Example of level one cuspidal Hecke Algebra T_k^0 such that p divides its index in its normalization, and p≥k-1? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:03:25Z http://mathoverflow.net/feeds/question/21323 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21323/example-of-level-one-cuspidal-hecke-algebra-t-k0-such-that-p-divides-its-index-i Example of level one cuspidal Hecke Algebra T_k^0 such that p divides its index in its normalization, and p≥k-1? Tommaso Centeleghe 2010-04-14T11:12:47Z 2010-04-14T13:52:43Z <p>The question is strongly focused on computations concerning modular forms and Hecke algebras. It is already in the title, but I will repeat it, adding a few details.</p> <p>Let $S_k$ be the complex vector space of classical, holomorphic cusp forms of weight $k$ and level one, and let $\mathbf{T}_k^0$ be the subring of the endomorphisms ring of $S_k$ generated, over $\mathbf{Z}$, by all the Hecke operators $T_\ell$. It is well known that $\mathbf{T}_k^0$ is finite free as $\mathbf{Z}$-module, and that $\mathbf{T}_k^0\otimes\mathbf{Q}$ is isomorphic to a product of finitely many number fields $K_i$. This is to say that $\mathbf{T}_k^0$ is isomorphic to a finite index subring of the product of the rings of integers $\mathcal{O}_i$'s of the $K_i$'s. Let this index be $d_k$.</p> <p>I would like to know if there are examples of weights $k$ so that $d_k$ is divisible by a prime $p$ with $p\geq k-1$ (which is to say, the perhaps more familiar, $k\leq p+1$).</p> <p>While performing computations with SAGE, I was not able to find any example of a pair $(p,k)$, with $k\leq p+1$, and $p&lt;1600$, so that $p$ divides $d_k$.</p> <p>Thanks.</p> <p>[The motivation for asking this question is the study of $\overline{\mathbf{F}}_p$-valued points of the spectrum of $\mathbf{T}_k^0$,</p> <p>that parametrize certain two-dimensional mod $p$ representations of $G_\mathbf{Q}$. One reason for which the cardinality of $\mathbf{Spec}(\mathbf{T}_k^0)(\overline{\mathbf{F}}_p)$ might turned out to be strictly smaller than $\mathbf{Spec}(\mathbf{T}_k^0)(\mathbf{C})$ is that $p$ divide the index $d_k$, in the notation above.]</p>