# 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?

• What is $A[b]$? Dec 30, 2014 at 12:35

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.