Let $R$ be an $E\_{\infty}$-ring spectrum and $B$ be an $E\_\infty$-space. Suppose we have an $E\_\infty$-map $$ f: B \to BGL\_1(S^0)$$ such that the composite $$f\_R: B \to BGL\_1(S^0) \to BGL\_1(R) $$
is null, then a choice of null-homotopy produces Thom isomorphism which is a weak-equivalence 
$$u: Mf \wedge R \simeq B\_+ \wedge R,$$


where $Mf$ is the Thom spectrum associated to $f$. 


Note that, both sides of the Thom isomorphism are ring spectra. 


Q: I wonder when $u$ is a ring-map? 


A possible guess is that if $f\_R$ is homotoped to null via infinite-loop space maps, then maybe $u$ is a ring map. I am not quite sure if this is true or how to see this. 



When is Thom isomorphism a ring map?
One comment is that you have to be careful, because to be an $E\_{\infty}$-map is not a property but rather additional structure. 


You are exactly right in that what is needed is a null-homotopy of $B \rightarrow BGL\_{1}(R)$ as a map of $E\_{\infty}$-spaces. This is Proposition 3.16 in Omar and Toby's "[A simple universal property of Thom ring spectra](https://arxiv.org/abs/1411.7988)" when you set $A = R$. 


