# Semiprime (but not prime) ring whose center is a domain

The center of a prime ring is a domain and the center of a semiprime ring is reduced.

Now I have no evidence to believe that if the center of a semiprime ring R is a domain,

then R has to be a prime ring. So I'm looking for some examples of semiprime rings R

with these properties: R is not prime and the center of R is a domain.

This is not important but it'd be nice if the center of R is not too small.

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I find it preposterous to completely answer a technical question providing an explanation and references, and get exactly ZERO response. – Victor Protsak Aug 5 '10 at 16:42

A whole class of examples of this kind can be obtained from prime ideals in enveloping algebras with the same central character. I will sketch the construction in the case of primitive ideals in simple Lie algebras, but these conditions can be considerably relaxed.

Let $\mathfrak{g}$ be a complex simple Lie algebra of rank at least $2$ (i.e. not $\mathfrak{sl}_2$), $U(\mathfrak{g})$ its universal enveloping algebra, $I_1, I_2$ two incomparable primitive ideals with the same infinitesimal character, and $I=I_1\cap I_2$ their intersection. Then $A=U(\mathfrak{g})/I$ is semiprimitive, and hence semiprime. By the assumption, $I_1$ and $I_2$ intersect $Z(\mathfrak{g})$ at the same maximial ideal, so $Z(A)=\mathbb{C},$ which is a domain. To get a larger center, you can repeat this construction with incomparable prime ideals whose intersection with $Z(\mathfrak{g})$ is the same non-maximal prime ideal of the latter ring.

If you know representation theory of simple Lie algebras, here is an explicit construction of a pair of ideals with these properties: let $\lambda$ be an integral dominant weight, choose two different simple reflections $s_i, i=1, 2$ in the Weyl group, and let $I_i=\text{Ann}\ L(s_i*\lambda)$ be the annihilator of the simple highest weight module with highest weight $s_i(\lambda+\rho)-\rho.$ The ideals $I_1$ and $I_2$ have the same infinitesimal character by the Harish-Chandra isomorphism and they are incomparable by the theory of $\tau$-invariant: $\tau(I_i)=\{s_i\}$, but $\tau$-invariant is compatible with the containment of primitive ideals.

Everything except for the $\tau$-invariant is explained in Dixmier's "Enveloping algebras", and you can find the rest in Borho–Jantzen's or Vogan's old papers (you need the main property of the $\tau$-invariant stated above) or read Jantzen's book about the enveloping algebras (in German) for the whole story.

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For those less familiar with Lie algebras here is a somewhat more prosaic example, with the added benefit that the ring is reduced. Let $F$ be a field, and take the free algebra $R:=F\langle a,b,c\ :\ ac^nb=bc^na=0,n\in \mathbb{N}\rangle$. In other words, any monomial containing both $a$ and $b$ is zero. Thus $aRb=0$, so $R$ is not prime.

Next, assume $r\in R$ satisfies $r^2=0$. Thus $\overline{r}\in R/(b)\cong F\langle a,c\rangle$ is nilpotent; but this factor ring is reduced. Thus $r\in (b)$. By a symmetric argument $r\in (a)$. But $(a)\cap (b)=(0)$. Thus $R$ is reduced.

Finally, suppose $r\in R$ commutes with $c$. If $r\notin F[c]\subseteq R$, let $m$ be a monomial of maximal degree in $r$ not an $F$-multiple of a power of $c$. Then $cm\neq mc$, hence (by maximality) $cr\neq rc$. But if $r\in F[c]$, then it doesn't commute with $a$ unless $r$ is constant. Thus the center of $R$ is $F$, a field.

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