I was told: if $X$ is bdd below and $p$-complete spectra then $X^{tC\_q}$ vanishes for primes $q \not= p$.


I do not see how this holds.




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I am aware from [I.2.9](https://arxiv.org/abs/1707.01799) that if $X$ is bdd. below, then $X^{tC\_q} \simeq \left(X^{\hat{}}\_q \right)^{tC\_P}$.


Here $\hat{}\_p$ denotes the $p$-completion of $X$. I believe my problem would be resolved if: we show $q$-completion of a $p$-complete spectra is trivial - is this true?



Vanishing tate of a $p$-complete spectra
That is how you prove this.


Recall that $(-)^\wedge\_p \simeq L\_{\mathbb{S}/p}(-)$, where $L\_E(-)$ is the Bousfield localization with respect to $E$. Now, multiplication by $q$ mod $p$ is an equivalence, so multiplication by $q$ becomes a weak equivalence on $p$-complete spectra. Then $\mathbb{S}/q\wedge X\simeq X/q\simeq 0$ for $p$-complete spectra, and so $X^\wedge\_q\simeq 0$.


