A question about group ring of a group over matrix ring!

Question

Let $S$ be a ring and $G$ a group. We denote by $SG$ the group ring of $G$ over $S$. Let $S=M_n(R)$ be the set of all $n \times n$ matrices over a ring $R$. Is it true $SG\cong M_n(RG)$ (as rings)?

Thanks to anybody answering!

Answer 1

We have $M_n(R) G \cong M_n(RG)$ as rings. Indeed, define the map $$\psi: M_n(R)G \to M_n(RG): \sum_{g \in G} A^g g \mapsto \left(\sum_{g \in G} A_{ij}^g g\right)_{1\leq i,j \leq n}$$

Clearly this is well-defined. This is a ring morphism:

$$\psi\left(\sum_g A^g g + \sum_g B^g g\right) = \psi\left(\sum_g (A^g +B^g)g \right)$$ $$= \left(\sum_{g \in G}((A^g + B^g)_{ij}g\right)_{ij} $$ $$= \left(\sum_{g \in G}(A_{ij}^g + B_{ij}^g)g\right)_{ij}$$

$$= \left(\sum_{g \in G}A_{ij}^g +\sum_{g \in G} B_{ij}^gg\right)_{ij}$$ $$= \left(\sum_{g \in G}A_{ij}^g\right)_{ij} +\left(\sum_{g \in G} B_{ij}^gg\right)_{ij}$$ $$=\psi\left(\sum_g A^g g\right) + \psi\left(\sum_g B^g g\right)$$

and

$$\psi\left(\left(\sum_g A^g g \right) \left( \sum_h B^h h\right)\right)$$ $$= \psi\left(\left(\sum_g A^g g \right) \left( \sum_h B^h h\right)\right)$$

$$= \psi\left(\sum_{g,h} A^g B^h (gh)\right)$$ $$= \left(\sum_{g,h} (A^g B^h)_{ij} gh\right)_{ij}$$

$$=\left(\sum_{g,h} \sum_{k=1}^n (A^g_{ik} B^h_{kj}) gh\right)_{ij}$$

On the other hand,

$$\left(\psi\left(\sum_{g} A_g g\right) \psi\left(\sum_{h} B_h h\right)\right)_{ij}$$

$$= \left(\sum_{k=1}^n \left(\sum_{g} A_{ik}^g g\right)\left(\sum_h B_{kj}^h h\right)\right)_{ij}$$

$$= \left(\sum_{k=1}^n \sum_{g,h} A_{ik}^g B_{kj}^h (gh)\right)_{ij}$$

and $\psi$ preserves multiplication because we can interchange the last two sums.

Surjectivity is clear and injectivity should also be clear. You can also check that if $R$ has an identity, that $\psi$ preserves the identity.