Divisors

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

Dedekind Domains

First, let us make the following definition.

Definition 1 For a ring \(A\), an ideal \(\mathfrak{a}\) 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 \(A\) vacuously has codimension \(1\). Then the following holds.

Theorem 2 For a Noetherian integral domain \(A\), suppose that the localization \(A_\mathfrak{m}\) at every maximal ideal of \(A\) is always a UFD.

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

Proof

1. First, suppose that \(\mathfrak{a}\) is invertible. Then, first, localizing \(\mathfrak{a}\) at an arbitrary maximal ideal \(\mathfrak{m}\), the resulting \(\mathfrak{a}A_\mathfrak{m}\) is, viewing \(A_\mathfrak{m}\) as contained in \(\Frac(A)\), a principal ideal generated by a non-zerodivisor of \(K\). Meanwhile, for any maximal ideal \(\mathfrak{m}\subseteq A\), \(A_\mathfrak{m}\) is a normal domain by §Integral Extensions, ⁋Proposition 9, so applying §Regular Local Rings, ⁋Theorem 8, for the principal ideal \(\mathfrak{a}A_\mathfrak{m}\) and its associated prime ideal \(\mathfrak{p}A_\mathfrak{m}\),

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

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

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

   holds; but since \(A_\mathfrak{p}\) is an integral domain, we know from \((0)\subsetneq \mathfrak{p}A_\mathfrak{p}\) that \(\codim \mathfrak{p}=1\).

