Dedekind Domains and Ideal Classes

Because $R$ is a domain, $(0)⊊\mathfrak{p}$ is a strict prime chain. Since [F1] gives $\dim R=1$, no longer strict chain ending at $\mathfrak{p}$ exists. Therefore $\mathfrak{p}$ has height $1$.

The ring $R$ is Noetherian and integrally closed by [F1], and step 1.1 shows that $\mathfrak{p}$ is height one. Hence [L1] applies and gives that $R_{\mathfrak{p}}$ is a discrete valuation ring.

At the zero prime, localisation inverts every nonzero element of $R$, so $R_{(0)}=\mathrm{Frac}(R)$ by [F1]. A field is integrally closed, so $R_{(0)}$ is integrally closed.

If $\mathfrak{p}\mathrm{\ne }(0)$ is prime, the hypothesis makes $R_{\mathfrak{p}}$ a discrete valuation ring. By [L1], every discrete valuation ring is integrally closed. Hence each nonzero-prime localisation of $R$ is integrally closed.

Steps 1.1 and 1.2 show that every prime localisation of $R$ is integrally closed. Therefore [L2] gives that $R$ itself is integrally closed.

Let $\mathfrak{p}⊊\mathfrak{q}$ be prime ideals of $R$ with $\mathfrak{p}\mathrm{\ne }(0)$. By hypothesis, $R_{\mathfrak{q}}$ is a discrete valuation ring. Because $R$ is a domain, any nonzero element of $\mathfrak{p}$ remains nonzero in $R_{\mathfrak{q}}$, so $\mathfrak{p}{R}_{\mathfrak{q}}$ is a nonzero prime ideal of $R_{\mathfrak{q}}$. Its contraction is $\mathfrak{p}$, so $\mathfrak{p}{R}_{\mathfrak{q}}⊊\mathfrak{q}{R}_{\mathfrak{q}}$. This contradicts [L3], which allows only $(0)$ and $\mathfrak{q}{R}_{\mathfrak{q}}$ as primes in a DVR. Therefore no nonzero prime ideal of $R$ is properly contained in another prime ideal.

Let $\mathfrak{p}$ be a nonzero prime ideal of $R$. By [L4], choose a maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$. Since $\mathfrak{p}$ is nonzero, step 1.1 forbids a strict inclusion $\mathfrak{p}⊊\mathfrak{m}$. Hence $\mathfrak{p}=\mathfrak{m}$, so every nonzero prime ideal of $R$ is maximal.

Assume now that $R$ is not a field. Choose a nonzero nonunit $x\in R$. Then $(x)$ is a proper ideal, so [L4] gives a maximal ideal $\mathfrak{m}$ containing it. The ideal $\mathfrak{m}$ is nonzero because it contains $x$, and step 2.1 shows that every nonzero prime ideal is maximal. Thus every strict prime chain has length at most $1$, while $(0)⊊\mathfrak{m}$ has length $1$. Hence $\dim R=1$.

If $R$ is Dedekind, then [L1] gives that each localisation at a nonzero prime is a DVR. This proves $(1)\Rightarrow (2)$.

Assume (2). Then [L2] makes $R$ integrally closed, and [L3] gives $\dim R=1$ together with maximality of the nonzero primes. Because $R$ was assumed Noetherian and nonfield, [F1] now says that $R$ is Dedekind. This proves $(2)\Rightarrow (1)$.

Still assuming (2), let $I$ be a nonzero proper ideal and let $\mathfrak{m}$ be a maximal ideal. If $I⊈\mathfrak{m}$, then $I_{\mathfrak{m}}=R_{\mathfrak{m}}$ because some element of $I$ becomes a unit. If $I\subseteq \mathfrak{m}$, then $\mathfrak{m}$ is nonzero and [L4] applies in the DVR $R_{\mathfrak{m}}$ from (2), so $I_{\mathfrak{m}}$ is principal. Therefore (3) holds.

