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Notes on condensed mathematics

Kiran S. Kedlaya

Section 2 Solid abelian groups

We study in more the category of solid abelian groups, i.e., the category of solid \(R\)-modules for \(R = \ZZ\text{.}\) As a reminder of Remark 1.7.4, we recall that this is the only case of the construction we will use in condensed mathematics; for \(R \neq \ZZ\) we will rather work with \(R\)-module objects in the category of solid abelian groups.
To simplify notation (and to remind ourselves of the previous point), let us define \(\Ab_\solid\) as a synonym for \(\Mod_{\ZZ_\solid}\text{,}\) the category of solid abelian groups. Let us also write \(M_\solid\) as shorthand for \(M \otimes_{\ZZ} \ZZ_\solid\text{,}\) meaning the image of a discrete abelian group \(M\) in \(\Ab_\solid\) (Definition 1.5.1).

Reference.

Subsection 2.2 is taken from [8], Lecture 1. Subsection 2.3 and Subsection 2.4 are taken from [8], Lecture 6.

Subsection 2.1 Solid abelian groups as topological groups

We insert a remark to help justify why we set up the theory of solid abelian groups as we did.

Remark 2.1.1.

The category \(\Ab_\solid\) was custom-built to enforce the relation (1.3). However, there is another context in which such a relation occurs more naturally: in the category \(\TopAb\) of topological abelian groups.
Let us spell this out in some detail. Start with \(\ZZ\) viewed as a discrete topological abelian group. The product \(\prod_I \ZZ\) is topologized as follows: one gets a neighborhood basis of open subsets by choosing an arbitrary finite subset \(I_0 \subseteq I\) and taking the preimage of an open subset under the projection \(\prod_I \ZZ \to \prod_{I_0} \ZZ\text{.}\) Since a finite product of copies of \(\ZZ\) is again discrete, every singleton set in the finite product is open; hence the neighborhood basis consists of subsets consisting of tuples with a fixed value in some finite set of coordinates and arbitrary values elsewhere.
Now consider a continuous map \(\prod_I \ZZ \to \ZZ\) of topological abelian groups. The inverse image of 0 must be open in \(\prod_I \ZZ\text{,}\) and contains the zero tuple, so for some finite subset \(I_0 \subseteq I\) it must contain the whole kernel of \(\prod_I \ZZ \to \prod_{I_0} \ZZ\text{.}\) But that makes the map a finite sum of coordinate projections; we conclude that
\begin{equation*} \Hom_{\TopAb}\left(\prod_I \ZZ, \ZZ\right) = \bigoplus_I \ZZ \end{equation*}
and similarly
\begin{equation*} \Hom_{\TopAb}\left(\prod_I \ZZ, \prod_J \ZZ\right) = \prod_J \bigoplus_I \ZZ\text{.} \end{equation*}
This formally implies (by taking colimits) that there is a functor \(\Ab_\solid \to \TopAb\) taking \(\prod_I \ZZ_\solid\) to \(\prod_I \ZZ\text{.}\)
However, this functor is not fully faithful because of the deleterious effect of taking colimits on underlying topological spaces (e.g., see Example 2.1.2). This can be corrected by replacing the category of topological abelian groups with the category of condensed abelian groups, in which cokernels behave more like in \(\Ab_\solid\text{.}\) This also will make the construction compatible with tensor products, which do not make sense in \(\TopAb\text{.}\) We will reconcile the points of view of solid abelian groups versus condensed abelian groups in Theorem 7.4.1.

Example 2.1.2.

In \(\Ab_\solid\text{,}\) the object \(\ZZ_\solid[T]\) is the functor \(\prod_I \ZZ_\solid \mapsto \left( \bigoplus_I \ZZ \right)[T] = \bigoplus_I \ZZ[T]\text{.}\) We also have an object \(\ZZ_\solid \langle T \rangle\) which is the functor taking
\begin{equation*} \prod_I \ZZ_\solid \mapsto \left( \prod_I \ZZ[T] \right) \cap \left( \bigoplus_I \ZZ \right) \llbracket T \rrbracket \subset \left( \prod_I \ZZ \right) \llbracket T \rrbracket\text{;} \end{equation*}
that is, we have an \(I \times \NN\) matrix over \(\ZZ\) in which each row and column has finite support. The cokernel of the obvious map \(\ZZ_\solid[T] \to \ZZ_\solid \langle T \rangle\) is an object of \(\Ab_{\solid}\) whose restriction to \(\Ab\) is zero (because \(\bigoplus_I = \prod_I\) when \(I\) is finite), but which is not the zero object.

