# Grothendieck group of derived category

Let $A$ be an essentially small abelian category, and $D(A)$ it derived category. Does $K_{0}(D(A)) = 0$? Thank you!

• Suppose $A$ is the category of finitely generated modules over a regular ring.... – Steven Landsburg Feb 15 '17 at 5:53
• This is true if your category has infinite sums (by the Eilenberg swindle), but clearly not in general. – Tom Bachmann Feb 15 '17 at 9:10
• This example is probably too obvious, but let me state it anyway. Let $A$ is the category of finite dimensional vector spaces over a field. Let $D(A)=D^b(A)$ denote the bounded derived category. Then $K_0(D(A))=\mathbb{Z}$. – MKO Feb 15 '17 at 13:24

Yes, it's always zero, assuming $D(A)$ means the unbounded derived category.
My complexes will be cochain complexes, and $X[1]$ will be $X$ shifted down in degree.
First suppose $X$ is bounded below, and let $X'=\bigoplus_{n\geq0}X[-2n]$. Even if $A$ doesn't have infinite direct sums, this is still an object of $D(A)$, since in each degree it only involves a finite direct sum. Then there is a triangle $$X'[-2]\to X'\to X\stackrel{0}{\to} X'[-1]$$ and so the class $[X]$ of $X$ in the Grothendieck group is zero.
A similar argument, using shifts in the opposite direction, works if $X$ is bounded above.
Finally, any $X$ fits in a triangle with a bounded below and a bounded above complex, so $[X]=0$ for all $X$.