Assume (3), and let $\mathfrak{m}$ be a maximal ideal. By [L6] and [L7], the ring $R_{\mathfrak{m}}$ is a Noetherian local domain. Let $J\subset R_{\mathfrak{m}}$ be a nonzero proper ideal, and let $I:=\{r\in R:r\mathrm{/}1\in J\}$. Then $J=I{R}_{\mathfrak{m}}$, and $I$ is a nonzero proper ideal contained in $\mathfrak{m}$. Applying (3) at the maximal ideal $\mathfrak{m}$ shows that $J=I_{\mathfrak{m}}$ is principal. Hence every nonzero proper ideal of the local domain $R_{\mathfrak{m}}$ is principal, and because $R$ is not a field its maximal ideal $\mathfrak{m}{R}_{\mathfrak{m}}$ is nonzero. Therefore [L5] makes $R_{\mathfrak{m}}$ a DVR. Now let $\mathfrak{p}$ be a nonzero prime ideal and choose a maximal ideal $\mathfrak{m}\supseteq \mathfrak{p}$. In the DVR $R_{\mathfrak{m}}$, the extended ideal $\mathfrak{p}{R}_{\mathfrak{m}}$ is a nonzero prime ideal, so it equals the maximal ideal $\mathfrak{m}{R}_{\mathfrak{m}}$. Contracting to $R$ gives $\mathfrak{p}=\mathfrak{m}$, so every nonzero prime is maximal and therefore $R_{\mathfrak{p}}$ is a DVR by the first part applied with $\mathfrak{m}=\mathfrak{p}$. This proves $(3)\Rightarrow (2)$.

By [L3], any of the equivalent conditions forces every nonzero prime ideal of $R$ to be maximal.

By [F1] and [L1], the localisation $S^{-1}R$ is Noetherian. Because $R$ is a domain and $0\mathrm{\notin }S$, so is $S^{-1}R$. By [F1] and [L2], it is integrally closed.

Suppose $S^{-1}R$ has a nonzero prime ideal $\mathfrak{n}$. By [L3], it is the extension of a nonzero prime ideal $\mathfrak{p}\subset R$ disjoint from $S$. In a Dedekind domain every nonzero prime is maximal, so every nonzero prime of $S^{-1}R$ is maximal as well by the same correspondence [L3]. Therefore $S^{-1}R$ has dimension $1$, and step 1.1 makes it a Dedekind domain.

Suppose instead that $S^{-1}R$ has no nonzero prime ideals. Then [L3] says that the only prime ideal of $S^{-1}R$ is $(0)$. In a domain, $(0)$ is maximal exactly when the ring is a field. Hence $S^{-1}R$ is a field.

The displayed pairing is exactly the trace form of the extension $L\mathrm{/}F$.

Because $L\mathrm{/}F$ is separable, [L1] makes that trace form nondegenerate.

Choose an $F$-basis $e_1,\dots ,e_n$ of $L$. For each $i$, choose a monic polynomial for $e_i$ over $F$ and one nonzero denominator $d_i\in R$ clearing all of its coefficients. Then $(d_i{e}_i{)}^m+d_i{c}_{m-1}(d_i{e}_i{)}^{m-1}+\cdots +d_i^m{c}_0=0$ for suitable $c_j\in F$, so $d_i{e}_i$ is integral over $R$ and therefore lies in $A$. Replacing each $e_i$ by $d_i{e}_i$ preserves the $F$-basis, so we may assume from the start that $e_1,\dots ,e_n\in A$. By [L1], there is a trace-dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ with ${\mathrm{Tr}}_{L\mathrm{/}F}(e_i{e}_j^{\ast })={\delta }_{ij}$. Clearing denominators in the coordinates of the $e_j^{\ast }$ relative to the basis $e_i$, choose $0\mathrm{\ne }c\in R$ such that $c{e}_j^{\ast }\in {\sum }_{i}R{e}_i\subseteq A$ for every $j$.

