# Solving polynomial equations in spectra?

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.

• Integral over what subring? Jan 20 at 21:52
• @FernandoMuro over $\mathbb Z[\Sigma,\Sigma^{-1}]$. I'm sure there are other reasonable choices, but this seems like the "minimal reasonable one" Jan 20 at 21:53
• Sorry, I missed that. Jan 20 at 21:53
• 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? Jan 21 at 4:26
• 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. Jan 21 at 4:28

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 $$H^\ast(X)$$.

• Yep, I meant a Cartan formula argument. Jan 21 at 22:51
• Also, I was focused on using just the pth power operations as they generate all operations. Jan 22 at 16:15
• @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. Jan 22 at 16:43