# radical of sum of two ideals

$I$ and $J$ are ideals in $k[x_1,\cdots,x_n]$.

Show that $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$.

I have no idea how to prove it. Can someone help?

"$\subset$" is trivial. For the other inclusion, assume $x \in \sqrt{\sqrt{I} + \sqrt{J}}$. Then there exists $m$ such that $x^m \in \sqrt{I} + \sqrt{J}$. Say $x^m = y + z$ with $y^n \in I$ and $z^k \in J$. What can you say about $(x^m)^{n + k - 1}$?
• $(x^m)^{n+k-1}=(y+z)^{n+k-1}=c_0y^{n+k-1}z^0+\cdots+c_iy^{n+k-i-1}z^i+\cdots+c_{n+k-1}y^0z^{n+k-1}$. Terms that can be divided by $y^n$ are in $I$ and terms that can be divided by $z^k$ are in $J$. So $(x^m)^{n+k-1}$ can be written in form of $a+b$ where $a\in I,b\in J$. – Falang May 11 '13 at 23:11
You don't have to argue with elements. $\sqrt{I+J}$ is the smallest radical ideal which contains $I+J$, i.e. which contains $I$ and $J$. On the other side, $\sqrt{\sqrt{I}+\sqrt{J}}$ is the smallest radical ideal which contains $\sqrt{I}+\sqrt{J}$, i.e. which contains $\sqrt{I}$ and $\sqrt{J}$, i.e. which contains $I$ and $J$. Thus both radical ideals are equal.
In general, for every closure operator $C$ on a partial order, we have $C(A \vee B) = C(C(A) \vee C(B))$ for all elements $A,B$. This is purely formal. In the example above, we have the partial order of ideals and $C(A)=\sqrt{A}$. You can also take the partial order of subsets of a topological space and $C(A)=\overline{A}$. But actually here the union of two closed sets is already closed, so that the right hand side simplifies. The corresponding statement for radical ideals does not hold and this has actually an algebro-geometric interpretation. See here. Also note that even more generally $C(\vee_i A_i)=C(\vee_i C(A_i))$. In the case of subsets of a topological space the right hand side doesn't simplify.