Let $x\in A$, and write $x={\sum }_i{a}_i{e}_i$ with $a_i\in F$. The dual-basis relation gives $a_j={\mathrm{Tr}}_{L\mathrm{/}F}(x{e}_j^{\ast })$ and therefore $c{a}_j={\mathrm{Tr}}_{L\mathrm{/}F}(xc{e}_j^{\ast })$. Because $x$ and $c{e}_j^{\ast }$ lie in $A$, their product is integral over $R$. The field trace of an integral element is a finite sum of its conjugates, hence is integral over $R$, and it lies in $F$; therefore [F2] forces $c{a}_j\in R$. Hence every coefficient $a_j$ lies in $c^{-1}R$.

Step 2.1 shows that $A\subseteq c^{-1}(R{e}_1+\cdots +R{e}_n)$. Multiplication by $c$ identifies this containing module with the finitely generated module $R{e}_1+\cdots +R{e}_n$, so it is finite over $R$. Since $A$ is an $R$-submodule of a finite module and $R$ is Noetherian, $A$ is finite over $R$.

By [L1], the ring $A$ is a finite $R$-module, hence integral over $R$. Since $R$ is Noetherian by [F1], every ideal of $A$ is an $R$-submodule of the finite $R$-module $A$, so $A$ is Noetherian. If $x\in \mathrm{Frac}(A)\subseteq L$ is integral over $A$, then transitivity of integrality makes $x$ integral over $R$, so the defining property of $A$ forces $x\in A$. Thus $A$ is integrally closed.

The inclusion $R\subseteq A$ is an injective integral extension of nonzero domains. Since [F1] gives $\dim R=1$, [L2] gives $\dim A=1$.

Steps 1.1 and 1.2 show that $A$ is a Noetherian integrally closed domain of dimension $1$. Hence [F1] makes $A$ a Dedekind domain.

Choose nonzero $d,e\in R$ with $dI\subseteq R$ and $eJ\subseteq R$ by [F1]. Then $de(I+J)\subseteq R$, $de(IJ)\subseteq R$, and if $a=u\mathrm{/}v$ with $u,v\in R$ and $v\mathrm{\ne }0$, then $vd(aI)\subseteq R$. The modules $I+J$, $IJ$, and $aI$ are nonzero because they contain any fixed nonzero element of $I$, the product of nonzero elements of $I$ and $J$, and the nonzero element $ax$ for $0\mathrm{\ne }x\in I$, respectively. Thus $I+J$, $IJ$, and $aI$ are fractional ideals.

Choose $0\mathrm{\ne }i\in I$ and $0\mathrm{\ne }y\in J$. Since $eJ\subseteq R$, one has $(ei)J\subseteq I$, so $0\mathrm{\ne }ei\in (I:J)$. If $x\in (I:J)$, then $xy\in I$, hence $dxy\in R$. Since $ey\in R$ is nonzero, one gets $(dey)x=dx(ey)\in R$. Therefore every $x\in (I:J)$ is bounded by one nonzero denominator, so $(I:J)$ is a fractional ideal.

These constructions are defined by subset conditions and finite sums inside the fixed field $K$, so they do not depend on which denominators were chosen in step 1.1. Hence the operations are well defined.

Assume (1). Choose a finite relation $1={\sum }_{i=1}^n{x}_i{y}_i$ with $x_i\in I$ and $y_i\in (R:I)$. For any $x\in I$ one has $x={\sum }_i(x{y}_i)x_i$, so $x_1,\dots ,x_n$ generate $I$. Define $\varphi :R^n\to I$ by $e_i\mapsto x_i$ and $\psi :I\to R^n$ by $x\mapsto (x{y}_1,\dots ,x{y}_n)$. Then $\varphi \psi ={\mathrm{id}}_I$, so $I$ is a direct summand of a finite free module and hence finite projective by [L4].

