The title gives the result whose proof I'm looking at. The proof is an induction on Krull dimension n, which uses the fact that a commutative Noetherian integral domain is a UFD if every rank 1 prime is principal. It all flows fine for me apart from one part.

A regular local ring $ R $ is a commutative Noetherian domain in which the dimension of $ J / J^2 $, where $ J $ is the unique maximal ideal, as a vector space over $ R / J $ is equal to the rank of $ J $.

Base case is easy. We assume the result for dimensions less than $ n $ and consider a regular local ring $ R $ with dimension $ n \geq 1 $. We denote its maximal ideal by $ J $. We choose $ p \in J \setminus J^2 $ (we can do this by Nakayama's Lemma). By an earlier result we know that $ R / pR $ is also a regular local ring. Since a regular local ring is an integral domain, $ pR $ is a prime ideal and $ p $ is prime. We let $ S = \{ p^n \colon n \in \mathbb{N} \} $ and consider the ring $ R_S $, which has Krull dimension strictly less than that of $ R $.

We choose a rank 1 prime of $ R, A $ (the goal is to show that it is principal). We can show that $ A R_S $ is a projective $ R_S $ module. Then, since $ A $ is finitely generated over $ R $, it is finitely generated over $ R_S $, and hence has finite free resolution (we can find an exact sequence $ 0 \to F_n \to \dots \to F_0 \to A R_S $ where each $ F_i $ is free). By a previous result we now know that $ A R_S $ is stably free, i.e. there are finitely generated free $ R_S $ modules $ F $ and $ G $ such that $ G \bigoplus A R_S \cong F $.

Here is the one step in which the proof completely loses me. We want to conclude from what we've done so far that $ A R_S $ is free. I understand everything that comes after that conclusion. However, it is claimed that we can apply a theorem of Kaplanksy to see this:

Let $ R $ be a commutative integral domain and $ A $ a nonzero ideal of $ R $ such that $ A \bigoplus R^{n-1} \cong R^n $ as $ R $-modules. Then $ A $ is a principal ideal of $ R $.

The conclusion of this is that $ A $ is principal, not free, while the proof here only uses the fact that $ A R_S $ is free to conclude that it is principal, so immediately it seems like there's something wrong here. However it doesn't seem to be just a labeling issue as nothing in the immediate vicinity allows me to quickly conclude that $ A R_S $ is free either. Does anyone know what result should be used here?

  • $\begingroup$ For a domain $R$ we have an ideal $I$ is principal iff $I$ as a $R$ module is free of rank 1, that might help? This is just by taking the generator as a basis and vice versa. $\endgroup$ – user277182 Dec 12 '18 at 23:05

First of all, with a bit of googling, I found the claim here on the first page that for a commutative ring, ideals cannot be nontrivial stably free modules. This source cites Lam's Serre's Conjecture 4.11, but I don't have access to Springerlink to confirm this or look any further into it.

I think, however, that I see why we can apply Kaplansky's theorem, though I agree with you that it would seem you can just immediately conclude that the ideal is principal given that statement.

Let $R$ be a commutative domain, $K$ its fraction field, and $A$ a nonzero ideal of $R$. If $A\oplus R^{n-1}$ is free for some $n$, then the rank of $A\oplus R^{n-1}$ is $n$.

To see this, simply tensor with $K$, and we have that the rank of $A\oplus R^{n-1}$ is the dimension of $(A\otimes_R K) \oplus K^{n-1}$, which is $n-1 + \dim A \otimes_R K$, and $A\otimes_R K \simeq K$. Thus the rank of $A\oplus R^{n-1}$ is necessarily $n$ if $A\ne 0$ and $A\oplus R^{n-1}$ is free.

Thus you can apply Kaplansky's theorem.

It's possible that the author forgot the precise statement of the theorem that they were citing, since a principal ideal in a commutative domain is certainly free of rank 1, and a free ideal in a domain is also certainly of rank 1 (as we can see by e.g. tensoring with $K$). Thus it doesn't make much difference whether the conclusion is that the ideal is free or the ideal is principal.


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