12
$\begingroup$

An elementary fact of algebra is that the category of nonunital commutative rings is equivalent to that of $\mathbb{Z}$-augmented unital commutative rings, the equivalence being given by forming unitisation and, conversely, forming augmentation ideals.

What may one say about the relation between the $\infty$-categories of nonunital $E_\infty$-rings and that of $E_\infty$-rings augmented over the sphere spectrum?

(This may well be answered by a simple abstract argument. Nevertheless I seem to be stuck, please bear with me.)

$\endgroup$
2
  • 7
    $\begingroup$ This seems to be a special case of Proposition 5.2.3.15 in Higher Algebra. $\endgroup$ Commented Aug 21, 2014 at 22:01
  • $\begingroup$ Ah, right. Excellent, thanks! $\endgroup$ Commented Aug 21, 2014 at 22:46

1 Answer 1

3
$\begingroup$

In Higher Algebra Proposition 5.4.4.10 1, Lurie proves that for a coCartesian fibration of $\infty$-operads $q:\mathscr{C}^\otimes\to\mathscr{O}^\otimes$, where $\mathscr{O}^\otimes$, when viewed as an $\infty$-category, is pointed and $\mathscr{C}$ is a stable $\mathscr{O}$-monoidal $\infty$-category under the coCartesian fibration $q$. There is an equivalence of $\infty$-categories $\mathrm{Alg^{nu}}_\mathscr{O}(\mathscr{C})\to\mathrm{Alg^{aug}_{\mathscr{O}}(\mathscr{C})}$.

Let $\mathscr{O}^\otimes=\mathrm{N}(\mathscr{F}\mathrm{in}_{*})$, where $\mathscr{F}\mathrm{in}_{*}$ is Segal's category of pointed finite sets (the $n$lab denotes it as $\Gamma$, if I'm not wrong). A $\mathbb{E}_\infty$-ring is a commutative monoidal object, hence $\mathbb{E}_\infty$-monoidal object, of $\mathrm{Sp}$, and the sphere spectrum $S$ is naturally a $\mathbb{E}_\infty$-ring, so we can let $\mathscr{C}$ to be $\mathrm{Mod}_S$. Then, the $\infty$-category of nonunital $\mathbb{E}_\infty$-rings is equivalent to the $\infty$-category of augmented $\mathbb{E}_\infty$-rings over the sphere spectrum.


1 This is an expansion of Haugseng's comment above.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .