If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

In a reduced commutative ring with $\dim R=0$ every ideal is radical. (This follows easily from the fact that $R_{\mathfrak{m}}$ is a field for every maximal ideal $\mathfrak{m}$ of $R$.) In particular, for an element $a\in R$ we have that $aR=a^2R$, so $R$ is VNR.

(Thanks Dr. Lam's wonderful Lectures on Modules and Rings for this proof.)

Suppose $R$ is commutative semiprime and has dimension zero.

Then if $M$ is a maximal ideal, the localization $R_M$ is a local ring with dimension zero also. Since $R$ is reduced, so is $R_M$. But $Nil(R_M)=M=0$, so $R_M$ is a field.

For all maximal ideals $M$, $(aR/a^2R)_M\cong (aR)_M/(a^2R)_M\cong aR_M/a^2R_M$, but since $R_M$ is always a field, we can see that the last term is always zero.

Since then $(aR/a^2R)_M=0$ for all maximal ideals, it follows that $aR/a^2R=0$, that is, $aR=a^2R$.