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.

Solution 1:

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.