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

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