Subsection 2.2 Whitehead’s problem for solid abelian groups

Remark 2.2.1.

The Whitehead problem asks whether every abelian group \(A\) with \(\Ext^1_{\ZZ}(A, \ZZ) = 0\) is free. It was shown by Shelah that this statement is in fact independent of the axioms of set theory (ZFC, or even ZFC plus the continuum hypothesis).
One interesting side effect of the fact that the passage to solid modules gives us a meaningful duality between infinite products and coproducts is that the analogue of Whitehead’s problem is easily resolved!

Proof.

See [26], Theorem B-2.28.

Proof.

Since \(\ZZ\) is a principal ideal domain, using Lemma 2.2.2 we may construct a free resolution of \(A\) of length 1:
\begin{equation} 0 \to \bigoplus_I \ZZ \to \bigoplus_J \ZZ \to A \to 0\text{.}\tag{2.1} \end{equation}
We may further ensure that \(|I|, |J| \leq \kappa\text{.}\)
By Proposition 1.5.3, this remains exact under the passage from \(\Ab\) to \(\Ab_\solid\text{:}\)
\begin{equation} 0 \to \bigoplus_I \ZZ_\solid \to \bigoplus_J \ZZ_\solid\to A_{\solid} \to 0\text{.}\tag{2.2} \end{equation}
Applying the dual functor \(\iHom_{\ZZ_\solid}(\bullet, \ZZ_\solid)\) and using our hypothesis (and Proposition 1.6.2), we obtain the exact sequence
\begin{equation*} 0 \to \iHom_{\ZZ_\solid}(A_{\solid}, \ZZ_{\solid}) \to \prod_J \ZZ_{\solid} \to \prod_I \ZZ_{\solid} \to \iExt^1_{\ZZ_\solid}(A_{\solid}, \ZZ_{\solid}) = 0\text{.} \end{equation*}
This splits because \(\prod_I \ZZ_{\solid}\) is projective (here we diverge from the classical setting). Consequently, applying \(\iHom_{\ZZ_\solid}(\bullet, \ZZ_\solid)\) recovers the sequence (2.2), but with the additional information that this sequence is now split. Since the functor \(A \mapsto A_\solid\) is fully faithful (see Definition 1.5.1), we deduce that \(A\) is a submodule of a free module, and hence free by Lemma 2.2.2 again.

Subsection 2.3 Finitely presented solid abelian groups

To further illustrate that one can really work with the solid modules despite the somewhat abstract definition, we introduce and classify finitely presented solid abelian groups.

Definition 2.3.1.

An object \(M \in \Ab_\solid\) is finitely generated if it can be written as a quotient of \(\prod_I \ZZ_\solid\) for some \(I\) with \(|I| \leq \kappa\text{.}\) The object \(M\) is finitely presented if it occurs as the cokernel of some morphism \(\prod_J \ZZ_\solid \to \prod_I \ZZ_\solid\text{.}\)

Example 2.3.2.

For any finitely generated abelian group \(A\text{,}\) \(A_\solid\) is finitely presented (see Remark 2.3.4 for a converse assertion). Also, any product \(\prod_I \ZZ_\solid\) with \(|I| \leq \kappa\) is finitely presented; this accounts for all finitely generated projective objects as per Lemma 2.3.3.

Proof.

Let \(M\) be a finitely generated projective object of \(\Ab_\solid\text{.}\) By definition, there exists a surjective morphism \(\prod_I \ZZ_\solid \to M\text{,}\) which then can be split. The resulting splitting of \(\prod_I \ZZ_\solid\) corresponds to a splitting of its dual, which we identify with \(\bigoplus_I \ZZ_\solid = \left( \bigoplus_I \ZZ \right)_\solid\) using Proposition 1.6.2. By Lemma 2.2.2, any direct summand of \(\bigoplus_I \ZZ\) has the form \(\bigoplus_J \ZZ\) with \(|J| \leq |I|\leq \kappa\text{;}\) dualizing back yields the claim.

