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Let $R$ be a commutative ring with unity and $F$ a finite rank free $R$-module, and $M$ a finitely generated $R$-module, such that $$ M \oplus F \cong F $$ I want to conclude that $M = 0$, but I can't prove this, and I'm not entirely sure that this is true.

Partial results:

  1. If $R$ is finite, this is obvious by cardinality counting.
  2. If $R$ is a field, then this is immediate, because every $R$-module is free so by dimension counting, $\dim M = 0$.
  3. If $R$ is a PID, then this is true. $M$ is projective, and over a PID, projective $\iff$ free, so $M$ is free, and by the classification theorem of finitely generated modules over a PID, both sides are free of the same rank, so the rank of $M$ is zero.
  4. If $R$ is local, this is true. Let $m$ be the maximal ideal, and tensor both sides of the isomorphism with the residue field $R/m$ over $R$. Then we get an isomorphism of $R/m$-vector spaces. On the right hand side, we get a vector space of dimension $\operatorname{rank} F$, and on the right side, we get the same vector space direct sum with $M \otimes_R R/m$, which is some $R/m$ vector space, but it must be dimension zero, $M \otimes_R R/m = 0$. Then $M=mM$, and applying Nakayama's Lemma, we can conclude $M = 0$.
  5. In the case where $R$ is an integral domain, I suspect this may be proved by passing to the fraction field and again reducing to a vector space dimension counting sort of thing, but I'm not sure.

Does anyone know if this is actually true in the generality I want?

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    $\begingroup$ Why can't you use item 4 with the local-global principal that is M is zero iff it is zero locally at every maximal ideal? $\endgroup$ – Youngsu Sep 17 '18 at 2:00
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Yes, it is true and isn't necessary $F$ free. The hypothesis implies that there is an exact sequence $0\rightarrow M\rightarrow F\xrightarrow{f} F\rightarrow0$. Since $F$ is finitely generated and $f$ is surjective then $f$ is an isomorphism and $M=0$.

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