The necessity of defining the stable equivalence in the construction of the Grothendieck group $K_0$

I am confused about the process of the construction of the Grothendieck group $K_0$ in Murphy's $C^*$-algebras and operator theory section 7.1.

Let $A$ be a $*$-algebra and $P[A]=\bigcup_{n=1}^\infty\{p\in M_n(A):\text {$p$is a projection} \}$. By projection I mean $p=p^*=p^2$.

Define the an equivalence relation on $P[A]$ by $p \sim q \Leftrightarrow$ there is a rectangle matrix $u$ with entries in $A$ such that $p=u^*u$ and $q=uu^*$.

Two projections $p$, $q$ in $P[A]$ are said to be stably equivalent(denoted by $p\approx q$) if $1_n\oplus p{\sim}1_n\oplus q$ for some identity matrix $1_n$.($1_n\oplus p$ is the block diagonal matrix with blocks $1_n$ and $p$)

There is a well-defined operation on ${P[A]}/{\sim}$ by defining the sum of two square matrices to be $p\oplus q$(block diagonal matrix). This also induces an operation on ${P[A]}/{\approx}$.

Murphy proved that $K_0(A)^+:={P[A]}/{\approx}$ is a cancellative abelian group with zero. But I want to find an example of $A$ such that ${P[A]}/{\sim}$ is not cancellative which shows the necessity of defining stable equivalence.

Thanks for your help!

• The Cuntz algebra $A$ is such that the is an isomorphism of modules $A\cong A^n$ for some $n\geq2$. Such an algebra provides an example. – Mariano Suárez-Álvarez Apr 23 '16 at 18:36
• Thanks~But why isn't ${P[A]}/{\sim}$ cancellative? – No One Apr 23 '16 at 18:48
• $P[A]/\sim$ is isomorphic to the set of finitely generated projective $A$-modules with the operation given by direct sum. If $A\oplus A\cong A$, then of course $A\oplus A\cong 0\oplus A$ and you cannot cancel the $A$. – Mariano Suárez-Álvarez Apr 23 '16 at 19:03
• Are your $A$'s in the second line projection matrices? The sum of two matrices is defined to be the block diagonal matrix with two blocks as summands. – No One Apr 23 '16 at 19:10
• As I wrote, $P[A]/\sim$ has a decription in terms of f.g. projective modules, and I am using that. $A$ and $A\oplus A$ are f.g. projective modules, so they have classes in $P[A]/\sim$, etc. – Mariano Suárez-Álvarez Apr 23 '16 at 19:16