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Let $HR$ be the Eilenberg-Maclane spectrum for a commutative ring $R$ and $M$ be a module over $HR.$ Then I want to prove that $M$ is a product of Eilenberg-Mac Lane spectra.

Construction: Let $\pi_k(M)$ be generated by a set $F_k$ for each $k \geq 0.$ Then we have a map

$\vee_{k \geq 0} \vee_{a \in F_k} HR_a \to M$.

Out of this, I need to find a structure of $M.$

Any suggestion will be appreciated.

Thank you in advance.

Ben Steffan
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Surojit
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  • For product of E-ML spectra you mean categorical product or smash product? – N.B. Oct 31 '18 at 10:18
  • I have edited the post. – Surojit Oct 31 '18 at 11:22
  • You still say you want to prove that a generic module $M$ is a product of copies of $HR$. But then in the formula you wrote coproducts. Are you trying to show that every module for any E-ML spectrum is free? – N.B. Oct 31 '18 at 11:27
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    In spectra, finite products and coproducts coincide. But module spectra over Eilenberg-Mac Lane spectra need not be free: e.g., $H\mathbb{Z}/2$ is an $H\mathbb{Z}$-module, but isn't a product of $H\mathbb{Z}$'s. This is only true if $R$ is a field. – JHF Nov 01 '18 at 15:21

1 Answers1

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What you're trying to show is false in general.

Indeed, if you're happy to rely on heavy machinery then the category of modules over the $\mathbb{E}_\infty$-ring $HR$ is equivalent to the derived category $D(R)$ of $R$, and your question thus becomes whether every chain complex over $R$ is formal. This is true iff $R$ is hereditary and false otherwise, see this answer. Even if $R$ is hereditary there is of course no reason to expect that an $HR$-module split as a direct sum of suspensions of $HR$, as JHF has already pointed out in the comments. This strong requirement only holds when $R$ is a field.

Ben Steffan
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