Algebras whose subalgebras are finitely generated

Question

Let $k$ be a commutative ring. Is there a name for those commutative $k$-algebras with the property that every subalgebra is finitely generated? (Equivalently, the partial order of subalgebras is Noetherian.) Can we say something about their structure? Where can I read more about them? I am particularly interested in the cases $k=\mathbb{Z}$ and when $k$ is field.

Notice that if $k$ is a field then $k[x,y]$ does not have the property, but $k[x]$ does. One can also check directly that the subrings of $\mathbb{Z} \times \mathbb{Z}$ are finitely generated because they are given by $\mathbb{Z}[(n,0)] = \mathbb{Z} \times_{\mathbb{Z}/n} \mathbb{Z}$, where $n \geq 0$. I wonder if there is a more conceptual reason for this.

(Usually an object is called Noetherian if its partial order of subobjects is Noetherian, and this applies for example to (non-abelian) groups and modules. But a ring is usually called Noetherian if its partial order of quotients is Noetherian. This is somewhat confusing. Therefore, although it would be consistent to call a ring Noetherian if its partial order of subrings is Noetherian, this would contradict the usual terminology.)

Edit. If $k$ is a field, then a necessary condition is that the algebra is finitely generated of Krull dimension $\leq 1$. This follows directly from Noether normalization and the fact, already mentioned above, that $k[x,y]$ does not have the property. But this condition is not sufficient, as the example $k[x,y]/(x^2)$ shows.

Summary of the answers so far. Keith Kearnes suggests the terms "supernoetherian" (if $k$ is Noetherian) and "hereditary finitely generated" (which sounds very good). YCor has reduced the general classification to the case of finitely generated $k$-domains (if $k$ is Noetherian). The classification in this case is still open. If $k$ is a field, is it equivalent to Krull dimension $\leq 1$? Is there a characterization if $k$ is not a field?

YCor has also generalized my observation on $\mathbb{Z} \times \mathbb{Z}$: The $k$-algebra $k \times k$ is hereditarily finitely generated if and only if $k$ is Noetherian, because in fact there is an isomorphism of partial orders between ideals of $k$ and subalgebras of $k \times k$ given by $I \mapsto k \cdot (1,1) + I \times \{0\}$.

Answer 1

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.

Answer 2

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.

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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.