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Let $R$ be noncommutative unital ring and $M$ a projective (right) $M$-module. Assume that $R$ embedds into $M$ as a right -module.

A) If $R$ is a semisimple ring, then of course $R$ admits an $R$-module complement. But this is a very strong assumption. What are weaker but sufficient criteria for an $R$-module complement to exist.

B) What is a non-free example where $M$ does not admit an $R$-module complement in $M.

C) What is a non-free module example where $M$ does admit a complement?

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    $\begingroup$ The simplest example is the embedding $2:\mathbb{Z}\to \mathbb{Z}$. $\endgroup$ Sep 26 at 9:52
  • $\begingroup$ Why is it clear that it does not admit a complement? $\endgroup$ Sep 26 at 9:59
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    $\begingroup$ The complement should be a $\mathbb{Z}/2$ and $\mathbb{Z}$ has no 2-torsion. More generally this works for any commutative ring $R$ and $x\in R$ non invertible nonzerodivisor. $\endgroup$ Sep 26 at 10:04
  • $\begingroup$ @Denis: thanks for the example! However, I was looking for a noncommutative example. $\endgroup$ Sep 26 at 12:00
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    $\begingroup$ @DickJohnson Just view $\mathbf{Z}$ as quotient of a noncommutative ring $\mathbf{Z}\times R$ to make this obvious counterexample "noncommutative". $\endgroup$
    – YCor
    Sep 26 at 15:37
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For (A) a ring $R$ is selfinjective if it is injective as an $R$-module (on e should say left or right). By definition of injective this means $R$ has a complement in any module. Examples include Frobenius and quasi-frobenius rings.

If $R$ is von Neumann regular, then any finitely generated submodule of a projective module has a complement. So again if $R$ embeds in a projective module there is a complement.

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B) A standard situation is the injective envelope $I(R)$ of $R$. There is always an embedding $R \rightarrow I(R)$ and rings where $I(R)$ is additionally projective are called QF-3 rings (a generalisation of Frobenius rings).

C) Cant we just take $M=R \oplus N$ for a projective module $N$ so that $M$ is not free? That is very easy to obtain for finte dimensional algebras.

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  • $\begingroup$ Yes, of course . . . so this part of the question was a bit stupid. $\endgroup$ Sep 26 at 13:46

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