Show that a radical ideal has no embedded prime ideals. 
Let $A$ be a commutative ring and $I$ a decomposable ideal. Let $I=\bigcap_{k=1}^{n} I_k$ be a minimal primary decomposition. Show that if $I=\sqrt{I}$ then $I$ has no embedded prime ideals. 

(I noticed that $I=\bigcap_{k=1}^{n} P_k$ where $P_k=\sqrt{I_k}$, $\forall k\in\lbrace1,...,n\rbrace$. I have to show that $P_i \nsubseteq P_j, \forall  i\neq j $.)
 A: Since $I=\bigcap_1^nQ_i$ is a minimal primary decomposition, by uniqueness theorem of a primary decomposition, the number $n$ is fixed by $I$ in all primary decompositions, i.e., $I$ has no primary decomposition with less than or equal to $n-1$ primary ideals. Now let $I=\sqrt{I}$, then $$I=\sqrt{I}=\sqrt{\bigcap_1^nQ_i}=\bigcap_1^n\sqrt{Q_i}=\bigcap_1^n P_i$$ is also a primary decompositon for $I$, where each $Q_i$ is $P_i$-primary. Now if for some $i\neq j$, say 1 and 2,  $P_2\subset P_1$, then $I=\bigcap_2^nP_i$ which is a primary decomposition with $n-1$ elements, a contradiction.
A: From Atiyah and Macdonald, Theorem 4.5, one knows that $$\{P_1,\dots,P_n\}=\operatorname{Spec}A\cap\{\sqrt{(I:x)}\mid x\in A\}.$$ This shows that the set $\{P_1,\dots,P_n\}$ is independent of the decomposition of $I$. (As you probably know, the primes $P_i$ are called the associated primes of $I$.)
Now, from $I=\sqrt I$ we get $I=\bigcap_{i=1}^nP_i$. If one can get rid of some primes in the intersection, then the set of associated primes of $I$ is smaller than $\{P_1,\dots,P_n\}$, a contradiction.
A: Suppose $P_k\subsetneq P_l$ for some $k\ne l$. Then $\sqrt I=\bigcap_{i\ne l}P_i$. Since $I=\bigcap_{i=1}^nQ_i$ is a minimal primary decomposition we have $\bigcap_{i\ne l}Q_i\not\subseteq Q_l$. Then there is $x\in(\bigcap_{i\ne l}Q_i)\setminus Q_l$ and therefore $x\in\bigcap_{i\ne l}P_i$, hence $x\in\sqrt I$. This gives $x\in I$, in particular, $x\in Q_l$, a contradiction.