   Conversely, suppose \(\mathfrak{a}\) is pure codimension \(1\) and let us show that it is invertible.
   First, under the hypotheses of the claim, in \(A\), any codimension \(1\) prime ideal \(\mathfrak{p}\) is invertible. This is because, first, if \(\mathfrak{p}\not\subseteq \mathfrak{m}\), then \(\mathfrak{p}A_\mathfrak{m}=A_\mathfrak{m}\), and if \(\mathfrak{p}\subseteq \mathfrak{m}\), then by the calculation above \(\mathfrak{p}A_\mathfrak{m}\) is a prime ideal of \(A_\mathfrak{m}\) of codimension \(1\), and since \(A_\mathfrak{m}\) is a UFD by assumption, this is a principal ideal. In fact, if we choose a non-\(0\) element \(x\in \mathfrak{p}A_\mathfrak{m}\) and factor it into a product of irreducible elements, then since \(\mathfrak{p}A_\mathfrak{m}\) is prime, one of its factors \(\pi\) belongs to \(\mathfrak{p}A_\mathfrak{m}\), and since in a UFD the ideal generated by an irreducible element is prime, \((0)\subsetneq (\pi)\subseteq \mathfrak{p}A_\mathfrak{m}\) is a chain of prime ideals, so from \(\codim \mathfrak{p}A_\mathfrak{m}=1\) we obtain \((\pi)=\mathfrak{p}A_\mathfrak{m}\). Now, since \(\pi\) is a non-\(0\) element of a domain, it is a non-zerodivisor, and therefore \(\mathfrak{p}\) is invertible. Now, since products of invertible modules are again invertible, it suffices for us to show that \(\mathfrak{a}\) is a product of codimension \(1\) prime ideals.
   Suppose, contrary to the conclusion, that there exists an ideal that is pure codimension \(1\) but cannot be expressed as a product of codimension \(1\) prime ideals, and let \(\mathfrak{A}\) be maximal among such ideals. If \(\mathfrak{p}\) is a prime ideal containing \(\mathfrak{A}\) of codimension \(1\), then \(\mathfrak{p}^{-1}\mathfrak{p}=A\), and therefore \(A\subsetneq\mathfrak{p}^{-1}\). Now, if \(\mathfrak{p}^{-1}\mathfrak{A}=\mathfrak{A}\), then by §Integral Extensions, ⁋Lemma 5, the elements of \(\mathfrak{p}^{-1}\) are integral over \(A\). But each \(A_\mathfrak{m}\) is a UFD by assumption, hence a normal domain by §Integral Extensions, ⁋Proposition 9, and since an intersection of normal domains with a common quotient field is again a normal domain, \(A=\bigcap_\mathfrak{m}A_\mathfrak{m}\) is also normal. (Here, if \(x\in K\) belongs to every \(A_\mathfrak{m}\), then the ideal \(\{a\in A\mid ax\in A\}\) is not contained in any maximal ideal, so \(x\in A\).) Therefore \(\mathfrak{p}^{-1}\mathfrak{A}=\mathfrak{A}\) is impossible and we must have \(\mathfrak{A}\subsetneq \mathfrak{p}^{-1}\mathfrak{A}\). Meanwhile, from \(\mathfrak{A}\subseteq \mathfrak{p}\) we have \(\mathfrak{p}^{-1}\mathfrak{A}\subseteq \mathfrak{p}^{-1}\mathfrak{p}=A\), so \(\mathfrak{p}^{-1}\mathfrak{A}\) is again an ideal of \(A\), and from \(\mathfrak{p}\mathfrak{p}^{-1}=A\) we have \(\mathfrak{p}\otimes_A A/\mathfrak{p}^{-1}\mathfrak{A}\cong \mathfrak{p}/\mathfrak{A}\). At this point, by the third result of §Associated Primes, ⁋Theorem 7, associated primes are determined by localization, and since \(\mathfrak{p}\) is invertible, tensoring with \(\mathfrak{p}\) does not change the associated primes. Therefore, applying §Associated Primes, ⁋Lemma 5 to \(\mathfrak{p}/\mathfrak{A}\subseteq A/\mathfrak{A}\), we obtain \(\Ass A/\mathfrak{p}^{-1}\mathfrak{A}=\Ass \mathfrak{p}/\mathfrak{A}\subseteq \Ass A/\mathfrak{A}\), and since all elements of the right-hand side have codimension \(1\), \(\mathfrak{p}^{-1}\mathfrak{A}\) is also pure codimension \(1\). Now, by the maximality of \(\mathfrak{A}\), \(\mathfrak{p}^{-1}\mathfrak{A}\) can be expressed as a product of codimension \(1\) prime ideals, and multiplying both sides by \(\mathfrak{p}\) yields a contradiction, giving the desired result. Therefore, every ideal that is pure codimension \(1\) is a product of codimension \(1\) prime ideals, and since each codimension \(1\) prime ideal is invertible, these are all invertible.

2. Now let us prove the uniqueness of the expression. Given two expressions, by multiplying by the inverse of one side, it suffices to show that for distinct codimension \(1\) prime ideals \(\mathfrak{p}_1,\ldots, \mathfrak{p}_r\) and integers \(n_1,\ldots, n_r\), if \(\prod_i \mathfrak{p}_i^{n_i}=A\), then all \(n_i\) are \(0\). Set \(\mathfrak{P}=\prod_{n_i>0}\mathfrak{p}_i^{n_i}\) and \(\mathfrak{Q}=\prod_{n_i<0}\mathfrak{p}_i^{-n_i}\); these are both ideals of \(A\), and by assumption \(\mathfrak{P}=\mathfrak{Q}\). If \(n_i>0\) for some \(i\), then \(\mathfrak{Q}=\mathfrak{P}\subseteq \mathfrak{p}_i\); if \(n_j<0\) holds for no \(j\), then \(\mathfrak{Q}=A\), which is a contradiction, and otherwise, since \(\mathfrak{p}_i\) is prime, for some \(j\) we have \(\mathfrak{p}_j\subseteq \mathfrak{p}_i\), and since \(A\) is a domain, from \((0)\subsetneq \mathfrak{p}_j\subseteq \mathfrak{p}_i\) and \(\codim \mathfrak{p}_i=1\) we obtain \(\mathfrak{p}_j=\mathfrak{p}_i\), contradicting the fact that they are distinct. Therefore all \(n_i\) are at most \(0\), and applying the same argument to \(\mathfrak{Q}\) shows that all \(n_i\) are \(0\). Now, for an arbitrary invertible fractional ideal \(\mathfrak{A}\), applying §Fractional Ideals, ⁋Corollary 6, for a suitable non-zerodivisor of \(A\), \(a\), we have \(\mathfrak{A}=(a \mathfrak{A})(a)^{-1}\), and since \(a \mathfrak{A}\) and \((a)\) are both invertible ideals, by the result above they are products of codimension \(1\) prime ideals. That is, \(\mathfrak{A}\) is a product of integer powers of codimension \(1\) prime ideals, and from this we conclude that \(\CaDiv(A)\) is the free abelian group generated by the codimension \(1\) prime ideals.

