### George M. Bergman, Two Statements about Infinite Products that Are Not Quite True, Contemporary Mathematics, v.420, 2006.  MR 2007k:16008.

Two questions have been answered, Question 6 and Question 18:

Solution to Question 6, p.40.  For  R  a ring and  (Mi)iω  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)iω,  let    be their direct product, find a subring  R ⊆   such that    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)iω  and an ultrafilter  U  on  ω  such that, if we write    for the ultraproduct of the  Ri  determined by  U,  then    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   → ,  it is not hard to verify that    will be countably but not finitely generated as a left R-module.

To find such    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    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 ⊂  = ∏ Ri  be the inverse image in that ring of the "R" constructed above.  (And then, if we wish, replacing this    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  (Mi)iω  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 = (x)κ  of elements of  κ,

α(x) =   x0    if  x  has  < μ  distinct components,
α(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.