sO_{E_f}$ to $\Fln$ we obtain a map $\T\into\Fln$. Let $\m$ be the kernel. Then $\m$ is a maximal ideal of $\T$. Wiles' Hecke algebra is the completion $\T_{\m}$ of $\T$ at $\m$. [[For some reason]] there exists $$\tilde{\rho}:G=\galq\into\GL(2,\T_{\m})$$ such that $\tr(\tilde{\rho}(\Frob_r))=T_r$. What makes this useful is that $\tilde{\rho}$ is {\em universal} for lifts of type $\Sigma$. This means that given any lift $\tau$ of type $\Sigma$ there exists a map $$\varphi:\GL(2,\T_{\m})\into\GL(2,A)$$ such that $\tau=\varphi\tilde{\rho}$. Another key idea that the approximation $\T$ obtained by just adjoining those $T_n$ with $(n,\ell N_{\Sigma})=1$ is, after completing at certain primes, actually equal to the whole Hecke algebra. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 4/24/96, ribetnotes.tex \chapter{The Hecke Algebra $T_{\Sigma}$} \section{The Hecke Algebra} Throughout this lecture $\ell\neq p$ and $\ell\geq 3$. We are studying the representation $\rho:G\into\GL(2,\Fln)$. This is an irreducible representation, $\ell$ is odd, $\rho$ is semistable\index{semistable}, and $\det\rho=\chi$. To single out certain classes of liftings we let $\Sigma$ be a finite set of primes. Let $A$ be a complete local Noetherian ring with residue filed $\Fln$. We take liftings $\tilde{\rho}:G\into \GL(2,A)$ such that $\tilde{\rho}$ reduces down to $\rho$, $\det\tilde{\rho}=\tilde{\chi}$, and $\tilde{\rho}$ is ``like'' $\rho$ away from $\Sigma$. For example, if $\rho$ is unramified at $p$ we also want $\tilde{\rho}$ unramified at $p$, etc. Assume $\tilde{\rho}$ is modular. We guess the serious divisibility condition that $N(f)|N_{\Sigma}$. Recall that $$N_{\Sigma}=\prod_{\substack{p\neq\ell\\p\in\Sigma}} p^2 \cdot \prod_{\substack{p\neq\ell\\p\not\in\Sigma}} p^{\ord_p N(\rho)} \cdot \ell^{\delta}.$$ To define $\delta$ consider two cases. \begin{itemize} \item {\em level 1 case.} Take $\delta=1$ if $\ell\in\Sigma$ or if $\rho$ is not finite at $\ell$. Take $\delta=0$ otherwise. \item {\em level 2 case.} This is the case when $\rho|I_{\ell}$ has order $\ell^2-1$. Take $\delta=0$. \end{itemize} {\em A priori} nobody seems to know how to prove that $N(f)|N_{\Sigma}$ given only that $\tilde{\rho}$ is modular. In the end we will show that all modular $\tilde{\rho}$ in fact come from $$S_2(\Gamma_0(N_{\Sigma})).$$ This can be regarded as a proof that $N(\rho)|N_{\Sigma}$. Last time we tried to get things going by defining the {\em anemic Hecke algebra} $$\T=\Z[\ldots,T_n,\ldots]\subset \End(S_2(\Gamma_0(N_{\Sigma})))$$ where we only adjoin those $T_n$ for which $(n,\ell N_{\Sigma})=1$. By some level lowering theorem there exists an $f\in S_w(\Gamma_0(N_{\Sigma}))$ giving $\tilde{\rho}$ so we obtain a map $\T\into \F$. The map sends $T_n$ to the reduction modulo $\lambda$ of its eigenvalue on $f$. ($\lambda$ is a prime of the ring of integers of $E_f$ lying over $\ell$.) [[... something about needing an $f$ of the right level = $N(\rho)$.]] Let $\m\subset\T$ be the kernel of the above defined map $\T\into\F$. Then $\m$ is a maximal ideal. Let $\T_{\m}$ be the completion of $\T$ at $\m$ and note that $$\T_{\m}\hookrightarrow \T\tensor_{\Z}\Zl.$$ We need to know that $\T_{\m}$ is Gorenstein. This is done by comparing $\T_{\m}$ to some full Hecke ring. Thus $$\T\subset R=\Z[\ldots,T_n,\ldots]\subset\End(S_2(\Gamma_0(N_{\Sigma})))$$ where the full Hecke ring $R$ is obtained by adjoining the $T_n$ for {\em all} integers $n$. We should think of $R$ as $$R=\T[T_{\ell}, \{U_P : p|N_{\Sigma}\}].$$ Note that $T_{\ell}$ may or may not be a $U_{\ell}$ depending on if $\ell|N_{\Sigma}$. \begin{lem} If $\ell\nd N_{\Sigma}$ then $R=\T[U_p,\ldots]$. Thus if $T_{\ell}$ is not a $U_{\ell}$ then we do not need $T_{\ell}$. \end{lem} \begin{proof} This lemma is an interesting thing and the proof goes as follows. Oooh. Sorry, this is not true. Ummm.... hmm. The ring $\T[U_p,\ldots]$ is clearly of finite index in $R$ since: if $q$ is a random prime number consider $R\tensor_{\Z}\Q_q$ compared to $\T[U_p,\ldots]\tensor_{\Z}\Q_q$. [[I do not know how to do this argument. The lemma as stated above probably isn't really true. The point is that the following lemma is the one we need and it is true.]] \end{proof} Let $\T[U_p,\ldots]$ be the ring obtained by adjoining to $\T$ just those $U_p$ with $p|N_{\Sigma}$. \begin{lem} If $\ell\nd N_{\Sigma}$ then the index $(R:\T[U_p,\ldots])$ is prime to $\ell$. Note that we assume $\ell\geq 3$. \end{lem} \begin{proof} We must show that the map $\T[U_p,\ldots]\into R/\ell R$ is surjective. [[I thought about it for a minute and did not see why this suffices. Am I being stupid?]] Let $A=\F_{\ell}[T_n : (n,\ell)=1]$ be the image of $\T[U_p,\ldots]$ in $R/\ell R$ so we have a diagram $$\begin{array}{ccc} \T[U_p,\ldots]&\longrightarrow&R/\ell R\\ \searrow & &\cup\\ & &A \end{array}$$ We must show that $A=R/\ell R$. There is a beautiful duality $$R/\ell R\cross S_2(\Gamma_0(\ns);\Fl)\into\Fl \quad \text{(perfect pairing)}.