Fork me on GitHub
Math for the people, by the people.

User login

invertible ideal is finitely generated

Type of Math Object: 
Major Section: 

Mathematics Subject Classification

13B30 no label found


Can anyone show this?

An invertible fractional ideal in an integral domain that is a local ring is principal.

Let I be an invertible fractional ideal of an integral domain D. Let J be its inverse. So IJ=D.

Write 1=sum of (a_k * b_k) , where a_k are in I and b_k are in J, k ranges from 1 to n.

If r is any element in I, multiply r on both sides of the equation in the previous paragraph, and rearrange the terms so that r = sum of (a_k *(r*b_k)). Since b_k are in J, r*b_k are in D. So I is generated by a_k, k ranging from 1 to n. I is finitely generated.

If D is local, it has a unique maximal ideal M, which contains all the non-units. If all of a_k are non-units, then 1, which is the sum of a_k*b_k, must also be a non-unit. This is impossible. So one of the a_k must be a unit. Being a unit, it must generate all of I. So I is principal.

What I mean in the last paragraph of the proof is that if all of a_k are non-units, they must all be in M. But then, 1, being the sum of a_k*b_k, must be in M as well. This means that M=D, an impossibility.

Subscribe to Comments for "invertible ideal is finitely generated"