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

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