= Solution
Take a nonzero eigenform of positive weight, and put $h(\tau)=f(p\tau)$ and $C=\chi(p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_0(Np)$, the conjugate $\begin{pmatrix}a&pb\\c/p&d\end{pmatrix}$ lies in $\Gamma_0(N)$, so
$$
h(\gamma\tau)=\chi(d)(c\tau+d)^kh(\tau).
$$
Its <Dirichlet character> is therefore the specified reduced <Dirichlet character>; <modular cusp> holomorphy is permitted in the question. This is the <Dirichlet character> version of an <oldform by argument dilation>.
A nonzero positive-weight <modular form> cannot be constant, because the <matrix> $\begin{pmatrix}1&0\\N&1\end{pmatrix}$ would force a nonzero constant to equal $(N\tau+1)^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.
At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) give
$$
U_pf=\lambda f-Ch,\qquad U_ph=f.
$$
Thus the <matrix> in the ordered basis $(f,h)$ is
$$
\boxed{\begin{pmatrix}\lambda&1\\-C&0\end{pmatrix},}
$$
with <characteristic polynomial> $X^2-\lambda X+C$. For its distinct roots,
$$
\boxed{f-\beta f(p\tau)\text{ has eigenvalue }\alpha,\qquad f-\alpha f(p\tau)\text{ has eigenvalue }\beta.}
$$
This is <prime stabilization of an oldform>.
The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial <Dirichlet character> has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau)=f(p\tau)$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.
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