Derived $\ell$-completion of $\mathbf{Q}_\ell$ sheaf?

Question

I came across some notation that I’m having trouble understanding in Hansen-Scholze’s preprint ‘Relative Perversity.’ In the last paragraph of Proposition 3.4 there is the notation $A\widehat{\otimes^{\mathbf L}}Rg_*\mathbf Z_\ell$. Here, $A$ is in $D_{\mathrm{cons}}(X,\mathbf Q_\ell)$ on a qcqs scheme $X$, separated and of finite presentation over $S$ (say $f:X\to S$) and $g:T\to S$ is a pro-étale map. (The notation should probably read $A\widehat{\otimes^{\mathbf L}}f^\ast Rg_\ast\mathbf Z_\ell$.) The authors say $\widehat{\otimes^{\mathbf L}}$ means the $\ell$-adically completed tensor product, using an integral structure on $A$; i.e. we suppose $A=A_0\otimes_{\mathbf Z_\ell}\mathbf Q_\ell$ where $A_0\in D_{\mathrm{cons}}(X,\mathbf Z_\ell)$. But what does that mean when $A$ has field coefficients? Dropping the $R$ and $\mathbf L$s, I would normally write $A\widehat{\otimes}f^\ast g_\ast\mathbf Z_\ell=\lim_n(A\otimes f^\ast g_\ast\mathbf Z_\ell\otimes\mathbf Z/\ell^n)$ (everything is derived) but this is null. If what is meant is rather $(A_0\widehat\otimes f^\ast g_\ast\mathbf Z_\ell)\otimes\mathbf Q_\ell$, this is nonzero but then I have trouble understanding their ensuing argument (namely, why this would be independent of the choice of integral model $A_0$).

Separately, in that same paragraph the authors claim an integral structure on $A$ exists ‘at least up to direct factors.’ Their Corollary 2.4 claims an equivalence $D_{\mathrm{cons}}(X,\mathbf Z_\ell)\otimes_{\mathbf Z_\ell}\mathbf Q_\ell\to D_{\mathrm{cons}}(X,\mathbf Q_\ell)$. (The functor sends $A_0\mapsto A_0\otimes_{\mathbf Z_\ell}\mathbf Q_\ell$.) I believe the $\infty$-category on the left-hand side is obtained by inverting $\ell$ and taking the idempotent completion. The idempotent completion is still a little mysterious to me. I suppose it’s a matter of finding some $\mathcal F\in D_{\mathrm{cons}}(X,\mathbf Z_\ell)$ which has no nontrivial direct summand but does after extending scalars. I suppose this can be seen on the level of triangulated categories but I don’t yet have an example.

Answer 1

Scholze explained to me that he meant by this is the following: consider $-\widehat\otimes f^*g_*\mathbf Z_\ell$ as a functor $D_{\mathrm{cons}}(X,\mathbf Z_\ell)\to D_{\mathrm{comp}}(X,\mathbf Z_\ell)$; then inverting $\ell$ yields a functor $D_{\mathrm{cons}}(X,\mathbf Z_\ell)\to D(X,\mathbf Q_\ell)$ and it factors via the Verdier quotient $D_{\mathrm{cons}}(X,\mathbf Q_\ell)=D_{\mathrm{cons}}(X,\mathbf Z_\ell)[\ell^{-1}]$ (this by their Corollary 2.4) if $(-\widehat\otimes f^*g_*\mathbf Z_\ell)[\ell^{-1}]$ sends torsion sheaves to zero, which it clearly does. Therefore in particular this functor is independent (up to isomorphism) of the integral model chosen.