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Nov 30 at 5:51 history edited Tim Campion CC BY-SA 4.0
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Nov 30 at 5:47 comment added Tim Campion I feel the need to put in a reminder that if we look at $Spectra$ instead of $D(\mathbb Z)$, then we should note that $p = 0$ on $\mathbb S / p$ iff $p$ is odd. On $\mathbb S / 2$ we have $4 = 0$ but $2 \neq 0$. This is related to the fact that $\mathbb S / 2$ does not admit a unital multiplication. In general in a symmetric monoidal stable category with unit $S$, if $\phi : \Sigma^n S \to S$ is an endomorphism-up-to-shift, then $\phi^2 = 0$ on $S / \phi$, but it's often the case that $\phi \neq 0$ on $S / \phi$.
Nov 30 at 5:43 history answered Tim Campion CC BY-SA 4.0