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When a submodule N of a R-module M is a direct summand of M (or of R-module R) ?.

For instance, if M is semisimple then N is semisimple, does this say that N is direct summand of M (or of R-module R)?, or, What ways to know that a submodule of a module M is direct summand of M (or of R-module R)?

Javier Esneider Mendez Alfonso
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2 Answers2

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When a submodule $N$ of a $R$-module $M$ is a direct summand of $M$?

There is not really any criterion (one significantly simpler than the definition) to spot summands in general. It all depends on the module structure of $M$. There is an important equivalent condition, though: $N$ is a summand of $M$ if and only if there is an idempotent homomorphism $e:M\to M$ such that $e(M)=N$.

or of $R$-module $R$?

At this point things are easier since the idempotent homomorphisms $R\to R$ are given by idempotent elements of $R$. So a right ideal $T$ is a summand of $R_R$ iff there is an idempotent element $e\in R$ such that $eR=T$.

For instance, if $M$ is semisimple then $N$ is semisimple, does this say that $N$ is direct summand of $M$ (or of $R$-module $R$)?

Well, yes, because a characterization of semisimple modules is that every submodule is a direct summand. I'm not sure which definition of "semisimple" you are using, so I don't know if you need help seeing this.

rschwieb
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  • Na verdade, eu estou com um problema particular, muito especifico. Porém, queria saber se há alguma generalização. – Javier Esneider Mendez Alfonso Aug 07 '17 at 00:04
  • Actually, I have a very specific, very specific problem. But I wanted to know if there is any generalization. – Javier Esneider Mendez Alfonso Aug 07 '17 at 00:04
  • An Artin R algebra, an R-module M and f: P \longrightarrow M is an epimorphism of R-modules (essencial epimorphism). The following is an exact short sequence 0 \longrightarrow ker (f') \longrightarrow P / rad ^ 2 (P) \longrightarrow M \longrightarrow 0 Where f' : P / rad ^ 2 (P) \longrightarrow M is defined by f' (x') = f (x) with x' = x + rad ^ 2 (P) and x \in P. – Javier Esneider Mendez Alfonso Aug 07 '17 at 00:14
  • I have been able to show, with the hypotheses of M having finite projective dimension and Loewy size (Loewy exactly 2) that ker (f ') is contained in rad (P) / rad ^ 2 (P) and since the latter is semisimple, – Javier Esneider Mendez Alfonso Aug 07 '17 at 00:16
  • I need to show that ker (f ') is direct summand from R / rad (R), but the only one I have is this up. I do not know which way to go. I have problems. – Javier Esneider Mendez Alfonso Aug 07 '17 at 00:16
  • Dear @rschwieb, could you give me a hint to prove the part $\Leftarrow$ in the "important equivalent condition" you pointed in the first paragraph? – rgm Jan 26 '18 at 16:24
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    @rgm Given the idempotent, $M=e(M)\oplus (1-e)M=N\oplus (1-e)M$. What else is there to say? – rschwieb Jan 26 '18 at 22:18
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I'm not sure if this answer will satisfy you, but I'm pretty sure, to the full extent of generality that you stated the problem, it is the best we can say.

Let $N\subset M$ and $L=M/N$ be the quotient module. Consider the short exact sequence ($\imath$ is inclusion, $\kappa$ quotient map) $$ 0\to N\xrightarrow{\imath}M \xrightarrow{\kappa}L\to 0 $$ This exact sequence is called a split exact sequence one of the following equivalent conditions hold:

  1. There exists $r: M\to N$ such that $r\circ \imath=\mathrm{id}$. In other words $r$ is a retraction.
  2. There exists $s: L\to M$ such that $\kappa \circ s=\mathrm{id}$. In other words $s$ is a section.
  3. $M$ is isomorphic to $N\oplus L$. In which case the natural map $N\to N\oplus L$ the direct sum is equipped with is the inclusion (composed with the isomorphism) and $L\to N\oplus L$ is the section of part 2. Treating $N\oplus L$ as $N\times L$, the natural map $N\times L\to L$ is the quotient map, while $N\times L\to N$ is the retraction of part 1.

Now note that if $N$ is a summand of $M$, then the other summand is necessarily isomorphic to $L$ (exercise). So basically you are asking when is the above sequence split. The answer is whenever condition 1 or 2 (if one holds, both do).

Hamed
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