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

Proof.  Let a be a central element of R and let x \in R be such that a=axa=a^2x. So a^2x is central. Let z \in R. Then a^2xz=za^2x and hence xa^2z=a^2zx, i.e. a^2z commutes with x and so it commutes with x^3. Therefore a^2x^3z=za^2x^3, i.e. y=a^2x^3 is central. But, since a^2x^2=ax, we have y=ax^2 and clearly aya=a^2x^2a=axa=a. \Box

Remark. In a commutative regular ring R every prime ideal P is maximal. To see this, let a \in R \setminus P and r \in R be such that a=ara. Then a(1-ra)=0 \in P and so 1-ra \in P. Therefore P+Ra=R and hence P is maximal.

Notation. Let M be a maximal ideal of Z(R), the center of R. The localization of R at M is denoted by R_M.

Lemma. If R is a commutative regular ring and M is a maximal ideal of R, then R_M is a field.

Proof. The unique maximal ideal of R_M is M_M. So, to prove that R_M is a field, we only need to show that M_M=\{0\}, i.e. for every a \in M, there exists some s \notin M such that sa=0. This is easy to see: we have a=xa^2, for some x \in R, because R is a commutative regular ring, and thus (1-xa)a=0 and clearly s=1-xa \notin M because a \in M. \ \Box

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 R is regular if and only if R_M is a field for any maximal ideal M of R is due to Kaplansky.

Theorem 2. (Armendariz, 1974) Let R be a ring with the center Z(R). If Z(R) is regular, then R/MR \cong R_M for any maximal ideal M of Z(R).

Proof. Let f: R \longrightarrow R_M be the natural homomorphism defined by f(x)=1^{-1}x, for all x \in R. Let S=Z(R) \setminus M.

1) f is surjective. To see this, let s^{-1}x \in R_M. Since Z(R) is regular, there exists some c \in Z(R) such that s=cs^2. Hence s(1-cs)=0 and therefore f(cx)=1^{-1}cx = s^{-1}x.

2) \ker f = MR. To see this, let f(x)=1^{-1}x=0. That means sx = 0 for some s \in S. Since Z(R) is regular, there exists some c \in Z(R) such that s(1-cs)=0 \in M. Thus 1-cs \in M because s \notin M. Therefore x=(1-cs)x \in MR. This proves \ker f \subseteq MR. For the other side of the inclusion, we first apply the above lemma to Z(R) to get 1^{-1}a=0 for all a \in M. Thus for every a \in M and x \in R we have f(ax)=1^{-1}ax = 1^{-1}a 1^{-1}x = 0. So ax \in \ker f and therefore MR \subseteq \ker f. \ \Box

  1. 23051982 says:

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

    • Yaghoub says:

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

      0 \longrightarrow I \longrightarrow R \longrightarrow aR \longrightarrow 0.

      since aR is projective, the sequence is split and hence R \cong aR \oplus I.

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

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