$$ Thus $A^{\perp}=0$ iff $A=R/\ell R$. Suppose $0\neq f\in A^{\perp}$, then $a_n(f)=0$ for all $n$ such that $(n,\ell)=1$. Thus $f=\sum a_{n\ell}q^{n\ell}$. Let $\theta=q\frac{d}{dq}$ be the theta operator. Since the characteristic is $\ell$, $\theta(f)=0$. On the other hand $w(f)=2$ and since $\ell\geq 3$, $\ell\nd 2$. Thus $w(\theta f)=w(f)+\ell+1=2+\ell+1$ which is a contradiction since $\theta f=0$ and $w(0)=0\neq 3+\ell$. [[The weight is an integer {\em not} a number mod $\ell$, right?]] \end{proof} \begin{example} The lemma only applies if $\ell\geq 3$. Suppose $\ell=2$ and consider $S_2(\Gamma_0(23))$. Then $$\T[U_p,\ldots]=\Z[\sqrt{5}]\subset R=\Z[\frac{1+\sqrt{5}}{2}]$$ so $(R:\T[U_p,\ldots])$ is not prime to $2$. \end{example} \begin{remark} If $N=\ns$ then $$\rank_{\Z}\T=\sum_{M|N}S_2(\Gamma_0(M))^{\new}$$ and $$\rank_{\Z} R = \dim S_2(\Gamma_0(N)).$$ There is an injection $$\bigoplus_{M|N} S_2(\Gamma_0(M))^{\new}\hookrightarrow S_2(\Gamma_0(N))$$ but $\oplus S_2(\Gamma_0(M))^{\new}$ is typically much smaller than $S_2(\Gamma_0(N))$. \end{remark} \section{The maximal ideal in $R$} The plan is to find a special maximal ideal $\m_R$ $$\begin{array}{ccc} \m_R&\subset& R\\ \cup& & \cup\\ \m &\subset& \T\end{array}.$$ Once we finally found the correct $\m$ and $\m_R$ then $\T_{\m}\iso R_{\m_R}$ will be an isomorphism. In finding $\m_R$ we will not invoke some abstract going up theorem but we will ``produce'' $\m_R$ by some other process. The ideal $\m$ was defined by a newform $f$ level $M|\ns$. The coefficients of $f$ lie in $$\sO_{\lambda}=(\sO_{E_f})_{\lambda}$$ where $E_f$ is the coefficient ring of $f$ and $\lambda$ is a prime lying over $\ell$. Thus $\sO_{\lambda}$ is an $\ell$-adic integer ring. Composing the residue class map $\sO_{\lambda}\into\Fbar$ with the eigenvalue map $\T\into\sO_{\lambda}$ we obtain the map $\T\into\Fbar$. In order to obtain the correct $\m_{R}$ we will make a sequence of changes to $f$ to make some good newform. [[Is the motivation for all this the fact that the lemma will not apply if $\ell|\ns$?]] \subsection{Strip away certain Euler factors} Write $f=\sum a_n q^n$. Replace $f$ by $$h=\sum_{\text{certain $n$}} a_n q^n$$ where the sum is over those $n$ which are prime to each $p\in\Sigma$. What does this mean? If we think about the $L$-function $L(f,s)=\prod_p L_p(f,s)$, then $h$ has $L$-function $$L(h,s)=\prod_{p\not\in\Sigma} L_p(f,s).$$ Furthermore making this change does not take us out of $S_2(\Gamma_0(\ns))$, i.e., $h\in S_2(\Gamma_0(\ns))$. [[He explained why but my notes are very incomplete. They say: Why. Because $h=(f\tensor\varepsilon)\tensor\varepsilon$, ($\varepsilon$ Dirichlet character ramified at primes in $\ns$). Get a form of level $\lcm(\prod_{p\in\Sigma}p^2,N(f))|\ns$. Can strip and stay in space since $\ns$ has correct squares built into it.]] \begin{remark} Suppose that $p||\ns$. Then $p\not\in\Sigma$ and $p||N(\rho)$. The level of $f$ is $$N(f)=\begin{cases}N(\rho),&\text{if $k(\rho)=2$}\\ N(\rho)\ell,&\text{if $k(\rho)=\ell+1$}\end{cases}.$$ If $p||N(\rho)$ then $f|T_p=a_p(f)f$. Thus $h$ is already an eigenform for $T_p$ unless $p\in\Sigma$ in which the eigenvalue is $0$. [[I do not understand this remark. Why would the eigenvalue being 0 mean that $T_p$ is not an eigenform? Furthermore, what is the point of this remark in the wider context of transforming $f$ into a good newform.]] \end{remark} \subsection{Make into an eigenform for $U_{\ell}$} We perform this operation to $f$ to obtain a form $g$ then apply the above operation to get the ultimate $h$ having the desired properties. Do this if $\ell|\ns$ but $\ell\nd N(f)$. This happens precisely if $\ell\in\Sigma$ and $\rho$ is good and ordinary at $\ell$. Then $f|U_{\ell}$ is just some random junk. Consider $g=f+*f(q^{\ell})$ where $*$ is some coefficient. We see that if $*$ is chosen correctly then $g|U_{\ell}=Cg$ for some constant $C$. There are 2 possible choices for $*$ which lead to 2 choices for $C$. Let $a_{\ell}=a_{\ell}(f)$, then $C$ can be either root of $$X^2+a_{\ell}X+\ell=0.$$ This equation has exactly one unit root in $\sO_{\lambda}$. The reason is because we are in the ordinary situation so $a_{\ell}$ is a unit. But $\ell$ is not a unit. The sum of the roots is a unit but the product is not. [[Even this is not clear to me right now. Definitely check this later.]] Make the choice of $*$ so that real root is $C$. Then we obtain a $g$ such that $$g|U_{\ell}=\text{(unit)}\cdot g.$$ Next apply the above procedure to strip $g$ and end up with an $h$ such that \begin{itemize} \item $h|U_{\ell}=\text{(unit)}\cdot h$, \item $h|U_p=0$ for $p\in\Sigma$ ($p\neq\ell$), and \item $h|T_p=a_p(f)\cdot h$ for $p\not\in\Sigma$. \end{itemize} Now take the form $h$. It gives a map $R\into\sO_{\lambda}$ which extends $\T\into\sO_{\lambda}$. Let $\m_R$ be the kernel of the map $R\into\Fbar$ obtained by composing $R\into\sO_{\lambda}$ with $\sO_{\lambda}\into\Fbar$. A lot of further analysis shows that $\T_{\m}\into R_{\m_r}$ is an isomorphism. We end up having to show separately that the map is injective and surjective. [[In this whole lecture $p\neq\ell$.]] [[Wiles' notation: His $\T_{\m_R}$ is my $\R$ and his $\T'$ is my $\T_{\m}$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 4/26/96, ribetnotes.tex \section{The Galois Representation} We started with a representation $\rho$, chose a finite set of primes $\Sigma$ and then made the