The Brown-Comenetz dualizing spectrum $I\_{\mathbf{Q/Z}}$ is not detected by very many spectra: it is $BP, \mathbf{Z}, \mathbf{F}\_2, X(n)$ for $n\geq 2$, and even $I\_{\mathbf{Q/Z}}$-acyclic. However, if $X$ is any nontrivial finite spectrum, then $X\wedge I\_{\mathbf{Q/Z}}$ is not contractible. This motivates a natural question: let $E$ be any spectrum such that $E\wedge I\_{\mathbf{Q/Z}}$ is not contractible. Then, is it true that $\langle E\rangle \geq \langle X\rangle$ for some nontrivial finite spectrum $X$?



Detecting the Brown-Comenetz dualizing spectrum
Your question appears to be equivalent to the 'dichotomy conjecture' of Hovey, which I believe is still open. 


First, note that any finite spectrum has a type, and all finite spectrum of type $n$ have the same Bousfield class, usually denoted $F(n)$. In Hovey and Strickland's memoir (Appendix B) they conjecture that if $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$. This is precisely your question. 


In his paper on the chromatic splitting conjecture, Hovey conjectured that every spectrum has either a finite acyclic or a finite local (the dichotomy conjecture). With Palmieri (The structure of the Bousfield lattice) he proved that the following conjectures are equivalent:



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> > (1) If $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$.
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> > (2) The dichotomy conjecture.
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> > (3) If $E$ has no finite acyclics, then $\langle E \rangle \ge \langle I \rangle$. 
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The analog of the dichotomy conjecture is known to hold in other categories with a good theory of support satisfying the tensor product property - see 'The Bousfield lattice of a triangulated category and stratification' by Iyengar and Krause. 


