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A class $\mathcal{C}$ of $R$-modules is called

-separative if $A \oplus A \simeq A \oplus B \simeq B \oplus B$ implies $A \simeq B$ for each $A,B \in \mathcal{C}$

-cancellative if $A \oplus C \simeq B \oplus C$ implies $A \simeq B$ for all $A,B,C \in \mathcal{C}$.

According to literature, if $R$ is commutative then the class of finitely generated projectives over $R$ is separative iff it is cancelative. Even though I keep finding it as 'easy to see' in literature I seem unable to prove separative => cancelative. I would be grateful for any hint.

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I'm not sure this is the intended "easy" argument, but here's what I could come up with. Fix a commutative ring $R$; the variables $A$, $B$, $C$, etc. will denote finitely generated projective $R$-modules. First, note that if $A$ locally has rank $>0$ on all of $\operatorname{Spec}(R)$, then for some $N\in\mathbb{N}$ there is a surjection $A^N\to R$. Indeed, the evaluation map (aka trace) $\operatorname{End}(A)=A\otimes \operatorname{Hom}(A,R)\to R$ is surjective, since locally it looks like the trace map $M_n(R)\to R$ for $n>0$. Choosing finitely many generators of $\operatorname{Hom}(A,R)$, we then get a surjection $A^N\to R$ for some $N$.

Now suppose $A\oplus C\simeq B\oplus C$. Note that $C$ is a direct summand of $R^n$ for some $n$; adding $R^n/C$ to both sides, we may assume $C=R^n$. Now let $n$ be minimal such that $A\oplus R^n\simeq B\oplus R^n$. Replacing $A$ and $B$ by $A\oplus R^{n-1}$ and $B\oplus R^{n-1}$, we may assume $n=1$, so we have $A\oplus R\simeq B\oplus R$ and wish to conclude $A\simeq B$. Note that the rank of $A$ is locally constant function on $\operatorname{Spec}(R)$, as is that of $B$, and these two functions must be equal. We lose no generality by restricting everything to the clopen subset of $\operatorname{Spec}(R)$ on which $A$ and $B$ have rank $>0$, since $A$ and $B$ are trivially isomorphic on the clopen set where they have rank $0$.

We thus may assume that $A$ and $B$ have rank $>0$ everywhere; let $N$ be such that there are surjections $A^N\to R$ and $B^N\to R$. These surjections split, so from the isomorphism $A\oplus R\simeq B\oplus R$ we get $A\oplus A^N\simeq B\oplus A^N$ and $A\oplus B^N\simeq B\oplus B^N$. We thus get that $$A^{2N}= A^N\oplus A^N\simeq A^{N-1}\oplus B\oplus A^N\simeq A^{N-2}\oplus B^2\oplus A^N\simeq\dots\simeq B^N\oplus A^N,$$ where at each step we combine one of the copies of $A$ in the first term with the $A^N$ in the last term and use $A\oplus A^N\simeq B\oplus A^N$. Swapping the roles of $A$ and $B$, we also have $B^N\oplus A^N\simeq B^{2N}$, so $A^{2N}\simeq B^{2N}$. By a similar argument, we have $A^n\simeq B^n$ for any $n\geq 2N$.

Now assume the finitely generated projective modules are separative, and suppose that $A\not\simeq B$. Let $k$ be the largest integer such that $A^k\not\simeq B^k$ (by the previous paragraph, such a largest integer exists and is $<2N$). Then by maximality of $k$, $A^{2k}\simeq B^{2k}$ and $A^{3k}\simeq B^{3k}$, so $$A^{2k}\oplus A^{2k}\simeq A^{2k}\oplus B^{2k}=(A^k\oplus B^k)\oplus (A^k\oplus B^k),$$ and also $$A^{2k}\oplus A^{2k}\simeq B^{2k}\oplus B^{2k}=B^{3k}\oplus B^k\simeq A^{3k}\oplus B^k=A^{2k}\oplus (A^k\oplus B^k).$$ Applying separativity to $A^{2k}$ and $A^k\oplus B^k$, we get $A^{2k}\simeq A^k\oplus B^k$. We can now apply separativity to $A^k$ and $B^k$ to get $A^k\simeq B^k$. This is a contradiction. Thus if the finitely generated projective modules are separative, we must have $A\simeq B$.

(The second half of this argument is adapted from the proof of Lemma 2.1 from this paper.)

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