Can the Picard-graded homotopy of a nonzero object be nilpotent? 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.
 A: 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.)
