In Rudyak's 'On Thom Spectra, Orientability, and Cobordism', the following fact is used:

Let $\mathbb{Z}_{(p)}$ be $\mathbb{Z}$ localized at the ideal $(p)$. Let $\pi,\tau$ be two cyclic $\mathbb{Z}_{(p)}$-modules, that is, $\pi,\tau=\mathbb{Z}/p^m$ or $\mathbb{Z}_{(p)}$. Then the stable cohomology groups of the Eilenberg Maclane spaces $\varinjlim H^{i+N}(K(\pi,N);\tau)$ have exponent $p$, i.e. $px=0$ for every element $x$, for $i>0$.

Rudyak says this fact is well-known and references Cartan's Seminaire in 1959. But I have skimmed through the articles in both Seminaire 1958-1959 and Seminaire 1959-1960, yet have not found this fact proved anywhere. Now it is perfectly possible that I have simply overlooked a proposition, since I can barely read French, but I would like to ask if anyone can provide a more specific reference that proves the above fact?

Alternatively, it would be appreciated if anyone can provide a sketch of a proof. Since Rudyak references Cartan Seminaire, it is likely that a proof using Serre's spectral sequence, which is heavily used in the Seminaire, can be constructed. Serre's mod $\mathcal{C}$ theory does not apply here since the groups with exponent $p$ do not form a Serre class. I believe a proof for the case $\pi=\mathbb{Z}/p$ would require a close inspection on how the order of the group dies down. And I don't have any idea how to start with $\pi=\mathbb{Z}_{(p)}$.

Any help is appreciated!

up vote 9 down vote accepted

Here is a sketch proof.

Step 1: For sensible spaces or spectra (connected, finite type) $X$, $H^*(X;\tau) $ will have exponent $p$ for all the coefficient groups $\tau$ you list exactly when the image of the Bockstein $\beta: H^*(X;\mathbb Z/p) \rightarrow H^{*+1}(X;\mathbb Z/p)$ equals the kernel of $\beta$. This can be proved by fooling around with the various coefficient sequences defining the Bockstein. Bill Browder developed this idea into the `Bockstein spectral sequence', which he made good use of.

Step 2: So now we need to look at $H^*(H\pi;\mathbb Z/p)$, viewed graded vector space with the Bockstein action, and we want to show that this module is `$\beta$--acylic'. The Kunneth theorem applies, and we can reduce to the cases when $\pi = \mathbb Z$ or $\pi=\mathbb Z/p^m$.

Now we use the calculations of Serre, et. al. First of all, $H^*(H\mathbb Z/p;\mathbb Z/p) = A$, the mod $p$ Steenrod algebra. When written with Serre's admissible basis, one can see that $A$ is $\beta$--acyclic. (In fact, $A$ is forced by general Hopf algebra theory to be free as a module over the exterior algebra on $\beta$.) Then $H^*(H\mathbb Z;\mathbb Z/p) = A/A\beta$, which one checks is also $\beta$--acyclic ($\beta$ acts on the left). Finally, if $m>1$, then $H^*(H\mathbb Z/p^m;\mathbb Z/p) = A/A\beta \oplus \Sigma A/A\beta$, and so is also $\beta$--acyclic.

So this is what Rudyak meant.

The canonical mod 2 reference is Serre, Jean-Pierre Groupes d'homotopie et classes de groupes abéliens. (French) Ann. of Math. (2) 58, (1953). 258–294. Odd primes would have been later, but not by much.

A little bit more added later $\dots$

The cofibration sequence $H\mathbb Z \rightarrow H\mathbb Z \rightarrow H\mathbb Z/p$ induces a short exact sequence $0 \rightarrow \Sigma H^*(H\mathbb Z; \mathbb Z/p) \rightarrow A \rightarrow H^*(H\mathbb Z; \mathbb Z/p) \rightarrow 0$. It is not hard to see that the short exact sequence $0 \rightarrow im \beta \rightarrow A \rightarrow coker \beta \rightarrow 0$ maps to this (with the identity in the middle slot). The fact that $A$ is $\beta$--acyclic ($im \beta = ker \beta$) implies that $im \beta$ and $coker \beta$ are the `same size', as graded vector spaces (up to a suspension). This forces the map between the short exact sequences to be an isomorphism, and so one has $H^*(H\mathbb Z; \mathbb Z/p) = coker \beta = A/A\beta$.

  • Great answer! Just one thing: can you please explain how we get $H^*(H \mathbb{Z};\mathbb{Z}/p)$ and $H^*(H \mathbb{Z}/p^m;\mathbb{Z}/p)$ from $H^*(H \mathbb{Z}/p;\mathbb{Z}/p)=A$? I tried using a cofibre sequence like $H\mathbb{Z} \to H\mathbb{Z} \to H\mathbb{Z}/p$ but I couldn't get it to work. – Tsang Mar 16 at 7:09
  • 1
    @Tsang You can find computations of this kind in Adams' blue book (Generalized cohomology and stable homotopy). – Denis Nardin Mar 16 at 8:12

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