Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$I$ 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?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

"$\subset$" 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 $(x^m)^{n + k - 1}$?

share|improve this answer
$(x^m)^{n+k-1}=(y+z)^{n+k-1}=c_0y^{n+k-1}z^0+\cdots+c_iy^{n+k-i-1}z^i+\cdots+c_{‌​n+k-1}y^0z^{n+k-1}$. 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
Thank you, Serkan. Your proof is beautiful! –  Falang May 11 '13 at 23:12
Essentially this proof reproves well-known standard facts instead of just using them (for example that the radical ideal is an ideal). When you replace element calculations by objects and their universal properties (i.e. their modern definition), the claim becomes much clearer and easier to prove. As always ... –  Martin Brandenburg May 11 '13 at 23:55

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.

share|improve this answer
Thank you for the concise proof and further explanation. –  Falang May 12 '13 at 3:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.