Remark 2.3.4.

If \(A\) is an abelian group which is not finitely generated, then \(A_\solid\) is not finitely generated. Namely, by (1.5), the maps from \(\prod_I \ZZ_\solid\) to \(A_\solid\) correspond to elements of \(\bigoplus_I A\text{,}\) so the image of any such map is (the image in \(\Ab_\solid\) of) a finitely generated subgroup of \(A\text{.}\)
In particular, if \(I\) is an infinite index set, then \(\bigoplus_I \ZZ_\solid = \left( \bigoplus_I \ZZ \right)_\solid\) is a projective object of \(\Ab_\solid\) which is not finitely generated. This implies in turn that the quotient \(\left( \prod_I \ZZ_\solid \right) / \left( \bigoplus_I \ZZ_\solid \right)\) is not finitely presented by a standard “two out of three” argument (compare [28], tag 064R).
We may generate more interesting examples of finitely presented objects of \(\Ab_\solid\) using the derived version of the dual functor \(\iHom_{\ZZ_\solid}(\bullet, \ZZ_\solid)\text{.}\)

Proof.

Choose a surjection \(\bigoplus_I \ZZ \to Q\) with \(|I| \leq \kappa\) (e.g., by taking \(I = \QQ\) and the map to be the counit of the adjunction between sets and abelian groups). By Lemma 2.2.2, the kernel of this map is again a free module of rank at most \(\kappa\text{;}\) that is, we have an exact sequence of abelian groups
\begin{equation} 0 \to \bigoplus_I \ZZ \stackrel{h}{\to} \bigoplus_J \ZZ \to Q \to 0\text{.}\tag{2.3} \end{equation}
Passing from \(\Ab\) to \(\Ab_\solid\) and then taking duals (Proposition 1.6.2) yields a new exact sequence
\begin{equation} 0 \to \iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid) \to \prod_J \ZZ_\solid \stackrel{g}{\to} \prod_I \ZZ_\solid \to \iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid) \to 0\tag{2.4} \end{equation}
where the final \(0\) occurs because \(\prod_J \ZZ_\solid\) is a projective object in \(\Ab_\solid\text{.}\)
We can say a bit more about the construction in Lemma 2.3.5 when starting with a torsion group.

Proof.

It suffices to compute the evaluation on \(\prod_I \ZZ_\solid\) for \(I\) at most countable: using Hom-tensor adjunction plus (1.8), we obtain
\begin{align*} \iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\left( \prod_I \ZZ_\solid \right) &= \Hom_{\ZZ_\solid}\left(\prod_I \ZZ_\solid, \iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\right)\\ & = \Hom_{\ZZ_\solid}\left(\left(\prod_I \ZZ_\solid\right) \otimes_{\ZZ_\solid} (Q_\solid), \ZZ_\solid\right)\\ & = \Hom_{\ZZ_{\solid}}\left(Q_\solid, \iHom_{\ZZ_\solid}\left(\prod_I \ZZ_\solid, \ZZ_\solid\right)\right)\\ & = \Hom_{\ZZ_{\solid}}\left(Q_\solid, \bigoplus_I \ZZ_\solid\right)\\ & = \Hom_{\ZZ}\left(Q, \bigoplus_I \ZZ\right)\text{.} \end{align*}
The last term injects into \(\prod_I \Hom_{\ZZ}(Q, \ZZ) = 0\text{.}\) (We will see another proof later in the language of condensed abelian groups; see Remark 7.4.3.)

Remark 2.3.7.

In Lemma 2.3.5, if \(\Hom{\ZZ}(Q, \ZZ) = 0\) (e.g., if \(Q\) is torsion), then Lemma 2.3.6 shows that the first term of (2.4) vanishes; that is, \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) admits a projective resolution of length 1 with terms of the form \(\prod_I \ZZ_\solid\) with \(|I| \leq \kappa\text{.}\) We will pick up on this point in Proposition 2.4.6.
For a countable torsion group, we can make the construction from Lemma 2.3.5 a bit more explicit. (We do not know whether the cardinality hypothesis is needed here, but see Remark 2.3.14.)

