**Theorem 1**. (von Neumann, 1936) The center of a regular ring is regular.

*Proof*. Let be a central element of and let be such that So is central. Let Then and hence i.e. commutes with and so it commutes with Therefore i.e. is central. But, since we have and clearly

**Remark**. In a commutative regular ring every prime ideal is maximal. To see this, let and be such that Then and so Therefore and hence is maximal.

**Notation**. Let be a maximal ideal of the center of The localization of at is denoted by

**Lemma**. If is a commutative regular ring and is a maximal ideal of then is a field.

*Proof*. The unique maximal ideal of is So, to prove that is a field, we only need to show that i.e. for every there exists some such that This is easy to see: we have for some because is a commutative regular ring, and thus and clearly because

The converse of the lemma is also true and we will prove it in a more general setting in part (3). This result that a commutative ring is regular if and only if is a field for any maximal ideal of is due to Kaplansky.

**Theorem 2**. (Armendariz, 1974) Let be a ring with the center If is regular, then for any maximal ideal of

*Proof*. Let be the natural homomorphism defined by for all Let

1) is surjective. To see this, let Since is regular, there exists some such that Hence and therefore

2) To see this, let That means for some Since is regular, there exists some such that Thus because Therefore This proves For the other side of the inclusion, we first apply the above lemma to to get for all Thus for every and we have So and therefore

“if aR is projective then aR is direct summand of R” can you help me?

let be the right annihilator of in and consider the following short exact sequence of right -modules:

since is projective, the sequence is split and hence

the above is true for any ring if is regular, then is always projective because it can be generated by an idempotent (see von Neumann regular rings (1)).