Help with proof : $Cl_F$ with only one equivalance class implies $D_F$ is a PID

Question

I am self-studying a set of legitimately downloaded notes on algebraic number theory. They are somewhat akin to "Ireland and Rosen," Ch. 12.

I would appreciate help in understanding a proof (in the direction) that if $Cl_F$ consists of one equivalence class, then $D_F$, the ring of integers of $F$, is a PID. $F$ is a number field.

For convenience, I've numbered the four questions I would like help with.

Using the definition of equivalence in this context, if there is only one equivalence class, any two ideals of $D_F$ are equivalent and there exist $\alpha, \beta \in D_F$ such that

$\alpha I = \beta D_F$,

where $I$ is any ideal in $D_F$. We want to show $I$ is a principal ideal. So far so good. (1) Although the notes has $\alpha, \beta \in I$, I think they ought to be in $D_F$. Is this correct?

Thus $\alpha I = (\beta)$.

The proof goes on: Let $\omega = \beta \alpha^{- 1} \in F$.

Now I get stuck:

Then (2) $\omega I = \beta I \subseteq I$. I don't see how to get the equality.

Based on this, we know $\omega \in D_F$. This is clear since this type of assertion, based on the inclusion, was previously proved.

Then I also can't get the final assertion, (3) that this implies $(\omega) = I$.

Lastly, I would appreciate any guidance or references as to (4) in general what the product, e.g., $\gamma J$ means, where $J$ is an ideal.

As always, thanks for your patience and help.

Answer 1

Hint $\rm\ \ a I = (b)\:\Rightarrow\: b\in aI\:\Rightarrow\: b = ai\:$ thus $\rm\:aI = a(i)\:\Rightarrow\: I = (i)\:$ by cancelling $\rm\:a\ne 0$.

Principal ideals are invertible, so cancellable, thus we can cancel $\rm\,(a)\ne 0\:$ from $\rm\,(a)I = (a)(i)\,$ above. If you're not familiar with invertible or fractional ideals then you can easily prove this directly: suppose $\rm\:aI = aJ\ne 0.\:$ To show $\rm\:I\subseteq J\:$ note $\rm\:i\in I\:\Rightarrow\:ai\in aI\subseteq aJ,\:$ so $\rm\:ai = aj\:$ for some $\rm\:j\in J.\:$ Cancelling $\rm\:a\ne 0\:$ yields $\rm\:i = j\in J,\:$ so $\rm\:I\subseteq J.\:$ By symmetry, $\rm\:J\subseteq I,\:$ therefore $\rm\:I = J.$