3
$\begingroup$

Suppose I am given two $E_2$-ring spectra $A$ and $B$ and a morphism of $E_2$-rings $\phi:A\to B$. Then I have $E_1$-monoidal categories of modules $LMod_A$ and $LMod_B$. Moreover I have morphisms $F:LMod_A\to LMod_B$ and $U:LMod_B\to LMod_A$ where by $F$ I mean the extension of scalars functor given by $-\otimes_A B$ and by $U$ I mean the functor that for a given $B$-module $M$ imbues it with the $A$-module structure given by $M\otimes A\overset{1\otimes\phi}\to M\otimes B\to M$. Since $LMod_A$ and $LMod_B$ are monoidal, I can talk about algebras therein. Is it the case that the functors $F$ and $U$ preserve this monoidal structure? If so, is there a reference for this fact within Lurie's $\infty$-categorical framework?

I should maybe mention that Lurie remarks that the functor $F$ defined above is monoidal in Remark 6.5 of DAG VII.

Improve this question
$\endgroup$
4
  • 6
    $\begingroup$ I'm confused. Suppose we work with just rings instead of ring spectra, so let $A, B$ be two commutative rings and let $f : A \to B$ be a morphism between them. Then $F$ is monoidal, but $U$ isn't. It's easiest to see this when $f$ is a field extension; $F$ preserves dimensions but $U$ multiplies them by the dimension of the extension. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 22, 2015 at 21:05
  • 1
    $\begingroup$ Fair point. So perhaps we should only ask $U$ to be lax monoidal? $\endgroup$
    – Jonathan Beardsley
    Commented Apr 22, 2015 at 21:50
  • 1
    $\begingroup$ There is Fausk-Hu-May which addresses this type of situation. It may be worth looking at to see what you could expect. $\endgroup$
    – Sean Tilson
    Commented Apr 23, 2015 at 12:53
  • 3
    $\begingroup$ @JonBeardsley I would bet indeed my money on $U$ being lax monoidal. $M\otimes_B N$ should be the coequalizer of the two arrows $M\otimes B \otimes N \to M\otimes N$ and likewise for $M\otimes_A N$. The map $A\to B$ should induce then a map $M\otimes_A N \to M\otimes_B N$. (If you want you can work with bimodules and strictly associative symmetric spectra to check that; then one does not have to say $\infty$ so often). $\endgroup$
    – Lennart Meier
    Commented Apr 23, 2015 at 14:50

1 Answer 1

Reset to default
5
$\begingroup$

Just so this doesn't go unanswered: the fact that $\mathrm{LMod}$ is a functor from $E_n$-ring spectra to the $\infty$-category of $E_{n-1}$-monoidal presentable $\infty$-categories is part of Proposition 7.1.2.6 in Higher Algebra (the other part of the proposition characterizes the image of $\mathrm{LMod}$ as consisting of those $E_{n-1}$-monoidal presentable $\infty$-categories in which the monoidal unit is a compact generator). This tells you that the functor $F$ is $E_{n-1}$-monoidal and a left adjoint. That $U$ is lax $E_{n-1}$-monoidal follows from Corollary 7.3.2.7 in Higher Algebra, which basically says that the right adjoint (relative to $\mathcal{O}^\otimes$) of an $\mathcal{O}$-monoidal functor is lax $\mathcal{O}$-monoidal.

Improve this answer
$\endgroup$

You must log in to answer this question.

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