completed Hecke algebra $\T_{\m}$. Our goal is to construct the universal deformation\index{deformations} of $\rho$ of type $\Sigma$. The universal deformation is a representation $$\tilde{\rho}:G\into\GL(2,\T_{\m})$$ such that for all primes $p$ with $p\nd\ell\ns$, $\tilde{\rho}(\Frobp)$ has trace $T_p\in\T_{\m}$ and determinent $p$. We now proceed with the construction of $\tilde{\rho}$. Let $$\T=\Z[\ldots,T_n,\ldots], \quad (n,\ell\ns)=1$$ be the anemic Hecke algebra. Then $\T\tensor\Q$ decomposes as a product of fields $$\T\tensor\Q=\prod_{f} E_f$$ where the product is over a set of representatives for the Galois conjugacy classes of newforms of weight 2, trivial character, and level dividing $\ns$. Since $\T$ is integral (it is for example a finite rank $\Z$-module), $\T\hookrightarrow\prod_{f}\sO_{f}$. Since $\Zl$ is a flat $\Z$-module, $$\T\tensor\Zl\hookrightarrow \prod_{f}\sO_f\tensor\Zl =\prod_{f,\lambda}\sO_{f,\lambda}$$ where the product is over a set of representatives $f$ and all $\lambda|\ell$. $\T_{\m}$ is a direct factor of $\T\tensor\Zl$. [[This is definitely not the assertion that $\T_{\m}$ is an $\sO_{f,\lambda}$. What exactly is it the assertion of really?]] We can restrict the product to a certain finite set $S$ and still obtain an injection $$\T_{\m}\hookrightarrow \prod_{(f,\lambda)\in S} \sO_{f,\lambda}.$$ The finite set $S$ consists of those $(f,\lambda)$ such that the prime $\lambda$ of $\sO_f$ pulls back to $\m$ under the map $\T\into\sO_f$ obtained by composing $\T\into\prod_f\sO_f$ with the projection onto $\sO_f$. [[Why is this enough so that $\T_{\m}$ still injects in?]] Restricting to a finite product is needed so that $$[\prod_{(f,\lambda)\in S}\sO_{f,\lambda}:\Tm]<\infty.$$ Given $f$ and $\lambda$ there exists a representation $$\rho_{f,\lambda}:G\into\GL(2,\sO_{f,\lambda}).$$ It is such that $\tr \rho_{f,\lambda}(\Frob_p)=a_p$ is the image of $T_p$ under the inclusion $$\T\tensor\Zl\hookrightarrow\prod_{(f,\lambda)\in S}\sO_{f,\lambda}.$$ Put some of these $\rho_{f,\lambda}$ together to create a new representation $$\rho'=\prod_{(f,\lambda)\in S}\rho_{f,\lambda}: G\into\GL(2,\prod\sO_{f,\lambda}) \subset \GL(2,\Tm\tensor\Q).$$ The sought after universal deformation\index{deformations} $\tilde{\rho}$ is a map making the following diagram commute $$\begin{array}{ccc} G& \xrightarrow{\rho'} & \GL(2,\Tm\tensor\Q)\\ & \tilde{\rho} \searrow & \cup\\ & & \GL(2,\Tm) \end{array}$$ \begin{thm} $\rho'$ is equivalent to a representation taking values in $\GL(2,\Tm)$. \end{thm} One way to [[try to]] prove this theorem is by invoking a general theorem of Carayol. [[and then what? does this way work? why is it not a good way?]] But the right way to prove the theorem is Wiles' way. \subsection{The structure of $\Tm$} Just as an aside let us review the structure of $\Tm$. \begin{itemize} \item $\Tm$ is local. \item $\Tm$ is not necessarily a discrete valuation ring. \item $\Tm\tensor\Q$ is a product of finite extensions of $\Ql$. \item $\Tm$ is not necessarily a product of rings $\sO_{f,\lambda}$. \item $\Tm$ need not be integral. [[or can I say, ``IS not integral.'' ?]] \end{itemize} \subsection{The philosophy in this picture} Choose $c$ to be a complex conjugation in $G=\galq$. Since $\ell$ is odd $\det(c)=-1$ is a very strong condition which rigidifies the situation. \subsection{Massage $\rho$} Choose two 1-dimensional subspaces so that $$\rho(c)=\Bigl(\begin{matrix}-1&0\\0&1\end{matrix}\Bigr)$$ with respect to any basis consisting of one vector from each subspace. For any $\sigma\in G$ write $$\rho(\sigma)= \Bigl( \begin{matrix} a_{\sigma} & b_{\sigma}\\ c_{\sigma} & d_{\sigma} \end{matrix} \Bigr).$$ Then $a_{\sigma}d_{\sigma}$ and $b_{\sigma}c_{\sigma}$ are somehow intrinsically defined. This is because $$\rho(\sigma c)= \Bigl( \begin{matrix} -a_{\sigma} & ?\\ ? & d_{\sigma} \end{matrix} \Bigr)$$ so $$a_{\sigma}=\frac{\tr(\rho(\sigma))-\tr(\rho(\sigma{}c))}{2}$$ and $$d_{\sigma}=\frac{\tr(\rho(\sigma))+\tr(\rho(\sigma{}c))}{2}.$$ Since we know the determinent it follows that $b_{\sigma}c_{\sigma}$ is also intrinsically known. [[The point is that we know certain things about these matrices in terms of their traces and determinants.]] \begin{prop} There exists $g\in G$ such that $b_g c_g\neq 0$. \end{prop} \begin{proof} Since $\rho$ is irreducible there exists $\sigma_1$ such that $b_{\sigma_1}\neq 0$ and there exists $\sigma_2$ such that $c_{\sigma_2}\neq 0$. If $b_{\sigma_2}\neq 0$ or $c_{\sigma_1}\neq 0$ then we are done. So the only problem case is when $b_{\sigma_1}=0$ and $c_{\sigma_2}=0$. Easy linear algebra shows that in this situation $g=\sigma_1 \sigma_2$ has the required property. \end{proof} Now rigidify by choosing a basis so that $b_g=1$. Doing this does not fix a basis because they are many ways to choose such a basis. \subsection{Massage $\rho'$} Choose a basis of $(\Tm\tensor\Q)^2$ so that $$\rho'(c)=\Bigl(\begin{matrix}-1&0\\0&1\end{matrix}\Bigr).