1
$\begingroup$

Let $I, J$ be ideals in a commutative ring with identity $R$. Define the quotient ideal $(I : J)$ by $$(I : J)=\{x\in R : xJ\subseteq I\}.$$

Define the radical $r(A)$, of an ideal $A$ of $R$ by

$$r(A)=\{x\in R\;:\;x^n\in A, \;\;for\;some\;n>0\}.$$

My question is: There is some link between $r[(I : J)]$ and the two radicals $r(I)$ and $r(J)$?

$\endgroup$

1 Answer 1

3
$\begingroup$
  1. We have the following relations: $$(I:r(J))\subseteq(I:J)\subseteq r(I:J)\subseteq(r(I):J)=(r(I):r(J)).$$ The first and the second of these inclusions are immediately clear. For the third one, consider $x\in r(I:J)$ and $y\in J$. There exists $n\in\mathbb{N}^*$ with $x^ny\in I$, implying $(xy)^n=x^ny\cdot y^{n-1}\in I$, hence $xy\in r(I)$ and therefore $x\in(r(I):J)$. Finally, for the equality it suffices to show that the left hand side is contained in the right hand side. So, let $x\in(r(I):J)$ and let $y\in r(J)$. There exists $n\in\mathbb{N}^*$ with $y^n\in J$, hence $xy^n\in r(I)$, and thus there exists $m\in\mathbb{N}^*$ with $x^my^{m+n}\in I$. It follows $(xy)^{m+n}=x^nx^my^{m+n}\in I$, therefore $xy\in r(I)$, and thus $x\in(r(I):J)$.

  2. None of the above inclusions needs to be an equality. For a counterexample, consider $R=A[X,Y]/\langle X^3,Y^2\rangle$ for some reduced ring $A\neq 0$, and the ideals $I=0$ and $J=\langle \overline{X}^2\rangle$ of $R$ (where $\overline{X}$ denotes the canonical image of $X$ in $R$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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