Invertible fractional ideal

A fractional ideal $I$ of an integral domain $R$ is invertible if another fractional ideal $J$ satisfies $IJ=R$ , using fractional ideal multiplication. The inverse is unique and equals

I^{-1}=(R:I)=\{x\in\operatorname{Frac}(R):xI\subseteq R\}.

Principal examples

Every nonzero principal fractional ideal is invertible, but arbitrary fractional ideals need not be.

Why the colon ideal is the inverse

If $IJ=R$ , then $J\subseteq(R:I)$ . Conversely write $1=\sum a_jb_j$ with $a_j\in I$ and $b_j\in J$ . For $xI\subseteq R$ , the equality $x=\sum(xa_j)b_j$ puts $x$ in $J$ .

The same finite expression proves that $I$ is finitely generated: $a=\sum a_j(b_ja)$ for every $a\in I$ . Invertible ideals are invertible modules, even when the domain is not Noetherian.

A noninvertible ideal of a quadratic order

Let $R=\mathbb Z[\sqrt{-3}]$ and $I=(2,1+\sqrt{-3})$ . Writing $u=1+\sqrt{-3}$ , one has $u^2=2u-4$ , so

I^2=(4,2u,u^2)=(4,2u)=2I.

If $I$ were invertible, cancellation would give $I=2R$ ; this is false because $u\notin2R$ . This example belongs to a nonmaximal order.

Dedekind domains

Every fractional ideal of a Dedekind domain is invertible. Conversely, a domain that is not a field and whose every nonzero fractional ideal is invertible is Dedekind. The exclusion of fields matches the dimension-one convention used here.

References

J. S. Milne, Algebraic Number Theory, author's text, §3, Theorem 3.20 and the following discussion of fractional-ideal inverses.
The displayed quadratic-order computation proves noninvertibility directly.