It is well known that the space of cuspforms of weight $k$ on a congruence subgroup is spanned by forms with integral Fourier coefficients. (This can be proven at least for $k \geq 2$ with a cohomological argument.)


Is the same statement true if the whole space of forms is replaced by the subspace of newforms?



Integral basis for newforms
Yes, this is true. The set of newforms is stable under $\operatorname{Aut}(\mathbf{C})$ (this is due to Shimura). So if you consider a newform $f$, and let $K\_f$ be the field generated by its Fourier coefficients, then $f^\sigma$ is a newform for any embedding $\sigma : K\_f \to \mathbf{C}$. Now if $(a\_1,\ldots,a\_d)$ is any $\mathbf{Z}$-basis of $\mathcal{O}\_{K\_f}$, then $f\_i := \sum\_{\sigma : K\_f \to \mathbf{C}} \sigma(a\_i) f^\sigma = \operatorname{Tr}\_{K\_f/\mathbf{Q}} (a\_i f)$ belongs to the new subspace and has coefficients in $\mathbf{Z}$. Moreover the $\mathbf{C}$-span of $(f\_1,\ldots,f\_d)$ coincides with the $\mathbf{C}$-span of $(f^\sigma)\_{\sigma}$ because the matrix $(\sigma(a\_i))\_{\sigma,i}$ is invertible.


