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Let $C$ be a compactly generated triangulated category. Can it contain a non-zero object $M$ such that there are no non-zero morphisms FROM $M$ into compact objects? I would be grateful for any example; can one obtain a certain "description" for these "left phantom" objects? For example, what happens in the unbounded derived category of a ring of infinite cohomological dimension?

Note that the subcategory of objects satisfying this conditions is triangulated and closed with respect to coproducts. I am willing to localize by this category (if it is non-zero; I would actually prefer to impose certain extra "orthogonality" conditions on the subcategory I kill). Yet I do not know whether the hom-classes in the quotient are sets. The localization functor should respect coproducts; yet it does not seems to respect the compactness of objects.

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This happens in spectra. By a theorem of Lin, there are no maps from the Eilenberg-MacLane spectra $H\mathbb{F}_p$ to finite spectra; the proof is an Adams spectral sequence computation using the fact that the Steenrod algebra is self-injective.

For a more algebraic example, let $A$ be ring containing an infinite regular sequence $(x_0,x_1,\dots)$ and let $M=A/(x_0,x_1,\dots)$. We can resolve $M$ by an infinite Koszul complex and compute that $\operatorname{Ext}^*(M,A)=0$. It follows that in the derived category of $A$, there are no maps from $M$ to compact objects.

As for getting some kind of control on these objects, I don't really know much, but I know Luke Wolcott has thought a lot about pathology in derived categories of non-Noetherian rings. You might try taking a look at his work and seeing if you can find anything useful.

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    $\begingroup$ Great answer. It made me wonder that this probably couldn't happen in a derived category of a Noetherian ring? $\endgroup$
    – Adam Przeździecki
    Commented Jun 18, 2015 at 18:34
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    $\begingroup$ @AdamPrzeździecki: Good question! I believe that's correct. By Neeman's classification of localizing subcategories, it suffices to show that $\operatorname{Ext}^*(k(p),A)\neq0$ for any residue field $k(p)$ at a prime $p$ of $A$. By comparison with a Koszul complex for a generating set of $p$, I think it can be shown that for some $n$, $\operatorname{Ext}^n(A/p,A)$ is a nontorsion $A/p$-module. When $p$ is maximal, we're done. When $p$ isn't maximal, I think you can then make a $\lim^1$ argument to show that $\operatorname{Ext}^{n+1}(k(p),A)$ cannot be finitely generated. $\endgroup$
    – Eric Wofsey
    Commented Jun 19, 2015 at 5:28

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