\begin{lemma}\label{algphi}
Let $\vphi_1,\ldots, \vphi_n$ be a $\Q$-basis for 
$\Hom_\Q(\sM_k(N,\Q),\Q)[I_f]$ and set 
$$\Psi=\vphi_1\cross\cdots\cross\vphi_n : \sM_k(N,\Q) \ra \Q^n.$$
Then $n=2d$ and $\ker(\Psi)=\ker(\Phi_f)$. 
\end{lemma}
\begin{proof}
This sort of rationality result is typically attributed to
Shimura\index{Shimura} \cite{shimura:onperiods}.
To compute the dimension of $\Hom_\Q(\sM_k(N,\Q),\Q)[I_f]$, we first tensor
with~$\C$.  Let $\Sbar_2$ denote the weight 2 anti-holomorphic cusp
forms and $E_2$ the weight $2$ Eisenstein series for $\Gamma_0(N)$.
Then $\sM_k(N,\C)$ is isomorphic as a $\T$-module to $S_2\oplus
\Sbar_2\oplus E_2$ (prop. 9 of \cite{merel:1585} and the
Eichler-Shimura\index{Eichler-Shimura} 
embedding).  Because of the Petersson\index{Petersson} inner product,
the dual $\Hom_\C(\sM_k(N,\C),\C)$ is also isomorphic as a $\T$-module to
$S_2\oplus \Sbar_2\oplus E_2$.  Since $f$ is new, by the Atkin-Lehner
\index{Atkin-Lehner}
theory,
$$(S_2\oplus \Sbar_2\oplus E_2)[I_f] = S_2[I_f]\oplus \Sbar_2[I_f]$$
has complex dimension $2d$, which gives the first assertion.

To see that $\ker(\Phi_f)\tensor\C\subset\ker(\Psi)\tensor\C$,
note that each of the maps $x\mapsto \langle f_i, x \rangle$ lies in 
$\Hom_\Q(\sM_k(N,\Q),\C)[I_f]$ and 
$\ker(\Psi)\tensor\C$ is the intersection of
the kernels of {\em all} maps in $\Hom_\Q(\sM_k(N,\Q),\C)[I_f]$. 
By Theorem~\ref{Af} the image of $\Phi_f$ is a lattice, so
$\dim_\Q \ker(\Phi_f)=\dim_{\Q} \sM_k(N,\Q) - 2d$. Since $\Psi$ is the
intersection of the kernels of $n=2d$ independent
linear functionals $\vphi_1,\ldots, \vphi_n$,
$\ker(\Psi)$ also has dimension $\dim\sM_k(N,\Q)-2d$.  Since
the dimensions are the same and there is an inclusion,
we have an equality
$\ker(\Phi_f)\tensor\C = \ker(\Psi)\tensor\C$ which forces
$\ker(\Phi_f)=\ker(\Psi)$.
\end{proof}