For my own convenience I'll work in $\infty$-categories, feel free to answer in whatever framework best suits you. My question is essentially how to show, given an $E_\infty$-ring object $R$ in an $\infty$-category $C$ and another object $M$ with a map $R\otimes M\to M$, that the pair $(R,M)$ is in fact an algebra, as Lurie calls it in Higher Algebra, for the $\infty$-operad $LM^\otimes$. In other words, how to show that $M$ is a module over $R$ with respect to all of $R$'s structure (not just the structure of being "homotopy associative and commutative").

One idea I have is the following: I'd like to produce a map of $\infty$-operads $F:LM^\otimes\to C^\otimes$, where $C^\otimes\to Fin_\ast$ is the coCartesian fibration determining the symmetric monoidal structure on $C$. To produce $F$ I need to have that it takes inert morphisms of $LM^\otimes$ to inter morphisms of $C^\otimes$. Now, recall that the inert morphisms of $LM^\otimes$ are precisely morphisms which map to inert morphisms in $Fin_\ast$ along the forgetful map $LM^\otimes\to Fin_\ast$, which are morphisms $\langle n\rangle\to\langle m\rangle$ that represent $\langle m\rangle$ as a "quotient" of $\langle n\rangle$, where everything we quotient out by gets sent to $\ast\in\langle m\rangle=\{\ast,1,\ldots,m\}$.

However, to describe an $R$-action on $M$, we certainly never send $M$ to $\ast$. As such, the inert morphisms in $LM^\otimes$ shouldn't have anything to do with $M$ (it's just coming along for the ride, so-to-speak). In other words, we need that the morphisms describing the monoidal structure on $R$ (inside of $LM^\otimes$) go to the right place, but we don't have to check anything involving inert morphisms with respect to $M$. Hence a functor $LM^\otimes\to C^\otimes$ which picks out an associative algebra object $R$ of $C$ also picks out an object of $LMod(R)$. If $R$ is in fact $E_\infty$ then we also know that $LMod(R)\simeq Mod^{Comm}(R)$, giving us what we need to know.

Does this seem valid? I have very little experience working with $\infty$-operads, and can't find this sort of thing anywhere anyway, so I might be making some very silly mistakes.

Assuming my argument above doesn't make any sense, does anyone have any other ideas?

Thanks!

For my own convenience I'll work in $\infty$-categories, feel free to answer in whatever framework best suits you. My question is essentially how to show, given an $E_\infty$-ring object $R$ in an $\infty$-category $C$ and another object $M$ with a map $R\otimes M\to M$, that the pair $(R,M)$ is in fact an algebra, as Lurie calls it in Higher Algebra, for the $\infty$-operad $LM^\otimes$. In other words, how to show that $M$ is a module over $R$ with respect to all of $R$'s structure (not just the structure of being "homotopy associative and commutative").

One idea I have is the following: I'd like to produce a map of $\infty$-operads $F:LM^\otimes\to C^\otimes$, where $C^\otimes\to Fin_\ast$ is the coCartesian fibration determining the symmetric monoidal structure on $C$. To produce $F$ I need to have that it takes inert morphisms of $LM^\otimes$ to inert morphisms of $C^\otimes$. Now, recall that the inert morphisms of $LM^\otimes$ are precisely morphisms which map to inert morphisms in $Fin_\ast$ along the forgetful map $LM^\otimes\to Fin_\ast$, which are morphisms $\langle n\rangle\to\langle m\rangle$ that represent $\langle m\rangle$ as a "quotient" of $\langle n\rangle$, where everything we quotient out by gets sent to $\ast\in\langle m\rangle=\{\ast,1,\ldots,m\}$.

However, to describe an $R$-action on $M$, we certainly never send $M$ to $\ast$. As such, the inert morphisms in $LM^\otimes$ shouldn't have anything to do with $M$ (it's just coming along for the ride, so-to-speak). In other words, we need that the morphisms describing the monoidal structure on $R$ (inside of $LM^\otimes$) go to the right place, but we don't have to check anything involving inert morphisms with respect to $M$. Hence a functor $LM^\otimes\to C^\otimes$ which picks out an associative algebra object $R$ of $C$ also picks out an object of $LMod(R)$. If $R$ is in fact $E_\infty$ then we also know that $LMod(R)\simeq Mod^{Comm}(R)$, giving us what we need to know.

Does this seem valid? I have very little experience working with $\infty$-operads, and can't find this sort of thing anywhere anyway, so I might be making some very silly mistakes.

Assuming my argument above doesn't make any sense, does anyone have any other ideas?

Thanks!