Proof.

When \(Q = T\) is finite, we may choose an exact sequence as in (2.3) with \(I\) and \(J\) finite, then compare (2.4) with the corresponding sequence obtained by taking algebraic duals. In the general case, write \(Q\) as a sequential union of torsion subgroups \(T_n\text{;}\) we can then choose presentations
\begin{equation*} 0 \to \bigoplus_{I_n} \ZZ \to \bigoplus_{J_n} \ZZ \to T_n \to 0 \end{equation*}
in which the inclusions \(T_n \to T_{n+1}\) are induced by inclusions \(I_n \to I_{n+1}\text{,}\) \(J_n \to J_{n+1}\text{.}\) When we dualize the resulting sequences to obtain
\begin{equation*} 0 \to \prod_{J_n} \ZZ_\solid \to \prod_{I_n} \ZZ_\solid \to \Ext^1_\ZZ((T_n)_\solid, \ZZ_\solid) \to 0, \end{equation*}
we get an inverse system of these sequences in which the transition maps in the left and middle columns are surjective. In particular there is no \(R^1\) term on the left, so for \(I = \bigcup_n I_n\text{,}\) \(J = \bigcup_n J_n\) we get an exact sequence
\begin{equation*} 0 \to \prod_J \ZZ_\solid \to \prod_I \ZZ_\solid \to \lim_n \Ext^1_\ZZ((T_n)_\solid, \ZZ_\solid) \to 0\text{.} \end{equation*}
Comparing with (2.4) again yields the desired comparison.

Example 2.3.9.

Let us spell out Lemma 2.3.5 in the case \(Q = \QQ/\ZZ\text{.}\) We first write \(Q\) as the colimit of \(\ZZ/n\ZZ\) over all positive integers \(n\text{;}\) by Proposition 2.3.8, \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) is the limit of \(\iExt^1_{\ZZ_\solid}((\ZZ/n\ZZ)_\solid, \ZZ_\solid)\) over all \(n\text{.}\)
For each \(n\text{,}\) use the exact sequence
\begin{equation} 0 \to \ZZ \stackrel{\times n}{\to} \ZZ \to \ZZ/n\ZZ \to 0\tag{2.5} \end{equation}
to obtain the exact sequence
\begin{equation} 0 \to \ZZ_\solid \stackrel{\times n}{\to} \ZZ_\solid \to \iExt^1_{\ZZ_\solid}((\ZZ/n\ZZ)_\solid, \ZZ_\solid) \to 0\tag{2.6} \end{equation}
to compute that \(\iExt^1_{\ZZ_\solid}((\ZZ/n\ZZ)_\solid, \ZZ_\solid) = \cong (\ZZ/n\ZZ)_\solid\text{.}\) Now note that if \(n'\) is a multiple of \(n\text{,}\) then the inclusion of \(\ZZ/n\ZZ\) into \(\ZZ/n' \ZZ\) via multiplication by \(n'/n\) corresponds to the map between exact sequences as in (2.5) which acts on the first copy of \(\ZZ\) as the identity and on the second copy as multiplication by \(n'/n\text{.}\) This then corresponds to the map between exact sequences as in (2.6) which acts on the first copy of \(\ZZ_\solid\) as multiplication by \(n'/n\) and on the second copy as the identity. We conclude that
\begin{equation*} \iExt^1_{\ZZ_\solid}((\QQ/\ZZ)_\solid, \ZZ_\solid) \cong \lim_n (\ZZ/n\ZZ)_\solid, \end{equation*}
which we recognize as the solid analogue of the profinite completion of the integers and therefore denote also as \(\widehat{\ZZ}_\solid\text{.}\)
We can also make the construction more explicit for \(\QQ\)-vector spaces.

Example 2.3.10.

