Examples of solid abelian groups

Question

I am reading through Clausen's and Scholze's Lectures on condensed mathematics. I am struggling to understand the concept of solid abelian groups so I am looking for some examples. Is the underlying condensed abelian group of a finitely generated module (with the unique induced topology) over a Banach ring solid? Do you know of other examples?

Answer 1

Here's a rule of thumb: As long as the construction is nonarchimedean and does not involve noncompleted tensor products, it's solid.

More precisely, anything you can build from discrete abelian groups by repeatedly forming limits (in particular, kernels), colimits (in particular, cokernels), extensions, and internal Hom's, is solid. This includes any nonarchimedean Banach ring: These are (usually) of the form $A=A_0[\tfrac 1t]$ where $A_0\subset A$ is the unit ball and $t\in A$ is some pseudouniformizer, and $A_0$ is $t$-adically complete, $A_0=\varprojlim_n A_0/t^n$. Here, all $A_0/t^n$ are discrete abelian groups, thus solid; thus the limit $A_0$ is solid; thus the colimit $A=A_0[\tfrac 1t]$ is solid.

Now if $M$ is a finitely generated $A$-module, you can write it as a cokernel of a map $f: A^n\to A^m$, and this makes $M$ naturally into a solid $A$-module, again by the above principles. (Beware that if you first do this topologically, endowing $M$ with the quotient topology, you may run into the issue that the quotient topology may not be separated. In that case, the corresponding condensed $A$-module may not agree with the above cokernel of $f$ taken in condensed $A$-modules. But I'd simply argue that here the condensed perspective gives you a more sensible thing to do, namely just take the cokernel of $f$ in condensed $A$-modules.)

Also beware that if $A$ is a Banach ring over the real numbers, then it's not solid (unless $A=0$).