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Let $n=p^k$ be a prime power.

When $k=1$, the algebra of stable operations in mod $p$ cohomology is the Steenrod algebra $\mathcal{A}_p$. It has a nice description in terms of generators and relations. Its dual (as a Hopf algebra) is also well understood by work of Milnor.

What about when $k>1$? Has any work been done on trying to understand the algebra of stable operations in mod $n$ cohomology? For instance, is there a description of $\mathcal{A}_4$ in terms of generators and relations? What about the dual Hopf algebra?

Of course I am just asking about the cohomology ring $H^\ast (H\mathbb{Z}/n;\mathbb{Z}/n)$, where $H\mathbb{Z}/n$ denotes the mod $n$ Eilenberg--Mac Lane spectrum. So I fully expect the answer to be either one of "yes, this is well-known and classical" or "no, this is known to be a big mess".

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This is not directly related, but it is not hard to compute the Steenrod algebra corresponding a finite field. I am guessing you are not interested in that though. –  Sean Tilson Sep 7 '12 at 14:30
@Sean: I am perhaps not so interested in the answer for finite fields, but I would be interested to know why it is easier (presumably issues with the Kunneth Theorems and duality)? –  Mark Grant Sep 7 '12 at 16:13
If I remember correctly, this can be calculated because the mod-$p$ Steenrod algebra has a relatively simple structure when considered as a bimodule over the subalgebra generated by the Bockstein. (I'm less sure at odd primes.) For $p=2$: in degrees zero and one you get $\mathbb{Z}/2^k$, generated by the Bockstein; I believe that in higher degrees it's abstractly isomorphic to the Steenrod algebra as a module except that admissible monomials starting with $Sq^1$ start with the Bockstein $\beta$ and admissible monomials ending with $Sq^r$, $r$ odd, end with $\beta Sq^{r-1}$. A lot of the mul –  Tyler Lawson Sep 7 '12 at 17:04
Err, a lot of the multiplications become zero. –  Tyler Lawson Sep 7 '12 at 17:04
@Tyler: This sounds promising. Was this a self-made calculation? If not, is there any hope that you could remember where you are remembering it from? Thanks. –  Mark Grant Sep 9 '12 at 7:34
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