## Ore domains (1)

Posted: January 3, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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Definition. Let $R$ be a domain and $S=R \setminus \{0\}.$ If $S$ is left (resp. right) Ore, then $R$ is called a left (resp. right) Ore domain.

Remark 1. Note that a domain $R$ is a left Ore domain if and only if $Rr_1 \cap Rr_2 \neq (0),$ for all non-zero elements $r_1,r_2 \in R.$ Similarly $R$ is a right Ore domain if and only if $r_1R \cap r_2R \neq (0),$ for all non-zero elements $r_1,r_2 \in R.$

Remark 2. The left (resp. right) quotient ring of a left (resp. right) Ore domain is a division ring. The reason is that every element of $Q(R)$ is in the form $x = s^{-1}r,$ where $0 \neq s \in R, \ r \in R.$ Now if $x \neq 0,$ then $r \neq 0$ and then $x^{-1}=r^{-1}s.$

Theorem. Every left (resp. right) Noetherian domain is left (resp. right) Ore.

Proof. We will prove the theorem for left Noetherian domains only. The “right” version, as usual, has the same argument. Let $r_1, r_2 \in R$ be non-zero. We need to show that $Rr_1 \cap Rr_2 \neq (0).$ So suppose, on the contrary, that $Rr_1 \cap Rr_2 = (0).$ We will show that the sum $\sum_{n=0}^{\infty} R r_1r_2^n$ is direct and thus $R$ cannot be left Noetherian because then we would have the non-stopping increasing chain of left ideals

$Rr_1 \subset Rr_1 \oplus Rr_1r_2 \subset Rr_1 \oplus Rr_1r_2 \oplus Rr_1r_2^2 \subset \cdots.$

To prove that the sum is direct, suppose that the sum is not direct and choose $n$ to be the smallest postive integer for which there exist $a_j \in R, \ j=0, \cdots , n,$ not all zero, such that $\sum_{j=0}^n a_jr_1r_2^j = 0.$ Clearly $n \geq 1.$ Thus

$-a_0r_1=\left (\sum_{j=0}^{n-1} a_{j+1}r_1r_2^{j} \right)r_2 \in Rr_1 \cap Rr_2 = (0).$

Therefore, since $r_1, r_2$ are non-zero and $R$ is a domain, we must have $a_0=0$ and $\sum_{j=0}^{n-1} a_{j+1}r_1r_2^j = 0,$ contradicting the minimality of $n. \ \Box$