Hecke algebra generated by a single element - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:27:42Z http://mathoverflow.net/feeds/question/67538 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67538/hecke-algebra-generated-by-a-single-element Hecke algebra generated by a single element B. Naskrecki 2011-06-11T20:28:47Z 2011-06-13T07:49:37Z <p>Let $\mathbb{T}_{\mathbb{Z}}$ be a $\mathbb{Z}$-module generated by Hecke operators $T_n$ acting on the space of cups forms $S_{k}(\Gamma,\mathbb{C})$ for the congruent subgroup satisfying $\Gamma_1(N)\subset\Gamma$ for some positive $N$. We know that the $\mathbb{Z}$-module $\mathbb{T}_{\mathbb{Z}}$ is finitely generated and the upper bound for the number of generators is given by the Sturm bound. When we compute some examples like for $\Gamma=\Gamma_0(33)$ and $k=2$ we get $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_3]/(T_3+1)(T_3^2+T_3+3)$. In all other cases I have computed it always happens that the algebra looks similarly,i.e. $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_k]/f(T_k)$ for some polynomial with integer coefficients. Is it always the case or are there any know examples that $\mathbb{T}_{\mathbb{Z}}$ as a ring is generated minimally by more than one Hecke operator ?</p> http://mathoverflow.net/questions/67538/hecke-algebra-generated-by-a-single-element/67545#67545 Answer by Kevin Buzzard for Hecke algebra generated by a single element Kevin Buzzard 2011-06-11T22:00:30Z 2011-06-11T22:34:04Z <p>[I took the time to chase this up so may as well post it as an answer.]</p> <p>There is a (cuspidal) modular (eigen)form of level $\Gamma_0(512)$ and weight 2, which if I remember correctly was shown to me by Luis Dieulefait, and which has the following property: its coefficient field is the biquadratic extension $\mathbf{Q}(\sqrt{2},\sqrt{3})$ of $\mathbf{Q}$, but writing $f=\sum_n a_n q^n=q+\ldots$, each $a_n$ has the property that $a_n^2\in\mathbf{Z}$!</p> <p>Hence the Hecke algebra at this level, even with $\mathbf{Q}$ coefficients, is not generated by one Hecke operator.</p> http://mathoverflow.net/questions/67538/hecke-algebra-generated-by-a-single-element/67555#67555 Answer by Tommaso Centeleghe for Hecke algebra generated by a single element Tommaso Centeleghe 2011-06-12T01:01:58Z 2011-06-13T07:49:37Z <p>In certain cases a reason that causes the subring generated by a single Hecke operator $T_\ell$ in $\mathbb{T}_\mathbb{Z}$ to have index greater than one is the existence of exceptional primes for the normalised eigenforms $f\in S_k(\Gamma,\mathbb{C})$ (we shall say that $p$ is exceptional for $f$ if (one of) the $p$-adic Galois representation $\rho_{f,\lambda}$ associated to $f$ has small residual image, i.e., the image of $\bar\rho_{f,\lambda}$ does not contain $SL_2(F)$, for any finite extension $F/F_p$).</p> <p>The special cases that I have in mind are instances of the inner twist phenomena that Kevin Buzzard has already mentioned in one of his comments, and you can see this also in level one, if I am right. Let me elaborate on that.</p> <p>Let $p$ be a prime $\equiv 3$ mod $4$, and set $k=(p+1)/2$. Let $h$ be the class number of $Q(\sqrt{-p})$. The space $S_k(\Gamma(1))$ gives rise to $n=(h-1)/2$ distinct systems of mod $p$ Hecke eigenvalues so that the associated mod $p$ Galois representations $\rho_1,\ldots,\rho_n$ (which are unramified outside $p$) have dihedral image. This implies that if $\ell$ is a prime that is not a quadratic residue mod $p$, then the trace of $\rho_i$ at a Frobenius element at $\ell$ is zero.</p> <p>In particular, if $n>1$ (i.e. $h>3$) then there are at least two DISTINCT systems of mod $p$ Hecke eigenvalues $(a_q)$ and $(b_q)$ such that their $\ell$-th members are equal. This implies that the integral Hecke ring $\mathbb{T}$ has more ring homomorphisms valued in $\bar{F_p}$ than the subring generated by $T_\ell$ alone does. In this case one can show that $p$ divides the index of the latter in the former.