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Let $bo$ be the connective cover of the real $K$-theory spectrum $KO$. This is a ring spectrum, and so one can look at its Adams spectral sequence. Mahowald does this in "$bo$-resolutions", and starts working from the $E_1$-page instead of passing to the $E_2$-page, presumably because $bo$ is not a flat ring spectrum. Why is this the case? Mahowald explicitly computes the cooperation algebra $\pi_*(bo \otimes bo)$ to show a lack of flatness, but even before this he seems to understand that $bo$ isn't flat: in the introduction of the paper, he phrases everything in terms of the $E_1$-page. Is there any intuition for why $bo$ should not be flat beyond looking at the cooperations?

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    $\begingroup$ Is there any reason for expecting a random ring spectrum to be flat? $\endgroup$
    – Denis T
    Commented Oct 2 at 17:55

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Let me me write $ko$ for connective $KO$, since I'm more used to that. If $ko_*ko$ were a flat $ko_*$-module, then by basechange to $H\mathbb{Z}$, $H\mathbb{Z}_*ko$ would be a flat $\mathbb{Z}$-module. But there is a ton of torsion, for example witnessed by the fact that the Bockstein $\mathrm{Sq}^1$ acts nontrivially on the mod $2$ homology of $ko$.

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    $\begingroup$ (but I agree with Denis T's comment in that if you're looking for intuition, it should be "flatness is a rare and special phenomenon") $\endgroup$
    – Achim Krause
    Commented Oct 2 at 18:13

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