Nov 18, 2013 04:22
Hi @Drew, I think the statement that the Picard group of the stable homotopy category is $\mathbb{Z}$ can be proved as follows: first, you argue that an invertible spectrum is compact, e.g., because the sphere is compact and tensoring with an invertible spectrum (as an autoequivalence) preserves compact objects. Now you need to show that the homology of your spectrum is $\mathbb{Z}$ concentrated in one dimension. So "tensor up" over $S^0 \to H \mathbb{Z}/p$ (i.e., consider mod $p$ homology).
Invertible objects in $H \mathbb{Z}/p$-modules are concentrated in one dimension since you have a K\"unneth theorem; now all that's left is to see that everything fits together at each prime. More generally, this argument can be used to show that the Picard group of modules over a connective $E_\infty$-ring (with $\pi_0 $ noetherian and connected, say) is given by the Picard group of $\pi_0$ times $\mathbb{Z}$ (for suspension).
One way to streamline this argument is to use the theory of flatness for modules over connective structured ring spectra. There's a nice criterion for flatness: if $A$ is an $A_\infty$-ring with $\pi_0 A$ noetherian commutative, then an $A$-module $M$ is flat if for every field $k$ equipped with a map $\pi_0 A \to k$, the relative tensor product $M \otimes_A k$ is discrete.
