Show that a radical ideal has no embedded prime ideals. [closed]

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 $.)

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 $I=\bigcap_2^n\sqrt{P_i}$ which is a primary decomposition with $n-1$ elements, a contradiction.