Definition 3 A Noetherian normal domain of dimension \(1\) is called a Dedekind domain.

Then the following holds.

Corollary 4 Every nonzero ideal of a Dedekind domain \(A\) is invertible and can be expressed uniquely as a product of prime ideals, and the same holds for the fractional ideals of \(A\). Consequently, \(\CaDiv(A)\) is the free abelian group generated by the maximal ideals of \(A\).

By §Fractional Ideals, ⁋Corollary 6, we know that \(\Pic(A)\cong \CaDiv(A)/K^\times\). Dedekind domains are used mostly in algebraic number theory, and following the terminology of that field, when \(A\) is a Dedekind domain we call \(\Pic(A)\) the class group.

Discrete valuation ring

DVRs arise naturally as localizations of Dedekind domains and play a central role in understanding divisors.

Definition 5 For a field \(K\), a subring \(A\) is called a valuation ring if, for every \(x \in K^\times\), either \(x \in A\) or \(x^{-1} \in A\).

A valuation ring \(A\) is called a discrete valuation ring (DVR) if \(A\) is Noetherian and the maximal ideal of \(A\), \(\mathfrak{m}\), is a non-\(0\) principal ideal.

In §Regular Local Rings, ⁋Definition 6, we first defined a valuation \(\nu:\Frac(A)^\times \rightarrow G\) and then defined the valuation ring as the set of elements whose value under \(\nu\) is nonnegative; a ring obtained in this way satisfies the condition above since \(\nu(x^{-1})=-\nu(x)\). Conversely, for an \(A\) satisfying the condition above, if on \(G=K^\times/A^\times\) we define \(\bar{x}\geq \bar{y}\) by \(x/y\in A\), then \(G\) becomes a totally ordered abelian group, the quotient map is a valuation, and the valuation ring of this valuation is again \(A\). Also, when \(G=\mathbb{Z}\), that is, for the valuation ring of a discrete valuation, as we saw after §Regular Local Rings, ⁋Definition 6, it is Noetherian and its maximal ideal is a non-\(0\) principal ideal, so it is a DVR in the above sense; conversely, for a DVR in the above sense, from the expression \(f=\pi^n u\) in Proposition 8, the map defined by \(f\mapsto n\) is a surjective valuation onto \(\mathbb{Z}\), so the two definitions agree.

Proposition 6 For a valuation ring \(A\), the following hold.

1. \(A\) is a local ring.
2. \(A\) is an integrally closed domain.
3. The ideals of \(A\) are totally ordered.

