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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?

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  • $\begingroup$ The intuition is right, but talking about the "number of $R$" and their sum may not sound legit if there are infinitely many copies of $R$ for $M$ and/or $N$. $\endgroup$ – Tunococ Sep 16 '13 at 9:45
  • $\begingroup$ @Tunococ: I don't agree. The (cardinal) numbers really add, also in the infinite case. $\endgroup$ – Martin Brandenburg Sep 16 '13 at 9:49
  • $\begingroup$ @MartinBrandenburg You are right. I was too narrow-minded with what "number" means. After all, the proof essentially has to go through the process of adding cardinal numbers. $\endgroup$ – Tunococ Sep 16 '13 at 9:58
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Your proof is correct. However, it would be better if you indicate the index sets in your proof. So I would write $M \cong \bigoplus_{i \in S} R$, $N \cong \bigoplus_{i \in T} R$, then $M \oplus N \cong \bigoplus_{i \in S \sqcup T} R$.

You can also show directly: If $S$ is a basis of $M$ and $T$ is a basis of $N$, then $S \cup T$ is a basis of $M \oplus N$.

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