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[Statement] If $A$ is Noetherian, then every fractional ideal is of the form $x^{-1} \frak{a}$ for some ideal $\frak{a}$ of $A$, $x \in A$.

[Attempt]

I find this in Atiyah Macdonald Commutative algebra, Chapter 9 ,page 96 , Fractional ideals.

They say if $A$ is Noetherian, then every fractional ideal is of the form $x^{-1} \frak{a}$ for some ideal $\frak{a}$ of $A$, $x \in A$ so every fractional ideal is finitely generated.

It is okay "so ever fractional ideal is finitely generated" because $A$ is noetherian so ideal $\frak{a}$ is finitely generated.

However, how to show above statement?

Let $M$ be fractional ideal. Then by definition, there is $\frac{b}{a} \in K:=\text{Frac}(A)$ such that $\frac{a}{b} M \subseteq A $, so $M \subseteq \frac{b}{a}A$.

What is next step?

hew
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1 Answers1

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Let $\{m_i\}_{i\in I}$ generate $M$ as an $A$-module. Then as $m_i A\subseteq M\subseteq\frac{b}{a}A,$ it follows that $m_i\in\frac{b}{a}A.$ Thus, for each $i,$ we may write $$m_i = \frac{b_i}{a},$$ with $b_i\in A.$ This implies that \begin{align*} M &= \sum_{i\in I} m_i A\\ &=\sum_{i\in I}\frac{b_i}{a}A\\ & = \frac{1}{a}\sum_{i\in I}b_i A. \end{align*} But now $\sum_{i\in I}b_i A$ is simply the ideal of $A$ generated by the $b_i,$ so we are done.

Stahl
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  • Awesome, it is generally true if $A$ is not noetherian. – hew Aug 18 '20 at 06:59
  • @hew It will still be true that $M = \frac1a\mathfrak{a}$ for some ideal $\mathfrak{a}\subseteq A$ and element $a\in A,$ but it need not be true that $M$ is finitely generated. As an example, take any ideal which is not finitely generated and consider it as a fractional ideal. – Stahl Aug 18 '20 at 07:23
  • Yes, you're right. That's what i'm saying. Every fractional ideal is of the form $\frac{1}{a} \frak{a}$... very nice. Thank you so much! – hew Aug 18 '20 at 07:31
  • @hew You're welcome! I misread your last comment as "is it" instead of "it is." – Stahl Aug 18 '20 at 07:32