More specifically, I was wondering if there are well-known conditions to put on $X$ in order to make $K\_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?



What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes?
Imposing that you can resolve by a length $2$ sequence of vector bundles is too strong. What you want is that there is some $N$ so that you can resolve by a length $N$ sequence of vector bundles. By Hilbert's syzygy theorem, this follows from requiring that the scheme be regular. (Specifically, if the scheme is regular of dimension $d$, then every coherent sheaf has a resolution by projectives of length $d+1$.)


Here is a simple example of what goes wrong on singular schemes. Let $X = \mathrm{Spec} \ A$ where $A$ is the ring $k[x,y,z]/(xz-y^2)$. Let $k$ be the $A$-module on which $x$, $y$ and $z$ act by $0$. I claim that $k$ has no finite free resolution. I will actually only show that $A$ has no *graded* finite free resolution. Proof: The hilbert series of $A$ is $(1-t^2)/(1-t)^3 = (1+t)/(1-t)^2$. So every graded free $A$-module has hilbert series of the form $p(t) (1+t)/(1-t)^2$ for some polynomial $p$; and the hilbert series of anything which has a finite resolution by such modules also has hilbert series of the form $p(t) (1+t)/(1-t)^2$. In particular, it must vanish at $t=-1$. But $k$ has hilbert series $1$, which does not.


There is, of course, a resolution of $k$ which is not finite. If I am not mistaken, it looks like


$$\cdots \to a[-4]^4 \to A[-3]^4 \to A[-2]^4 \to A[-1]^3 \to A \to k$$


