Multiplier ring of a fractional ideal

Let $I$ be a nonzero fractional ideal of a domain $R$ , with fraction field $K$ . Its multiplier ring is

(I:I)=\{x\in K:xI\subseteq I\}.

It is a subring of $K$ containing $R$ . Multiplication identifies it with $\operatorname{End}_R(I)$ : an $R$ -linear endomorphism extends to the one-dimensional $K$ -vector space $K\otimes_R I\cong K$ , and hence is scalar multiplication.

Orders

If $R=\mathcal O$ is an order in a number field $K$ , then $(I:I)$ is another order containing $\mathcal O$ . To see finiteness, choose $0\ne a\in I$ ; then $(I:I)\subseteq a^{-1}I$ , a finite integer lattice. It spans $K$ because it contains $\mathcal O$ .

If $I$ is invertible, multiplying $xI\subseteq I$ by $I^{-1}$ gives $x\mathcal O\subseteq\mathcal O$ , so $(I:I)=\mathcal O$ .

A larger multiplier ring

Put $s=\sqrt{-3}$ , $\omega=(1+s)/2$ , $R=\mathbb Z[s]$ , and $I=(2,1+s)=2\mathbb Z[\omega]$ . Then

(I:I)=\mathbb Z[\omega]\supsetneq R.

The identity follows by cancelling the nonzero scalar $2$ : a scalar preserves $\mathbb Z[\omega]$ precisely when it belongs to that ring.

References

Andrew Sutherland, MIT 18.783 Lecture 18, §18.3, the multiplier order of a fractional ideal.
The localization and lattice arguments above verify the endomorphism and order claims directly.