Divisors

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In this post we examine $\CaDiv(A)$ and $\Pic(A)$, which were defined in §Fractional Ideals, ⁋Definition 5, in somewhat greater detail.

Dedekind Domains

First, we 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 $A$ vacuously has 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 of $A$ is a UFD.

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

Proof

First assume that $\mathfrak{a}$ is invertible. Then for any maximal ideal $\mathfrak{m}$, the localization $\mathfrak{a}A_\mathfrak{m}$ is, viewing $A_\mathfrak{m}$ as a subring of $\Frac(A)$, a principal ideal generated by a non-zerodivisor in $K$. On the other hand, for any maximal ideal $\mathfrak{m}\subseteq A$, the ring $\mathfrak{A}_\mathfrak{m}$ is a normal domain by §Integral Extensions, ⁋Proposition 9; applying §Regular Local Rings, ⁋Theorem 9 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
\[\codim \mathfrak{p}=\dim A_\mathfrak{p}=\codim \mathfrak{p}A_\mathfrak{p}\leq 1\]
Since $A_\mathfrak{p}$ is an integral domain, $(0)\subseteq \mathfrak{p}A_\mathfrak{p}$ implies $\codim \mathfrak{p}=1$.

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 $\mathfrak{p}A_\mathfrak{m}=A_\mathfrak{m}$; if $\mathfrak{p}\subseteq \mathfrak{m}$ then by the computation above $\mathfrak{p}A_\mathfrak{m}$ is a prime ideal of codimension $1$ in $A_\mathfrak{m}$, hence a minimal prime ideal since $A_\mathfrak{m}$ is a domain, and thus applying §Dimension, ⁋Corollary 8 it becomes a principal ideal, hence 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 contradiction that there exists an ideal of pure codimension $1$ which is not a product of codimension $1$ prime ideals, and let $\mathfrak{A}$ be a maximal such ideal. If $\mathfrak{p}$ is a codimension $1$ prime ideal containing $\mathfrak{A}$, then $\mathfrak{p}^{-1}\mathfrak{p}=A$ and hence $A\subsetneq\mathfrak{p}^{-1}$. 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 by §Integral Extensions, ⁋Proposition 9 $A$ is normal, so this is impossible; hence $\mathfrak{A}\subsetneq \mathfrak{p}^{-1}\mathfrak{A}$. By maximality of $\mathfrak{A}$, the ideal $\mathfrak{p}^{-1}\mathfrak{A}$ can be expressed as a product of codimension $1$ prime ideals, and multiplying both sides by $\mathfrak{p}$ yields a contradiction. The uniqueness of this expression can be shown by induction, and applying §Fractional Ideals, ⁋Corollary 6 we see that $\CaDiv(A)$ is 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 ideal of a Dedekind domain $A$ is invertible, and can be uniquely expressed as a product of prime ideals; the same holds for fractional ideals of $A$. Hence $\CaDiv(A)$ is the free abelian group generated by the maximal ideals of $A$.

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

Discrete Valuation Rings

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

Definition 5 A subring $A$ of a field $K$ 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 its maximal ideal $\mathfrak{m}$ is principal.

Proposition 6 For a valuation ring $A$, the following hold.

$A$ is a local ring.
$A$ is an integrally closed domain.
The ideals of $A$ are totally ordered.

Proof

Let $\mathfrak{m}$ be a maximal ideal of $A$. It suffices to show that any $x \in A \setminus \mathfrak{m}$ is a unit. If $x$ were not a unit, then $x^{-1} \notin A$, and by the definition of a valuation ring we would have $x^{-1} \in A$. But since $x \in A$, this gives $x \cdot x^{-1} = 1 \in \mathfrak{m}$, a contradiction. Hence $x$ is a unit and $A$ is a local ring with unique maximal ideal $\mathfrak{m}$.
Let $x \in K = \Frac(A)$ be integral over $A$. Then there exists a monic polynomial $f(t) \in A[t]$ with $f(x) = 0$. 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 $x^{-(n-1)}$ yields $x = -a_{n-1} - \cdots - a_0 x^{-(n-1)} \in A$, a contradiction. Therefore $x \in A$ and $A$ is integrally closed.
Let $\mathfrak{a}, \mathfrak{b}$ be two proper ideals. If $\mathfrak{a} \not\subseteq \mathfrak{b}$, there exists $a \in \mathfrak{a} \setminus \mathfrak{b}$. For any $b \in \mathfrak{b}$, we have $a/b \in K$, and by the definition of a valuation ring either $a/b \in A$ or $b/a \in A$. If $b/a \in A$, then $b = (b/a) \cdot a \in \mathfrak{a}$. If $a/b \in A$ and $b/a \notin A$, then $a/b$ must be a unit. (Otherwise $a/b \in \mathfrak{m}$ and since $b/a$ is the inverse of $a/b$, we would have $b/a \in A$, a contradiction.) Hence $b/a \in A$ and $b \in \mathfrak{a}$.

