More specifically, I was wondering if there are wellknown 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 $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?
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 $k[x,y,z]/(xzy^2)$. 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 $(1t^2)/(1t)^3 = (1+t)/(1t)^2$. So every graded free $A$module has hilbert series of the form $p(t) (1+t)/(1t)^2$ 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 $p(t) (1+t)/(1t)^2$. 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
$$\cdots \to a[4]^4 \to A[3]^4 \to A[2]^4 \to A[1]^3 \to A \to k$$

1$\begingroup$ To finish and show that k can not have even have finite projective resolution, we can use the resolution in the last sentence to compute $Tor_i^A(k,k)$ (the resolution is periodic, so one only needs to do a few steps) and see that they are all nonzero. But if k has finite projective dimension one must have $Tor_i(k,k)=0$ for $i>>0$. $\endgroup$ – Hailong Dao Apr 22 '10 at 4:08

$\begingroup$ Oh, of course. The homological definition of regular of dimension $d$ is that $Ext^{d+1}$ vanishes which implies projective dimension less than or equal to $d+1$. I'm just learning this stuff, so I wanted a resolution of length 2 because from that it was clear where to send the sheaf. I'll need to think what the map is in this case. Probably just an alternating sum of the vector bundle associated to the free sheaf in each part of the resolution? I still feel like regular is strange. $\endgroup$ – Matt Apr 22 '10 at 4:14

$\begingroup$ @hilbertthm90: That's right, one takes alternating sum. We probably can't avoid regular, because finite projective dimension of the residue field characterize regular local rings. $\endgroup$ – Hailong Dao Apr 22 '10 at 4:26
You want coherent sheaves to have finite global resolutions by locally free sheaves. So definitely you need the regularity of $X$ to ensure that a locally free resolution stops at a finite stage. You also need a global condition such as quasiprojectivity over an affine base to guarantee that you can start the process. (The last condition is not optimal.)
Edit: In reading the follow up comments, I realize my answer was a bit cryptic. The inverse map $K_0(X)\to K^0(X)$ would send the class of a coherent sheaf to the alternating sum of the classes in a resolution. In general, these groups behave quite differently. $K^0(X)$ is contravariant like cohomology and $K_0(X)$ is covariant for proper maps like (BorelMoore) homology. That they coincide for regular schemes is reminiscent of Poincaré duality.

$\begingroup$ Faltings proves that, for a noetherian regular scheme $X$ which is separated over $\Z$, the natural morphism $K^0(X)\longrightarrow K_0(X)$ is an isomorphism on page 1 of his book "Lectures on the Arithmetic RiemannRoch theorem", I believe $\endgroup$ – Ariyan Javanpeykar Apr 22 '10 at 7:40
Asking $K^0(X)$ to be isomorphic to $K_0(X)$ is not always "good enough". Of course, it will allow you to carry over constructions for $K_0(X)$ to $K^0(X)$, but not canonically. And it can happen that $K^0(X)\cong K_0(X)$ without $X$ being regular. For example, take $X= \textrm{Spec} A$, $A=k[x]/(x^n)$ with $n\geq 2$. Then you have an infinite resolution as given in David's answer for $k$. Computing $Tor^A_i(k,k)$ shows that $k$ has no finite resolution. (In fact, $Tor_i^A(k,k) = k^2$ for all $i>0$.) Now, although the above "existence of finite resolution" fails, it is not hard to see that $K^0(X)\cong \mathbf{Z}\cong K_0(X)$ in this case. (Use that $A$ is a local ring and the length map on $A$.) Of course, the natural map $K^0(X) \longrightarrow K_0(X)$ is not an isomorphism. (It is given by $1\mapsto n$.)
[Edit: I added another example]
[Edit 2: There was something wrong with the example below as noted by Michael. I fixed the problem]
Let me also add to my answer the following "snake in the grass". If you work with general schemes, even if regular, one requires the extra assumption of "finitedimensionality". For example, take the scheme $X=\textrm{Spec} (k \times k[t_1]\times k[t_1,t_2] \times \ldots)$. Now, even though $A = k\times k[t_1]\times\ldots$ is regular, there is an infinite resolution for $k$ of the form $$\ldots \longrightarrow A\longrightarrow A\longrightarrow A \longrightarrow k \longrightarrow 0$$ which corresponds geometrically to taking a point, then adding a line, then adding a plane, etc. Again, take the Tor's to see that $k$ has no finite resolution. Do note that $X$ is not noetherian.
[Edit 3: I added the following for completeness]
Let $X$ be a regular finitedimensional scheme. Assume that $X$ has enough locally frees. (This notion also arose in Are schemes that "have enough locally frees" necessarily separated ). Then the canonical morphism $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. In the second example, $X=\textrm{Spec} \ A$ is regular, but not finitedimensional. Does $X$ have enough locally frees?

$\begingroup$ Something is slightly wrong here  $X$ is not quasicompact (as you say), but all affine schemes $\mathrm{Spec}(A)$ are quasicompact, so $X \ne \mathrm{Spec}(A)$. $\endgroup$ – Michael Sep 14 '11 at 6:56

$\begingroup$ Thank you for your comment. I edited my answer accordingly. $\endgroup$ – Ariyan Javanpeykar Sep 14 '11 at 20:38