Proof

1. If we show that the set of non-unit elements of \(A\), \(\mathfrak{m}\), is an ideal, then every proper ideal of \(A\) contains no unit and hence is contained in \(\mathfrak{m}\), so \(\mathfrak{m}\) is the unique maximal ideal. First, let \(x, y \in \mathfrak{m}\). If one of them is \(0\), it is clear that \(x+y \in \mathfrak{m}\), so we may assume that \(x, y\) are both not \(0\); then by the definition of a valuation ring, either \(x/y \in A\) or \(y/x \in A\). By symmetry, suppose \(x/y \in A\); then \(x + y = y(1 + x/y) \in (y)\), and if \(x+y\) were a unit, then \(y\) would also be a unit, contradicting \(y \in \mathfrak{m}\). Therefore \(x + y \in \mathfrak{m}\). Also, for any \(a \in A\), if \(ax\) were a unit, then \(a(ax)^{-1}\) would be an inverse of \(x\), again a contradiction, so \(ax \in \mathfrak{m}\). Thus \(\mathfrak{m}\) is an ideal and \(A\) is a local ring.

2. Let an element of \(K = \Frac(A)\), \(x\), be integral over \(A\). Then there exists a monic polynomial \(f(t) \in A[t]\) such that \(f(x) = 0\). If \(x \notin A\), then \(x^{-1} \in A\). Writing \(f(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_0\), we have \(x^n = -a_{n-1}x^{n-1} - \cdots - a_0\), and multiplying both sides by \(x^{-(n-1)}\) gives \(x = -a_{n-1} - \cdots - a_0 x^{-(n-1)} \in A\), a contradiction. Therefore \(x \in A\), and \(A\) is integrally closed.

3. Suppose two proper ideals \(\mathfrak{a}, \mathfrak{b}\) are given. If \(\mathfrak{a} \not\subseteq \mathfrak{b}\), then there exists \(a \in \mathfrak{a} \setminus \mathfrak{b}\), and we show that \(\mathfrak{b} \subseteq \mathfrak{a}\). For any \(b \in \mathfrak{b}\), if \(b = 0\) then clearly \(b \in \mathfrak{a}\), so suppose \(b \neq 0\). Also, since \(0 \in \mathfrak{b}\), we have \(a \neq 0\), and hence \(a/b \in K^\times\); by the definition of a valuation ring, either \(a/b \in A\) or \(b/a \in A\). If \(a/b \in A\), then \(a = (a/b) \cdot b \in \mathfrak{b}\), contradicting \(a \notin \mathfrak{b}\). Therefore \(b/a \in A\), and \(b = (b/a) \cdot a \in \mathfrak{a}\).

Definition 7 For a DVR \(A\), an element generating the maximal ideal \(\mathfrak{m} = (\pi)\), \(\pi\), is called a uniformizer or a uniformizing parameter.

Proposition 8 For a DVR \(A\), its maximal ideal \(\mathfrak{m}\), and \(K = \Frac(A)\), the following hold.

1. For \(\mathfrak{m}\), any generator \(\pi\) is a uniformizer, and two uniformizers \(\pi, \pi'\) satisfy the relation \(\pi' = u\pi\) (with \(u \in A^\times\)). Therefore, a uniformizer is unique up to multiplication by a unit.
2. For any \(f \in K^\times\), there exist a unique integer \(n \in \mathbb{Z}\) and a unit \(u \in A^\times\) such that \(f = \pi^n u\).

Proof

1. If \(\mathfrak{m} = (\pi) = (\pi')\), then \(\pi' = a\pi\) and \(\pi = b\pi'\) for some \(a, b \in A\). From this, \(\pi = ab\pi\), and since \(\pi \neq 0\), we have \(ab = 1\), and therefore \(a, b \in A^\times\).

2. First consider the case \(f \in A \setminus \{0\}\). Since \(A\) is a Noetherian local ring and \(\mathfrak{m} = (\pi)\), by §Blowup Algebras, ⁋Corollary 8 (Krull intersection theorem) we have \(\bigcap_{n \geq 0} \mathfrak{m}^n = 0\). Therefore, for \(f \neq 0\), \(f \in \mathfrak{m}^n \setminus \mathfrak{m}^{n+1}\) holds for a unique nonnegative integer \(n\). Now since \(\mathfrak{m}^n = (\pi^n)\), we can write \(f = \pi^n u\). Here \(u \notin \mathfrak{m}\), so \(u \in A^\times\).

