The Clausen-Scholze theory of condensed mathematics offers an abelian category with enough projective objects that embraces the study of arbitrary locally compact (and Hausdorff) groups. The behaviour of the tensor product is managed by restricting to a subcategory of solid abelian groups within the category of condensed abelian groups and there is a solidification functor that is left adjoint to this inclusion. Putting my first toe in the water, I am restricting to F_p vector spaces where p is prime: in other words, condensed abelian groups of prime exponent. The Clausen-Scholze theory already provides sufficient in this special case to be able to solve problems that were inaccessible with classical Galois cohomology. My question: is every condensed vector space over a finite field automatically solid? And if not, exactly what would be the advantage of solidification in this context?
Category Theory – Are Condensed Vector Spaces Over Finite Fields Always Solid?
abelian-groupscondensed-mathematicsct.category-theoryfinite-fieldstensor-products
Related Solutions
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 $\mathbb Z$).
I will prove that the result is true if $F$ is a finitely generated field, but fails if $F$ is countably generated field that is not finitely generated.
Let me first discuss the case $F=\mathbb Q$. For $F=\mathbb Q$, one has the idempotent solid $\mathbb Z$-algebra $\hat{\mathbb Z}=\mathrm{lim}_n \mathbb Z/n\mathbb Z$ (where $n$ runs over nonzero integers). One can form the Bousfield localization of $D(\mathbb Z_\blacksquare)$ that kills all $\hat{\mathbb Z}$-modules. As a Bousfield localization, this can also be described as a full subcategory of $D(\mathbb Z_\blacksquare)$ in terms of the corresponding local objects; and one computes that the localization of $\mathbb Z_\blacksquare[S] = \mathrm{lim}_i \mathbb Z[S_i]$ is $\mathrm{lim}_i \mathbb Q[S_i]$ (as the quotient $\mathrm{lim}_i \mathbb Q/\mathbb Z[S_i]$ is a module over $\hat{\mathbb Z}$, while $\hat{\mathbb Z}$ has no maps to $\mathbb Q$, and thus no maps $\mathrm{lim}_i \mathbb Q[S_i]$). This easily implies that $\mathbb Q$ with $\mathbb Q_\blacksquare[S] = \mathrm{lim}_i \mathbb Q[S_i]$ defines an analytic ring.
If $F$ is any finitely generated field, write it as the field of fractions of some domain $R$ that is a finitely generated $\mathbb Z$-algebra. Then $R_\blacksquare$ exists as an analytic ring, and one can form the idempotent $R_\blacksquare$-algebra $\hat{R}=\mathrm{lim}_f R/f$, where $f$ runs over nonzero elements of $R$. Passing to the corresponding Bousfield localization, the localization of $\mathrm{lim}_i R[S_i]$ is $\mathrm{lim}_i K[S_i]$, from which one gets the desired result.
Now assume that $F$ is not finitely generated, but still countably generated. I claim that $$\mathrm{Ext}^1_F(\prod_{\mathbb N} F,F)\neq 0,$$ where the $\mathrm{Ext}^1$ is computed in condensed $F$-modules; but if $F_\blacksquare$ was an analytic ring then such Ext-groups would have to be zero.
Write $F$ as a sequential colimit of finitely generated fields $F_n$. By writing $R\mathrm{Hom}_F$ as $R\mathrm{lim}_n R\mathrm{Hom}_{F_n}$ and a lim-lim^1-sequence, it suffices to show that $$ \mathrm{lim}^1 \mathrm{Hom}_{F_n}(\prod_{\mathbb N} F,F)\neq 0.$$ But one can show that any morphism of condensed abelian groups $\prod_{\mathbb N} F\to F$ factors over a finite product, hence it suffices that $$ \mathrm{lim}^1 (\bigoplus_{\mathbb N} \mathrm{Hom}_{F_n}(F,F))\neq 0.$$
But for any tower $(X_n)_n$ of abelian groups, one has $\mathrm{lim}^1(\bigoplus_{\mathbb N} X_n)\neq 0$ as long as $(X_n)_n$ does not satisfy the Mittag-Leffler condition (see I. Emmanouil. Mittag-Leffler condition and the vanishing of lim^1. Topology, 35(1):267–271, 1996). But if the $F_n$ are strictly increasing, $\mathrm{Hom}_{F_n}(F,F)$ forms a strictly decreasing chain of abelian groups.
It is likely that the result fails for any $F$ that is not finitely generated; I did not try to think about this.
Best Answer
Peter Scholze's comment gives a good answer to the main direct question and tells us that the condensed mod p vector space with basis a compact Hausdorff space S is solid if and only if S is finite. The advantages of solidification are going to become more apparent as we use the condensed maths more widely. Solidification is particularly important when using tensor product. One situation where I believe there is no need to invoke solidification is when tensoring a solid vector space (over F_p) with a finite vector space: such a tensor is automatically solid from the get-go.