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Let $R$ be a ring and $M$ be an $R$-module. Also assume that $N$ is a complement in $M$ (recall that a submodule $K$ of $M$ is called a complement in $M$ if there exists a submodule $L$ of $M$ such that $K\cap L=\{ 0\}$ and $M=K+L$. In other words, $M=K\oplus L$).

Assume that $M$ is a module. Is there a non-negative integer, say $n(M)$, with the following conditions:

(1) $n(M)<\infty$;

(2) if $M\cong N$, then $n(M)=n(N)$;

(3) if $K$ is a proper complement in $M$, then $n(K)\lneq n(M)$.

In particular, I'm interested in the class of Noetherian moduels, but I don't know this class of modules is countable.

Can $n(G)$ be the number of direct summands of $M$? Can $n(G)$ be the uniform (or Goldie) dimension of $M$?

My try: I define $n(G)$ inductively: $n(\{0\})=0$ and $$n(M)=1+\sup\{ n(N): N\; \text{is a proper complement in}\; M\}.$$ But my problem is that whether $\sup\{ n(N): N\; \text{is a proper complement in}\; M\}$ exist or not.

M.Ramana
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  • Is (3) suppposed to have nothing to do with $n(-)$? – Randall Jul 23 '23 at 14:15
  • @Randall Sorry, I've just edited. – M.Ramana Jul 23 '23 at 14:16
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    I'm afraid you can always construct some module such that $n$ is infinite, by considering things like $\bigoplus_{k \in \Bbb N} N$ – Crostul Jul 23 '23 at 14:30
  • @Crostul You are right, like the cardinality of $M$. But I need $n(M)$ to be finite and minimum in comparison with other $n$'s. For example, for the class of finitely generated $\mathbb{Z}$-modules, we can define $n(M)$ to be the number of direct summands of $M$. – M.Ramana Jul 23 '23 at 14:34

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