Let us spell out Lemma 2.3.5 in the case \(Q = \QQ\text{.}\) Using the exact sequence
\begin{equation*} 0 \to \ZZ \to \QQ \to \QQ/\ZZ \to 0\text{,} \end{equation*}
we obtain an exact sequence
\begin{equation*} 0 = \iHom_{\ZZ_\solid}(\QQ_\solid, \ZZ_\solid) \to \ZZ_\solid \to \iExt^1_{\ZZ_\solid}((\QQ/\ZZ)_\solid, \ZZ_\solid) \to \iExt^1_{\ZZ_\solid}(\QQ_\solid, \ZZ_\solid) \to 0 \end{equation*}
where the first term vanishes by Lemma 2.3.6. Using Example 2.3.9 to identify the middle term, this yields an isomorphism
\begin{equation*} \iExt^1_{\ZZ_\solid}(\QQ_\solid, \ZZ_\solid) \cong \widehat{\ZZ}_\solid/\ZZ_\solid\text{.} \end{equation*}
For \(Q\) a \(\QQ\)-vector space with \(|Q| \leq \kappa\text{,}\) by choosing a basis we can write \(Q = \bigoplus_I \QQ\) with \(|I| \leq \kappa\text{.}\) We then obtain
\begin{equation*} \iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid) \cong \prod_I \left(\widehat{\ZZ}_\solid/\ZZ_\solid \right)\text{.} \end{equation*}

Remark 2.3.11.

For a general abelian group \(Q\) with \(|Q| \leq \kappa\text{,}\) we have a canonical exact sequence
\begin{equation*} 0 \to T_0 \to Q \to Q \otimes_{\ZZ} \QQ \to T_1 \to 0 \end{equation*}
in which \(T_0\) and \(T_1\) are both torsion groups. This allows us to reduce the computation of \(\iExt^1_{\ZZ_\solid}(Q, \ZZ_\solid)\) to cases where \(Q\) is either a torsion group or a \(\QQ\)-vector space.
We next show that the construction of Lemma 2.3.5 is essentially complete.

Proof.

The “if” assertion follows from Lemma 2.3.5; we prove the “only if” assertion as follows. Let \(M\) be a finitely presented object of \(\Ab_\solid\text{.}\) By definition, we can find an exact sequence of the form
\begin{equation} 0 \to \prod_J \ZZ_\solid \stackrel{g}{\to} \prod_I \ZZ_\solid \to M \to 0\text{.}\tag{2.7} \end{equation}
By taking duals (Proposition 1.6.2) and noting that everything in sight is discrete, we get a map \(h \colon \bigoplus_I \ZZ \to \bigoplus_J \ZZ\text{.}\)
Suppose first that \(h\) is injective. We can then set \(Q := \coker(h)\) and carry out Lemma 2.3.5; comparing (2.7) with the resulting exact sequence (2.4) shows that \(M \cong \iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\text{.}\)
To reduce the general case to the previous case, apply Lemma 2.2.2 to see that \(\ima(h)\) is also a free \(\ZZ\)-module. This means that the source \(\bigoplus_I \ZZ\) splits as \(\ker(h)\) plus a submodule isomorphic to \(\ima(h)\text{,}\) and both factors are free. This splitting transposes to a decomposition of the target \(\prod_I \ZZ_\solid\) as a product of two factors, each itself a product of copies of \(\ZZ_\solid\) (as in Lemma 2.3.3). Now note that the dual of \(\ima(h)\text{,}\) viewed as a subobject of \(\prod_I \ZZ_\solid\text{,}\) is precisely the image of \(g\text{;}\) consequently, our splitting of \(\prod_I \ZZ_\solid\) projects to a splitting of \(M\) as the direct sum of the dual of \(\ker(h)\) with another module. We can now simultaneously discard the dual of \(\ker(h)\) as a summand of both \(\prod_I \ZZ_\solid\) and \(M\) (leaving \(\prod_J \ZZ_\solid\) unchanged) leaving us in a situation where now the map \(h\) is injective. Applying the previous paragraph yields the desired result.

Remark 2.3.13.

In Proposition 2.3.12, the group \(Q\) cannot in general be taken to be torsion. This can be seen for instance from Example 2.3.10: taking \(\kappa = \aleph_0\text{,}\) \(Q = \QQ\) yields the object \(\widehat{\ZZ}_\solid/\ZZ_\solid\) which is not profinite, and therefore by Proposition 2.3.8 cannot arise from a torsion group.

