Lemma of Harada and Sai on sums of modules with a "chain" of monomorphisms between them

Question

I am trying to get a contradiction from the following set of hypotheses:

Let $R$ be a ring. Let $M$ be a direct sum of non-zero $R$-modules $M_1$, $M_2$, $\dotsc$. For each $i\ge1$, let $f_i:M_i\to M_{i+1}$ be an $R$-monomorphism that is not onto and let $M'_i=\{x+f_i(x)\mid x\in M_i\}$.

Assume that $M=M_1\oplus M'_2\oplus M_3\oplus M'_4\oplus\dotsb$. For $n\ge1$, let $\psi_{2n}$ be an $R$-isomorphism from $M_{2n}$ onto one of the summands in this direct sum. Assume that $M=M'_1\oplus M'_3\oplus\dotsb\oplus\psi_2(M_2)\oplus \psi_4(M_4)\oplus\dotsb$. Assume that for all $n\ge1$, there exists $m\ge1$ such that $\psi(M_{2m})=M'_{2n}$. Also assume that, for some $n_0\ge1$, for all $n\ge n_0$ we have $\psi_{2n}(M_{2n})=M'_{2m}$ for some $m\ge1$.

Lemma 9 on p. 336 of Harada and Sai, "On categories of indecomposable modules. I," Osaka Math. J. 7 (1970), 323–344 is different than this, but I hope I have extracted what I need above.

Answer 1

Note that $M_i'\oplus M_{i+1}=M_i\oplus M_{i+1}$. Thus, $$ \left(\bigoplus_{i=1}^{n}M_i'\right)\oplus M_{n+1}=\bigoplus_{i=1}^{n+1}M_i. $$

Case 1: Suppose that for every integer $n\geq 1$ that $\psi_{2n}(M_{2n})$ does not equal $M_{2m-1}$ (for any integer $m\geq 1$). Then your assumptions yield $$ M=\bigoplus_{i=1}^{\infty}M_i' $$ which is impossible (say, since the sum on the right contains no nonzero element of $M_1$).

Case 2: Suppose that for some integer $n\geq 1$ that $\psi_{2n}(M_{2n})=M_{2m-1}$ (for some integer $m\geq 1$). The condition about $n_0$ guarantees that there are only finitely many such $n$ (each less than $n_0$). Let $m_0$ be the biggest of all the corresponding $m$'s.

Now, by the computation in the first paragraph I wrote above, and from your assumptions, we have $$ M=\bigoplus_{i=1}^{2m_0-1}M_i\oplus \bigoplus_{i=2m_0}^{\infty}M_i'. $$ (Moreover, with a little more work we see that there was exactly one such corresponding $m$, namely $m_0$.) This is again impossible, since the sum on the right contains none of the elements in $M_{2m_0}-f_{2m_0-1}(M_{2m_0-1})$.