Let $\newcommand{\Sp}{\mathrm{Sp}}\Sp$ denote the $\infty$-category of spectra and $\newcommand{\Z}{\mathbb{Z}} H\Z$ the Eilenberg-Mac Lane spectrum of the integers. Given any spectrum $X \in \Sp$, I am interested in understanding its Bousfield localisation $\newcommand{\LZ}{L_{H\Z}}l\colon X \to \LZ X$ at $H\Z$, i.e. the unique (up to equivalence) pair $(l, \LZ X)$ such that $\LZ X$ is $H\Z$-local and $l$ is an $H\Z$-equivalence.
$H\Z$ seems to not fall into any of the classes commonly discussed when discussing Bousfield localisation (rationalisation, localisation at or away from a prime, $p$-completion / localisation at $\mathbb{S} / p$), all of which can be concisely described, so I'm wondering if one can give a concrete expression to $\LZ X$.
My understanding of localisations is fairly basic and I don't have a good way to tackle this question, but I can at least offer a few basic observations towards this goal:
- Any spectrum whose homotopy groups are bounded below in dimension is already $H\Z$-local, so that it is its own $H\Z$-localisation; this can be proved by induction over the Whitehead tower, see e.g. Bousfield's The Localization of Spectra with Respect to Homology, Lemma 3.3.
- The extension of this fact to general spectra is false: Take for instance $\newcommand{\KU}{\mathrm{KU}} \KU / n$ for some $n > 1$. This is $H\Z$-acyclic by virtue of $H_*(\KU; \Z)$ being rational, so its localisation is equivalent to the zero spectrum, but $\KU / n$ is not contractible (consider its homotopy groups).
- This rules out that $\LZ$ is given by smashing with a fixed spectrum $X_{H\Z}$ since it shows that $\LZ$ does not in general commute with colimits (every spectrum can be written as a colimit of spectra whose homotopy groups are bounded below).
- I've also tried ruling out that $\LZ$ is induced from a map $Y \to \mathbb{S}$ of spectra via $$ X \simeq X^{\mathbb{S}} \to X^Y = \LZ X $$ (I'm using superscripts to denote mapping spectra) but didn't manage to do so. The best I could do is to show that $\KU^Y \simeq L_{H\mathbb{Q}} \KU$: To see this, note that we have fibre sequences $$ (\KU / n[-1])^Y \to \KU^Y \overset{\cdot n}{\to} \KU^Y $$ for all $n > 1$, using that fibres are cofibres up to a shift, that mapping spectra commute with limits in the first variable, and that the multiplication by $n$-map on any spectrum $X$ agrees with the multiplication by $n$-map on any mapping spectrum $X^Z$. But $\KU / n[-1]$ is $H\Z$-acyclic, so the spectrum on the left is trivial, which is to say that all $n \in \mathbb{N}_{> 1}$ act invertibly on $\KU^Y$ and therefore that $\KU^Y$ is already rational, i.e. $\KU^Y \simeq L_{H\mathbb{Q}} \KU^Y$. Since the map $X \to \KU^Y$ is an $H\Z$-equivalence, it is also a $H\mathbb{Q}$-equivalence, but this is to say that $\KU^Y$ is already the $H\mathbb{Q}$-localization of $\KU$ and thus that $\KU^Y \simeq L_{H \mathbb{Q}} \KU \simeq \KU \otimes L_{H \mathbb{Q}} \mathbb{S}$.
Unfortunately I cannot see how to push this argument further. Certainly this implies that $\KU^*(Y) \cong \mathbb{Q}[u^\pm]$ for $u$ a class in degree 2, so one should be able to derive some information about $Y$ to hopefully be able to compare it to $X^Y$ for other spectra $X$, but I'm constrained by the fact that I don't understand $H_*(X)$ for spectra whose homotopy groups are not bounded below for pretty much any $X$ that aren't $\KU$ (or derived therefrom); all of this supposing that the hypothesis of point 4 is even true.
Thus, to repeat my question: Does the Bousfield localisation at $H\Z$ have a concrete expression (for spectra whose homotopy groups are not bounded below)?
Also, although I'll be perfectly happy to receive an answer to just this first question, I'd much appreciate if someone can tell me how to finish the incomplete argument above (or, alternatively, why it won't work).
