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Let $\mathcal C$ be a symmetric monoidal stable category such that the thick subcategory generated by the unit is all of $\mathcal C$ -- in particular, every object is dualizable (I'm particularly interested in the case where $\mathcal C = Spt_{T(h)}\langle S \rangle$ is the thick subcategory of $Spt_{T(h)}$ generated by the sphere). Let $Pic(\mathcal C)$ denote the Picard group, i.e. the group of $\otimes$-invertible elements in $\mathcal C$.

Question: Let $X \in \mathcal C$. Suppose that every map $P \to X$ or $X \to P$ is $\otimes$-nilpotent, for $P \in Pic(\mathcal C)$. Then is $X = 0$?

Here, I say that $f : X \to Y$ is "$\otimes$-nilpotent" if there is $n \in \mathbb N$ such that $f^{\otimes n} : X^{\otimes n} \to Y^{\otimes n}$ is null. So I'm asking whether the Picard-graded homotopy of $X$ must contain a non-nilpotent element.

I suspect the answer to my question is no, and probably it is specifically no in the case $\mathcal C = Spt_{T(h)}\langle S \rangle$ which I'm interested in. It would be nice to see a counterexample.

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    $\begingroup$ Are you asking for one of (maps in) or (maps out) to consist of nilpotents, or are you asking for both of them to simultaneously consist of nilpotents? $\endgroup$
    – Tyler Lawson
    Commented Dec 14, 2022 at 22:36
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    $\begingroup$ @TylerLawson I'm asking for both of them to simultaneously consist of nilpotents, although I guess it might already be interesting to consider just one side at a time. $\endgroup$
    – Tim Campion
    Commented Dec 14, 2022 at 23:36
  • $\begingroup$ So if $M$ is nilpotent as in Tyler's answer, and dualizable, then $M=0$. So maybe one can try to argue as follows : the collection of $P$'s that satisfy your hypothesis is thick, therefore contains $M$, and therefore $M=0$. The fact that the $P$'s are closed under cofiber sequences seems to be the hard part (if it's even true...) $\endgroup$
    – Maxime Ramzi
    Commented Dec 15, 2022 at 13:07
  • $\begingroup$ @MaximeRamzi Hmm... if this form argument should work, I'd expect the argument to equally show "if every map $S \to X$ is nilpotent, then $X = 0$", since I'm assuming every object is perfect. But this would be too strong: for instance, if $P$ is invertible in $Sp_{K(n)}$ but not a shift of spheres, then since $K(n)_\ast$ is 1-dimensional, every map $S \to P$ must be zero on $K(n)_\ast$. So if $P$ is perfect, then every such map is nilpotent. I'm told the "upside-down question mark" is an example of such a perfect element of $Pic$. $\endgroup$
    – Tim Campion
    Commented Dec 15, 2022 at 17:23
  • $\begingroup$ yes, I was expecting this to maybe be true with shifts of spheres - if there's a counterexample though... $\endgroup$
    – Maxime Ramzi
    Commented Dec 15, 2022 at 21:39

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For a polynomial algebra $k[x]$ in one variable, there is a module that I'll write $M(x)$, sometimes called $k[x] / (x^\infty)$, which sits in an exact sequence $$ 0 \to k[x] \to k[x^{\pm 1}] \to M(x) \to 0. $$ One can use this sequence to compute Tor. We find that the derived tensor product of $M(x)$ with itself over $k[x]$ is equivalent to the shift $M(x)[1]$. The element $x^{-1}$ also generates an inclusion $k \to M(x)$ of modules.

Now consider an infinite polynomial ring $A = k[x_1, x_2, \dots] = \bigotimes^k k[x_i]$, which has a module $M = \bigotimes^k M(x_i)$. (To take the implicit colimit involved in this infinite tensor, we use the maps of modules $k \to M(x_i)$ to assemble them into a directed system.) The fact that Tor commutes with filtered colimits in the ring and module variable, and the component-by-component calculation, allows us to conclude that $$ M \otimes^L_A M \simeq 0$$ in the derived category $D(A)$. This means that $M$ itself is nilpotent, and so certainly maps into or out of $M$ from elements of Pic are nilpotent as well.

(Something like this might work in stable homotopy theory if you use the analogous Brown-Comenetz dual of the Brown-Peterson spectrum.)

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    $\begingroup$ $M$ is not perfect though, is it ? $\endgroup$
    – Maxime Ramzi
    Commented Dec 15, 2022 at 10:34
  • $\begingroup$ @MaximeRamzi Ah, darn it. I misread and thought that the unit should be a generator, not that all objects were perfect. My apologies. $\endgroup$
    – Tyler Lawson
    Commented Dec 15, 2022 at 14:43

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