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 $\mathbb Z[\Sigma,\Sigma^{-1}]$.

  • 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.

  • $\begingroup$ Integral over what subring? $\endgroup$ Jan 20, 2021 at 21:52
  • $\begingroup$ @FernandoMuro over $\mathbb Z[\Sigma,\Sigma^{-1}]$. I'm sure there are other reasonable choices, but this seems like the "minimal reasonable one" $\endgroup$
    – Tim Campion
    Jan 20, 2021 at 21:53
  • $\begingroup$ Sorry, I missed that. $\endgroup$ Jan 20, 2021 at 21:53
  • 4
    $\begingroup$ Doesn't the Brown-Comenetz dual of the sphere, $I$, have the property that $I\wedge I\simeq 0$? So I guess it's integral? $\endgroup$ Jan 21, 2021 at 4:26
  • 1
    $\begingroup$ The same statement holds for any spectrum $E$ such that $\langle E\rangle < \langle H\mathbb{F}_p\rangle$ (these are the Bousfield classes) I believe. So you have a whole slew of spectra that satisfy the equation $x^2=0$, basically by taking the Brown-Comenetz dual of any connective spectrum with finitely generated homotopy groups. $\endgroup$ Jan 21, 2021 at 4:28

2 Answers 2


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.


Here's a slightly fleshed-out version of Nicholas Kuhn's argument. I wasn't able to verify the exact statement he uses, but a variant thereof.

For a spectrum $X$, let $f(X)$ be the maximal $k \in \mathbb N$ such that $Sq^k$ acts nontrivially on $H^\ast(X)$ (note that we're not using $Sq^{2^k}$). If $X$ is finite, then $f(X) < \infty$. Then clearly

  • $f(X \oplus Y) = \max(f(X),f(Y))$

  • $f(\Sigma X) = f(X)$

Let us show that:

  • $f(X_1\wedge X_2) = f(X_1) + f(X_2)$

By the Cartan formula, $Sq^k(x_1 \otimes x_2) = \sum_{k_1 + k_2 = k} S^{k_1}(x_1) \otimes Sq^{k_2}(x_2)$. If $k > f(X_1) + f(X_2)$, then in each summand, at least one tensor factor has $Sq^{k_i}$ acting for $k_i > f(X_i)$, so $Sq^k$ vanishes. Thus $f(X_1 \wedge X_2) \leq f(X_1)+f(X_2)$. If $k = f(X_1) + f(X_2)$, then every summand has a tensor factor for which $Sq^{k_i}$ acts with $k_i > f(X_i)$ except for $k_i = f(X_i)$, so $Sq^k = Sq^{f(X_1)} \otimes Sq^{f(X_2)}$. So if $x_i \in H^\ast(X_i)$ are such that $Sq^{f(X_i)}(x_i) \neq 0$, then $Sq^k(x_1 \otimes x_2) \neq 0$. Thus $f(X_1 \wedge X_2) \geq f(X_1) + f(X_2)$.

Now suppose that $X$ is integral, i.e. $X^n \cong \oplus_{i=0}^{n-1}\oplus_j a_{ij} \Sigma^j X^i$. Then by the above three formulas, we have $n f(X) = f(X^n) = f(\oplus_{i=0}^{n-1} \oplus_j a_{ij} \Sigma^j X^i) = \max_{i=0}^{n-1} i f(X) = (n-1) f(X)$. Therefore $f(X) = 0$, i.e. the Steenrod algebra acts trivially on $H^\ast(X)$.

  • $\begingroup$ Yep, I meant a Cartan formula argument. $\endgroup$ Jan 21, 2021 at 22:51
  • $\begingroup$ Also, I was focused on using just the pth power operations as they generate all operations. $\endgroup$ Jan 22, 2021 at 16:15
  • $\begingroup$ @NicholasKuhn I figured as much. But with just the assumption that $Sq^{2^k}(x) \neq 0$ (even assuming that $Sq^{2^{\geq k+1}}$ acts trivially on $H^\ast(X)$ as well), I couldn't get the Cartan argument to work to show that $Sq^{2^{k+1}}(x \otimes x) \neq 0$. For instance, it might be the case that $Sq^{2^k + 1}(x), Sq^{2^k - 1}(x) \neq 0$ as well, and then there could be cancellation in the Cartan formula. I tried showing that, for $l \geq 1$, $Sq^{2^k + l}$ is in the 2-sided ideal of $\mathcal A^\ast$ generated by $Sq^{2^{k+1}},Sq^{2^{k+2}},\dots$, but I got lost in Adem relations. $\endgroup$
    – Tim Campion
    Jan 22, 2021 at 16:43

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