The following question is extracted from this question on MSE, which got no answer so far, probably because it was a bit hidden by another question which a posteriori was totally obvious.

Let $A$ be a ring with $1$, that I am happy to suppose commutative if necessary. Let $\mathrm{Proj}(A)$ be the commutative monoid of isomorphism classes $\langle M\rangle$ of finitely generated projective modules, the internal law being given by: $\langle M\rangle+\langle N\rangle =\langle M\times N\rangle$.

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 $ \langle M_2\rangle =\langle M_1\rangle +\langle M_3\rangle $ 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 $\langle M \rangle $ under the canonical projection, then any element may be written as $[M]-[N]$, and we have

$[M]-[N]=0$ if and only if there exists $r,s\geq 0$ such that $M\times A^r\simeq N\times A^s.$

Def 2.$K_0(A)$ is the Grothendieck group (i.e.symmetrization) of the monoid $\mathrm{Proj}(A)$, that is the quotient set of $\mathrm{Proj}(A)\times \mathrm{Proj}(A)$ wrt to the equivalence relation $$(\langle M_1\rangle,\langle N_1\rangle)\sim (\langle M_2\rangle ,\langle N_2\rangle)\\ \iff \exists \ \langle P\rangle\in \mathrm{Proj}(A), \langle M_1\rangle + \langle N_2\rangle +\langle P\rangle=\langle M_2\rangle+\langle N_1\rangle +\langle P\rangle.$$

If $[M]$ denotes this time the class of $(\langle M\rangle,0)$, then any element may be written as $[M]-[N]$, and we have $[M]-[N]=0$ if and only if there exists $n\geq 0$ such that $M\times A^n\simeq N\times A^n$.

Question .Do these two constructions yield isomorphic groups ?

I do not know any counterexample though, at least amongst all the very few examples of $K_0$ I am aware of.

Thanks in advance of any enlightning thoughts.

Greg

**Edit 1** I switched the equivalences between the definitions, since they were misplaced.

**Edit 2.** the equivalence in Def.1 is false. I read this in some reference i found on the web, and believed it without doublechecking at the time being. Thanks to the answers, I realize know that the right equivalence is the same as in Def.2, which solves immediately the question.