Two questions have been answered, Question 6 and Question 18:
Solution to Question 6, p.40. For R a ring and (Mi)i∈ω a family of cyclic left R-modules, that question asks whether the R-module ∏I Mi must either be finitely generated, or require uncountably many generators. We shall answer this in the negative.
The idea will be to take a family of rings (Ri)i∈ω, let R´ be their direct product, find a subring R ⊆ R´ such that R´ is countably but not finitely generated as a left R-module, and also projects surjectively to each Ri, and let each Mi be Ri regarded as a left R-module. Of course, saying this simply reduces the problem to that of finding such an inclusion R ⊆ R´.
The discussion following the question shows that the difficulty is somehow located in the vicinity of finitely many nonprincipal ultrafilters on the index-set ω, without loss of generality, just one; so let us push the problem back one more level: Let us try to find a family of rings (Ri)i∈ω and an ultrafilter U on ω such that, if we write S´ for the ultraproduct of the Ri determined by U, then S´ has a subring S over which it is countably but not finitely generated as a left module. If we then let R be the inverse image of S under the natural map R´ → S´, it is not hard to verify that R´ will be countably but not finitely generated as a left R-module.
To find such S´ and S, we use the fact that it is easy to get such an example for vector spaces, since any infinite-dimensional vector space has a subspace of countably infinite codimension. So let k be any field, let V be an infinite-dimensional vector space over k, make the direct sum k ⊕ V into a commutative ring in the standard way (using the multiplication of k, the scalar multiplication relating k and V, and the zero multiplication on V ), and use this ring for each of the Ri. The ultraproduct of these rings with respect to any nonprincipal ultrafilter on ω will be a ring of the form S´ = k* ⊕ V*, and if we let W ⊂ V* be any k*-subspace of countably infinite codimension, then S = k* ⊕ W will be a subring of S´ over which the latter is countably but not finitely generated as a module, completing the construction.
If we would like our ring R to be an integral domain, we can achieve this by replacing the R obtained above by any integral domain that can be mapped surjectively to it; e.g., a polynomial ring in sufficiently many indeterminates. We can make our module-structures "nicer" in the same way, replacing, in each of the ring-and-module pairs Ri , Mi , the ring Ri with an integral domain that maps surjectively to it and the module Mi with a free module of rank 1 over that domain, and letting R ⊂ R´ = ∏ Ri be the inverse image in that ring of the "R" constructed above. (And then, if we wish, replacing this R´ with an integral domain that maps surjectively to it.)
Something it is not clear is whether we can do is indicated in part (1) of the following question. An affirmative answer to part (2) would imply an affirmative answer to part (1).
Question 6´. (1) If R is a ring and
a family of simple
left R-modules, can their direct product be countably but
not finitely generated as an R-module?
(2) Can a nonprincipal ultraproduct k of fields have a subfield over which it is countably but not finitely generated as a vector space? (If not, does there at least exist such an example for division rings and generation as a left vector space?)
Solution to Question 18, p.47. This was answered by Adam Megacz. The question concerns an algebra M all of whose operations have arities < λ for a regular infinite cardinal λ. Theorem 16 shows that for any κ ≥ λ, Mκ requires either > κ or < λ generators; and Question 18 asks whether in this statement < λ can be replaced by < ℵ0. Megacz shows that when λ is the successor μ+ of a cardinal μ < κ, the bound of that theorem cannot be improved at all. The following somewhat simplifies and generalizes his construction.
Let κκ denote the set of all functions κ → κ, and for each f ∈ κκ, let αf be the μ-ary operation on κ such that for any μ-tuple x = (xi )i ∈κ of elements of κ,
αf (x) =
if x has < μ distinct components,
αf (x) = f (x0) if x has μ distinct components.
Let A be the algebra with underlying set κ and the above operations αf , which all have arities μ < λ. Applied to any subset of cardinality < μ, these operations give nothing new, so A cannot be generated by < μ elements, hence neither can Aκ. However, Aκ can be generated by μ elements, namely the identity map κ → κ, and any μ distinct constants (or indeed, any family of μ elements which together achieve μ distinct values at each coordinate i ∈κ). For, given an element f ∈ Aκ = κκ that one wants to obtain, one takes the corresponding operation αf and applies it to the μ-tuple having first entry the identity map, and remaining entries the μ other elements described above.
Hence, though A can be generated by ≤ κ elements, and therefore, as shown in Theorem 16, by < λ elements, namely, by μ elements, it cannot be generated by a set of any smaller cardinality.
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