</p> <p>With the previous argument you see that for almost all primes $p\equiv 3$ mod $4$ in weight $(p+1)/2$ half of the primes $\ell$ are such that $T_\ell$ does not generate $\mathbb{T}$. But what about the other half of the primes $\ell$?</p> <p>By relating the traces of $\rho_i$ at Frobenius elements over primes $\ell$ that are split in $Q(\sqrt{-p})$ to the characters of the class group of $Q(\sqrt{-p})$ valued in $\bar{F_p}$ we get the following:</p> <p>PROPOSITION: Let $A$ be the class group of $Q(\sqrt{-p})$. Assume that for every $a\in A$ there exists a pair of non trivial characters $\chi_i:A\rightarrow\bar{F_p}^*$, with $\chi_1\neq\chi_2$ and $\chi_1\neq \chi_2^{-1}$, such that $\chi_1(a)+\chi_1(a^{-1}) =\chi_2(a)+\chi_2(a^{-1})$. Then the integral Hecke ring $\mathbb{T}$ in weight $(p+1)/2$ and level one cannot be generated by a single $T_\ell$, for $\ell$ prime.</p> <p>Few remarks: 1) The above does not say anything about the possibility of having a $T_n$ (with $n$ not prime) generating $\mathbb{T}$. In fact I think we cannot rule this out a priori.</p> <p>2) Probably a little more group theoretic work can be done to reformulate the condition on the mod $p$ characters of $A$, and turn it into something nicer. I think if $A$ is not cyclic and $3$ does not divide its order then the assumption is satisfied.</p> <p>3) Notice that we do not need to worry about $T_p$ generating $\mathbb{T}$ when $n>1$. In fact, since $P$ splits in $H/Q({\sqrt{-p}})$, where $H$ is the Hilbert Class Field of $Q(\sqrt{-p})$, all the $\rho_i$'s are the same locally at $p$ and one can show that the eigenvalue $a_p$ is $1$ for all of them.</p> <p>An example. The cuspidal, integral Hecke ring of level one and weight $k=246=(491+1)/2$ is so that $p=491$ does not divide its discriminant. However every Hecke operators $T_n$, for $2\leq n\leq 153$ generates a subring of $\mathbb{T}$ of index divisible by $p$.</p> <p>May be it could be nice to see whether the class group of $Q(\sqrt{491})$ satisfies the assumption of the proposition. I suspect it does!</p> <p>[EDIT: The assumption of the proposition is NOT satisfied for $p=491$, unless I am wrong to an infinite amount. Therefore my suspicion is not confirmed. The proposition above, as far as I can tell, remains valid, although I have no example of a prime $p\equiv 3$ mod $4$ such that the class group $A$ of $Q(\sqrt{-p})$ satisfies the assumption of the proposition.</p> <p>There still remains to give an explanation of the fact that $p=491$ seems to divide the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of level $1$ and weight $246$. The arguments above explain such divisibility for primes $\ell$ that are non-quadratic residue mod $p$. For what concerns the other primes $\ell$, I am tempted to say that there should be a mod $491$ Galois representation arising from weight $246$ and level $1$ that is tamely ramified, reducible at $p$, and non-dihedral (may be with small image?): this in fact would cause the existence of a system of mod $p$ eigenvalues $(a_\ell)$, with $\ell\neq p$, arising from $S_{246} (\Gamma(1))$ such that its quadratic twist $(a_\ell \ell^{(p-1)/2})$ also arises from the same space.</p> <p>Summarising</p> <p>PROPOSITION: Let $p\equiv 3$ mod $4$. Assume that</p> <p>1) the class number of $Q(\sqrt{-p})$ is > 3;</p> <p>2) there exists a mod $p$ representation $\rho$ of $G_Q$ arising from $S_{(p+1)/2}(\Gamma(1))$ that is tamely ramified and reducible at $p$ and it is non-dihedral;</p> <p>then for every prime $\ell$, $p$ divides the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of weight $(p+1)/2$ and level $1$.</p> <p>When $p=2083$ I learnt from some tables that there exists an odd, $A_5$ extension of $Q$ unramified outside $p$ that gives rise to a representation of the type we want in 2). Since condition 1) is also satisfied we see that the Hecke ring in weight $1044$ and level $1$ cannot be generated by a single $T_\ell$, for $\ell$ prime.]</p>