Remark 2.3.14.

When \(Q\) is a torsion group that is not countable, we may still write it as a countable direct sum of Sylow subgroups:
\begin{equation*} Q \cong \bigoplus_p Q[p^\infty] \end{equation*}
where \(p\) runs over primes. We may correspondingly split \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) as a direct sum of \(\iExt^1_{\ZZ_\solid}(Q[p^\infty]_\solid, \ZZ_\solid)\text{.}\)
Suppose now that \(Q\) is \(p^\infty\)-torsion for some prime \(p\text{.}\) We may write \(Q\) as the sequential colimit of \(Q[p^n] = \ker(Q \stackrel{\times p^n}{\to} Q)\) over \(n \in \NN\text{,}\) so we can still view \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) as the limit of \(\iExt^1_{\ZZ_\solid}(Q[p^n]_\solid, \ZZ_\solid)\text{.}\)
However, when we write \(Q[p^n]\) as a non-sequential colimit of finite torsion groups, it is not clear whether this colimit transforms into a limit via \(\iExt^1\text{.}\) This does hold for \(n=1\text{,}\) though, as in this case \(Q[p]\) is a free module over \(\ZZ/p\ZZ\) and so the coproduct transforms into a product.

Subsection 2.4 The category of finitely presented solid abelian groups

When \(\kappa = \aleph_0\text{,}\) we can say even more about the category of finitely presented solid abelian groups.

Remark 2.4.1.

In Lemma 2.3.5, the group \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) may be further analyzed as follows. We have already seen in Proposition 2.4.6 that if \(\Hom_{\ZZ}(Q, \ZZ) = 0\text{,}\) then \(\iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) admits a projective resolution of length 1. If \(\Hom_{\ZZ}(Q, \ZZ) = \ZZ\text{,}\) then we obtain an exact sequence of the form
\begin{equation*} 0 \to Q_0 \to Q \to \ZZ \to 0 \end{equation*}
which we can split to obtain a summand \(Q_0\) of \(Q\) with \(\Hom_{\ZZ}(Q_0, \ZZ) = 0\text{;}\) we then have \(\iExt^1_{\ZZ_\solid}((Q_0)_\solid, \ZZ_\solid) = \iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\text{.}\) By the same token, if \(\Hom_{\ZZ}(Q, \ZZ)\) is finitely generated, we can similarly split \(Q\) as a direct sum \(Q_0 \oplus \ZZ^n\) with \(\Hom_\ZZ(Q_0, \ZZ) = 0\text{.}\)
However, if \(\Hom_{\ZZ}(Q, \ZZ)\) is infinitely generated then no such splitting exists in general. For instance, if \(Q = \prod_\NN \ZZ\text{,}\) then the projection maps form a countable family of homomorphisms \(Q \to \ZZ\) with zero kernel, so we would have to take \(Q_0 = 0\text{;}\) but this is not possible because \(Q\) is not itself a free abelian group (see Remark 2.4.2). Notably, this example is ruled out by taking \(\kappa = \aleph_0\text{,}\) and indeed this process can be pushed through under this restriction.

Remark 2.4.2.

We verify that the product \(\prod_\NN \ZZ\) is not a free abelian group. Suppose to the contrary that \(\prod_\NN \ZZ \cong \bigoplus_I \ZZ\) for some \(I\text{,}\) necessarily with \(|I| = 2^{\aleph_0}\text{.}\) The image of \(\bigoplus_\NN \ZZ \to \prod_\NN \ZZ\) would then be contained in \(\bigoplus_J \ZZ\) for some countable subset \(J\) of \(I\) (namely, the union of the supports of the images of each generator of \(\bigoplus_\NN \ZZ\) in \(\bigoplus_I \ZZ\)). This would in turn imply that the divisible subgroup of \(\left( \prod_\NN \ZZ \right) / \left( \bigoplus_\NN \ZZ \right)\) is countable (as this quotient would be contained in the product of the torsion-free group \(\bigoplus_{I \setminus J} \ZZ\) with a countable group), but this is manifestly false: the map
\begin{equation*} \left( \prod_\NN \ZZ \right) / \left( \bigoplus_\NN \ZZ \right) \to \left( \prod_\NN \ZZ \right) / \left( \bigoplus_\NN \ZZ \right), \qquad (a_n)_n \mapsto (n! a_n)_n \end{equation*}
is injective, its domain is uncountable, and its image is contained in the divisible subgroup of the codomain.

