relation between finite generating set and ascending chain condition

Question

A commutative ring satisfies the ascending chain condition (ACC) if and only if every ideal is finitely-generated. Are there algebraic structures for which this equivalence is not true? For example, is ACC $\Leftrightarrow$ finitely-generated true for modules? If not, what are some examples of modules that satisfy one but not the other condition?

Answer 1

If by "algebra" you mean "a set with a collection of finitary operations on it" (a finitary operation on $A$ means a function $A^n\to A$, where $n$ is a natural number, possibly $0$), and by "subalgebra" we mean "a subset which is closed under the operations", then we have three statements about an algebra $A$:

1. Every subalgebra of $A$ is finitely generated.
2. Every ascending chain of subalgebras of $A$ stabilizes.
3. Every nonempty collection of subalgebras of $A$, partially ordered by inclusion, has maximal elements.

Theorem. For an algebra $A$:

1. $(1)\implies (2)$.
2. Assuming the Axiom of Dependent Choice, $(2)\implies (1)$.
3. $(3)\implies (1)$ and $(3)\implies (2)$.
4. Assuming the Axiom of Choice, $(1)\implies(3)$ and $(2)\implies(3)$.

(Cf. with this previous post )

Proof. $(1)\implies(2)$: It is easy to verify that the union of an increasing chain of subalgebras of $A$ is itself a subalgebra of $A$ (this uses the fact that operations are finitary, btw). If $A_1\subseteq A_2\subseteq\cdots$ is an ascending chain of subalgebras of $A$, then let $B=\cup A_i$; then this is a subalgebra. By $(1)$, it is finitely generated. Let $b_1,\ldots,b_k\in B$ be such that $\langle b_1,\cdots,b_n\rangle = B$. For each $j$, $j=1,\ldots,k$, there exists an index $i_j$ such that $b_j\in A_{i_j}$. Letting $i_0=\max\{i_1,\ldots,i_k\}$, we have that $b_j\in A_{i_0}$ for $j=1,\ldots,k$. Therefore, for all $n\geq i_0$, $$ B = \langle b_1,\ldots,b_k\rangle \subseteq A_{i_0}\subseteq A_n \subseteq B,$$ so $A_{i_0}=A_n$ for all $n\geq i_0$. Thus, the chain stabilizes, as claimed.

$(2)\implies (1)$, using Dependent Choice. Let $B$ be a subalgebra of $A$. If $B$ is empty, then it is generated by the empty set, so we may assume $B$ is not empty. Assume $B$ is not finitely generated. Let $x_0\in B$; then $\langle x_0\rangle\neq B$, so we can choose $x_1\in B\setminus\langle x_0\rangle$. Continuing this way, using the Axiom of Dependent Choice, we obtain as sequence $x_0,x_1,x_2,\ldots,x_n,\ldots$ of elements of $B$ such that $$\langle x_0\rangle\subsetneq \langle x_0,x_1\rangle \subsetneq \cdots \subsetneq \langle x_0,x_1,\ldots,x_n\rangle\subsetneq \cdots$$ so that $A$ does not satisfy ACC on substructures.

$(3)\implies(1)$. Let $B$ be a subalgebra of $A$. Consider the set of all finitely generated subalgebras of $B$, ordered by inclusion. The set is nonempty (it contains the subalgebra generated by the empty set), so by $(3)$ it has maximal elements. Let $C$ be a maximal element of this collection. Then $C\subseteq B$. For each $b\in B$, we have that $\langle C,b\rangle$ is a finitely generated subalgebra of $B$, and $C\leq\langle C,b\rangle$, so the maximality of $C$ gives $C=\langle C,b\rangle$; hence $b\in C$. Thus, $B\subseteq C$, and we have equality. In particular, $B=C$ is finitely generated.

$(3)\implies(2)$. We can obtain the implication by "going through $(1)$", or directly: if $B_1\subseteq B_2\subseteq\cdots$ is an ascending chain of subalgebras of $A$, then the set $\{B_i\}_{i\in I}$ has maximal elements. If $B_k$ is a maximal element of the chain, then for all $n\geq k$ we have $B_k\subseteq B_n$, hence $B_k=B_n$ by maximality of $B_k$. So the chain stabilizes.

$(1)\implies(3)$: this is Zorn's Lemma: if $\mathcal{S}$ is a collection of subalgebras of $A$ ordered by inclusion, let $\mathcal{C}$ be a chain in $\mathcal{S}$, and let $C=\cup\mathcal{C}$. Then $C$ is finitely generated, and so we can prove as in the implication from $(1)$ to $(2)$ that $C$ is an element of $\mathcal{C}$, and therefore that $\mathcal{C}$ is bounded above in $\mathcal{S}$. Therefore, $\mathcal{S}$ has maximal elements.

$(2)\implies(3)$: Using AC, we can go from $(2)$ to $(1)$ to $(3)$. $\Box$

The result for ideals follows by considering an enriched algebraic structure on $A$, in which we add additional unary operations $\ell_r$ and $\rho_r$ for each $r\in R$, corresponding to left multiplication by $r$ and right multiplication by $r$, respectively. Then the ideals are precisely the "subalgebras" of $R$.

So to get examples where $(1)$ and $(2)$ are not equivalent, you would need a structure with infinitary operations in which an ascending union of subalgebras is not necessarily a subalgebra.