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Can someone give a precise definition of a direct summand of an $R$-module $M$. (You can assume $R$ is commutative with unity).

Here is what I thought till date:

"We say an $R$-module $N$ is a direct summand of $M$ if there exists an $R$-module $N'$ such that $M$ is isomorphic to $N \oplus N'$".

But while going through some articles on commutative algebra/ homological algebra, I don't think this is taken as a defintion.

For example, in my definition, $2\mathbb{Z}$ is a direct summand of $\mathbb{Z}$. But is it really so according to standard literature? Can someone point me to a definition of direct summand in some popular textbook?

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  • $\begingroup$ A submodule $N$ is a direct summand of $M$ if there exists a submodule $N'$ of $M$ such that $M$ is the (internal) direct sum $N\oplus N'$, i.e., for every $m\in M$ there exists unique $(n,n')\in N\times N'$ such that $m=n+n'$. $\endgroup$
    – user10354138
    Apr 11, 2021 at 14:53
  • $\begingroup$ Your definition is good. To be extremely precise, one should probably distinguish between internal direct summands and external direct summands but since this is usually/always clear from the context, it isn't done. In your case, yes, $2 \mathbb{Z}$ is an (external) direct summand but I wouldn't phrase it that way as the $2$ can be confusing and the important thing here is rather that $2 \mathbb{Z}$ is embedded via $2 \mathbb{Z} \cong \mathbb{Z}$ in $\mathbb{Z}$. $\endgroup$
    – Qi Zhu
    Apr 12, 2021 at 6:10

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A submodule $A$ of $B$ is a direct summand if there exists a $C$ such that $B\cong A\oplus C$ where the canonical map of the first summand is the inclusion map.

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  • $\begingroup$ So does direct summand always refer to submodules? Can you give a reference to this definition? $\endgroup$
    – Anupam
    Apr 11, 2021 at 14:57
  • $\begingroup$ @Anupam That depends on the context, for most purposes it suffices to say that it is any module $A$ for which there exists a module $C$ such that $B \cong A \oplus C$. (And certainly, one could then identify $A$ with a submodule of $B$.) Ok, I just saw that this was exactly your textbook definition as well. $\endgroup$
    – Qi Zhu
    Apr 12, 2021 at 5:39
  • $\begingroup$ @Qi In Rotman's "Advanced Modern Algebra", the definition (or rather one of the equivalent criteria) for a submodule to be a direct summand is if there exists a retract onto the submodule (i.e., if the inclusion map splits). So in that setup, I believe we should better talk in terms of internal direct sums. $\endgroup$
    – Anupam
    Apr 12, 2021 at 8:32

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