# Have mod $p^k$ Dyer Lashof operations been studied?

Here is one of the motivations for my question, when $$p=2$$. The homology of the spectrum $$H\mathbb F_2$$ as an algebra is generated by the Dyer Lashof operations on the single generator $$\xi_1$$ (and it is enough to consider the $$E_2$$ operation), and for $$H\mathbb Z/2^k$$ ($$k>1$$), we have to take more generators, i.e. $$x$$ the dual of the higher Bockstein, $$\xi_1^2$$ and $$\overline\xi_2$$. A reason for that is that $$x$$ vanishes under the Dyer Lashof operations, because the higher bockstein vanishes under the Steenrod operations.

Somehow I have the feeling that looking at the mod $$2$$ homology of the spectra $$H\mathbb Z/2^k$$ gives very degenerate information (it doesn't even depend on $$k$$), and that some statement like the very informal following one should be true:

$$H_*(H\mathbb Z/2^k;\mathbb Z/2^k)$$ is generated as an algebra over the mod $$2^k$$ Dyer Lashof operations, by a single generator $$\xi^{(k)}_1$$

Here $$\xi^{(k)}_1$$ would be dual to the Bockstein associated to the exact sequence $$\mathbb Z/2^k\to \mathbb Z/2^{k+1}\to\mathbb Z/2^k$$ in some sense.

Can all that make sense ? Is there such a thing as mod $$p^k$$ Dyer Lashof operations ? Of course there is such a thing but I am wondering to which extent this notion is relevant and has been studied.

• the whole Bockstein spectral sequence for the homology of free E_infty-algebras (and I think also free E_n-algebras) is determined in May's 'general algebraic approach'; so this includes the mod p^k homology of free algebras, hence all the operations. I don't know the answer to your other question though. (But, whereas the dual Steenrod algebra is actually a free E_2-algebra, the same is not true for the Z/p^k-dual Steenrod algebra) – Dylan Wilson Nov 24 '20 at 20:16
• Thanks for the reference. Do you have a reference for your remark between brackets ? – elidiot Dec 14 '20 at 10:44
• this is a consequence of theorem 1.5 here: arxiv.org/pdf/1707.00956.pdf – Dylan Wilson Dec 14 '20 at 19:25