It is known that for any spectrum $X$, $H\mathbb{Z}^*(X)=0$ implies that $H\mathbb{Z} \wedge X =0$.

Also, for the case $HF_p,$ if we consider $ HF_p^*(X) =0.$ This gives $[HF _ p \wedge X , \sum^i HF _ p] _{ HF_p−module}=0$ for all i. So, we get $HF_p \wedge X =0$ , since $HF _p \wedge X$ is an $HF_p$-module therefore we can take $HF _p \wedge X \cong \bigvee_{j \in J} \sum^j HF _ p $, then $ [HF _p \wedge X, \sum^i HF _ p]=[ \bigvee_{j \in J} \sum^j HF _ p , \sum^ iHF _ p]= \bigoplus_{j \in J} [ \sum^ j HF _ p, \sum^i HF_p] =0$ for all i. So, if $ HF _p \wedge X$ is non zero then considering $ i=j $ for some $ j $ gives a contradiction.( This is due to a discussion with Eric Wolfsey)

Also, this thing holds for the spectrum $HR$ when $R$ is a field.

Can we expect a similar thing for $HR,$ where $R$ is any subring of rationals? More generally, Can we classify ring spectrum $E$ for which $E^*(X)=0$ implies $E \wedge X =0$ for all spectra $X?$

Thank you so much in advance.

  • $\begingroup$ See the comments on Eric Wofsey's answer to this question, for the aforementioned discussion. $\endgroup$ Mar 19 '19 at 16:03

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