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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 $\mathfrak{F}_A$ 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.

Improve this question
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    $\begingroup$ I may be misunderstanding your question, but isn't $\mathbb{C}[x^2,x^3]\subseteq\mathbb{C}[x]$ a counterexample? $\endgroup$
    – Jeremy Rickard
    Commented Jun 6, 2014 at 18:57
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    $\begingroup$ I think your edit has several typos, but if $\pi$ is surjective, how can $\pi'$ be surjective if $A\neq B$? $\endgroup$
    – Jeremy Rickard
    Commented Jun 8, 2014 at 17:16
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    $\begingroup$ You call $\iota'$ an inclusion, but it's not automatically injective (if it is, then so is $j$, obviously). And $\pi_A$ and $\pi_B$ are not uniquely determined, so it's conceivable that the injectivity depends on the choice. Is the hypothesis that there is some choice of $\pi_A$ and $\pi_B$ such that $\iota'$ is injective? Also, I think some of the maps in your diagram are mislabelled, and is "$A$ is some unit of $B$" in the last paragraph a typo? $\endgroup$
    – Jeremy Rickard
    Commented Jun 10, 2014 at 15:22
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    $\begingroup$ Did you mean to drop the "smallest possible" condition on $\mathfrak{F}_A$ and $\mathfrak{F}_B$ in your last edit? If so, then you can make $\mathbb{C}[x^2,x^3]\subseteq\mathbb{C}[x]$ into a counterexample by choosing a reduntantly large set of generators for $\mathbb{C}[x]$. $\endgroup$
    – Jeremy Rickard
    Commented Jun 11, 2014 at 8:59
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    $\begingroup$ There's still the problem that the hypotheses allow non-trivial choice in the selection of $\pi_A$, $\pi_B$ and $\iota'$ which can affect whether or not $\iota'$ and $j$ are injective. For example, for a field $k$, take $A=k[x^2,x^3]$, $\mathfrak{F}_A=k\langle u,v\rangle$ with $\pi_A(u)=x^2$, $\pi_A(v)=x^3$, $B=k[x,s_1,s_2,t_1,t_2]$, $\mathfrak{F}_B=k\langle x,s_1,s_2,t_1,t_2\rangle$ with the obvious $\pi_B$. Then the obvious choice of $\iota'$ isn't injective, but there are choices that are: e.g., $\iota'(u)=x^2+s_1s_2-s_2s_1$ and $\iota'(v)=x^3+t_1t_2-t_2t_1$. $\endgroup$
    – Jeremy Rickard
    Commented Jun 11, 2014 at 21:09

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