Conditions for $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$ Let $A$ be a commutative ring with identity and, $\mathfrak{a}$ and $\mathfrak{b}$ ideals.I'm trying to find sufficient and necessary conditions for  $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$ holds. I think that it holds for any UFD. I could not find any counter-example (for any ring) nor prove the reciprocal for the UFD assertion above. 
Thanks in advance.
 A: The question is whether the sum of two radical ideals is radical. In general this is far from being true, for example we have $(y)+(x^2-y)=(x^2,y)$ in $k[x,y]$.
There is an algebro-geometric explanation for this: If $I,J$ are radical ideals of a commutative ring $A$, this means that we have reduced subschemes $V(I)$ and $V(J)$ of the affine scheme $\mathrm{Spec}(A)$, but their intersection $V(I) \times_{\mathrm{Spec}(A)} V(J) = V(I+J)$ doesn't have to be reduced. In the above example we intersect the parabola $y=x^2$ with the axis $y=0$, this gives a point of multiplicity $2$.
It seems plausible that $V(I+J)$ is reduced iff $V(I)$ are $V(J)$ are transversal iff the intersection multiplicities are $\leq 1$ (whenever these notions are well-defined, for example for smooth curves).
Some observations in the positive direction:
Lemma. Let $A$ be a commutative ring. If every localization of $A$ has the property that the sum of two radical ideals is radical, then this also holds for $A$.
Proof. Let $\mathfrak{a},\mathfrak{b} \subseteq A$ be radical ideals, we have to show that $A/(\mathfrak{a}+\mathfrak{b})$ is reduced. This is well-known to be a local property. Quotients and sums of ideals commute with localizations. Besides, the localization of a radical ideal is easily seen to be a radical ideal. QED
Proposition. In a $1$-dimensional integral domain the sum of two radical ideals is radical.
Proof. By the Lemma we may assume that $A$ is local, say with maximal ideal $\mathfrak{m}$. The only prime ideals are $0$ and $\mathfrak{m}$, so these are only the only radical ideals. But these are obviously closed under sum. QED
For rings with zero divisors this fails.
Example. Let $k$ be a field and $A = k[x,y]/(y(x^2  - y))$. We have $\dim(A)=1$. The quotients $A/(y)=k[x]$ and $A/(x^2-y)=k[x,y]/(x^2-y)$ are reduced, but $A/(y,x^2-y)=k[x]/(x^2)$ is not.
I doubt (but cannot prove) that there is any $2$-dimensional finitely generated $k$-algebra with the property.
A: If $A$ is a commutative ring, one clearly has $\sqrt{\mathfrak{a + b}} = \sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$ for all ideals $\mathfrak{a},\mathfrak{b}\leq A$ if and only if $\sqrt{aA + bA} = \sqrt{aA} + \sqrt{bA}$ for all $a,b\in A$. For, assuming the latter, if $z\in\sqrt{\mathfrak{a + b}}$, then $z^{n}=a+b$ for some $a\in\mathfrak{a}$, $b\in\mathfrak{b}$,  $n\in\mathbb{N}$, so that $z\in\sqrt{aA + bA}=\sqrt{aA} + \sqrt{bA}\subseteq\sqrt{\mathfrak{a}} + \sqrt{\mathfrak{b}}$.
It follows that any $A$ can be embedded in a ring in which the radical ideals are closed under taking sums. Indeed, let $\widetilde{A}$ be the quotient of the polynomial ring $A[X_{\tau},Y_{\tau},S_{\tau},T_{\tau}]$, where $\tau$ runs over all triplets $(a,b,z)\in(A-\{0\})^{3}$ with $z\in\sqrt{aA+bA}$, modulo the ideal generated by the elements $z-X_{\tau}-Y_{\tau},X_{\tau}^{2}-aS_{\tau},Y_{\tau}^{2}-bT_{\tau}$ for all $\tau$. Then $A\subseteq\widetilde{A}$, and for all $\tau=(a,b,z)$ in $(A-\{0\})^{3}$ with $z\in\sqrt{aA+bA}$ in $A$, we have $z=X_{\tau}+Y_{\tau}$ with $X_{\tau}\in\sqrt{a\widetilde{A}}$ and $Y_{\tau}\in\sqrt{b\widetilde{A}}$ in $\widetilde{A}$. Letting $A_{0}:=A$ and $A_{n+1}:=\widetilde{A_{n}}$ for all $n$, we obtain a ring $B:=\bigcup_{n}A_{n}$ containing $A$, in which the set of radical ideals is closed under addition. Moreover, $B$ is a domain iff $A$ is. 
Additivity of radicals holds for Prüfer domains $A$, but these are generally not Noetherian (if $A$ is Prüfer and Noetherian, then dim$(A)\leq 1$, so $A$ is just a Dedekind domain or a field). However, for domains, being Prüfer is much stronger than having all sums of radical ideals being radical: simply consider rings such as the $B$ described just now.
If $A$, domain or not, is of finite type over a field $k$ and dim$(A)\geq 2$, then by Noether Normalization there exist $x,y\in A$ which are algebraically independent over $k$. With $x$ and $y$ carefully chosen, the counterexample given by Martin can be replicated - that is, we can have $yA$ and $(x^{2}-y)A$ radical ideals of $A$ with $x\notin yA+(x^{2}-y)A=(x^{2},y)$.
