Let $M$ be the mod-$p$ Moore spectrum where $p \geq 3$ is a (power of) a prime. Then $M$ satisfies the "polynomial equation" $M \wedge M \cong M \oplus \Sigma M$. Is this a general phenomenon, or is it something very special about Moore spectra?


More formally, let $K = K\_0^\oplus(\mathbb S\_{(p)})$ be the direct-sum $K$-theory of the $p$-local sphere spectrum, i.e. the free commutative ring on equivalence classes of finite $p$-local spectra, mod the equations $[0] = 0$, $[A] + [B] = [A \oplus B]$, $[\mathbb S] = 1$, and $[A][B] = [A \wedge B]$. Then $K$ is an algebra over the ring $\mathbb Z[\Sigma,\Sigma^{-1}]$, where the action of $\Sigma$ is given by suspension. The above observation says that when $M$ is a Moore spectrum [1], the element $[M] \in K$ is integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$.


**Question 1:** What are some other examples of integral elements of $K = K^\oplus\_0(\mathbb S\_{(p)})$, considered as a $\mathbb Z[\Sigma,\Sigma^{-1}]$-algebra?


**Question 2:** Is every element of $K$ integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$? If not, is it perhaps the case that every finite type $n$, $p$-local spectrum is integral when $p$ is large compared to $n$?


**Question 3:** Are there other interesting "polynomial equations" involving not-necessarily-finite spectra, or not-necessarily-finite power series?


[1] When $p=2$, I believe I've read that the mod-2 Moore spectrum $M$ satisfies a polynomial equation involving powers $\leq 3$, but also the coefficent $\mathbb C \mathbb P^2$. This "reduces" the question of whether $M$ is integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$ to the question of whether $\mathbb C \mathbb P^2$ is integral, but I'm not sure this is an improvement. I think I'd tend to suspect that neither of these spectra are integral over $\mathbb Z[\Sigma,\Sigma^{-1}]$.




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* Via rather indirect means, I've convinced myself that if $A$ is a finite spectrum of type $\geq 1$, then any polynomial $f(X)$ satisfied by $A$ must lie in the ideal $(X,\Sigma-1)$. Perhaps there is a direct way to see this.
* Note that if we were working with exact-sequence $K$-theory, then the action of $\Sigma$ would be by $-1$, thanks to exact sequences of the form $X \to 0 \to \Sigma X$. But in direct sum $K$-theory, there's no reason for this to be the case.



Solving polynomial equations in spectra?
Here is a simple argument that would show many finite complexes can not be `integral' in your sense.


If $Sq^{2^k}$ acts nontrivially on $H^\*(X;\mathbb Z/2)$ then $Sq^{2^{k+1}}$ will act nontrivially on $H^\*(X \wedge X;\mathbb Z/2)$. But if $X$ were integral then there would be an upper bound on $k$ such that $Sq^{2^k}$ acts nontrivially on $H^\*(X^{\wedge d};\mathbb Z/2)$ for some $d$. Thus if $X$ were integral, all nontrivial Steenrod operations would have to vanish on its mod 2 cohomology.


A similar argument would work at odd primes, using the operations $\mathcal P^{p^k}$. And this answers your question 2 in the negative, as, for all primes $p$, and all $n$, there are certainly type $n$ complexes at $p$ with nontrivial Steenrod operations acting on their cohomology.


