Question 1.223: Let k be a positive integer. Prove that the following equali......
Let k be a positive integer. Prove that the following equalities are true for two ideals I and J from a ring R :
(a) I^{k} \subseteq J \Longrightarrow \operatorname{rad}(I) \subseteq \operatorname{rad}(J);
(b) I^{k} \subseteq J \subseteq I \Longrightarrow \operatorname{rad}(J)=\operatorname{rad}(I);
(c) \operatorname{rad}(I J)=\operatorname{rad}(I \cap J)=\operatorname{rad}(I) \cap \operatorname{rad}(J);
(d) \operatorname{rad}(\operatorname{rad}(I))=\operatorname{rad}(I);
(e) \operatorname{rad}(I)+\operatorname{rad}(J) \subseteq \operatorname{rad}(I+J)=\operatorname{rad}(\operatorname{rad}(I)+\operatorname{rad}(J)).
(a) We start with an easy observation: if I_{1} and I_{2} are two ideals such that I_{1} I_{2} \subset P for some prime ideal P, then either I_{1} \subset P, or I_{2} \subset P. Indeed, assuming otherwise we find two elements a \in P-I_{1} and b \in P-I_{2} such that a b \in P. Since P is prime, this produces a contradiction. This observation implies also that if P contains a power of an ideal I^{k}, then I \subset P. Since \operatorname{rad}(I) is the intersection of all primes containing
I (Problem 201), we see that \operatorname{rad}(I) \subset \operatorname{rad}\left(I^{k}\right). On the other hand, as I^{k} \subset J, any element x \in R such that x^{m} \in I^{k} satisfies x^{m} \in J. Equivalently, \operatorname{rad}\left(I^{k}\right) \subset \operatorname{rad}(J). Therefore, \operatorname{rad}(I) \subset \operatorname{rad}(J).
(b) By part (a) we know that \operatorname{rad}(I) \subset \operatorname{rad}(J). Since J \subset I \Longrightarrow \operatorname{rad}(J) \subset \operatorname{rad}(I) the equality follows.
(c) Since I J \subset I \cap J, we see that \operatorname{rad}(I J) \subset \operatorname{rad}(I \cap J). Conversely, let x \in R be an element of \operatorname{rad}(I \cap J) that is to say x^{m} \in I \cap J for some m \geq 1. Then x^{m} \in I and x^{m} \in J. Thus x^{2 m} \in I J. It follows that x \in \operatorname{rad}(I J). (Almost) the same argument shows that \operatorname{rad}(I) \cap \operatorname{rad}(J) \subset \operatorname{rad}(I \cap J). Since I \cap J \subset I, J we see that \operatorname{rad}(I \cap J) \subset \operatorname{rad}(I), \operatorname{rad}(J). Therefore, \operatorname{rad}(I \cap J)=\operatorname{rad}(I) \cap \operatorname{rad}(J).
(d) The inclusion \operatorname{rad}(\operatorname{rad}(I)) \subset \operatorname{rad}(I) follows from the definition of radical ideals. Conversely, I \subset \operatorname{rad}(I) implies that \operatorname{rad}(I) \subset \operatorname{rad}(\operatorname{rad}(I)).
(e) Let x \in \operatorname{rad}(I), y \in \operatorname{rad}(J) be two elements. Then x^{m} \in I and y^{n} for some positive integers m, n. Then terms of (x+y)^{m n} are of the form \left(\begin{array}{c}k+l \\ k\end{array}\right) x^{k} y^{l} with k+ l=m n. In this case, either k \geq m, or l \geq n. Therefore, either \left(\begin{array}{c}k+l \\ k\end{array}\right) x^{k} y^{l} \in I or \left(\begin{array}{c}k+l \\ k\end{array}\right) x^{k} y^{l} \in J. It follows that (x+y)^{m n} \in I+J. Finally, we prove the last equality. Since \operatorname{rad}(I)+\operatorname{rad}(J) \subset \operatorname{rad}(I+J), taking radicals of both sides implies that \operatorname{rad}(\operatorname{rad}(I)+\operatorname{rad}(J)) \subset \operatorname{rad}(\operatorname{rad}(I+J))=\operatorname{rad}(I+J) (by the previous part of the problem). Conversely, if x \in \operatorname{rad}(I+J), then there exists m \geq 1 such that x^{m} \in I+J. Since I+J \subset \operatorname{rad}(I)+\operatorname{rad}(J), we have x^{m} \in \operatorname{rad}(I)+\operatorname{rad}(J). In other words, x belongs to the radical of \operatorname{rad}(I)+\operatorname{rad}(J). This finishes the proof.