Prove that if $P\oplus R^m\cong R^\infty $ , then $P\cong R^\infty$

Question

$R$ is a ring with unit.(not necessary commutative) . $R^\infty$ is a fixed free R-module on a countably infinite basis

Prove that if $P\oplus R^m\cong R^\infty$, then $P\cong R^\infty$.

This problem come from the Weibel's K-book. And he gives the following hint.

Hint: The image of $R^m$ is contained in some $R^n\subset R^\infty$. Writing $R^\infty\cong R^n\oplus F $and $Q = P\cap R^n $, show that $P\cong Q\oplus F$ and $F\cong F\oplus R^m$

I have no ideal how to get the $P\cong Q\oplus F$ and $F\cong F\oplus R^m$. Could someone help me?

Answer 1

You can choose $R^m$ and $F$ in the hint to be free submodules spanned by canonical basis elements of $R^\infty$. In this case, the existence of an isomorphism $F\oplus R^m \cong F$ is clear, because $F$ is again free on countably many generators.

Now note that the composition $P\hookrightarrow R^\infty\to F$ is surjective and its kernel is $Q$. Since $F$ is projective (it is a direct summand of $R^\infty$), the short exact sequence $0\to Q\to P\to F\to 0$ splits. This proves $P\cong Q\oplus F$.

Now we have a chain of isomorphisms $P\cong Q\oplus F\cong Q\oplus F\oplus R^m\cong P\oplus R^m\cong R^\infty$.