Let $D$ be an integral domain with quotient field $K$ and let $\mathcal{ I}% (D) $ be the set of nonzero ideals of $D$. Call, for $I,J\in \mathcal{ I}(D)$ , the product $IJ$ of ideals condensed if $IJ=\{ ij|i\in I,j\in J\}.$ Call $D$ a condensed domain if for each pair $I,J$ the product $IJ$ is condensed. We show that if $a,b$ are elements of a condensed domain such that $aD\cap bD=abD,$ then $(a,b)=D.$ It was shown in [Comm. Algebra 15 (1987), 1895-1920] that a pre-Schreier domain is a $\ast $-domain, i.e., $D$ satisfies $\ast :$ For every pair $\{ a_{ i}\}_{ i=1}^{ m},\{ b_{ j}\}_{ j=1}^{ n}$ of sets of nonzero elements of $D$ we have $(\cap (a_{ i}))(\cap b_{ j})=\cap (a_{ i}b_{ j}).$ We show that a condensed domain $D$ is pre-Schreier if and only if $D$ is a $ \ast $-domain. We also show that if $A\subseteq B$ is an extension of domains and $A+XB[X]$ is condensed, then $B$ must be a field and $A$ must be condensed and in this case $[B:K]<4.$ In particular we study the necessary and sufficient conditions for $D+XL[X]$ to be condensed, where $D$ is a domain and $L$ an extension field of $K.$ It may be noted that if $D$ is not a field $D[X]$ is never condensed. So for $D$ condensed $D+XK[X]$ is a way of constructing new condensed domains from old
Categories: math.AC