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It is true for the unbounded derived category of a noetherian ring: the minimal Bousfield classes (which are not all of $D(R)$) are $\langle k(\mathfrak{p})\rangle$ and $$0 = \langle \coprod_{\mathfrak{p}} k(\mathfrak{p}) \rangle$$ as tensoring with the residue fields detects whether an object is non-zero. I am fairly sure it is not known what conditions would suffice in general; as Fernando points out this is probably a difficult issue.
Sorry for the dumb question but I am confused - why is this ring zero dimensional? For $i\in \mathbb{N}$ aren't the ideals $(x_j\vert j\neq i)$ prime with quotient $\mathbb{Q}[x_i]$? So this ring would have dimension 1... Even if I made a silly mistake here I don't see how it could be von Neumann regular.
@Martin: I would say a thick subcategory is one which is closed under summands and has the two out of three property for short exact sequences rather than being the same as a Serre subcategory.
cont'd: - $i_*(O_Z)$ is not actually an object of $K_{\mathrm{ac}}(\mathrm{Inj}\;X)$ but there is a functor from the derived category making this homotopy category an infinite completion of the singularity category which one can apply. The reference for this is Krause's paper "The Stable Derived Category of a Noetherian Scheme". Also, thanks Daniel and Kevin - I'll try to remember to edit in a link when this stuff is ready (which will be soon).
