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

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

 1. *if $u$ is a unit in $B$, then $u$ is in $A$.*
 2. If $\mathfrak{F}$ is a free algebra, $\pi:\mathfrak{F}\rightarrow A$ and $\pi'\rightarrow B$ are the corresponding quotient $R$-algebra maps and $\iota: A \rightarrow B$ is the inclusion of $A$ into $B$ then the morphism $j$ in the unique commutative diagram of $R$-modules: 


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

is an $R$-monomorphism.  



Can anything be deduced about *the global dimension $D(B)$ of $B$, with respect to $D(A)$?*
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 $k[x_1,..,x_n]$.  

Moreover, any "counter example" where $A$ is a subring of $B$ and $D(A)\not\leq D(B)$ is generated from an example where $A$ is some unit of $B$, for example any $\mathbb{Z}$-algebra as relating to any $\mathbb{R}$-algebra.