Dedekind Domains, Invertible Ideals

Invertible Ideals

Let H1 and H2 be R modules contained in the fraction field F, and assume H1*H2 = R (the base ring). In this case we say H1 and H2 are invertible, and each is the inverse of the other. Note that 0 is never an invertible module, because H*0 can never produce 1.

We need to show inverses are unique.

Start with H2 and let H2′ be the set of fractions x in F such that x*H2 lies in R. Show that H2′ is an R module.

At this point H2 has an inverse H1, and a possibly larger inverse H2′. Write the following equation.

H2′ = R*H2′ = (H1*H2)*H2′ = H1*R = H1

The inverse of H2 is unique, assuming H2 is invertible, and the inverse is precisely the set of elements that drive H2 into R. If H2 already lives in R, its inverse contains all of R.

Take any numerator n in H2 and note that n*H1 lies in R. Thus H1 is a fractional ideal. Reverse this to show H2 is a fractional ideal. If a submodule of F is invertible it is automatically a fractional ideal.

By definition, an "invertible ideal" is an invertible fractional ideal; it need not be an ideal of R. If you mean an ideal in R that happens to be invertible, you must say so.

If H2 is a fractional ideal that is not invertible, H2′′ may not equal H2. Let H2 be the ideal in K[x,y] generated by x and y. Here H2′ is the base ring K[x,y], and H2′′ is the same ring, which is larger than our ideal. This is a round about way to prove an ideal is not invertible.

Invertible Principal Ideals

Let H2 be principal, with generator g. Invert g in F and let this generate H1. Now H1H2 = R, and the inverse H1 is principal. In other words, an invertible ideal is principal iff its inverse is principal, and their generators are inverses in F.

Cancellation

Whenever H2 is invertible, one can multiply an equation through by H1 to cancel H2, like an integral domain. The invertible ideals form a group within the monoid of all fractional ideals.