1. Let $f$ be modular of level $p^nN$, $(p,n) = 1$, $p > 2$ with character $\chi\psi\eta$, where $\chi$ has conductor dividing $N$, $\psi$ conductor power of $p$ and order power of $p$, and $\eta$ conductor $p$ and order dividing $p-1$. Since $p$ is odd, one can write $\psi = \xi^{-2}$. (i) Why is the character of $f \otimes \xi$ equal to $\chi\eta$; (ii) why is the reduction mod $p$ of this equal to the reduction of $f$; (iii) why is for $r \gg n$ $f \otimes \xi$ modular with respect to $\Gamma\_0(p^r) \cap \Gamma\_1(pN)$? Can this be proven using the converse theorem?


2.a Why is the twisted Eisenstein series $G = a\_0 + \sum\_{n=1}^\infty\sum\_{d \mid n}\eta^{-1}(d)d^{i-1}q^n$ of Nebentypus $\eta^{-1}$ with respect to $\Gamma\_0(p)$?


2.b Why is $fG$ modular with respect to $\Gamma\_0(p^r) \cap \Gamma\_1(N)$, wenn $f \in S\_k(\Gamma\_0(p^r) \cap\ \Gamma\_1(pN), \chi\eta)$ ist?


(The article is here: math.berkeley.edu/~ribet/Articles/motives.pdf )



questions regarding modular forms
William Stein has answered your question (i). As for your question (ii), since $\xi$ has $p$-power order, and since any $p$-power root of unity is congruent to $1$ modulo the unique prime ideal lying over $p$ in $\mathbb Q$ adjoin the $p$-power roots of unity, we see
that $f\otimes \xi$ is congruent to $f$ modulo any prime ideal lying over $p$ in the
field of definition of $f$ adjoin the $p$-power roots of unity.


Finally, you have already noted in (i) that the character of $f\otimes \xi$ is just
$\chi\eta$. Since the conductor of $\xi$ divides $N$ and the conductor of $\eta$ divides $p$,
we see that the conductor of $\chi\eta$ divides $p N$. This gives (iii). (Note that (iii)
is simply a statement about the conductor of the character of $f\otimes \xi$: In general, a modular form $g$ of level $C$ is modular on $\Gamma\_0(C)\cap \Gamma\_1(D)$, rather than just $\Gamma\_1(C)$, for some integer $D$ dividing $C$, 
if the character of $g$, a priori a character of $(\mathbb Z/C)^{\times}$, actually
factors through $(\mathbb Z/D)^{\times}$. In your particular case, $f$ has conductor a power of $p$ times $N$, and $\xi$ has conductor a power of $p$, so $f\otimes \xi$ has 
conductor a power of $p$ times $N$. The conductor of its character divide $p N$, as
already noted, and so (iii) follows.)


