1
$\begingroup$

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?

EDIT:

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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