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.
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.
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 $x\notin yA+(x^{2}-y)A=(x^{2},y)$.