Algebras whose subalgebras are finitely generated
Let $k$ be a commutative ring. Is there a name for those commutative $k$-algebras with the property that every subalgebra is finitely generated? $\ldots$ Where can I read more about them?
The paper
Rogalski, D.; Sierra, S. J.; Stafford, J. T., Algebras in which every subalgebra is Noetherian. Proc. Amer. Math. Soc. 142 (2014), no. 9, 2983-2990.
introduces the term supernoetherian for a not-necessarily-commutative $k$-algebra $A$ that has the property that all subalgebras of $A$ are both (i) finitely generated and (ii) Noetherian. In the commutative case, when $k$ is Noetherian, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here when $k$ is Noetherian. The authors of this paper consider only the case where $k$ is an algebraically closed field, and in this case they do observe that the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.
I'll assume that $k$ is noetherian. I'll just write $k$-ACC, or ACC if no ambiguity, to mean the ascending chain condition on $k$-subalgebra.
(For $k$ arbitrary, $A=k$ is the only $k$-subalgebra of itself so satisfies the property, so can be arbitrarily bad.)
Here's an equivalence which then boils down to the case of a domain:
A $k$-algebra $A$ ($k$ is noetherian) has ACC iff the following 3 condition hold:
(i) $A$ is noetherian
(ii) the nilradical $N_A$ of $A$ is a finitely generated $k$-module
(iii) $A/P$ has ACC for every (minimal) prime ideal $P$ of $A$.
Indeed suppose that $A$ has ACC. (iii) immediately follows.
Let $(I_n)$ be an increasing sequence of ideals; so $(I_n\cap k1_A)$ is an ascending sequence of ideals of $k1_A$ and hence is stationary, say for $n\ge n_0$. Also $(k1_A+I_n)$ is an ascending sequence of $k$-subalgebras of $A$. So there exists $n_1$ (say $\ge n_0$) such that for every $n\ge n_1$ and $x\in I_n$, one can write $x=x'+t1_A$ with $x'\in I_{n_1}$ and $t\in k$. Then $x'-x\in I_n\cap k1_A$ and hence $x'-x\in I_{n_0}$. Thus $x\in I_{n_1}$; whence $I_n=I_{n_1}$ for all $n\ge n_1$. This proves (i).
To prove (ii), use that $A$ is noetherian to write a nested sequence of submodules $0\le N_1\le \dots N_k=N_A$ with each $N_i$ isomorphic as an $A$-module to $A/P_i$ for some prime $i$. Suppose by contradiction that some $N_i$ is an infinitely generated $k$-module. Since ACC passes to quotients, we can suppose that $i=1$. Since $P_i$ annihilates $N_1$ and contains the nilradical, we see that $xy=0$ for all $x,y\in N_1$. Therefore, for every $k$-submodule $V$ of $N_1$, the $k$-subalgebra generated by $V$ is reduced to $k1_A+V$. So if $(V_n)$ is an increasing sequence of submodules, from ACC we deduce that for large $n$, $k1_A+V_n=k1_A+V_{n+1}$ (in other words, the canonical map $V_n/(k1_A\cap V_n)\to V_{n+1}/(k1_A\cap V_{n+1})$ is an isomorphism). Since $k$ is noetherian, $(k1_A\cap V_n)$, as an ascending sequence of $k$-submodule of $k1_A$, is also stationary, say with union $W$. Hence the above canonical map is the inclusion $V_n/W\to V_{n+1}/W$; since it is an isomorphism it means that $V_{n+1}=V_n$.
Conversely suppose that (i),(ii),(iii) hold. It is easy to check (see below) that a finite direct product of $k$-algebras with ACC has ACC. By (i), $A$ has finitely many minimal primes $P_i$, so $A/N_A$ embeds as subalgebra in the finite product $\prod A/P_i$, which has ACC using (iii), hence $A/N_A$ has ACC. Also it is immediate that if an algebra $A$ has an ideal $I$ that is a f.g. $k$-module and $A/I$ has ACC then $A$ has ACC. Then using (ii) $A$ has ACC.
Fact: ($k$ noetherian) the $k$-ACC condition passes to finite direct products.
Proof: it's enough to do it for a product $A\times B$. Let $(H_n)$ be an ascending sequence of subalgebras of $A\times B$. The projections being stationary, we can suppose that the projections of $H_n$ on both $A$ and $B$ are surjective. It follows that the intersection $H_n\cap (A\times\{0\})$ is an ideal in $A$. Since $A$ is noetherian (as I first checked: this didn't use this finite product claim), this intersection is stationary and we're done.