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Let $I$ be an invertible ideal in an integral domain $R$. I claim that it is a maximal ideal. Please tell my i am correct or not.

Here is my attempt: If $I\subseteq J$, for some proper ideal $J$ of $R$. Then $R=II^{-1}\subseteq JI^{-1}$ and hence $JI^{-1}=R$. It implies that $J=I$.

waqas mahmood
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  • How are you concluding $JI^{-1}=R$? – Eric Wofsey Apr 04 '19 at 01:05
  • In an integral domain, every principal ideal is invertible. So certainly not every invertible ideal is maximal. In the other direction, maximal ideals are often not invertible. For example, the maximal ideal of a valuation domain is invertible iff it is principal. – Badam Baplan Apr 04 '19 at 04:13
  • Dear, Wofsey. Thanks for pint out the mistake in my proof. – waqas mahmood Apr 04 '19 at 04:53

1 Answers1

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The issue is that you seem to be assuming that $JI^{-1}\subseteq R$, which need not be the case.

Let's look at what happens with a specific example; the simplest thing we could possibly look at is $R=\Bbb Z$. Let's take $I=(4)$ and $J=(2)$; then $JI^{-1}=(\frac 12)$ (i.e. this is the additive subgroup of $\mathbb Q$ generated by the element $\frac12$), and clearly this is not actually contained in $\Bbb Z$.

Alex Mathers
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