In the case \(f \in K^\times \setminus A\), we have \(f^{-1} \in \mathfrak{m}\) by the definition of a valuation ring. Applying the argument above to \(f^{-1}\), we get \(f^{-1} = \pi^m v\) (with \(m \geq 1\), \(v \in A^\times\)), and hence \(f = \pi^{-m} v^{-1}\). Here \(v^{-1} \in A^\times\), so we obtain the desired expression.

Let us show the uniqueness of the expression. Suppose \(\pi^n u = \pi^{n'} u'\) and \(n \geq n'\). Then \(\pi^{n-n'} u = u'\). If \(n > n'\), the left-hand side belongs to \(\mathfrak{m}\) while the right-hand side is a unit, a contradiction. Therefore \(n = n'\) and \(u = u'\).

There is a close relationship between DVRs and Dedekind domains.

Proposition 9 For a Noetherian integral domain \(A\), the following are equivalent.

1. \(A\) is a Dedekind domain.
2. In \(A\), for every maximal ideal \(\mathfrak{m}\), the localization \(A_\mathfrak{m}\) is a DVR.

Proof

(1) ⇒ (2): If \(A\) is a Dedekind domain, then \(A\) is a Noetherian normal domain of dimension 1. For any maximal ideal \(\mathfrak{m}\), \(A_\mathfrak{m}\) is a 1-dimensional Noetherian local domain and is normal. By §Regular Local Rings, ⁋Theorem 8, \(A_\mathfrak{m}\) is a regular local ring, and hence \(\mathfrak{m}A_\mathfrak{m}\) is a principal ideal. That is, \(A_\mathfrak{m}\) is a DVR.

(2) ⇒ (1): If \(A_\mathfrak{m}\) is a DVR, then it is 1-dimensional, so \(\dim A = 1\). Also, since DVRs are integrally closed, \(A_\mathfrak{m}\) is integrally closed for every maximal ideal \(\mathfrak{m}\), and therefore \(A\) is also integrally closed. That the localizations are Noetherian does not by itself imply that the original ring is Noetherian, so that \(A\) is Noetherian must be additionally assumed. Indeed, if \(A\) is Noetherian and every localization \(A_\mathfrak{m}\) is a DVR, then \(A\) is a Dedekind domain.

Divisors

Corollary 4 shows that, just as every element of a UFD can be expressed uniquely as a product of irreducible elements, in a Dedekind domain every ideal can be expressed uniquely as a product of codimension \(1\) prime ideals, and fractional ideals can be expressed by additionally allowing negative powers. Generalizing this further, we make the following definition for an arbitrary ring \(A\).

Definition 10 For a ring \(A\), a Weil divisor of \(A\) is an element of the free abelian group generated by its codimension \(1\) prime ideals, \(\Div(A)\).

It is customary to write the operation of \(\Div(A)\) as \(+\). Thus an element of \(\Div(A)\) is a formal linear combination of codimension \(1\) prime ideals.

Theorem 11 Fix a Noetherian ring \(A\). Then, satisfying for every invertible ideal \(\mathfrak{a}\)

\[\mathfrak{a}\mapsto \sum_\text{\scriptsize$\mathfrak{p}\subseteq A$ a codimension $1$ prime} \length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}) \mathfrak{p}\]

there exists a group homomorphism \(\Phi: \CaDiv(A) \rightarrow \Div(A)\).

Proof

From §Fractional Ideals, ⁋Corollary 6 and the universal property of free abelian groups, it suffices to show that the above formula is well-defined and preserves products of ideals.

