Proposition 3.1
Suppose that
![$ d_k(\Gamma)$](img34.png)
is finite.
Then the discriminant valuation
![$ d_k(\Gamma)$](img34.png)
is nonzero
if and only if there is a mod-
![$ p$](img1.png)
congruence
between two Hecke eigenforms in
![$ S_k(\Gamma)$](img19.png)
(note that the two congruent eigenforms might
be Galois conjugate).
Proof.
It follows from Proposition
2.2 that
![$ d_k(\Gamma)>0$](img36.png)
if and only if
![$ \mathbb{T}\otimes \overline{\mathbb{F}}_p$](img37.png)
is not
separable. The Artinian ring
![$ \mathbb{T}\otimes \overline{\mathbb{F}}_p$](img37.png)
is
not separable if and only if the number of ring
homomorphisms
![$ \mathbb{T}\otimes \overline{\mathbb{F}}_p\rightarrow \overline{\mathbb{F}}_p$](img38.png)
is
less than
Since
![$ d_k(\Gamma)$](img34.png)
is finite, the number of ring
homomorphisms
![$ \mathbb{T}\otimes \overline{\mathbb{Q}}_p\rightarrow \overline{\mathbb{Q}}_p$](img40.png)
equals
![$ \dim_\mathbb{C}S_k(\Gamma)$](img41.png)
. Using the standard bijection between
congruences and normalized eigenforms, we see that
![$ \mathbb{T}\otimes \overline{\mathbb{F}}_p$](img37.png)
is not separable if and only
if there is a mod-
![$ p$](img1.png)
congruence between two eigenforms.
Example 3.2
If
![$ \Gamma=\Gamma_0(389)$](img42.png)
and
![$ k=2$](img43.png)
,
then
![$ \dim_\mathbb{C}S_2(\Gamma) = 32$](img44.png)
.
Let
![$ f$](img45.png)
be the characteristic polynomial of
![$ T_2$](img46.png)
.
One can check that
![$ f$](img45.png)
is square free and
![$ 389$](img47.png)
exactly
divides the discriminant of
![$ f$](img45.png)
, so
![$ T_2$](img46.png)
generated
![$ \mathbb{T}\otimes \mathbb{Z}_{389}$](img48.png)
as a ring. (If it generated a subring of
![$ \mathbb{T}\otimes \mathbb{Z}_{389}$](img48.png)
of finite index, then the discriminant of
![$ f$](img45.png)
would be divisible
by
![$ 389^2$](img49.png)
.)
Modulo
the polynomial
is congruent to
The factor
![$ (x+175)^2$](img51.png)
indicates that
![$ \mathbb{T}\otimes \overline{\mathbb{F}}_{389}$](img52.png)
is not separable
since the image of
![$ T_2+175$](img53.png)
is nilpotent
(its square is 0). There are
![$ 32$](img54.png)
eigenforms over
![$ \mathbb{Q}_2$](img55.png)
but only
![$ 31$](img56.png)
mod-
![$ 389$](img47.png)
eigenforms, so there must be a congruence.
Let
![$ F$](img57.png)
be the
![$ 389$](img47.png)
-adic newform whose
![$ a_2$](img58.png)
term is a root of
Then the congruence is between
![$ F$](img57.png)
and its
![$ \Gal(\overline{\mathbb{Q}}_{389}/\mathbb{Q}_{389})$](img60.png)
-conjugate.
Example 3.3
The discriminant of the Hecke algebra
![$ \mathbb{T}$](img22.png)
associated
to
![$ S_2(\Gamma_0(389))$](img61.png)
is
I computed this using the following algorithm, which was suggested
by Hendrik Lenstra. Using the Sturm bound I found a
![$ b$](img63.png)
such that
![$ T_1,\ldots,T_b$](img64.png)
generate
![$ \mathbb{T}$](img22.png)
as a
![$ \mathbb{Z}$](img23.png)
-module. I then
found a subset
![$ B$](img65.png)
of the
![$ T_i$](img66.png)
that form a
![$ \mathbb{Q}$](img67.png)
-basis for
![$ \mathbb{T}\otimes _\mathbb{Z}\mathbb{Q}$](img68.png)
.
Next, viewing
![$ \mathbb{T}$](img22.png)
as a ring of matrices acting on
![$ \mathbb{Q}^{32}$](img69.png)
,
I found a random vector
![$ v\in\mathbb{Q}^{32}$](img70.png)
such that the set of
vectors
![$ C=\{T(v) : T \in B\}$](img71.png)
is linearly independent. Then
I wrote each of
![$ T_1(v),\ldots, T_b(v)$](img72.png)
as
![$ \mathbb{Q}$](img67.png)
-linear combinations
of the elements of
![$ C$](img73.png)
. Next I found a
![$ \mathbb{Z}$](img23.png)
-basis
![$ D$](img74.png)
for the
![$ \mathbb{Z}$](img23.png)
-span of these
![$ \mathbb{Q}$](img67.png)
-linear combinations of elements of
![$ C$](img73.png)
.
Tracing everything back, I find the trace pairing on
the elements of
![$ D$](img74.png)
, and deduce the discriminant by computing
the determinant of the trace pairing matrix. The most difficult
step is computing
![$ D$](img74.png)
from
![$ T_1(v),\ldots, T_b(v)$](img72.png)
expressed
in terms of
![$ C$](img73.png)
, and this explains why we embed
![$ \mathbb{T}$](img22.png)
in
![$ \mathbb{Q}^{32}$](img69.png)
instead of viewing the elements of
![$ \mathbb{T}$](img22.png)
as vectors in
![$ \mathbb{Q}^{32^2}$](img75.png)
.
This whole computation takes one
second on an Athlon 2000 processor.