$$ For any $\sigma\in G$ write $$\rho'(\sigma)=\Bigl( \begin{matrix} a_{\sigma} & b_{\sigma}\\ c_{\sigma} & d_{\sigma} \end{matrix}\Bigr)$$ Using an argument as above shows that $a_{\sigma}, d_{\sigma}\in\Tm$ since the traces live in $\Tm$. Furthermore $b_{\sigma}c_{\sigma}\in\Tm$ since the determinent is in $\Tm$. The {\em key observation} is that $b_{\sigma} c_{\sigma}$ reduces mod $\m$ to give the previous $b_{\sigma}c_{\sigma}\in\F$ corresponding to $\rho(\sigma)$. This is because the determinants and traces of $\rho'$ are lifts of the ones from $\rho$. Since $\m$ is the maximal ideal of a local ring and $b_g c_g$ reduces mod $\m$ to something nonzero it follows that $b_g c_g$ is a unit in $\Tm$. Choose a basis so that $$\rho'(c)= \Bigl(\begin{matrix}-1&0\\ 0& 1 \end{matrix}\Bigr)$$ and also so that $$ \rho'(g)= \Bigl(\begin{matrix}a_g&1\\ u& d_g \end{matrix}\Bigr) \in \GL(2,\Tm).$$ Here $u$ is a unit in $\Tm$. \begin{prop} Write $$\rho'(\sigma)= \Bigl(\begin{matrix}a_{\sigma} & b_{\sigma}\\ c_{\sigma} & d_{\sigma} \end{matrix}\Bigr)$$ with respect to the basis chosen above. Then $a_{\sigma}, b_{\sigma}, c_{\sigma}, d_{\sigma}\in\Tm$. \end{prop} \begin{proof} We already know that $a_{\sigma}, d_{\sigma}\in\Tm$. The question is how to show that $b_{\sigma}, c_{\sigma}\in \Tm$. Since $$\rho'(\sigma g)= \Bigl(\begin{matrix}a_{\sigma}a_{g}+b_{\sigma}u & ?\\ ? & c_{\sigma}+d_{\sigma}d_g \end{matrix}\Bigr)$$ we see that $a_{\sigma}a_g+b_{\sigma}u\in\Tm.$ Since $a_{\sigma}a_g\in\Tm$ it follows that $b_{\sigma}u\in\Tm$. Since $u$ is a unit in $\Tm$ this implies $b_{\sigma}\in\Tm$. Similarly $c_{\sigma}+d_{\sigma} d_g\in\Tm$ so $c_{\sigma}\in\Tm$. \end{proof} As you now see, in this situation we can prove Carayol's theorem with just some matrix computations. [[This is basically a field lowering representation theorem. The thing that makes it easy is that there exists something (namely $c$) with distinct eigenvalues which is rational over the residue field. Schur's paper, models over smaller fields. ``Schur's method''.]] \subsection{Representations from modular forms mod $\ell$} If you remember back in the 70's people would take an $f\in S_2(\gon;\F)$ which is an eigenform for almost all the Hecke operators $$T_p f=c_p f\quad\text{ for almost all $p$, and $c_p\in\F$}.$$ The question is then: Can you find $$\rho:G\into\GL(2,\F)$$ such that $$\tr(\rho(\Frobp))=c_p\text{ and }\det(\rho(\Frobp))=p$$ for all but finitely many $p$? The answer is yes. The idea is to find $\rho$ by taking $\rho_{f,\lambda}$ [[which was constructed by Shimura?]] for some $f$ and reducing mod $\lambda$. The only special thing that we need is a lemma saying that the eigenvalues in characteristic $\ell$ lift to eigenvalues in characteristic $0$. \subsection{Representations from modular forms mod $\ell^n$} Serre and Deligne asked: ``What happens mod $\ell^n$?'' More precisely, let $R$ be a local finite Artin ring such that $\ell^n R=0$ for some $n$. Take $f\in S_2(\gon;R)$ satisfying the hypothesis $$\{r\in R : rf=0 \} = \{0\}.$$ This is done to insure that certain eigenvalues are unique. Assume that for almost all $p$ one has $T_p f = c_p f$ with $c_p\in R$. The problem is to find $\rho:G\into\GL(2,R)$ such that $$\tr(\rho(\Frobp))=c_p\text{ and }\det(\rho(\Frobp))=p$$ for almost all $p$. The big stumbling block is that $\rho$ need not be the reduction of some $\rho_{g,\lambda}$ for any $g, \lambda$. [[I couldn't understand why -- I wrote ``can mix up $f$'s from characteristic 0 so can not get one which reduces correctly.'']] Let $\T=\Z[\ldots,T_p,\ldots]$ where we only adjoin those $T_p$ for which $f$ is an eigenvector [[I made this last part up, but it seems very reasonable]]. Then $f$ obviously gives a rise to a map $$\T\into R : T\mapsto \text{ eigenvalue of $T$ on $f$ }.$$ The strange hypothesis on $f$ insures that the eigenvalue is unique. Indeed, suppose $Tf=a f$ and $Tf=bf$, then $0=Tf-Tf=af-bf=(a-b)f$ so by the hypothesis $a-b=0$ so $a=b$. Since the pullback of the maximal ideal of $R$ is a maximal ideal of $\T$ we get a map $\Tm\into R$ for some $\m$. [[I do not understand why we suddenly get this map and I do not know why the pullback of the maximal ideal is maximal.]] Now the problem is solved. Take $\rho':G\into\GL(2,\Tm)$ with the sought after trace and determinent properties. Then let $\rho$ be the map obtained by composing with the map $\GL(2,\Tm)\into\GL(2,R)$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 4/29/96, ribetnotes.tex \section{$\rho'$ is of type $\Sigma$} Let $\rho$ be modular irreducible and semistable\index{semistable} mod $\ell$ representation with $\ell>2$. Let $\Sigma$ be a finite set of primes. Then $N(\rho)|N_{\Sigma}$. We constructed the anemic Hecke algebra $\T$ which contains a certain maximal ideal $\m$. We then consider the completion $\Tm$ of $\T$ at $\m$. Next we constructed $$\rho':G\into\GL(2,\Tm)$$ lifting $\rho$. Thus the diagram $$\triCD{G}{\rho'}{GL(2,\Tm)}{}{GL(2,\F)}{\rho}$$ commutes. Some defining properties of $\rho'$ are \begin{itemize} \item $\det \rho'=\tilde{\chi}$. \item $\tr \rho'(\Frob_r)=T_r$. Since topologically $\Tm$ is generated by the $\Frob_r$ this is a tight condition. \item $\rho'$ is a lift of type $\Sigma$. \end{itemize} To say $\rho'$ is a lift of type $\Sigma$ entails that $\rho'$ is unramified outside primes $p|\ns$. This is true because $\rho'$ is constructed from various $\rho_{f,\lambda}$ with $N(f)|\ns$. If $p\neq\ell$ and $p|N(\rho)$ then $p||\ns$. Recall that $$\rho|D_p\sim\mat{\alpha}{*}{0}{\beta}$$ where $\alpha=\beta\chi$ and $\alpha$ and $\beta$ are unramified. For $\rho'$ to be a lift of type $\Sigma$ we require that $$\rho'|D_p\sim\mat{\tilde{\alpha}}{*}{0}{\tilde{\beta}}$$ where $\tilde{\alpha}$ and $\tilde{\beta}$ are unramified lifts and $\tilde{\alpha}=\tilde{\beta}\chi$. [[I find it mighty odd that $\alpha$ is $\chi$ times an unramified character and yet $\alpha$ is not ramified! How can that be? Restricted to inertia $\beta$ and $\alpha$ would be trivial but $\chi$ would not be.]] Is this true of $\rho'$? Yes since by a theorem of Langlands the factors $$\rho_{f,\lambda}|D_p\sim\mat{\tilde{\alpha}}{*}{0}{\tilde{\beta}}.