# 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? (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 $I \mapsto k \cdot (1,1) + I \times \{0\}$.

• About terminology: the confusion, if any, is in the other direction: Noetherian was originally defined for finite generation of ideals in rings and then extended to other contexts.
– YCor
Oct 28, 2016 at 5:52
• For non-fields $k$, even for $k = \mathbb{Z}$, this seems to be a very restrictive property. For example, $\mathbb{Z}[X]$ does not have the property since the ring $\mathbb{Z}[2X,2X^2,2X^3,\ldots]$ is not finitely generated. I wonder (just a guess) if $\mathbb{Z}$-algebras with this property have to be subrings of a finite direct product of localizations of $\mathbb{Z}$ at finitely generated submonoids of the monoid of all positive integers under multiplication, or something like that. Oct 28, 2016 at 6:52
• It seems reasonable to conjecture that when $k$ is a field, these are exactly the finitely generated algebras of Krull dimension $\leq 1$. Oct 28, 2016 at 7:19
• @EricWofsey: I think that $k[x,y]/(x^2)$ is a counterexample to this claim. Oct 28, 2016 at 8:53
• @JesseElliott: It will not be that easy. For instance, finite commutative rings have the property. Oct 28, 2016 at 9:41

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.

• Thank you. This answers the terminology question. I would call this then "super finitely generated", since my question is not primarily about the property of being Noetherian. Notice that (i) -> (ii) needs that $k$ is Noetherian, but in their paper $k$ is just an alg. closed field. Oct 30, 2016 at 19:44
• I found a paper: Hereditarily Finitely Generated Commutative Monoids by J. C. Rosales and J. I. Garcı́a-Garcı́a, Journal of Algebra 221, 723-732 (1999). They use "hereditarily finitely generated" to mean a monoid whose submonoids are all finitely generated. This term might work for you. Oct 30, 2016 at 20:11
• Probably there's no need for new terminology. For instance, Philip Hall (Finiteness conditions for soluble groups, 1954) refers to "the maximal condition for right ideals" (in a ring), "the maximal condition for subgroups" "the maximal condition for normal subgroups", etc. "The maximal condition for subalgebras" can also be found in old papers, e.g. this one: archive.numdam.org/ARCHIVE/CM/CM_1975__31_1/CM_1975__31_1_31_0/…
– YCor
Oct 30, 2016 at 20:38
• @YCor: Thank you. But "ascending chain condition" (ACC) seems to be more standard and refers to arbitrary partial orders. Oct 30, 2016 at 20:45
• @KeithKearnes: This sounds good, especially because it is an adjective in contrast to ACC or maximal condition. Oct 30, 2016 at 20:46

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.

• Thank you. Why is ACC preserved by finite products? This seems to be a nontrivial statement. A subalgebra of a product is not determined by its projections. Oct 30, 2016 at 20:49
• You have to play with both intersections and projections. If $H_1\subset H_2\subset A\times B$ and $H_1$ and $H_2$ have the same projections and intersections then $H_1=H_2$.
– YCor
Oct 30, 2016 at 20:55
• What do you mean by intersection? $A$ is not a subalgebra of $A \times B$. Oct 30, 2016 at 20:55
• Here is a comment about the Noetherian assumption on $k$: A commutative ring $k$ is Noetherian iff the $k$-algebra $k\times k$ is hereditarily finitely generated. Oct 30, 2016 at 22:03
• @KeithKearnes thanks for the remark (there's indeed a canonical poset isomorphism between the poset of ideals in $k$ and the poset of subalgebras of $k\times k$)
– YCor
Oct 30, 2016 at 22:20