I would like to hear what's known about the homotopy type of smash products of mod-$p^j$ Moore spectra, for $p$ an odd prime.


First, here is what I'm specifically interested in: there is a short exact sequence $$0 \to \mathbb{Z} \xrightarrow{p^j} \mathbb{Z} \to \mathbb{Z}/p^j \to 0.$$ Tensoring this short exact sequence against your favorite group $G$ yields an exact sequence $$\cdots \to G \xrightarrow{p^j} G \to G \otimes \mathbb{Z}/p^j \to 0,$$ which exhibits $G \otimes \mathbb{Z}/p^j \cong G / p^j G$ as $G$ with its $p^j$-divisible part stripped out.


Moore spectra play a related role in homotopy theory: they are defined by the cofiber sequence $$S \xrightarrow{p^j} S \to M(p^j).$$ Smashing through with any spectrum $X$ gives the new cofiber sequence $$X \xrightarrow{p^j} X \to X \wedge M(p^j),$$ and chasing this around shows that the homotopy group $\pi\_n X \wedge M(p^j)$ is a mix of $\pi\_n X / p^j(\pi\_n X)$, as one would expect, together with the $p^j$-torsion of $\pi\_{n-1} X$, which is new and different. So, though $X \wedge M(p^j)$ is sometimes written $X / p^j$, and though this notation suggests a useful analogy, this isn't exactly true, and we have to be careful about things we expect to follow from the algebraic setting.


The specific algebraic fact I'm interested in is that the composition of the tensor functors $- \otimes \mathbb{Z}/p^j$ and $- \otimes \mathbb{Z}/p^i$ for $j > i$ has a reduction: $$- \otimes \mathbb{Z}/p^j \otimes \mathbb{Z}/p^i \cong - \otimes \mathbb{Z}/p^i.$$ The exact translation of this statement to Moore spectra and the smash product is not true --- one can, for instance, compute the reduced integral homology of $M(p^i) \wedge M(p^j)$ to see that there are too many cells around for it to be equivalent to $M(p^i)$ alone. However, the same homology calculation suggests something related: there is an abstract isomorphism between the reduced homology groups of $M(p^j) \wedge M(p^i)$ and those of $M(p^i) \wedge M(p^i)$. This is, of course, also true for groups; it is indeed the case that $\mathbb{Z}/p^i \otimes \mathbb{Z}/p^j \cong \mathbb{Z}/p^i \otimes \mathbb{Z}/p^i$. This is what I want to know:



> 
> For $j > i$, is $M(p^j) \wedge M(p^i)$ homotopy equivalent to $M(p^i) \wedge M(p^i)$?
> 
> 
> 


If this is not true, I'm willing to throw in some extra qualifiers. For instance, is the situation improved if we work $K(n)$-locally? Is it true only when $j \gg i$? What if we additionally restrict to $j \gg i \gg 0$?


This specific question aside, I am also interested in any and all known features of the homotopy type of $M(p^i) \wedge M(p^j)$ --- any favorite fact you have that would help me get a grip on them. I'm also specifically interested in variants of the above question for generalized Moore spectra: can anything similar be said about those?



Homotopy type of tensors of Moore spectra
For $p\neq 2$ the answer is yes. It's an easy computation usinghomology decompositions in the sense of Eckmann–Hilton the well known exact sequence


$$\operatorname{Ext}(A,\pi\_{n+1}(X))\hookrightarrow [M(A,n),X]\twoheadrightarrow \operatorname{Hom}(A,\pi\_n(X)),$$


and the computation


$$\pi\_{n+1}(M(A,n))=A\otimes\mathbb{Z}/2.$$


Here $M(A,n)$ is the Moore spectrum whose homology is the abelian group $A$ concentrated in degree $n$.


As you point out, it is easy to check that


$$H\_n(M(A,s)\wedge M(B,t))=
\begin{cases}
A\otimes B,&n=s+t,\\
\operatorname{Tor}\_1(A,B),&n=s+t+1,\\
0,&\text{otherwise}.
\end{cases}$$


Therefore, $M(A,s)\wedge M(B,t)$ can be obtained as the homotopy cofiber of a map
$$f\colon M(\operatorname{Tor}\_1(A,B),s+t)\longrightarrow M(A\otimes B,s+t)$$
which is trivial in homology $H \_{\*}(f)=0$.


Suppose for instance that $A$ and $B$ are finite and do not have $2$-torsion. Then the previous short exact sequence shows that homology induces an isomorphism


$$H \_{s+t}\colon [M(\operatorname{Tor}\_1(A,B),s+t), M(A\otimes B,s+t)]\cong \operatorname{Hom}(\operatorname{Tor}\_1(A,B),A\otimes B).$$


Therefore $f$ must be null-homotopic, so


$$M(A,s)\wedge M(B,t) \simeq
M(A\otimes B,s+t)\vee M(\operatorname{Tor}\_1(A,B),s+t+1).$$


If you take $A=\mathbb{Z}/p^i$ and either $B=\mathbb{Z}/p^j$ or $B=\mathbb{Z}/p^i$ you always obtain the same thing on the right.


For $p= 2$ the answer is no. The spectrum $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is the mapping cone of
$$4\colon M(\mathbb{Z}/2,0)\longrightarrow M(\mathbb{Z}/2,0)$$
which is knonw to be null-homotopic, actually $[M(\mathbb{Z}/2,0),M(\mathbb{Z}/2,0)]\cong\mathbb{Z}/4$ generated by the identity. Hence


$$M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)\simeq M(\mathbb{Z}/2,0)\vee M(\mathbb{Z}/2,1).$$


In particular the action of the Steenrod algebra on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is trivial.


On the other hand, it is known that the Steenrod algebra mode 2 acts on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ in a non-trivial way, i.e. the first Steenrod operation sends the 0-dimensional generator to the 1-dimensional generator. Therefore $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ and $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ cannot be homotopy equivalent.


As for references, see Hatcher's book. This book doesn't deal with spectra but since we are working with finite-dimensional CW-complexes you can assume that we are in the stable range.


