4
$\begingroup$

Hi!

I'm wanting to see why the following is true: Given 2 finitely generated ideals $B$ and $C$ in a Prufer domain $D$ with bases of $n$ and $m$ generators respectively, $B\cap C$ has a basis of $m+n$ generators, and $B:C$ has a basis of $m(m+n)$ generators.

This is a result used in Gilmer and Heinzer's paper Overrings of Prufer domains II, Lemma 2, and they use the following facts in a way that is unclear to me: $B:C=B\cap C: C$ (this is clear), so $B\cap C=(B:C)C$, and $BC=(B\cap C)(B+C)$ (I understand the proof for this latter statement), and so since $BC$ is invertible, $B\cap C$ is, and so $(B:C)$ is invertible too.

I understand that if an ideal $A$ has $k$ generators, its inverse has $k$ generators as well, so I think I see how the result on the number of generators of $B:C$ follows from the calculation of the number of generators of $B\cap C$, but I'm not seeing where the number for $B\cap C$ is coming from. How can we make this calculation without knowing the exact number of generators for $BC$ and $B+C$ (we don't know these numbers since we could have redundant generators, I think)?

Does anyone here see what's going on? Thanks!

Improve this question
$\endgroup$
4
  • $\begingroup$ Hi. If it is not much of trouble to you, would you tell me the definition of a Prufer domain, base, and inverse of an ideal? $\endgroup$
    – Youngsu
    Commented Apr 14, 2013 at 21:50
  • $\begingroup$ Sure. When I say "bases," I mean the plural of basis (of an ideal) since I'm considering multiple ideals. A Prufer domain is a domain such that every finitely generated ideal admits an inverse. By an inverse of an ideal $I$, I mean a fractional ideal $I^{-1}$ such that $II^{-1}=R$. $\endgroup$
    – rmg512
    Commented Apr 14, 2013 at 23:48
  • $\begingroup$ Isn't a PID a Prufer domain? If so, how can you have $B \cap C$ minimally generated by more than 1 element? $\endgroup$
    – Youngsu
    Commented Apr 15, 2013 at 1:21
  • $\begingroup$ Yes. PID's are a class of Prufer domains since principal ideals are always invertible. I am not assuming the ring we're working over here is a principal ideal domain, though. Just one where all finitely generated ideals are invertible (not necessarily Noetherian). $\endgroup$
    – rmg512
    Commented Apr 15, 2013 at 2:23

1 Answer 1

Reset to default
4
$\begingroup$

(For the following we can obviously assume $B$ and $C$ are nonzero.) The equation $BC = (B \cap C)(B+C)$ can be seen by localization. (One of $(B \cap C)_M$ and $(B+C)_M$ is $B_M$ and the other is $C_M$, depending on which contains which in $D_M$.) Multiply it by $(BC)^{-1}$ to obtain $D = (B \cap C)(B^{-1}+C^{-1})$. So $(B \cap C)^{-1} = B^{-1} + C^{-1}$ and $B \cap C$ is $(m+n)$-generated. The equation $B \cap C = (B : C)C$ can be seen by localization. (Since $C_M$ is a principal ideal containing $(B \cap C)_M$, we have $(B \cap C)_M = ((B \cap C)_M : C_M)C_M = (B : C)_MC_M$. The last equality is using the fact that $C$ is finitely generated.) For each generator of $B \cap C$, write $a = \sum b_ic_i$ with the $b_i$'s in $(B : C)$ and the $c_i$'s distinct elements from a minimal generating set for $C$. Let $B_0$ be the ideal generated by the $b_i$'s, and note that it is $m(m+n)$-generated, since there are $m+n$ sums and at most $m$ $b_i$'s in each sum. So $B_0C = B \cap C = (B : C)C$ and $B_0 = (B : C)$.

I suppose you have probably answered this problem long ago, but I stumbled across this today, and it's related to things I am working on.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .