Divisors

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

In this post we take a closer look at \(\CaDiv(A)\) and \(\Pic(A)\), which were defined in §Fractional Ideals, ⁋Definition 5.

Dedekind Domains

We first make the following definition.

Definition 1 An ideal \(\mathfrak{a}\) of a ring \(A\) is said to be of pure codimension \(1\) if every associated prime ideal of \(\mathfrak{a}\) has codimension \(1\).

In particular, when \(\mathfrak{a}=A\) there are no associated prime ideals, so in this case \(A\) is vacuously of pure codimension \(1\). Then the following holds.

Theorem 2 Let \(A\) be a Noetherian integral domain such that the localization \(A_\mathfrak{m}\) at every maximal ideal \(\mathfrak{m}\) of \(A\) is a UFD.

1. An ideal \(\mathfrak{a}\subseteq A\) is invertible if and only if it is of pure codimension \(1\).
2. Every invertible fractional ideal \(\mathfrak{A}\subseteq K\) can be written uniquely as a finite product of prime ideals of codimension \(1\); hence \(\CaDiv(A)\) is the free abelian group generated by the prime ideals of codimension \(1\).

Proof

1. First assume that \(\mathfrak{a}\) is invertible. Then for any maximal ideal \(\mathfrak{m}\), the localization \(\mathfrak{a}A_\mathfrak{m}\) is a principal ideal generated by a non-zerodivisor of \(K\), viewing \(A_\mathfrak{m}\) as a subring of \(\Frac(A)\). On the other hand, for any maximal ideal \(\mathfrak{m}\subseteq A\), \(\mathfrak{A}_\mathfrak{m}\) is a normal domain by §Integral Extensions, ⁋Proposition 9; applying §Regular Local Rings, ⁋Theorem 8 to the principal ideal \(\mathfrak{a}A_\mathfrak{m}\), for its associated prime ideal \(\mathfrak{p}A_\mathfrak{m}\) we have

   \[(\mathfrak{p}A_\mathfrak{m})(A_\mathfrak{m})_{\mathfrak{p}A_\mathfrak{m}}\cong \mathfrak{p}A_\mathfrak{p}\]

   is a principal ideal in \((A_\mathfrak{m})_{\mathfrak{p}A_\mathfrak{m}}\cong A_\mathfrak{p}\), and therefore

   \[\codim \mathfrak{p}=\dim A_\mathfrak{p}=\codim \mathfrak{p}A_\mathfrak{p}\leq 1\]

   and since \(A_\mathfrak{p}\) is an integral domain, from \((0)\subseteq \mathfrak{p}A_\mathfrak{p}\) we see that \(\codim \mathfrak{p}=1\).

   Conversely, assume that \(\mathfrak{a}\) is of pure codimension \(1\) and let us show that it is invertible.
   Under the hypothesis of the claim, any codimension \(1\) prime ideal \(\mathfrak{p}\) of \(A\) is invertible. Indeed, if \(\mathfrak{p}\not\subset \mathfrak{m}\) then \(\mathfrak{p}A_\mathfrak{m}=A_\mathfrak{m}\), while if \(\mathfrak{p}\subseteq \mathfrak{m}\) then by the computation above \(\mathfrak{p}A_\mathfrak{m}\) is a prime ideal of codimension \(1\) in \(A_\mathfrak{m}\), and since \(A_\mathfrak{m}\) is a domain it is a minimal prime ideal; hence by §Dimension, ⁋Corollary 8 it is a principal ideal and therefore invertible. Since the product of invertible modules is again invertible, it suffices to show that \(\mathfrak{A}\) is a product of codimension \(1\) prime ideals.
   Suppose, for a contradiction, that there exists an ideal of pure codimension \(1\) which cannot