Prove that if $M$ and $N$ are free $R$-module, then $M \bigoplus N$ is also a free $R$-module.
I have this result 'Every free module is the direct sum of isomoprhic copies of underlying ring $R$'. Hence, my proof goes like this:
Since $M$ is free and $N$ is free, we have $M=\bigoplus M_i \cong \bigoplus R$ and $N=\bigoplus N_i \cong \bigoplus R$, the direct sum $M \bigoplus N=\bigoplus R$ where the number of $R$ in $M \bigoplus N$ is the sum of $R$ in $M$ and $N$.
Is my proof correct?