Proof.

Choose a basis of \(M\) and record the elements of this basis as a semi-infinite matrix of integers (with finitely many columns and countably many rows). We can then perform elementary row and column operations on this matrix without changing the isomorphism class of either \(M\) or its divisible closure. In particular, we can put the matrix into Hermite normal form, from which the claim is evident.

Proof.

By hypothesis, we can find a surjection \(\bigoplus_J \ZZ \to M\) with \(|J| \leq \aleph_0\text{.}\) There is nothing to check if \(J\) is finite, as then we can project from \(I\) onto some finite index set and then apply Lemma 2.2.2. We may thus assume that \(J = \NN\text{.}\)
Let \(F_n\) be the divisible closure in \(M\) of the image of \(\bigoplus_{\{0,\dots,n-1\}} \ZZ\text{.}\) Since \(F_n\) is contained in the divisible closure of the image of \(\bigoplus_{\{0,\dots,n-1\}} \ZZ\) in \(\prod_I \ZZ\text{,}\) which by Lemma 2.4.3 is finitely generated, so then is \(F_n\text{.}\)
For each \(n \gt 0\text{,}\) \(F_n/F_{n-1}\) is finitely generated (by the previous paragraph) and torsion-free (since \(M/F_{n-1}\) is by construction), hence free. for each \(n \gt 0\text{.}\) In particular, we can choose a splitting \(F_n \cong F_{n-1} \oplus (F_n/F_{n-1})\text{.}\) Combining these splittings yields a coherent sequence of isomorphisms
\begin{equation*} F_n \cong F_0 \oplus (F_1/F_0) \oplus \cdots \oplus (F_n/F_{n-1})\text{,} \end{equation*}
and taking colimits yields an isomorphism
\begin{equation*} M \cong F_0 \oplus (F_1/F_0) \oplus \cdots \end{equation*}
which shows that \(M\) is free.

Proof.

Note that \(Q^{\vee \vee}\) embeds into \(\prod_{Q^\vee} \ZZ\text{;}\) this restricts to an embedding of \(Q/Q_0\) into \(\prod_{Q^\vee} \ZZ\text{.}\) We may thus apply Lemma 2.4.4 to deduce that \(Q/Q_0\) is free. Using any resulting splitting of \(Q\text{,}\) we see that any nonzero homomorphism \(Q_0 \to \ZZ\) extends to a homomorphism \(Q \to \ZZ\) which does not vanish identically on \(Q_0\text{;}\) from this we conclude that in fact \(\Hom_\ZZ(Q_0, \ZZ) = 0\text{.}\)

Proof.

With notation as in Proposition 2.3.12, it suffices to check the claim assuming \(K = \emptyset\text{,}\) as we can just insert a copy of \(K\) into \(I\) at the end. That is, it suffices to treat the case of the object \(M = \iExt^1_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) for \(Q\) an abelian group with \(|Q| \leq \kappa\text{.}\)
By Lemma 2.4.5, we may split \(Q\) as the kernel \(Q_0\) of the evaluation map plus a free summand. As in Remark 2.4.1, we may discard the free summand without changing \(M\text{.}\) That is, we may assume \(Q = Q_0\text{,}\) which means that \(\Hom_{\ZZ}(Q, \ZZ) = 0\) (again by Lemma 2.4.5). This case was already treated in Remark 2.3.7.

Proof.

Let \(M\) be a finitely generated subobject of \(\prod_K \ZZ_\solid\) in \(\Ab_\solid\text{.}\) Then \(\coker(M \to \prod_K \ZZ_\solid)\) is finitely presented, and hence by Proposition 2.4.6 has projective dimension at most 1. Since \(\prod_K \ZZ_\solid\) is projective, we deduce that \(M\) is a projective object of \(\Ab_\solid\) (this is again a “two out of three” argument as in [28], tag 064R).

