4
$\begingroup$

In Nilpotence in Stable Homotopy Theory II, Hopkins–Smith define a field spectrum to be a ring spectrum $E$ such that $E_*X$ is a free $E_*$-module for all spectra $X$. They then show that any field spectrum $E$ has the homotopy type of a wedge of suspensions of a Morava K-theory $K(p,n)$ for some $n\in\mathbb{N}\cup\{\infty\}$ and some prime $p$. In this sense, the Morava K-theories are the prime fields in stable homotopy theory.

This is very different from the first definition of fields we usually see in commutative algebra, i.e. that $k$ is a field if $k^\times=k\setminus\{0\}$, however.

Question. Is it possible to define field spectra by means of some kind of invertibility condition, similar to the classical definition of fields?

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ It's a bit odd to cite "Maximilien-Shipley" : one of them is a first name and the other one a last name. Also note that $\mathbb Z$ (and friends) is not the free ring on one element, but the free ring on $0$ element (i.e. the initial ring). I don't know about your questions - note that unlike in the Set-context, not every pointed space is of the form $X_+$, so not every pointed space admits a canonical $E_\infty$-comonoid structure - in particular $\Omega^\infty E$ is never of this form, except for discrete $E$ (because all components are equivalent) $\endgroup$
    – Maxime Ramzi
    Commented Aug 11, 2021 at 22:08
  • $\begingroup$ @MaximeRamzi Thanks! I corrected the two points you mentioned. Also, I wonder if the problem with $\Omega^{\infty}E$ would go away if we relaxed the requirement for it to be an $\mathbb{E}_\infty$-Hopf algebra to it being isomorphic (in some appropriate weak sense) to some $\mathbb{E}_\infty$-monoid in $\mathcal{S}_*$, which then is an $\mathbb{E}_{\infty}$-Hopf algebra there? $\endgroup$
    – Emily
    Commented Aug 11, 2021 at 22:16
  • 1
    $\begingroup$ I don't get what you mean by that, but let me say it in a different way : if the Péroux-Shipley result is indeed true for $S_*$, then all $E_\infty$-comonoids are of the form $X_+$ - this is never the case for $E_\infty$-monoids of the form $\Omega^\infty E$ - in particular your "to some $E_\infty$-monoid, which is then a Hopf algebra" doesn't make sense, because there is no such "then" $\endgroup$
    – Maxime Ramzi
    Commented Aug 11, 2021 at 22:24
  • 2
    $\begingroup$ if you fix it - the underlying $E_\infty$ space only depends on the connective cover. Connective ring spectra are only field spectra if they are ordinary fields, so whatever you define in terms of underlying $E_\infty$ spaces won't be related to field spectra. $\endgroup$
    – Achim Krause
    Commented Aug 12, 2021 at 22:47
  • 1
    $\begingroup$ In the cited result (saying that any field spectrum is a $K(p,n)$-module for some $p,n$), one must allow $n = \infty$ and not just $n \in \mathbb N$ to include the case of $H \mathbb F_p$. Regarding the question, I would probably start first by considering the question in $D(\mathbb Z)$ -- the derived category of abelian groups -- rather than in Spectra. Already there, it seems unclear to me how to characterize "differential graded fields" via "invertibility conditions", but if you work out such a characterization, you could then try to generalize it to spectra. $\endgroup$
    – Tim Campion
    Commented Aug 23, 2021 at 16:18

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .