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Suppose $A$ is a commutative ring with unity and an ideal $q$ of $A$ is $p$-primary, i.e. $\sqrt{q}=p$. It is known that for $x \in A $, we have

  • if $x \not\in q$, then $(q:x)$ is $p$-primary.
  • if $x\not\in p$, then $(q:x)=q$.

I was wondering whether the following is true:

if $(q:x)=q$, then $x$ is not in $p$.

Thank you for your time in advance.

Bingyan Liu
  • 181

1 Answers1

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Suppose $(q:x)=q$ and $x \in p$, then we have $$x \not \in q \quad\text{ and }\quad x^n \in q \ \text{ for some $n>1$,}$$ where the first statement is true since otherwise $(q:x)=(1)$.

Choose $m$ to be the minimal positive integer such that $x^m \in q$ and $x^{m-1} \not \in q$. Such $m$ exists since in the sequence $x,x^2,x^3,....$, we have $x \not \in q$ and $x^k \in q$ for $k\geq n>1$.

So by definition we have $x^{m-1} \in (q:x)$, which contradict $(q:x)=q$.

Bingyan Liu
  • 181