Solid tensor product of pro-discrete space with Laurent series

Consider the category $$\operatorname{Solid}_{\mathbf{Z}}$$ of solid abelian groups in the sense of Clausen-Scholze. This category is a full subcategory of condensed abelian groups, $$\operatorname{Cond}_{\mathbf{Z}}$$. These are, modulo set theoretical technicalities, abelian sheaves on the site of profinite sets, with finite families of jointly surjective maps as coverings. There is a left adjoint to the inclusion, and thus one can define a tensor product by tensoring in $$\operatorname{Cond}_{\mathbf{Z}}$$ and then solidifying.

Question. Let $$V$$ be a pro-discrete topological abelian group. Note that $$\mathbf{Z}((T))$$ is solid, as it is the limit along open inclusions of pro-discrete spaces, and any map from a compact space must factor through some pro-discrete subspace. What is the solid tensor product of $$V$$ with $$\mathbf{Z}((T))$$? I presume it is $$V\{T\}$$, the module of two way infinite Laurent series over $$V$$, whose Laurent tail coefficients tend to $$0$$ in $$V$$? To prove this it would seem I need to prove that the solid tensor product commutes with certain cofiltered limits, which I have not been able to do.

• Prop 3.14 of arxiv.org/abs/2105.12591 tells you the result after $\mathbb Z[T]$-solidification.
– Z. M
Mar 26, 2022 at 1:52
• @Z.M thank you, that is perfect!!
– EBz
Mar 26, 2022 at 11:16

1 Answer

This is not true in general, the most important observation being that it fails already when $$V$$ is discrete. In that case $$V\otimes^{\blacksquare} \mathbb Z((T))$$ is just the usual algebraic tensor product. This agrees with $$V((T))$$ only if $$V$$ is finitely generated.

In a different direction, for those pro-discrete abelian groups $$V$$ that are limits of finitely generated abelian groups, a variant of the claim is true; namely, one gets $$V\otimes^\blacksquare \mathbb Z((T)) = V((T))$$ (but the Laurent tail is finite). Indeed, in that case $$V$$ is a compact object of the category of solid abelian groups, so has a finite resolution by product of copies of $$\mathbb Z$$, reducing on to that case; and then it follows from $$\prod_I \mathbb Z\otimes^\blacksquare \prod_J \mathbb Z=\prod_{I\times J} \mathbb Z$$.

As Z. M observes in the comment, a related true statement is that $$V\otimes_{\mathbb Z_\blacksquare} \mathbb Z[T]_\blacksquare$$ is given by $$V\langle T\rangle$$ (those series $$\sum_{i\geq 0} v_i T^i$$ for which $$v_i\to 0$$ as $$i\to \infty$$). The tensor product here is not a solid tensor product, but a base change from solid $$\mathbb Z$$-modules to solid $$\mathbb Z[T]$$-modules (where being $$\mathbb Z[T]$$-solid is stronger than being solid over $$\mathbb Z$$).