Assume (3), and choose generators $x_1,\dots ,x_n$ of $I$. For a maximal ideal $\mathfrak{m}$, write $I_{\mathfrak{m}}=a{R}_{\mathfrak{m}}$ with $a\in K^{\times }$. For each $i$ there is $u_i\in R_{\mathfrak{m}}$ such that $x_i\mathrm{/}1=a{u}_i$, and we may choose $s_i\mathrm{\notin }\mathfrak{m}$ with $s_i{u}_i\in R$. Putting $t:=s_1\cdots s_n$, we get $(t{a}^{-1})x_i\in R$ for every $i$, so $t{a}^{-1}\in (R:I)$ and $a^{-1}\mathrm{/}1=(t{a}^{-1})\mathrm{/}t\in (R:I{)}_{\mathfrak{m}}$. Hence $1=(a\mathrm{/}1)(a^{-1}\mathrm{/}1)\in I_{\mathfrak{m}}(R:I{)}_{\mathfrak{m}}\subseteq R_{\mathfrak{m}}$, so in fact $I_{\mathfrak{m}}(R:I{)}_{\mathfrak{m}}=R_{\mathfrak{m}}$. By [L1], localising the quotient $R\mathrm{/}I(R:I)$ and using [L2] gives $(R\mathrm{/}I(R:I){)}_{\mathfrak{m}}=0$ for every maximal ideal $\mathfrak{m}$. Therefore [L3] gives $I(R:I)=R$, so $I$ is invertible.

Still under (1), localise at a maximal ideal $\mathfrak{m}$. From $1=\sum x_i{y}_i$, one summand $x_j{y}_j$ is a unit in the local ring $R_{\mathfrak{m}}$, so $x_j$ generates $I_{\mathfrak{m}}$. Thus (1) implies (3), and also yields the local rank-one part of (2).

Condition (2) implies (3) because finite projective modules are finitely generated, and a free rank-one module over $R_{\mathfrak{m}}$ is principal. Steps 1.1 and 2.1 prove $(1)\Rightarrow (2)$ and $(1)\Rightarrow (3)$, step 1.2 proves $(3)\Rightarrow (1)$, and the present step proves $(2)\Rightarrow (3)$. Therefore the three conditions are equivalent.

Choose $0\mathrm{\ne }d\in R$ with $J:=dI\subseteq R$. Then $J$ is a nonzero integral ideal of $R$, so [F1] makes it finitely generated; hence the fractional ideal $I=d^{-1}J$ is finitely generated as well. Let $\mathfrak{m}$ be a maximal ideal. If $J⊈\mathfrak{m}$, then $J_{\mathfrak{m}}=R_{\mathfrak{m}}$, so $I_{\mathfrak{m}}=d^{-1}{R}_{\mathfrak{m}}$ is principal. If $J\subseteq \mathfrak{m}$, then $\mathfrak{m}$ is a nonzero prime, [L1] makes $R_{\mathfrak{m}}$ a DVR, and [L2] makes the integral ideal $J_{\mathfrak{m}}$ principal. Hence $I_{\mathfrak{m}}=d^{-1}{J}_{\mathfrak{m}}$ is principal in either case. Thus $I$ is finitely generated and every maximal localisation of $I$ is principal.

Applying [L3] to step 1.1 shows that $I$ is invertible.

Choose $0\mathrm{\ne }x\in I$ and $0\mathrm{\ne }d\in R$ with $dI\subseteq R$, and put $a:=dx\in R$. If a nonzero prime ideal $\mathfrak{p}$ contains neither $a$ nor $d$, then both $a\mathrm{/}1$ and $d\mathrm{/}1$ are units in $R_{\mathfrak{p}}$, so $x=(a\mathrm{/}1)(d\mathrm{/}1{)}^{-1}$ is a unit of $R_{\mathfrak{p}}$. Hence $R_{\mathfrak{p}}=x{R}_{\mathfrak{p}}\subseteq I_{\mathfrak{p}}\subseteq d^{-1}{R}_{\mathfrak{p}}=R_{\mathfrak{p}}$, so $I_{\mathfrak{p}}=R_{\mathfrak{p}}$ and therefore $v_{\mathfrak{p}}(I)=0$.