$$ [[Ribet said more about this but it does not form a cohesive whole in my mind. Here is what I have got. Since $\rho_{f,\lambda}$ obviously ramified at $p$, $p||N(f)|\ns$. $\rho_{f,\lambda}|D_p$ is like an elliptic curve with bad multiplicative reduction at $p$. That $\rho_{f,\lambda}|D_p\sim\smat{\tilde{\alpha}}{*}{0}{\tilde{\beta}}$ really comes down to Deligne-Rapaport. If write $f=\sum a_n q^n$, then $a_p\neq 0$ and $\tilde{\beta}(\Frobp)=a_p$, $\tilde{\alpha}(\Frobp)=pa_p$. Thus $a_p^2=1$ since $\tilde{\alpha}\tilde{\beta}=\chi$. Thus $a_p=\pm 1$ and $a_p \mod\lambda = \beta(\Frobp)=\pm 1$ independent of $(f,\lambda)$. So we have these numbers $a_p=a_p(f)=\pm 1$, independent of $\lambda$. ]] \section{Isomorphism between $\Tm$ and $\rmr$} Let $\T\subset R=\Z[\ldots,T_n,\ldots]$ be the anemic Hecke algebra with maximal ideal $\m$. The difference between $\T$ and $R$ is that $R$ contains all the Hecke operators whereas $\T$ only contains the $T_p$ with $p\nd\ell\ns$. Wiles proved that the map $\Tm\into \rmr$ is an isomorphism. Which Hecke operators are going to hit the missing $T_p$? If we do the analysis in $\rmr$ we see that [[I think for $p\neq \ell$!]] $$T_p=\begin{cases} \pm 1, & \text{for $p|\ns$, $p\not\in\Sigma$}\\ 0, & \text{for $p\in\Sigma$}\end{cases}.$$ This takes care of everything except $T_{\ell}$. In proving the surjectivity of $\Tm\into\rmr$ we are quite happy to know that $T_p=\pm 1$ or $0$. Proving this is a bit unpalatable. It is described in \cite{ddt}. Consider the commuting diagram $$\begin{array}{ccc} \Tm&\hookrightarrow&\prod\sO_{f,\lambda}\\ &\searrow&\uparrow\\ & &\rmr \end{array}.$$ The map $\rmr\into\prod\sO_{f,\lambda}$ is constructed by massaging $f$ by stripping away certain Euler factors so as to obtain an eigenvector for all the Hecke operators. This diagram forces $\Tm\into\rmr$ to be injective. [[Ogus\index{Ogus}: Is it clearly surjective on the residue field? Ribet: Yes. Ogus: OK, then we just need to prove it is \`{e}tale.]] From the theory of the $\theta$ operator we already know two-thirds of the times that $\T$ contains $T_{\ell}$. Suppose $\ell|\ns$. This entails that we are in the ordinary case, $\rho$ is not finite at $\ell$, or $\ell\in\Sigma$. We did not prove in this situation that $T_{\ell}\in\Z[\ldots, T_n,\ldots : (n,\ell)=1]$. Using generators and relations and brute force one shows that $\rmr\into\prod\sO_{f,\lambda}$ is an injection. Then we can compare everything in $\prod \sO_{f,\lambda}$. Now $$\rholamf|D_{\ell}\sim\mat{\tilde{\alpha}}{*}{0}{\tilde{\beta}}$$ and $$\tilde{\beta}(\Frob_{\ell})=T_{\ell}\in\prod\sofl.$$ Using arguments like last time one shows that $\tilde{\beta}(\Frob_{\ell})$ can be expressed in terms of the traces of various operators. This proves surjectivity in this case. Ultimately we have $\Tm\isom\rmr$. The virtue of $\Tm$ is that it is generated by traces. The virtue of $\rmr$ is that it is Gorenstein. We have seen this if $\ell\nd\ns$. In fact it is Gorenstein even if $\ell||\ns$. [[Ribet: As I stand here today I do not know how to prove this last assertion in exactly one case. Ogus\index{Ogus}: You mean there is another gap in Wiles' proof. Ribet: No, it is just something I need to work out.]] When $\ell||\ns$ there are 2 cases. Either $\rho$ is not finite at $\ell$ or it is. The case when $\rho$ is not finite at $\ell$ was taken care of in [[cite a paper by Mazur-Ribet, Asterisque, reference?]]. A proof that $\rmr$ is Gorenstein when $\rho$ is finite at $\ell$ ($\ell\in\Sigma$) is not in the literature. Now forget $\rmr$ and just think of $\Tm$ in both ways: trace generated and Gorenstein. \section{Deformations} \index{deformations} Fix an absolutely irreducible modular mod $\ell$ representation $\rho$ and a finite set of primes $\Sigma$. Consider the category $\sC$ of complete local Noetherian $W(\F)$-algebras $A$ (with $A/\m=\F$). Here $\F=\Fln$ and $W(\F)$ is the ring of Witt vectors, i.e., the ring of integers of an unramified extension of $\Ql$ of degree $\nu$. Define a functor $\sF:\sC\into\Set$ by sending $A$ in $\sC$ to the set of equivalence classes of lifts $$\tilde{\rho}:G\into\GL(2,A)$$ of $\rho$ of type $\Sigma$. The equivalence relation is that $\tilde{\rho_1}\sim\tilde{\rho_2}$ iff there exists $M\in\GL(2,A)$ with $M\equiv\mat{1}{0}{0}{1}\pmod{\m}$ such that $\tilde{\rho_1}=M^{-1}\tilde{\rho_2}M.