(Sorry about my poor english...)
Let $k,N$ be integers and $\chi$ be a Dirichlet character modulo $N$. I know that if $F\in S^{new}_{k}(\Gamma_0(N),\chi)$ be a normalized newform then all coefficients of $F$ are algebraic integer. My question is..
Q. If $F=\sum_{n=1}^{\infty} a(n)q^n \in S_{k}(\Gamma_0(N),\chi)$ be a modular form with $a(1)=1$. For prime $p\nmid N$, there exists a complex number $\lambda_p$ such that
\begin{equation}
F|T(p,k,\chi)=\lambda_p F.
\end{equation}
Are all coefficients of $F$ algebraic integer?
Why or Why not..?
(Thanks for reading..)
Let $$\Delta(\tau) = q \prod_{n=1}^{\infty} (1 - q^n)^{24} = q + \ldots \in S_{12}(\Gamma(1)).$$ Let $N > 1$ be any integer. Then $\Delta(\tau)$ is also a form in $S_{12}(\Gamma_0(N))$, as is $$\Delta(N \tau) = q^N \prod_{n=1}^{\infty} (1 - q^{nN}) = q^N + \ldots$$ Both $\Delta(\tau)$ and $\Delta(N \tau)$ are eigenforms for all the Hecke operators $T_p$ with $p \nmid N$. But then so is:
$$F = \Delta(\tau) + \pi \cdot \Delta(N \tau) = q + \ldots $$
But $F$ does not have algebraic integer coefficients, even though it is an eigenform for all $p \nmid N$.
If you want to force old eigenforms to have algebraic coefficients, you should insist that they are still eigenvalues for the new Hecke operators $U_p$ for $p | N$. (The point is that newforms automatically are eigenforms for $U_p$ with eigenvalue either $0$ or a root of unity times $p^{(k-2)/2}$, so this extra condition is superfluous.)