# Does trivial cohomology of spectra imply trivial homology?

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.

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