This question is related to the MO questions [What is the difference between Grothendieck groups K\_0(X) vs K^0(X) on schemes?](https://mathoverflow.net/questions/22120/what-is-the-difference-between-grothendieck-groups-k-0x-vs-k0x-on-schemes) and [Does a fully faithful functor between triangulated categories induce embedding of their Grothendieck groups?](https://mathoverflow.net/questions/203106/does-a-fully-faithful-functor-between-triangulated-categories-induce-embedding-o)


Let $X$ be a noetherian scheme over a field $k$. We have the derived category of perfect complexes on $X$, $Perf(X)$, and the derived category of bounded complex of coherent sheaves on $X$, $D^b(coh(X))$. It is clear that $Perf(X)$ is a strictly full triangulated subcategory of $D^b(coh(X))$ and for non-regular $X$, these two categories are not equivalent.


Now we define $K^0(X)$ be the Grothendieck group of $Perf(X)$ and $K\_0(X)$ be the Grothendieck group of $D^b(coh(X))$. Then the embedding $Perf(X)\to D^b(coh(X))$ induces a homomorphism
$$
K^0(X)\to K\_0(X).
$$


$\textbf{My question}$ is: is the above homomorphism an momomorphism? (under some assumptions on $X$ or the base field $k$, if necessary) If not, is there any counter-example?



Is $K^0(X)\to K_0(X)$ monomorphic for a noetherian scheme $X$?
No, this is not always a monomorphism. For the underlying reduced scheme $X\_{\text{red}}$ of $X$, the pushforward homomorphism 
$$
K\_0(X\_{\text{red}})\to K\_0(X)
$$ 
is an isomorphism (via devissage). If you read Manin's "Lectures on the K-functor", you will see that the natural map 
$$
\text{Pic}(X) \to K^0(X)
$$ 
is an injection. Yet the composition to $K\_0(X)$ factors through the pullback homomorphism $$
\text{Pic}(X)\to \text{Pic}(X\_{\text{red}}),
$$ which may easily have a kernel, cf. Hartshorne, Chapter III, Exercise 4.6.