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

The reason this element $\pi$ is called a uniformizer is as follows. In a general ring, elements appear only in numerators, but in a DVR the element $\pi^{-1}$ exists in the field of fractions $K$. Hence any $f \in K^\times$ can be expressed uniformly in the form

\[f = \pi^n \cdot u\]

Here $n \in \mathbb{Z}$ and $u$ is a unit. If $n > 0$, there are $n$ copies of $\pi$ in the numerator; if $n < 0$, there are $\lvert n\rvert$ copies of $\pi$ in the denominator; and if $n = 0$, there is no $\pi$ at all. That is, $\pi$ serves as a uniform standard by which all elements are expressed in the same form.

Proposition 8 Let $\mathfrak{m}$ be the maximal ideal of a DVR $A$ and let $K = \Frac(A)$. Then the following hold.

Any generator $\pi$ of $\mathfrak{m}$ is a uniformizer, and two uniformizers $\pi, \pi'$ satisfy $\pi' = u\pi$ for some $u \in A^\times$. Thus the uniformizer is unique up to multiplication by a unit.
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

If $\mathfrak{m} = (\pi) = (\pi')$, then there exist $a, b \in A$ with $\pi' = a\pi$ and $\pi = b\pi'$. This gives $\pi = ab\pi$, and since $\pi \neq 0$ we have $ab = 1$; hence $a, b \in A^\times$.
First consider the case $f \in A \setminus \{0\}$. Since $A$ is a Noetherian local ring with $\mathfrak{m} = (\pi)$, by §Blowup Algebras, ⁋Corollary 8 (Krull intersection theorem) we have $\bigcap_{n \geq 0} \mathfrak{m}^n = 0$. Hence for $f \neq 0$ there exists a unique non-negative integer $n$ such that $f \in \mathfrak{m}^n \setminus \mathfrak{m}^{n+1}$. Since $\mathfrak{m}^n = (\pi^n)$, we can write $f = \pi^n u$. Here $u \notin \mathfrak{m}$, so $u \in A^\times$.

If $f \in K^\times \setminus A$, then by the definition of a valuation ring we have $f^{-1} \in \mathfrak{m}$. Applying the above argument to $f^{-1}$ gives $f^{-1} = \pi^m v$ with $m \geq 1$ and $v \in A^\times$, and therefore $f = \pi^{-m} v^{-1}$. Since $v^{-1} \in A^\times$, we obtain the desired expression.

To show uniqueness, suppose $\pi^n u = \pi^{n'} u'$ with $n \geq n'$. Then $\pi^{n-n'} u = u'$. If $n > n'$, the left side lies in $\mathfrak{m}$ while the right side is a unit, a contradiction. Hence $n = n'$ and $u = u'$.

There is a close relationship between DVRs and Dedekind domains.

Proposition 9 For an integral domain $A$, the following are equivalent.

$A$ is a Dedekind domain.
For every maximal ideal $\mathfrak{m}$ of $A$, 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 $\mathfrak{m}$, the localization $A_\mathfrak{m}$ is a one-dimensional Noetherian local domain which is normal. By §Regular Local Rings, ⁋Theorem 9, $A_\mathfrak{m}$ is a regular local ring, and hence $\mathfrak{m}A_\mathfrak{m}$ is a principal ideal. Thus $A_\mathfrak{m}$ is a DVR.

(2) ⇒ (1): If $A_\mathfrak{m}$ is a DVR then it has dimension 1, so $\dim A = 1$. Moreover, since a DVR is integrally closed, $A_\mathfrak{m}$ is integrally closed for every maximal ideal $\mathfrak{m}$, and therefore $A$ is integrally closed. The Noetherian property is not preserved under localization, so we must additionally assume that $A$ is Noetherian. 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 in a UFD every element can be uniquely expressed as a product of irreducible elements, in a Dedekind domain every ideal can be uniquely expressed as a product of codimension $1$ prime ideals, and fractional ideals can be expressed by including negative exponents. We generalize this further and make the following definition for an arbitrary ring $A$.

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

The operation in $\Div(A)$ is customarily denoted by $+$. Thus an element of $\Div(A)$ is a formal linear combination of codimension $1$ prime ideals.

