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So, I believe that, given a ring spectrum $R$ and an $R$-module $A$, we say that $A$ is "free" if $A\simeq \vee_IR$, i.e. some indexed wedge of copies of $R$. Now, as far as I understand, we can consider a ring spectrum $F$ a "field" if every $F$-module is free (do we need the wedge to be finite?).

I believe that a large class of examples of such fields are the Morava K-theories for a given prime, $K(n)$. Are the Morava $K$-theories the only "fields"? Specifically, do the Morava K-theories contain every Eilenberg-MacLane spectrum $\mathbb{H}F$ for $F$ a field?

If not, is there some other general classification of the fields in brave new algebra?

Many thanks to you!

Edit: One other thought. Is there a correspondence between $K(n)$ at a prime $p$ and $\mathbb{F}_{p^n}$? This is a total guess and completely unsubstantiated, but it would make sense I guess....

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$H\mathbb R$ is a field but not a Morava k-theory. Moreover, The coefficient ring of Eilenberg-MacLane spaces is concentrated in degree zero whereas the coefficient ring of Morava K-theories $K(n)$ is not even bounded below if $n\neq 0$ –  user17786 Dec 17 '11 at 1:21
Thankyou! Those are good counterexamples. –  Jon Beardsley Dec 17 '11 at 22:21

1 Answer 1

up vote 4 down vote accepted

The commented reply is essentially the only counterexample. In fact, if you set $K(0) = H\mathbb{Q}$ and $K(\infty) = H\mathbb{F}_p$ (which is quite reasonable if you view the $K(n)$ as built from the $E(n)$ and $P(n)$ spectra), then you get the following:

[Hopkins-Smith, Nilpotence in Stable Homotopy Theory II, Prop. 1.9]: For $E$ a field (in your sense), there is a prime $p$ and $n \in [0, \infty]$ for which $E$ is a wedge of suspensions of $K(n)$.

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