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The question you raise doesn't come up often enough to be dealt with explicitly in textbooks and such, I think. One way to extract an answer (possibly overkill) is to look more closely at the cde-triangle in the formulation of Serre or Curtis & Reiner. Changing your notation a bit, take $p$ not dividing $|G|$ and then form a triple $(K,A,k)$ with $A$ a complete d.v.r. (such as $\mathbb{Z}_p$) having $K$ as fraction field and $k$ as the finite residue field of characteristic $p$. Without assuming that these fields are "large enough", one knows that the decomposition homomorphism $d: \mathrm{R}_K(G) \rightarrow \mathrm{R}_k(G)$ is surjective: see my older question here.

EDIT: Looking back at what I wrote next, it seems too superficial. Maybe a more careful comparison of the behavior under field extensions is really needed. Back to the drawing board.

ADDED: Maybe I've missed something, but I think what Serre does in his Section 15.5 avoids any use of the assumption that the fields are large enough. So this should dispose of the original question asked, while equating dimensions of correlated simple modules over $K$ and $k$. (Serre tends to be careful about specifying where it matters that fields are large enough.) Even though simple modules may decompose further over field extensions, working with any fixed $p$-modular system seems to yield trivial Cartan and decomposition matrices. (In this situation, the fact that $d$ is surjective follows from the assumption that $p$ doesn't divide the group order.)

In any case Alex has addressed the modified questions well.

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The question you raise doesn't come up often enough to be dealt with explicitly in textbooks and such, I think. One way to extract an answer (possibly overkill) is to look more closely at the cde-triangle in the formulation of Serre or Curtis & Reiner. Changing your notation a bit, take $p$ not dividing $|G|$ and then form a triple $(K,A,k)$ with $A$ a complete d.v.r. (such as $\mathbb{Z}_p$) having $K$ as fraction field and $k$ as the finite residue field of characteristic $p$. Without assuming that these fields are "large enough", one knows that the decomposition homomorphism $d: \mathrm{R}_K(G) \rightarrow \mathrm{R}_k(G)$ is surjective: see my older question here. Moreover, the kernel of $d$ is trivial, from the viewpoint of ordinary and Brauer characters

EDIT: Looking back at what I wrote next, since all conjugacy classes of $G$ are $p$-regular.

From this comparison one sees that there is it seems too superficial. Maybe a natural bijection between the natural bases more careful comparison of the two Grothendieck groups; so in this sense the simple modules over $K$ and $k$ are "the same". (And there is no distinction over either behavior under field between simple modules and their projective covers. One need not worry about comparing projectives indirectly, via "lifting of idempotents" or equivalent arguments.) Now the decomposition map extensions is just an identity map on Grothendieck groups, so in particular the "reduction mod $p$" makes dimensions of simple modules coincide over really needed. Back to the two fieldsdrawing board.

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The question you raise doesn't come up often enough to be dealt with explicitly in textbooks and such, I think. One way to extract an answer (possibly overkill) is to look more closely at the cde-triangle in the formulation of Serre or Curtis & Reiner. Changing your notation a bit, take $p$ not dividing $|G|$ and then form a triple $(K,A,k)$ with $A$ a complete d.v.r. (such as $\mathbb{Z}_p$) having $K$ as fraction field and $k$ as the finite residue field of characteristic $p$ Without assuming that these fields are "large enough", one knows that the decomposition homomorphism $d: \mathrm{R}_K(G) \rightarrow \mathrm{R}_k(G)$ is surjective: see my older question here. Moreover, the kernel of $d$ is trivial, from the viewpoint of ordinary and Brauer characters, since all conjugacy classes of $G$ are $p$-regular.

From this comparison one sees that there is a natural bijection between the natural bases of the two Grothendieck groups; so in this sense the simple modules over $K$ and $k$ are "the same". (And there is no distinction over either field between simple modules and their projective covers. One need not worry about comparing projectives indirectly, via "lifting of idempotents" or equivalent arguments.) Now the decomposition map is just an identity map on Grothendieck groups, so in particular the "reduction mod $p$" makes dimensions of simple modules coincide over the two fields.