Theorem 11 Fix a Noetherian ring $A$. Then for every invertible ideal $\mathfrak{a}$, there exists a group homomorphism $\Phi: \CaDiv(A) \rightarrow \Div(A)$ satisfying

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

Proof

By §Fractional Ideals, ⁋Corollary 6 and the universal property of the free abelian group, it suffices to show that the above expression is well defined and preserves ideal products.

First we show that the expression is well defined. That is, for any invertible ideal $\mathfrak{a}$ we must show that the given sum is finite. For any codimension $1$ prime ideal $\mathfrak{p}$, consider the one-dimensional local ring $A_\mathfrak{p}$. By §Fractional Ideals, ⁋Theorem 3, $\mathfrak{a}$ contains a non-zerodivisor of $A$, and hence $\mathfrak{a}A_\mathfrak{p}$ also contains a non-zerodivisor $a$ of $A_\mathfrak{p}$. On the other hand, since $\dim A_\mathfrak{p}=\codim \mathfrak{p}A_\mathfrak{p}$ is always at least as large as the codimension of any ideal contained in $\mathfrak{p}A_\mathfrak{p}$, we have

\[\dim \mathfrak{a}A_\mathfrak{p}\leq\dim \mathfrak{p}A_\mathfrak{p}\leq \dim A_\mathfrak{p}-\codim \mathfrak{p}A_\mathfrak{p}\]

and since $\mathfrak{p}A_\mathfrak{p}$ contains a non-zerodivisor we have $\codim \mathfrak{p}A_\mathfrak{p}=1$, so $\dim \mathfrak{a}A_\mathfrak{p}=0$. From this, by §Dimension, ⁋Corollary 3 and §The Jordan-Hölder Theorem, ⁋Theorem 3, we know that $\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})<\infty$.

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 hence $\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})=0$. 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 for a given $\mathfrak{a}$ there are only finitely many such $\mathfrak{p}$. The above discussion confirms 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 need to show

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

For convenience of notation, let $A$ denote the local ring obtained by localization, let $\mathfrak{p}$ be its maximal ideal, and let the $\mathfrak{a}_i$ be ideals of $A$. Since each $\mathfrak{a}_i$ is an invertible ideal, they are each generated by a single non-zerodivisor $a_i$, and then we have the filtration

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

To prove the desired formula it suffices to show $(\prod_{j< i}a_j)/(\prod_{j\leq i} a_j)\cong A/(a_i)$. 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 map

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

its kernel is exactly $(a_i\prod_{j< i}a_j)=(\prod_{j\leq i}a_j)$, which is obvious.

In particular, if $A$ is one-dimensional, then by the second part of the computation in §The Jordan-Hölder Theorem, ⁋Theorem 3,

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

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

On the other hand, we were able to write $\Pic(A)$ as the set of equivalence classes of elements of $\CaDiv(A)$ under a suitable equivalence relation. In a similar vein we make the following definition.

Definition 12 Let $A$ be a Noetherian ring and let $K$ be its total ring of fractions. 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 $\Prin(A)\subseteq \Div(A)$ of principal divisors of $A$, the quotient group $\Div(A)/\Prin(A)$ is called the codimension $1$ Chow group of $A$ and is denoted $\Chow(A)$.

Then by 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 induced group homomorphism $\Psi:\Pic(A) \rightarrow \Chow(A)$.

Definition 13 For a reduced ring $A$, the normalization of $A$ is the integral closure of $A$ in its total ring of fractions $K$. If the normalization of $A$ is $A$ itself, then $A$ is called a normal ring.

If $A$ is an integral domain, then $K=\Frac(A)$, and hence 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 $A$, the maps $\Phi,\Psi$ defined above are both injective.

Proof

Consider the following diagram

$relations_between_divisors$

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

So let $\mathfrak{a}, \mathfrak{b}$ be two invertible ideals satisfying $\Phi(\mathfrak{a})=\Phi(\mathfrak{b})$, and let us show $\mathfrak{a}=\mathfrak{b}$. By symmetry it suffices to show $\mathfrak{a}\subseteq \mathfrak{b}$, and by the result of §Associated Primes, ⁋Corollary 4 this in turn reduces to showing $\mathfrak{a}A_\mathfrak{p}\subseteq \mathfrak{b}A_\mathfrak{p}$ 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 11 (Serre) we have $\codim \mathfrak{p}=1$ and $A_\mathfrak{p}$ is a DVR. From $\Phi(\mathfrak{a})=\Phi(\mathfrak{b})$ we obtain $\length(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p})=\length(A_\mathfrak{p}/\mathfrak{b}A_\mathfrak{p})$, and from this we know $\mathfrak{a}A_\mathfrak{p}=\mathfrak{b}A_\mathfrak{p}$.

References

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

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Dedekind Domains

Discrete Valuation Rings

Divisors

Dimension and Normalization

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