Skip to main content
added 6 characters in body
Source Link
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86

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 a fieldNoetherian, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here when $k$ is a fieldNoetherian. The authors of this paper do observe that, whenconsider only the case where $k$ is aan 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.

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 a field, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here when $k$ is a field. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

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.

added 21 characters in body
Source Link
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86

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 a field, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here when $k$ is a field. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

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, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

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 a field, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here when $k$ is a field. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

added 22 characters in body
Source Link
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86

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, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

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, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here. The authors of this paper do observe that the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

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, (ii) follows from (i) by the Hilbert Basis Theorem, so these are exactly the $k$-algebras asked about here. The authors of this paper do observe that, when $k$ is a field, the commutative supernoetherian algebras have Krull dimension at most $1$, but they do not classify them.

Source Link
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
Loading