## von Neumann regular rings (2)

Posted: October 22, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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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 $MR \subseteq \ker f. \ \Box$

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)).