First, let us show that the above formula is well-defined. That is, we must show that for any invertible ideal \(\mathfrak{a}\), the given expression is a finite sum. For an arbitrary codimension \(1\) prime ideal \(\mathfrak{p}\), consider the 1-dimensional local ring \(A_\mathfrak{p}\). Then by §Fractional Ideals, ⁋Theorem 3, \(\mathfrak{a}\) contains a non-zerodivisor of \(A\), and therefore \(\mathfrak{a}A_\mathfrak{p}\) also contains a non-zerodivisor of \(A_\mathfrak{p}\), \(a\). Meanwhile, by §Associated Primes, ⁋Theorem 7, the minimal prime ideals of \(A_\mathfrak{p}\) consist only of zerodivisors, so since it contains the non-zerodivisor \(a\), \(\mathfrak{a}A_\mathfrak{p}\) is not contained in any minimal prime ideal. But since \(\dim A_\mathfrak{p}=1\), a prime ideal of \(A_\mathfrak{p}\) is either minimal or the maximal ideal \(\mathfrak{p}A_\mathfrak{p}\) itself, and hence the only prime ideal containing \(\mathfrak{a}A_\mathfrak{p}\) is \(\mathfrak{p}A_\mathfrak{p}\), so \(\dim A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}=0\). From this, by §Krull Dimension, ⁋Corollary 3 and §The Jordan-Hölder Theorem, ⁋Theorem 3, we know that \(\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})<\infty\).

Now, if \(\mathfrak{a}\not\subseteq \mathfrak{p}\), then \(\mathfrak{a}A_\mathfrak{p}\) contains a unit of \(A_\mathfrak{p}\), so \(\mathfrak{a}A_\mathfrak{p}=A_\mathfrak{p}\) and therefore \(\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})=0\). Also, if \(\mathfrak{a}\subseteq \mathfrak{p}\), then \(\mathfrak{p}A_\mathfrak{p}\) must be one of the minimal prime ideals containing \(\mathfrak{a}A_\mathfrak{p}\), and therefore only finitely many such \(\mathfrak{p}\) are fixed for each given \(\mathfrak{a}\). From the discussion above, we can confirm that the given expression is well-defined.

Now we must show that \(\Phi\) preserves multiplication. That is, if \(\mathfrak{a}=\prod \mathfrak{a}_i\), we must show that the equation

\[\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})=\sum \length(A_\mathfrak{p}/\mathfrak{a}_iA_\mathfrak{p})\]

holds. For convenience of notation, let \(A\) denote the local ring obtained by localization, \(\mathfrak{p}\) its maximal ideal, and let the \(\mathfrak{a}_i\) be ideals of \(A\). Then, since the \(\mathfrak{a}_i\) are invertible ideals, they are each generated by a single non-zerodivisor \(a_i\), and then the following filtration

\[A\supseteq \mathfrak{a}_1=(a_1)\supseteq \mathfrak{a}_1 \mathfrak{a}_2=(a_1a_2)\supseteq\cdots\supseteq \prod_{i\in I} \mathfrak{a}_i=\left(\prod_{i\in I} a_i\right)\]

exists. Now, to show the given equation, it suffices to show that \((\prod_{j< i}a_j)/(\prod_{j\leq i} a_j)\cong A/(a_i)\). This is clear: since each \(a_i\) is a non-zerodivisor, multiplication by \(\prod_{j< i}a_j\) induces an isomorphism from \(A\) to \((\prod_{j< i}a_j)\), and considering the kernel of the following map

\[(\prod_{j< i}a_j) \rightarrow A \rightarrow A/(a_i)\]

this is precisely \((a_i\prod_{j< i}a_j)=(\prod_{j\leq i}a_j)\).

In particular, if \(A\) was \(1\)-dimensional, then by the calculation in the second result of §The Jordan-Hölder Theorem, ⁋Theorem 3,

\[\length A/\mathfrak{a}=\sum_\text{\scriptsize$\mathfrak{a}\subseteq \mathfrak{p}$ a codimension $1$ prime of $A$} \length (A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})\]

holds, and therefore, defining \(\Div(A) \rightarrow \mathbb{Z}\) by \(\sum n_\mathfrak{p}\mathfrak{p}\mapsto \sum n_\mathfrak{p}\), composition with the map in Theorem 11 shows that there exists a map satisfying \(\mathfrak{a}\mapsto \length(A/\mathfrak{a})\) from \(\CaDiv(A)\) to \(\mathbb{Z}\).

