The ring $R$ is commutative with unit. An ideal $I$ is called primary, if it stands the following:

If $ab \in I$ then $a \in I$ or $b^n \in I$, for a natural number $n$.

Show that if $I$ is a primitive ideal of $R$, then $Rad(I)$ is a prime ideal of it.

Could you give me a hint how we could show this?


That's what I have tried:

$Rad(I)=\{ x \in R| \exists n \in \mathbb{N} \text{ such that } x^n \in I \}$

$P$ is a prime ideal iff $a,b \in R$, $a \cdot b \in \mathbb{P}$, then $a \in P$ or $b \in P$

Let $a \cdot b \in Rad(I) \Rightarrow \exists m \in \mathbb{N}$ such that $(a \cdot b)^m \in I \Rightarrow a^m \cdot b^m \in I \Rightarrow a^n \in I \text{ or } (b^{m})^n \in I \Rightarrow a \in Rad(I) \text{ or } b^{m \cdot n } \in I \Rightarrow a \in Rad(I) \text{ or } b \in Rad(I)$

Could you tell me if it is right?

  • $\begingroup$ Straightforward hint: Write down the definitions $x \in\mathrm{rad}(I) \iff \exists n : x^n\in I \dots$ and $\mathrm{rad}(I)$ is prime $\iff \forall a,b\in R : (ab\in \mathrm{rad}(I) \implies a\in \mathrm{rad}(I) \vee b\in \mathrm{rad}(I)\dots)$ $\endgroup$ – Myself Dec 4 '14 at 21:43
  • $\begingroup$ @Myself I edited my post... So, is it right? $\endgroup$ – evinda Dec 4 '14 at 21:52

$ab\in\sqrt I\Rightarrow\exists m\ (ab)^m\in I$; $(ab)^m=a^mb^m\in I\Rightarrow a^m\in I$ or $\exists n\ (b^m)^n\in I$ $\Rightarrow\cdots$

  • $\begingroup$ I edited my post... So, is it right? $\endgroup$ – evinda Dec 4 '14 at 21:53
  • 1
    $\begingroup$ @evinda Of course it is! $\endgroup$ – user26857 Dec 4 '14 at 21:55
  • $\begingroup$ Great... Thanks a lot...!!! $\endgroup$ – evinda Dec 4 '14 at 21:56

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