$ Mazur proved \cite{mazur2} that $\sF$ is representable. \begin{thm} There exists a lift $$\rhouniv:G\into\GL(2,R_{\Sigma})$$ of type $\Sigma$ such that given any lift $\tilde{\rho}:G\into\GL(2,A)$ there exists a unique homomorphism $\varphi:R_{\Sigma}\into A$ such that $\varphi\circ\rhouniv\sim\tilde{\rho}$ in the sense of the above equivalence relation. \end{thm} Lenstra\index{Lenstra} figured out how to concretely construct $R_{\Sigma}$. Back in the student days of Ribet and Ogus, Schlessinger wrote a widely quoted thesis which gives conditions under which a certain class of functors can be representable. Mazur checks these conditions in his paper. [[Buzzard\index{Buzzard}: What happened to Schlessinger anyways? Ribet: He ended up at University of North Carolina, Chapel Hill.]] We have constructed $\tilde{\rho}=\rho':G\into\GL(2,\Tm)$. By the theorem there exists a unique morphism $\varphi:R_{\Sigma}\into\Tm$ such that $\rho'=\varphi\circ\rhouniv$. \begin{thm} $\varphi$ is an isomorphism thus $\rho'$ is the universal deformation\index{deformations} and $\Tm$ is the universal deformation ring. \end{thm} This will imply that any lift of type $\Sigma$ is modular. The morphism $\varphi$ is surjective since $$T_p=\tr\tilde{\rho}(\Frobp)=\tr\varphi\circ\rhouniv(\Frobp) =\varphi(\tr(\rhouniv(\Frobp))).$$ We have two very abstractly defined local Noetherian rings. How would you prove they are isomorphic? Most people would be terrified by this question. Wiles dealt with it. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 5/1/96 ribetnotes.tex \section{Wiles Main Conjecture} \begin{quote} ``We are like a train which is trying to reach Fermat's Last Theorem. Of course it has not made all of its scheduled stops. But it is on its way.'' \end{quote} We have a representation $\rho:G\into\GL(2,\F)$. Take $\F=\Fl$ for our applications today. Then the ring of Witt vectors is $W(\F)=\Zl$. The Hecke algebra can be embedded as $$\Tm\subset\prod_{(g,\mu)\in \sA}\sO_{g,\mu}.$$ The Hecke algebra $\Tm$ has the following properties. \begin{itemize} \item The index of $\Tm$ in $\prod\sO_{g,\mu}$ is finite. \item Gorenstein as a $\Zl$-module, i.e., there exists an isomorphism $\Hom_{\Zl}(\Tm,\Zl)\isom\Tm$. \item $\Tm$ is generated by the $T_r$ with $r$ prime and $(r,\ell\ns)=1$. \end{itemize} We have constructed a representation $$\rho':G\into\GL(2,\Tm).$$ Composing appropriately with the map $\Tm\hookrightarrow \prod\sO_{g,\mu}$ gives a map $$G\into\prod_{(g,\mu)}\GL(2,\sO_{g,\mu}).$$ This is the product of representations $\prod\rho_{g,\mu}$. The triangle is $$\triCD{G}{\rho'}{\GL(2,\Tm)}{} {\prod\GL(2,\sO_{g,\mu})}{\prod\rho_{g,\mu}}$$ Moreover, $\rho'$ is a deformation\index{deformations} of $\rho$ of type $\Sigma$ so it lifts $\rho$ and satisfies certain ``nice as $\rho$'' properties at primes $p\not\in\Sigma$. Let $$\rhouniv:G\into\GL(2,\rs)$$ be the universal deformation\index{deformations} of $\rho$ of type $\Sigma$. Lenstra\index{Lenstra} gave a very concrete paper \cite{lenstra} constructing this $\rhouniv$. Before his paper there was only Schlesinger's thesis. By the definition of $\rhouniv$ there exists a unique map $\varphi:\rs\into\Tm=\T_{\Sigma}$ such that $\varphi\circ\rhouniv=\rho'$. By $\varphi\circ\rhouniv$ we mean the composition of $\rhouniv$ with the map $\GL(2,\rs)\into\GL(2,\Tm)$ induces by $\varphi$. As noted last time it is easy to see that $\varphi$ is surjective. \begin{thm}[Wiles main `conjecture'] $\varphi$ is an isomorphism (for each $\Sigma$). \end{thm} The theorem implies the following useful result. \begin{thm}Suppose $E$ is a semistable\index{semistable} elliptic curve over $\Q$ and that for some $\ell>2$ the representation $\rho=\rho_{\ell,E}$ on $E[\ell]$ is irreducible and modular. Then $E$ is modular. \end{thm} \begin{proof} The representation $$\tilde{\rho}=\rho_{\ell^{\infty},E}:G\into\GL(2,\Zl)$$ on the $\ell$-power torsion $E[\ell^{\infty}]=\union E[\ell^n]$ is a lift of $$\rho:G\into\Aut(E[\ell])=\GL(2,\Fl).$$ Furthermore, $\tilde{\rho}$ is a deformation\index{deformations} of $\rho$ of type $\Sigma$. Applying universality and using the theorem that $\R_{\Sigma}\isom\T_{\Sigma}$ we get a map $$\T_{\Sigma}\into\Zl: \quad T_r\mapsto a_r=a_r(E)=\trace \tilde{\rho}(\Frob_r).$$ The relevant diagram is $$\ltriCD{\rs}{} {\Zl}{}{\ts}{}$$ where the map $\rs\into\Zl$ is given by $$\trace \rhouniv (\Frob_r)\mapsto a_r.$$ Now the full Hecke algebra $\Z[\ldots,T_n,\ldots]$ embeds into the completion $\T_{\Sigma}$. Composing this with the map $\T_{\Sigma}\into\Zl$ above we obtain a map $$\alpha:\Z[\ldots,T_n,\ldots]\into\Zl.