Meanwhile, by endowing \(\CaDiv(A)\) with an appropriate equivalence relation, we could write the collection of equivalence classes of elements of \(\CaDiv(A)\) as \(\Pic(A)\). In a similar vein, we define the following.

Definition 12 Let a Noetherian ring \(A\) and \(A\)’s total ring of fractions \(K\) be given. Then for any \(a\in K^\times\), the image of the invertible ideal \((a)\) under \(\Phi:\CaDiv(A) \rightarrow \Div(A)\) is called a principal divisor.
For the group of principal divisors of \(A\), \(\Prin(A)\subseteq \Div(A)\), the quotient group \(\Div(A)/\Prin(A)\) is called \(A\)’s codimension \(1\) Chow group and is denoted \(\Chow(A)\).

Then from the definition, we know that \(\Phi:\CaDiv(A) \rightarrow \Div(A)\) induces \(\Psi: \Pic(A) \rightarrow \Chow(A)\).

Dimension and Normalization

Above, we defined the group homomorphism \(\Phi:\CaDiv(A) \rightarrow \Div(A)\) and the group homomorphism \(\Psi:\Pic(A) \rightarrow \Chow(A)\) obtained from it.

Definition 13 For a reduced ring \(A\), the normalization of \(A\) is the integral closure, in \(A\)’s total ring of fractions \(K\), of \(A\). If the normalization of \(A\) is itself, we call \(A\) a normal ring.

This is a restatement of the definition given in §Regular Local Rings, ⁋Definition 9. Also, if \(A\) is an integral domain, then \(K=\Frac(A)\), so in this case the notions of normalization and normal ring coincide with those defined in §Integral Extensions, ⁋Definition 3. Then the following holds.

Proposition 14 For a normal Noetherian ring \(A\), both \(\Phi,\Psi\) defined above are injective.

Proof

Considering the diagram

by [Homological Algebra] §Diagram chasing, ⁋Proposition 1 (The four lemma), it suffices to show that \(\Phi\) is injective. That is, it suffices to prove that if two invertible ideals of \(A\) are sent via \(\Phi\) to the same element of \(\Div(A)\), then they are equal.

Therefore, suppose two invertible ideals \(\mathfrak{a}, \mathfrak{b}\) satisfy \(\Phi(\mathfrak{a})=\Phi(\mathfrak{b})\), and let us show that \(\mathfrak{a}=\mathfrak{b}\). By symmetry, it suffices to show \(\mathfrak{a}\subseteq \mathfrak{b}\), and by the result of §Associated Primes, ⁋Corollary 4, it in turn suffices to show that \(\mathfrak{a}A_\mathfrak{p}\subseteq \mathfrak{b}A_\mathfrak{p}\) holds for every associated prime \(\mathfrak{p}\). But now \(\mathfrak{p}A_\mathfrak{p}\) is an associated prime of \(\mathfrak{b}A_\mathfrak{p}\), and since \(\mathfrak{b}A_\mathfrak{p}\) is a principal ideal generated by a non-zerodivisor, by the equivalence in §Regular Local Rings, ⁋Theorem 10 (Serre), \(\codim \mathfrak{p}=1\) and \(A_\mathfrak{p}\) is a DVR. Now, from \(\Phi(\mathfrak{a})=\Phi(\mathfrak{b})\) we have \(\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})=\length(A_\mathfrak{p}/\mathfrak{b}A_\mathfrak{p})\), and from this we conclude that \(\mathfrak{a}A_\mathfrak{p}=\mathfrak{b}A_\mathfrak{p}\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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