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.

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    $\begingroup$ Prop 3.14 of arxiv.org/abs/2105.12591 tells you the result after $\mathbb Z[T]$-solidification. $\endgroup$
    – Z. M
    Mar 26, 2022 at 1:52
  • $\begingroup$ @Z.M thank you, that is perfect!! $\endgroup$
    – user108998
    Mar 26, 2022 at 11:16

1 Answer 1


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 $\mathbb Z$).


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