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(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$,$\mathfrak{F}_B$ are the smallest free algebras such that there are canonical projection epimorphisms: $\pi_A:\mathfrak{F}_A\rightarrow A$ , $\pi_B:\mathfrak{F}_B\rightarrow B$ (of $R$-algebras), making the following diagram of $R$-modules commute:

\begin{array}{ccccccccc} 0 & \longrightarrow & Ker(\pi) & \overset{ker(\pi_A)}{\longrightarrow} & \mathfrak{F}_A & \overset{\pi_A}{\longrightarrow} & A & \longrightarrow & 0\\ \| & & j\downarrow & & \iota'\downarrow& & \iota\downarrow & & \|\\ 0 & \longrightarrow & Ker(\pi_B) & \overset{ker(\pi')}{\longrightarrow} & \mathfrak{F}_B & \overset{\pi_B}{\longrightarrow} & B & \longrightarrow & 0 \end{array}

(where $\iota: A \rightarrow B$, $\iota': \mathfrak{F}_A \rightarrow \mathfrak{F}_B$ are $R$-algebra inclusion (injective) morphisms of $A$ into $B$ and $\mathfrak{F}_A$ into $\mathfrak{F}_B$, respectivly and $j$ is the unique $R$-module monomorphism making the diagram commute).

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.

(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$ and $\mathfrak{F}_B$ are the smallest free algebras admitting canonical projection (R-algebra) morphisms $\pi_A:\mathfrak{F}_A \rightarrow A$ and $\pi_B: \mathfrak{F}_B \rightarrow B$ respectivly; then, $\mathfrak{F}_A$ must be a subalgebra of $\mathfrak{F}_B$.

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.

(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$,$\mathfrak{F}_B$ are free algebras such that there are canonical projection epimorphisms: $\pi_A:\mathfrak{F}_A\rightarrow A$ , $\pi_B:\mathfrak{F}_B\rightarrow B$ (of $R$-algebras), making the following diagram of $R$-modules commute:

\begin{array}{ccccccccc} 0 & \longrightarrow & Ker(\pi) & \overset{ker(\pi_A)}{\longrightarrow} & \mathfrak{F}_A & \overset{\pi_A}{\longrightarrow} & A & \longrightarrow & 0\\ \| & & j\downarrow & & \iota'\downarrow& & \iota\downarrow & & \|\\ 0 & \longrightarrow & Ker(\pi_B) & \overset{ker(\pi')}{\longrightarrow} & \mathfrak{F}_B & \overset{\pi_B}{\longrightarrow} & B & \longrightarrow & 0 \end{array}

(where $\iota: A \rightarrow B$, $\iota': \mathfrak{F}_A \rightarrow \mathfrak{F}_B$ are $R$-algebra inclusion (injective) morphisms of $A$ into $B$ and $\mathfrak{F}_A$ into $\mathfrak{F}_B$, respectivly and $j$ is the unique $R$-module monomorphism making the diagram commute).

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.

(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$,$\mathfrak{F}_B$ are the smallest free algebras such that there are canonical projection epimorphisms: $\pi_A:\mathfrak{F}_A\rightarrow A$ , $\pi_B:\mathfrak{F}_B\rightarrow B$ (of $R$-algebras), making the following diagram of $R$-modules commute:

\begin{array}{ccccccccc} 0 & \longrightarrow & Ker(\pi) & \overset{ker(\pi_A)}{\longrightarrow} & \mathfrak{F}_A & \overset{\pi_A}{\longrightarrow} & A & \longrightarrow & 0\\ \| & & j\downarrow & & \iota'\downarrow& & \iota\downarrow & & \|\\ 0 & \longrightarrow & Ker(\pi_B) & \overset{ker(\pi')}{\longrightarrow} & \mathfrak{F}_B & \overset{\pi_B}{\longrightarrow} & B & \longrightarrow & 0 \end{array}

(where $\iota: A \rightarrow B$, $\iota': \mathfrak{F}_A \rightarrow \mathfrak{F}_B$ are $R$-algebra inclusion (injective) morphisms of $A$ into $B$ and $\mathfrak{F}_A$ into $\mathfrak{F}_B$, respectivly and $j$ is the unique $R$-module monomorphism making the diagram commute).

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.

(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$,$\mathfrak{F}_B$ are the smallest free algebras such that there are canonical projection epimorphisms: $\pi_A:\mathfrak{F}_A\rightarrow A$ , $\pi_B:\mathfrak{F}_B\rightarrow B$ (of $R$-algebras), making the following diagram of $R$-modules commute:

\begin{array}{ccccccccc} 0 & \longrightarrow & Ker(\pi) & \overset{ker(\pi_A)}{\longrightarrow} & \mathfrak{F}_A & \overset{\pi_A}{\longrightarrow} & A & \longrightarrow & 0\\ & & j\downarrow & & \iota'\downarrow& & \iota\downarrow\\ 0 & \longrightarrow & Ker(\pi_B) & \overset{ker(\pi')}{\longrightarrow} & \mathfrak{F}_B & \overset{\pi_B}{\longrightarrow} & B & \longrightarrow & 0 \end{array}

(where $\iota: A \rightarrow B$, $\iota': \mathfrak{F}_A \rightarrow \mathfrak{F}_B$ are the $R$-algebra inclusion morphisms of $A$ into $B$ and $\mathfrak{F}_A$ into $\mathfrak{F}_B$, respectivly and $j$ is the unique $R$-module morphism making the diagram commute, )

• then: $j$ is a $R$-module monomorphism.

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.

(All rings here are always assumed to be unital and associative).

Setup

Let $R$ be a ring, and $A$ and $B$ be $R$-algebras, with $A$ a commutative subalgebra of $B$ satisfying:

1. If $u$ is a unit in $B$, then $u$ is in $A$.
2. If $\mathfrak{F}_A$,$\mathfrak{F}_B$ are free algebras such that there are canonical projection epimorphisms: $\pi_A:\mathfrak{F}_A\rightarrow A$ , $\pi_B:\mathfrak{F}_B\rightarrow B$ (of $R$-algebras), making the following diagram of $R$-modules commute:

\begin{array}{ccccccccc} 0 & \longrightarrow & Ker(\pi) & \overset{ker(\pi_A)}{\longrightarrow} & \mathfrak{F}_A & \overset{\pi_A}{\longrightarrow} & A & \longrightarrow & 0\\ \| & & j\downarrow & & \iota'\downarrow& & \iota\downarrow & & \|\\ 0 & \longrightarrow & Ker(\pi_B) & \overset{ker(\pi')}{\longrightarrow} & \mathfrak{F}_B & \overset{\pi_B}{\longrightarrow} & B & \longrightarrow & 0 \end{array}

(where $\iota: A \rightarrow B$, $\iota': \mathfrak{F}_A \rightarrow \mathfrak{F}_B$ are $R$-algebra inclusion (injective) morphisms of $A$ into $B$ and $\mathfrak{F}_A$ into $\mathfrak{F}_B$, respectivly and $j$ is the unique $R$-module monomorphism making the diagram commute).

Question:

Can anything be deduced about the global dimension $D(B)$ of $B$, with respect to $D(A)$?

Hypothesis & Some Test Results:

I'm strongly inclined to believe, that in such a situation $D(R)\geq D(A)$, for example this hold for the Weyl algebra $A_n(k)$ and $k[x_1,..,x_n]$.
For $R[x_1,..,x_n]$ and $Z(R)[x_1,...,x_n]$...

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ does not contain some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra. Or assumes $A$ to be a subalgebra of $B$ with "more relations" which is not possible by the assumptions $1$ and $2$, respectively.