## Regular submonoids

Posted: December 28, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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Throughout $S$ is a regular submonoid of the ring $R,$ i.e.$S$ is multiplicatively closed, $1_R \in S$ and every element of $S$ is regular.

Remarks:

1) $S^{-1}R$ (resp. $RS^{-1}$) is defined if and only if $S$ is left (resp. right) Ore. In this case, since every element of $S$ is regular, the corresponding map $\varphi : R \longrightarrow S^{-1}R$ (resp. $\varphi : R \longrightarrow RS^{-1}$) defined by  $\varphi (r) = 1_R^{-1}r$ (resp. $\varphi (r) = r1_R^{-1}$) is one-to-one and hence $S^{-1}R$ (resp. $RS^{-1}$) contains a copy of $R.$

2) If every element of $S$ is a unit, then $Rs = R,$ for all $s \in S$ and thus $Rs \cap Sr = Sr \neq \emptyset,$ for all $s \in S, \ r \in R.$ So, $S$ is left (and right) Ore and therefore $S^{-1}R$ (and $RS^{-1}$) both exist and are equal to $R.$ The reason that they are equal to $R$ is that $s^{-1} \in R,$ for all $s \in S$ and thus $s^{-1}r \in R,$ for all $s \in S, \ r \in R.$

3) An example of 2): let $R$ be a left Artinian ring and $s$ a regular element of $R.$ Then the chain $Rs \supseteq Rs^2 \supseteq \cdots$ must terminate, i.e. there exists $n \geq 1$ such that $Rs^n = Rs^{n+1}.$ Thus $s^n = rs^{n+1},$ for some $r \in R.$ Therefore $(1-rs)s^n = 0,$ which gives us $rs = 1.$ Obviously the same result holds for right Artinian rings.

4) If $S$ contains all the regular elements of $R$ and it is left (resp. right) Ore, then the left (resp. right) quotient ring of $R$ is denoted by $Q(R).$

5) If $S$ is central, i.e. it is contained in the center of $R$, then it’s clearly both left and right Ore and thus $S^{-1}R \cong RS^{-1}.$

Lemma. Suppose that $S$ is central and $A=\{ s^{-1}r_1, \cdots , s^{-1}r_n \} \subset S^{-1}R.$ Let $B=\{r_1, \cdots , r_n \}.$ Then $A^m = (s^{-1})^m B^m,$ for all $m \in \mathbb{N}.$

Proof. Since $S$ is central, we have $s^{-1}x = xs^{-1}$ for all $s \in S, \ x \in R.$ Therefore $s^m(s^{-1}x_1 s^{-1}x_2 \cdots s^{-1}x_m)=x_1x_2 \cdots x_m,$ for all $x_j \in R. \ \Box$

We will use the above lemma to prove that the GK-dimension of an algebra is invariant under localization with respect to a central submonoid:

Theorem. If $R$ is an algebra over a field $F$ and $S$ is central, then $\text{GK}(S^{-1}R) = \text{GK}(R).$

Proof. Let $W=\sum_{j=1}^n Fq_j,$ where $q_j \in S^{-1}R.$ As we’ve seen before, there exit $s \in S$ and $r_1, \cdots , r_n \in R$ such that $q_j = s^{-1}r_j,$ for all $j.$ Let $V=\sum_{j=1}^n F r_j.$ Then, by the lemma, $s^m W^m = V^m,$ for all integers $m \geq 0$ (note that for $m = 0$ both sides are equal to $F$). Therefore $W^m = (s^{-1})^m V^m$ and hence $\dim_F W^m = \dim_F (s^{-1})^m V^m \leq \dim_F V^m.$ Thus $\text{GK}(S^{-1}A) \leq \text{GK}(A).$ The other direction of the inequality is trivial because $A \subseteq S^{-1}A. \ \Box$