$$ Because of the duality between the Hecke algebra and modular forms there exists a modular form $h\in S_2(\Gamma_0(\ns),\Zl)$ corresponding to $\alpha$. Since $\alpha$ is a homomorphism $h$ is a normalized eigenform. Furthermore $a_r(h)=a_r(E)\in\Z$ for all primes $r\nd\ell\ns$. Since almost all coefficients of $h$ are integral it follows that $h$ is integral. Because we know a lot about eigenforms we can massage $h$ to an eigenform in $S_2(\Gamma_0(N_{E}),\Z)$. \end{proof} [[Some undigested comments follow.]] \begin{itemize} \item Once there is any connection between $\rho$ and a modular form one can prove Taniyama-Shimura in as strong a form as desired. See the article {\em Number theory as Gadfly}. \item Take the abelian variety $A_h$ attached to $h$. The $\lambda$-adic representation will have pieces with the same representations. Using Tate's conjecture we see that $E$ is isogenous to $A_h$. Use at some points Carayol's theorem: If $g$ is a form giving rising to the abelian variety $A$ then the conductor of $A$ is the same as the conductor of $g$. \item Tate proved that if two elliptic curves have isomorphic $\rho_{\ell^{\infty}}$ for some $\ell$ then they are isogenous. \end{itemize} \section{$\T_{\Sigma}$ is a complete intersection} \index{complete intersection} Recall the construction of $\ts=\Tm$. Let $\T$ be the anemic Hecke algebra. Then $$\T\tensor\Zl\hookrightarrow\prod \sO_{g,\mu}$$ where the product is over a complete set of representatives (for the action of Galois on eigenforms) $g$ and primes $\mu$ lying over $\ell$. We found a specific $(f,\lambda)$ for $\Sigma=\emptyset$ such that $\rholamfbar=\rho$. The maximal ideal $\m$ was defined as follows. The form $f$ induces a map $\T\into\sO_{f,\lambda}$. Taking the quotient of $\sO_{f,\lambda}$ by its maximal ideal we obtain a map $\T\into\Flbar$. Then $\m$ is the kernel of this map. The diagram is $$\ltriCD{\T}{}{\Flbar}{}{\sO_{f,\lambda}}{}$$ The map $\sO_{f,\lambda}\into\Flbar$ is $$a_r(f)\mapsto\trace\rho(\Frob_r).$$ To fix ideas we cheat and suppose $\sO_{f,\lambda}=\Zl$. [[In Wiles' optic\index{optic} this is OK since he can work this way then tensor everything at the end.]] Now $\rholamf$ is a distinguished deformation\index{deformations} of $\rho$ [[``Distinguished'' is not meant in a mathematical sense]]. The map $f$ gives rise to a map $\ts\into\sO=\Zl$ which we also denote by $f$ $$\triCD{\rs}{\varphi}{\ts}{f}{\sO=\Zl}{}$$ %Let $\wp_{\T}=\ker f$ and let $\wp_{R} = \ker f\circ\varphi$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% 5/6/96 \section{The inequality $\#\sO/\eta\leq\#\wpt/\wpt^2\leq\wpr/\wpr^2$} Let $$\rho:G=\galq\into\GL(2,\Fl)$$ be irreducible and modular with $\ell>2$. Let $\Sigma$ be a finite set of primes. We assume there is a modular form $f$ of weight 2 with coefficients in $\Zl$ which gives rise $\rho$. Let $$\rholamf:G\into\GL(2,\Zl)$$ be the representation coming from $f$, then $\rholamf$ reduces to $\rho$ modulo $\ell$. Let $\rs$ be the universal deformation\index{deformations} ring, so every deformation of $\rho$ of type $\Sigma$ factors through $\rs$ in an appropriate sense. Let $\ts$ be the Hecke ring associated to $\Sigma$. It is a $\Zl$-algebra which is free of finite rank. Furthermore $$\ts\subset\prod_{(g,\mu)\in \sA} \sO_{g,\mu} =\sO_{f,\mu}\cross \prod_{(g,\mu)\in \sA-\{(f,\mu)\}}\sO_{g,\mu}$$ where $\sA$ is as defined before. Define projections $\pr_1$ and $\pr_2$ onto the first and rest of the factors, respectively \begin{eqnarray*} \pr_1&:&\prod_{(g,\mu)\in\sA}\sO_{g,\mu}\into\sO=\sO_{f,\lambda}\\ \pr_2&:&\prod_{(g,\mu)\in\sA}\sO_{g,\mu}\into \prod_{(g,\mu)\neq(f,\lambda)}\sO_{g,\mu} \end{eqnarray*} Let $\varphi:\rs\into\ts$ be the map coming from the universal property of $\rs$. This map is surjective. The famous triangle which dominates all of the theory is $$\begin{array}{ccc}\rs&\stackrel{\varphi}{\onto}&\ts\\ & & \downarrow \pr_1\\ & &\sO=\Zl \end{array}$$ \subsection{The definitions of the ideals} We now define two ideals. View $\ts$ as sitting in the product $\prod \sO_{g,\mu}$. \begin{enumerate} \item The {\em congruence ideal} $\eta\subset\sO$ is $$\eta := \sO\intersect\ts = \ker \Bigl(\pr_2:\ts\into \prod_{(g,\mu)\neq(f,\lambda)}\sO_{g,\mu}\Bigr)$$ \item The prime ideal $\wpt\subset\ts$ is $$\wpt = \ker \Bigl(\pr_1:\ts\into\sO\Bigr)$$ \end{enumerate} It is true that $$\#\sO/\eta \leq \#\wpt/\wpt^2.$$ The condition for equality is a theorem of Wiles. \begin{thm} $\ts$ is a complete intersection\index{complete intersection} iff $\#\sO/\eta = \#\wpt/\wpt^2.$ \end{thm} There is an analogous construction for $$\psi=\pr_1\circ\varphi:\rs\into\sO.$$ The diagram is $$\triCD{\rs}{\varphi}{\ts}{\pr_1}{\sO}{\psi}.$$ Let $\wpr$ be the kernel of $\psi$. From the commutativity of the above diagram we see that $\psi$ maps $\wpr\onto\wpt$. Thus we have an induced map $\overline{\psi}$ on ``tangent spaces'' $$\overline{\psi}:\wpr/\wpr^2\onto\wpt/\wpt^2.$$ It follows that $$\#\sO/\eta\leq\#\wpt/\wpt^2\leq \#\wpr/\wpr^2.$$ There is an analogous theorem. \begin{thm} The above inequalities are all equalities iff \begin{itemize} \item $\varphi:\rs\into\ts$ is an isomorphism, and \item $\ts$ is a complete intersection ring. \end{itemize} \end{thm} \subsection{Aside: Selmer Groups} Let $M$ be the set of matrices in $M(2,\Ql/\Zl)$ which have trace $0$. Then $\GL(2,\Zl)$ operates on $M$ by conjugation. Thus $G$ acts on $M$ via the representation $\rho':G\into\GL(2,\Zl)$. To $\wpr/\wpr^2$ there corresponds the {\em Selmer group} which is a subgroup of $H^1(\galq,M)$. The subgroup is $$H^1_{\Sigma}(G,M)=\Hom_{\sO}(\wpr/\wpr^2,\Ql/\Zl) \subset H^1(\galq,M).$$ [[Since Flach's thesis there has been a problem of trying to get an upper bound for the Selmer group. Wiles converted it into the above problem.]] \subsection{Outline of some proofs} We outline the key steps in the proof that $\#\wpr/\wpr^2\leq \#\sO/\eta$. \subsubsection{Step 1: $\Sigma=\emptyset$} The key step is the minimal case when $\Sigma=\emptyset$. This is done in \cite{tw}. They claim to be proving the apparently weaker statement that $\ts$ a complete intersection\index{complete intersection} implies $$\#\wpt/\wpt^2=\#\sO/\eta.