Non-commutative normalization Let $A$ be a (non-commutative) associative algebra with 1. Assume that $A$ contains a cental subalgebra $Z$ such that 
a) $Z$ is a noetherian domain
b) $A$ is a finitely generated module over $Z$.
Let us denote by $K$ the field of fractions of $Z$ and let $B=A\underset{Z}\otimes K$.
Let us also set
$$
\overline{A}=\{ b\in B|\ A[b]\ \text{is a finitely generated module over $Z$}\}
$$
$\mathbf{Questions:}$
1) Is $\overline{A}$ a subalgebra of $B$?
2) If the answer to 1) is positive, then is $\overline{A}$ a finitely generated module over $Z$?
3) If the answer to either 1) or 2) is negative, are there any conditions that one can impose on $A$ so that the answer becomes positive?
 A: Let $Z=\mathbb{Z}$ be the ring of integers.  This is a noetherian domain, and so satisfies property (a).
Let $A=\mathbb{Z}\langle x,y\ :\ x^2=2x,y^2=2y,xyx=2x,yxy=2y\rangle$.  This is a (non-commutative) associative algebra with $1$.  Further, it contains $Z$ as a central subalgebra.  Moreover, $A$ is finitely generated as a $Z$-module, by the elements $\{1,x,y,xy,yx\}$.  Thus condition (b) is satisfied.
Letting $K=\mathbb{Q}$, this is the field of fractions of $Z$, and $B=A\otimes_Z K\cong \mathbb{Q}\langle x,y\ :\ x^2=2x,y^2=2y,xyx=2x,yxy=2y\rangle$.  We will write elements of $B$ using this isomorphism.
Since $A$ naturally embeds into $B$ (because $Z$ is regular in $A$) I'll take $A[b]$ to mean the subalgebra of $B$ generated by $A$ and $b$ (so we don't have to deal with Sasha's issue).
Consider the element $b_1=\frac{1}{2}x\in B$.  It turns out that $A[b_1]$ is a finitely generated $Z$-module, generated by $\{1,\frac{1}{2}x,y,\frac{1}{2}xy,\frac{1}{2}yx\}$.  Similarly, $A[b_2]$ is finitely generated, with $b_2=\frac{1}{2}y$.
However, $A[b_1,b_2]$ is not finitely generated, since alternating products $b_1b_2b_1b_2\cdots$ can give rise to arbitrarily large denominators.  So the answer to (1) is no.
The algebra $A$ is fairly simple.  We could even replace $\mathbb{Z}$ by the localization $\mathbb{Z}_{(2)}$, so $A$ would be a finitely generated module over a local ring.
