Let $A$ be a ring with $1$. I have encountered two definitions of stably isomorphic (left) $A$-modules.
Def. 1. Two finitely generated $A$-modules $M,N$ are stably isomorphic if there exists $r,s\geq 0$ such that $M\times A^r\simeq N\times A^s.$
Def. 2. Two finitely generated $A$-modules $M,N$ are stably isomorphic if there exists $n\geq 0$ such that $M\times A^n\simeq N\times A^n.$
Question 1. Are these two definitions equivalent ?
Theis question is motivated by the following. Let $Proj(A)$ be the commutative monoid of isomorphism classes $<M>$ of finitely generated projective modules, the internal law being given by: $<M>+<N>=<M\times N>$.
Once again, I know two possible definitions of the group $K_0(A)$.
Def 1'. $K_0(A)$ is the quotient of the free abelian group on $Proj(A)$ by the relations $ <M_2>=<M_1>+<M_3>$ whenever we have a short exact sequence $0\to M_1\to M_2\to M_3\to 0$. If we denote by $[M]$ the image of $<M>$ under the canonical projection, we have
$[M]=[N]$ if and only if $M$ and $N$ are stably isomorphic in the sense of Def 1.
Def 2'. $K_0(A)$ is the Grothendieck group (i.e. symmetrization) of the monoid $Proj(A)$, that is the quotient set of $Proj(A)\times Proj(A)$ wrt to the equivalence relation $$(<M_1>,<N_1>)\sim (<M_2>,<N_2>)\iff \exists \ <P>\in Proj(A), <M_1>+ <N_2>+<P>=<M_2>+<N_1>+<P>.$$
If $[M]$ denotes this time the class of $(<M>,0)$, we have $[M]=[N]$ if and only if $M$ and $N$ are stably isomorphic in the sense of Def. 2.
Question 2. Do these two constructions yield isomorphic groups ?
Of course, if Def 1 and Def 2 are equivalent, this is clear, but if not, it is not so clear to me.
I do not know any counterexample though, at least amongst all the very few examples of $K_0$ I am aware of.
EDIT. The answer of Question 2 is obviously NO (see the answer of A. Pongracz). So I guess the real interesting question is Question 2.
Thanks in advance of any enlightning thoughts.
Greg
Unless there are hidden conditions, the answer to question $1$ is no. It is already false for $A=\mathbb{Z}$. For finitely generated $\mathbb{Z}$-modules (i.e., finitely generated Abelian groups) the notion of rank makes sense. So $\mathbb{Z}^a\not \cong \mathbb{Z}^b$ for $a\neq b\in \mathbb{N}$. So two such modules satisfy definition $1$ but not definition $2$.
In fact, the same argument works for any PID due to https://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain