# Integral basis for newforms

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?

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## 1 Answer

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.

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Many thanks. I knew this result of Shimura, but I didn't know the second argument that allows one to descend to Z. Do know you of a reference for this second argument? – MF1 Oct 17 '12 at 15:31
I don't know a precise reference. I consider this result is essentially equivalent to Shimura's result, so you can simply refer to Shimura. – François Brunault Oct 18 '12 at 9:47
Well...okay I guess. – MF1 Oct 18 '12 at 17:36