# Why are isomorphic rings also isomorphic modules?

Let $G$ be a finite group and $F$ be an algebraically closed field such that $char(F)$ does not divide $|G|$.

Then, Artin-Weddenburn theorem implies that there exist positive integers $n_1,...,n_k$ such that $F[G]\cong \prod_{i=1}^k Mat_{n_i}(F)$ as rings. How do I conclude that they are isomorphic as $F$-modules?

Texts I am reading say that "since they are isomorphic as rings, they are isomorphic as $F$-modules, hence $|G|=dim_F(F[G])=dim_F(\prod_{i=1}^k Mat_{n_i}(F))=\sum n_i^2$." However, to imply this, we must show that there exists an $F$-linear isomorphism which preserves given $F$-space structures. Since $F[G]$ and $\prod_{i=1}^n Mat_{n_i}(F)$ are isomorphic as rings, of course we can embed an $F$-space structure from one to another naturally induced by a ring isomorphism. However, dimension may differ and this does not mean that $F[G]$ and $\prod_{i=1}^k Mat_{n_i}(F)$ are isomorphic as $F$-spaces where the $F$-space structures are already given.

How do I prove this? Thank you in advance.

• Would it be more accurate to say that your question is "If two $F$-algebras are isomorphic as rings, why are they isomorphic as $F$-vector spaces?" – rschwieb Mar 20 '17 at 10:41
• @rschwieb Yes! That's right – Rubertos Mar 20 '17 at 10:48
• It seems to me that these authors probably meant "since they are isomorphic as $F$-algebras they are isomorphic as $F$ vector spaces." That much is obvious, and would be the natural context. – rschwieb Mar 20 '17 at 10:53
• @rschwieb Okay, so you mean I have to go the bottom of the proof and check whether those ring isomorphisms used to prove Artin-Weddenburn theorem are indeed $F$-algebra isomorphisms in this case? – Rubertos Mar 20 '17 at 10:57
• Yes, I think you will find it is the case. – rschwieb Mar 20 '17 at 11:25