1

For a Dedekind domain $A$ such that $\operatorname{frac}(A)=K$, we define invertible ideals to be $I\hookrightarrow K$ and rank one projective modules over $A$. (I think this definition is motivated from the definition of Picard group).

I want to prove that if there exists $J \hookrightarrow K$ such that $IJ=A$ then $I$ is invertible.
I can localize and prove this locally and that would suffice because I know the fact:

Projectiveness is local property for finitely presented modules.

So I have $I_\mathfrak p J_\mathfrak p =A_\mathfrak p$. As $J_\mathfrak p$ is a module over PID, and $J_\mathfrak p \hookrightarrow K$ it is free of rank one. So principal. This also implies $I_\mathfrak p$ is is principal since if $J_\mathfrak p= (b)$ then $I_\mathfrak p =(1/b)$. So $I_p$ is projective of rank one and from the above result, we have that $I = \cap _\mathfrak p I_p \hookrightarrow K$ is invertible.
But the proof of the above result, that projectiveness is local, does not seem to be trivial and I wish to avoid it (or prove it in my special case).

user26857
  • 53,190
mathemather
  • 2,991
  • Can you clarify your context, i.e., what you're "allowing yourself" to use here? E.g., in Milnor's K-theory book, he proves "Steinitz' theorem", about structure theory of finitely-generated modules over Dedekind rings. I'd imagine that the several implications among nearly-equivalent assertions in such a discussion would include yours. But/and what do you choose as starting point? E.g., the local-ness of projectivity (in some circumstances) is not used in Milnor's discussion, if I recall correctly. Just direct arguments. – paul garrett Mar 10 '22 at 19:33
  • @paulgarrett Thank you for your comment. Maybe I will try to give a little context that will clarify some of the things. This is part of a course on algebraic number theory which takes a somewhat highbrow approach. Rather than defining invertible ideal $I$ as $IJ=A$ we define it as projective A-module of rank one contained in $K$. And then prove the two defintions are equivalent. – mathemather Mar 11 '22 at 06:03
  • 1
    Don't know if they cover as much ground as you would like to, but I have skirted this theme here, and in a very specific case here. Those are about showing how an invertible ideal is necessarily projective. The other direction I don't see at the moment, but may the argument can be reverse engineered. – Jyrki Lahtonen Mar 11 '22 at 06:44

0 Answers0