Solution
= Solution
\b[The equality is false.] In $A=k[x,y]$, take $I=(y)$ and $J=(y-x^2)$. Both are <prime ideals>, since their quotient rings are isomorphic to $k[x]$, so both are radical. But
$$
\sqrt I+\sqrt J=(y,x^2),\qquad\sqrt{I+J}=(x,y).
$$
The element $x$ belongs to the latter ideal and not the former. This also shows why taking the <radical of an ideal> can enlarge a sum of already radical ideals.