$$ But in Wiles' paper \cite{wiles} he obtains the inequality $$\#\wpr/\wpr^2 \leq \frac{(\#\wpt/\wpt^2)^2}{\#\sO/\eta}.$$ Combining these two shows that $$\#\wpr/\wpr^2\leq \#\sO/\eta.$$ In an appendix to \cite{tw} Faltings proves directly that $$\#\wpr/\wpr^2\leq\#\sO/\eta.$$ [[At this point there were some remarks about why Wiles might have taken a circular route in his Annals paper. Ribet replied, \begin{quote} ``As Serre says, it is sometimes better to leave out any psychological behavior related to how people did something but instead just report on what they did.'']] \end{quote} \subsubsection{Step 2: Passage from $\Sigma=\emptyset$ to $\sigma$ general} The second step is the induction step in which we must understand what happens as $\Sigma$ grows. Thus $\Sigma$ is replaced by $\Sigma'=\Sigma\union\{q\}$ where $q$ is some prime not in $\Sigma$. We will use the following notation. The object attached to $\Sigma'$ will be denoted the same way as the object attached to $\Sigma$ but with a $'$. Thus $(\wpr/\wpr^2)'$ denotes the Selmer\index{Selmer group} group for $\Sigma'$. The change in the Selmer group when $\Sigma$ is replaced by $\Sigma'$ is completely governed by some local cohomology group. There is a constant $c_q$ such that $$\#(\wpr/\wpr^2)'\leq c_q\#(\wpr/\wpr^2)\leq c_q\#\sO/\eta.$$ So we just need to know that $$\#\sO/\eta'\geq c_q \#\sO/\eta,$$ i.e., that $\eta'$ is {\em small} as an ideal in $\sO$. We need a formula for the ratio of the two orders. Let $\T$ be the anemic ring of Hecke operators on $S_2(\Gamma_0(\ns))$ obtained by adjoining to $\Z$ all the Hecke operators $\T_n$ with $n$ prime to $\ell\ns$. Let $\T'$ be the anemic Hecke ring of Hecke operators on $S_2(\Gamma_0(N_{\Sigma'}))$. Since $\ns|N_{\Sigma'}$ there is an inclusion $$S_2(\Gamma_0(N_{\Sigma}))\hookrightarrow S_2(\Gamma_0(N_{\Sigma'})).$$ There is one subtlety, this injection is not equivariant for all of the Hecke operators. But this is no problem because $\T$ and $\T'$ are anemic. So the inclusion induces a restriction map $r:\T'\into \T$. We now introduce a relative version of $\eta$ which is an ideal $I\subset\T$. One way to think of $I$ is as $\T\intersect\T'$ where $\T$ and $\T'$ are both viewed as subrings of $\prod \sO_{g,\mu}$ $$\begin{array}{ccc} \T' & \stackrel{r}{\longrightarrow} & \T\\ \cap & & \cap\\ \T'_{\m'} & & \Tm\\ \cap & & \cap\\ \prod_{(g,\mu)} \sO_{g,\mu} & \hookrightarrow & \prod_{\text{more }(g,\mu)} \sO_{g,\mu} \end{array}$$ The definition Lenstra\index{Lenstra} would give is that $$I:=r(\Ann_{\T'}(\ker(r))).$$ The amazing formula is $$\eta'=\eta\cdot f(I)$$ where $f:\T\into\sO$ is the map induced by the modular form $f$. [[After introducing this definition Ogus\index{Ogus} was very curious about how deep it is, in particular, about whether its proof uses the Gorensteiness of $\T$. Ribet said, ``somehow I do not think this formula can possibly be profound.'']] We pause with an aside to consider Wiles' original definition of $\eta$. By duality the map $f:\ts\into\Zl$ induces $$f^{\dual}:\Hom_{\Zl}(\Zl,\Zl)\into\Hom_{\Zl}(\ts,\Zl)\isom\ts.$$ Because $\ts$ is Gorenstein there is an isomorphism $\Hom_{\Zl}(\ts,\Zl)\isom\ts$. Now $f^{\dual}(\id)\in\ts$ so $f(f^{\dual}(\id))\in\sO=\Zl$. Wiles let $\eta=(f(f^{\dual}(\id)))$ be the ideal generated by $f(f^{\dual}(\id))$. To finish step 2 we must show that $\#\sO/f(I)\geq c_q$, i.e., that ``$I$ is small''. [[I do not see how this actually finishes step 2, but it is reasonable that it should. How does this index relate to the index of $f(I)\eta$ in $\sO$?]] Let $J=J_0(\ns)$ and $J'=J_0(N_{\Sigma'})$. Since $$S_2(\Gamma_0(\ns))\hookrightarrow S_2(\Gamma_0(N_{\Sigma'}))$$ functoriality of the Jacobian\index{Jacobian} induces a map $J\hookrightarrow J'$. By autoduality we also obtain an injection $J^{\dual}\hookrightarrow J'$ and $J\intersect J^{\dual}$ is a finite subgroup of $J'$. [[I definitely do not understand why $J$ is not just equal to $J^{\perp}$. Where does the other embedding $J^{\perp}\hookrightarrow J'$ come from?]] It can be seen that $J\intersect J^{\dual}=J[\delta]$ for some $\delta\in\T$. It turns out that $$\Ann_{\T}(J\intersect J^{\dual})=\T\intersect\delta\End(J) \supseteq\delta\T.$$ It is an observable fact that $f(\delta\T)$ is an ideal of $\sO$ of norm $c_q$. The {\em heart} of the whole matter is to see that the inclusion $$\delta\T\subseteq\T\intersect\delta\End(J)$$ is an equality after localization at $\m$. To do this we have to know that $\Tate_{\m}(J)\isom\Tm^2$. This is equivalent to the Gorenstein business. With this in hand one can just check this equality. [[Unfortunately we were now 10 minutes passed when the course should end. Realizing this Ribet brought everything to a close. As is the tradition at the end of a course the room erupted in applause.]] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %END \backmatter \begin{thebibliography}{HHHHHHH} \bibitem{antwerp} N. 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