Proof.

By emulating the case of pseudocoherent modules over a commutative ring ([28], tag 064N), we see that the full subcategory of \(\Ab_\solid\) consisting of objects admitting finite resolutions by objects of the form \(\prod_I \ZZ\) with \(|I| \leq \kappa\) is closed under kernels, cokernels, and extensions. By Proposition 2.4.6, these are precisely the finitely presented objects.
By the same token, since the objects \(\prod_I \ZZ\) with \(|I| \leq \kappa\) are compact generators of \(\Ab_\solid\text{,}\) we deduce that every object is a colimit of finitely presented objects.
Using Proposition 2.4.6, we also read off compatibility with tensor product and with products over index sets of cardinality \(\leq \kappa\text{.}\) Since we already have compatibility with kernels and cokernels, this implies compatibility with limits indexed by sets of cardinality \(\leq \kappa\) and with finite colimits (since finite products and coproducts coincide in \(\Ab_\solid\)).

Remark 2.4.9.

We show by way of contradiction that the conclusions of Proposition 2.4.8 cannot all hold when \(\kappa \geq 2^{\aleph_0}\text{.}\) Since we will have to make the restriction \(\kappa = \aleph_0\) later for other reasons, this is ultimately not so problematic for our purposes.
Set \(Q := \prod_\NN \ZZ\) and choose a presentation as in (2.3) with \(|I|, |J| \leq 2^{\aleph_0}\text{.}\) In (2.4), all of the terms except \(\iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) are finitely presented, so by the conclusions of Proposition 2.4.8, so must be \(\iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\text{.}\) In particular, the image of \(\iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\) in \(\prod_J \ZZ_\solid\) must be finitely generated, so we can apply the conclusion of Corollary 2.4.7 to deduce that it can be written as \(\prod_L \ZZ_\solid\) for some \(L\) with \(|L| \leq 2^{\aleph_0}\text{.}\)
Define \(M \in \Ab_\solid\) to make
\begin{equation*} 0 \to \iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid) \to \prod_J \ZZ_\solid \to M \to 0 \end{equation*}
an exact sequence. Since this sequence splits, it dualizes to an exact sequence
\begin{equation*} 0 \to \iHom_{\ZZ_\solid}(M, \ZZ_\solid) \to \bigoplus_J \ZZ \to R \to 0 \end{equation*}
where
\begin{equation*} R := \iHom_{\ZZ_\solid}(\iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid), \ZZ_\solid) \cong \iHom_{\ZZ_\solid}\left(\prod_L \ZZ_\solid, \ZZ_\solid\right) \cong \bigoplus_L \ZZ \end{equation*}
is the solid double dual of \(Q\text{.}\)
Although \(R\) is not a priori equal to the algebraic double dual of \(Q\text{,}\) we may argue as in Remark 2.4.1 that the projection maps \(\prod_\NN \ZZ \to \ZZ\) correspond to sections of \(\iHom_{\ZZ_\solid}(Q_\solid, \ZZ_\solid)\text{.}\) Since the projection maps have zero joint kernel, the evaluation map \(Q \to R\) must be injective.
On the other hand, since \(g\colon \prod_J \ZZ_\solid \to \prod_I \ZZ_\solid\) factors through the projection \(\prod_J \ZZ_\solid \to M\text{,}\) \(h \colon \bigoplus_I \ZZ \to \bigoplus_J \ZZ\) factors through the inclusion \(\iHom_{\ZZ_\solid}(M, \ZZ_\solid) \to \bigoplus_J \ZZ\text{.}\) From this, we see that the evaluation map \(Q \to R\) is also surjective. We conclude that \(Q \cong R \cong \bigoplus_L \ZZ\text{,}\) a contradiction against the fact that \(Q\) is known not to be a free abelian group (Remark 2.4.2).

Remark 2.4.10.

One can prove an analogue of Proposition 2.4.8 for the category of \(R\)-module objects in \(\Ab_\solid\) where \(R\) is an algebra of finite type over \(\ZZ\text{.}\) See [8], Lecture 11, timestamp 24:30.
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