Ore domains (1)

Posted: January 3, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , ,

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

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