Uncountable integral domain such that every countable subset is contained in a finitely generated $\mathbb{Z}$-algebra

Question

Is there an uncountable integral domain such its every countable subset is contained in a finitely generated $\mathbb{Z}$-algebra?

Answer 1

1- No. Indeed, let $A$ be an uncountable domain and $X$ a maximal algebraically independent subset (over the minimal subring $A_0=\mathbf{Z}$ or $\mathbf{F}_p$). Then $A$ is contained in an algebraic closure of the field of rational functions $\mathrm{Frac}(A_0)(x:x\in X)$. In particular, it follows that $X$ is uncountable. Then, for every infinite countable subset $Y$ of $X$ there is no finitely generated subalgebra containing it.

---

2- Note: I'm more familiar with the analogous property for groups (every countable subgroup is contained in a finitely generated subgroup). There are uncountable groups with this property (e.g., symmetric groups — Galvin 1995), but no abelian ones. Similarly, there are many non-commutative (associative) algebras with this property, for instance, group algebras $FG$ of such groups $G$ for $F$ finitely generated field. I guess infinite-dimensional matrix algebras over finite fields are good candidates too.

---

3- Added: the conclusion still holds assuming that $A$ is an arbitrary scalar (= associative commutative unital) ring.

Indeed, let $A$ satisfy the condition. The first consequence is that $A$ is noetherian. Indeed, noetherian is equivalent to the condition that for every sequence $(a_n)_{n\ge 0}$, there exists $N$ such that for every $n\ge N$ there exists $b_0,\dots,b_{n-1}$ such that $a_n=\sum_{i<n}a_ib_i$. (If $A$ is not noetherian choose $J$ infinitely generated, $a_0=0$ and choose by induction $a_n$ not in the ideal $\sum_{i<n}a_iA$.) Hence, if $A$ is not noetherian, let $(a_n)$ be such a sequence: it is contained in a finitely generated subring $B$, which is noetherian, which in turns yields a contradiction.

Now $A$ is noetherian. By the domain case, for every prime ideal $P$ of $A$, $A/P$ is countable. Since $A$ is noetherian, as $A$-module, $A$ is an iterated extension of various such $A/P$. It follows that $A$ is countable.

---

4- There is a property "uncountable cofinality" which means "cannot be written as union of an increasing sequence of proper subrings" (for a countable ring, it is equivalent to being finitely generated). The property "every countable subalgebra is contained in a finitely generated one" clearly implies uncountable cofinality. The argument in 1 above easily shows the stronger result that no uncountable domain has uncountable cofinality. (Similarly, every uncountable field has countable cofinality in the category of fields.)

However there exist uncountable scalar rings of uncountable cofinality. An example, due to Koppelberg and Tits, is the Boolean algebra $(\mathbf{Z}/2\mathbf{Z})^X$ for any infinite set $X$.