Let $A$ be a commutative ring. Let $I$ be an ideal of $A$. Assume that $\sqrt{I}=\sqrt{J}$ for some finitely generated ideal $J$. Does it follow that $\sqrt{I} = \sqrt{J'}$ for some finitely presented ideal $J'$? This is trivial if $A$ is Noetherian. In the general case, I think that there will be counterexamples.
Some definitions which are necessary here: An ideal of $A$ is just a $A$-submodule of $A$. An $A$-module $M$ is called finitely generated if there is an epimorphism of $A$-modules $A^r \to M$. An $A$-module $M$ is called finitely presented if there is an exact sequence $A^s \to A^r \to M \to 0$.
Let $k$ be a field and $R=k\langle x_1,x_2,\ldots\mid x_1x_i=0\text{ for }i>1\rangle$. Let $J=x_1R$ and $K=x_2R+x_3R+\ldots$, so $$ R/J\cong k[x_2,x_3,\ldots]. $$ In particular $J$ is prime, so $\sqrt{J}=J$. Also note that the ring homomorphism $i:k[x_1]\rightarrow R$ is injective and its image contains $J$.
Suppose $\sqrt{J'}=J$ for some ideal $J'$. In particular $J'\subseteq J\subseteq i(k[x_1])$, so $J'=i(i^{-1}(J'))$. Now $\sqrt{i^{-1}(J')}=i^{-1}(J)=x_1k[x_1]$, so $i^{-1}(J')=x_1^nk[x_1]$ for some $n>0$. Hence $J'=x_1^nR$. We have an exact sequence $$ K\hookrightarrow R\twoheadrightarrow J' $$ where the right hand map is $r\mapsto rx_1^n$. Since $R$ is finitely generated but $K$ is not (as an $R$-module), $J'$ is not finitely presented.