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!


1 Answer 1


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 $H^*(H\mathbb Z; \mathbb Z/p) = coker \beta = A/A\beta$.

  • $\begingroup$ 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. $\endgroup$ Mar 16, 2018 at 7:09
  • 1
    $\begingroup$ @Tsang You can find computations of this kind in Adams' blue book (Generalized cohomology and stable homotopy). $\endgroup$ Mar 16, 2018 at 8:12
  • 4
    $\begingroup$ I am Rudyak. I only recently saw this discussion and regret not seeing it sooner. Anyway, it's nice. Special thanks to Nick Kuhn. $\endgroup$ Feb 29 at 21:52
  • $\begingroup$ In a comment in the original posting, @Yuli (which was converted to a comment, as it wasn't an answer to the question), Dave Benson said "Welcome to Math Overflow!" $\endgroup$
    – David Roberts
    Mar 1 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.