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.


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.

  • 1
    $\begingroup$ The statement after Def. 2 (about $r$, $s$) seems to be wrong. The condition comes out the same as under Def. 1. This is fairly standard material. $\endgroup$ – მამუკა ჯიბლაძე Aug 21 '18 at 12:17
  • $\begingroup$ This is a misprint; The two equivalences should be switched. Thanks for pointing this out. $\endgroup$ – GreginGre Aug 21 '18 at 23:07

The two constructions give the same result, namely the universal group with a monoid homomorphism from $Proj(A)$ aka the Grothendieck group. The first definition gives this because a short exact sequence with $M_3$ projective always splits so that the relations are simply "$[M_2] = [M_1]+[M_3]$ in the group whenever $[M_2]=[M_1]+[M_3]$ in the monoid". The second definition is simply the explicit construction of the Grothendieck group. Because objects defined by universal properties are unique up to unique isomorphism, the two constructions are isomorphic.

  • $\begingroup$ Well, I realize thanks to your answer that the real problem was that the equivalence in def 1 (that i read in some reference found online) was false. I'm now convinced that both are the same, so they indeed give the same group ! $\endgroup$ – GreginGre Aug 21 '18 at 23:22

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