If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).
The converse, although not immediately apparent, can be proven quite easily. But I've been stuck on this one for awhile.
I found that Bourbaki has a question in his treatise on commutative algebra, namely: if all points in Spec(A) are closed (ie. prime ideals are maximal) then A/R (where R is the nilradical) is von Neumann regular (absolutely flat). If this can be proven, the above statement follows immediately since the nilradical of a semiprime ring 0?
Ps: this is my first time posting a question so apologies for any formatting issues.