Put $b:=ad$. If $b$ is a unit, then no nonzero prime ideal contains $b$. Otherwise every prime ideal containing $(b)$ is nonzero, hence maximal because [F2] gives $\dim R=1$. By [L2], every prime ideal of $R\mathrm{/}(b)$ is therefore maximal. The quotient $R\mathrm{/}(b)$ is Noetherian by [L3], so [L4] makes it Artinian, and then [L5] gives only finitely many maximal ideals. Translating back through [L2], only finitely many nonzero prime ideals of $R$ contain $b$.

Combining steps 1.1 and 2.1, only finitely many nonzero prime ideals can satisfy $v_{\mathfrak{p}}(I)\mathrm{\ne }0$.

Fix a nonzero prime ideal $\mathfrak{p}$. By [F1], write $I_{\mathfrak{p}}={\mathfrak{p}}^m{R}_{\mathfrak{p}}$ and $J_{\mathfrak{p}}={\mathfrak{p}}^n{R}_{\mathfrak{p}}$. Localizing the defining finite sums for a product gives $(IJ{)}_{\mathfrak{p}}=I_{\mathfrak{p}}{J}_{\mathfrak{p}}={\mathfrak{p}}^{m+n}{R}_{\mathfrak{p}}$, so $v_{\mathfrak{p}}(IJ)=m+n=v_{\mathfrak{p}}(I)+v_{\mathfrak{p}}(J)$.

By [L1], only finitely many integers $v_{\mathfrak{p}}(I)$ are nonzero, so $J:={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{v_{\mathfrak{p}}(I)}$ is a well-defined nonzero fractional ideal. Fix a nonzero prime ideal $\mathfrak{q}$. If $\mathfrak{p}\mathrm{\ne }\mathfrak{q}$, choose $x\in \mathfrak{p}\setminus \mathfrak{q}$; then $x\mathrm{/}1$ is a unit in $R_{\mathfrak{q}}$, so $({\mathfrak{p}}^{v_{\mathfrak{p}}(I)}{)}_{\mathfrak{q}}=R_{\mathfrak{q}}$. Hence only the $\mathfrak{q}$-factor survives after localizing, and $J_{\mathfrak{q}}={\mathfrak{q}}^{v_{\mathfrak{q}}(I)}{R}_{\mathfrak{q}}=I_{\mathfrak{q}}$.

Let $M:=(I+J)\mathrm{/}J$. Localizing the short exact sequence $0\to J\to I+J\to M\to 0$ and using [L2], we get $M_{\mathfrak{m}}\cong (I_{\mathfrak{m}}+J_{\mathfrak{m}})\mathrm{/}{J}_{\mathfrak{m}}$ for every maximal ideal $\mathfrak{m}$. Step 1.1 gives $I_{\mathfrak{m}}=J_{\mathfrak{m}}$, so $M_{\mathfrak{m}}=0$ for all $\mathfrak{m}$. Therefore [L3] gives $M=0$, hence $I\subseteq J$. The same argument with $(I+J)\mathrm{/}I$ gives $J\subseteq I$, so $I=J$. This proves existence of the factorization.

If also $I={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$, localising at a fixed nonzero prime $\mathfrak{q}$ gives ${\mathfrak{q}}^{v_{\mathfrak{q}}(I)}{R}_{\mathfrak{q}}=I_{\mathfrak{q}}={\mathfrak{q}}^{n_{\mathfrak{q}}}{R}_{\mathfrak{q}}$. Uniqueness of powers in the DVR $R_{\mathfrak{q}}$ forces $n_{\mathfrak{q}}=v_{\mathfrak{q}}(I)$, so the factorization is unique.

If $I\subseteq R$, then $I_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$ for every nonzero prime ideal $\mathfrak{p}$, so the exponent in the DVR equality $I_{\mathfrak{p}}={\mathfrak{p}}^{v_{\mathfrak{p}}(I)}{R}_{\mathfrak{p}}$ must be nonnegative. Hence integral ideals have only nonnegative exponents.

Write $I={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{a_{\mathfrak{p}}}$ and $J={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{b_{\mathfrak{p}}}$ with $a_{\mathfrak{p}},b_{\mathfrak{p}}\ge 0$ by [L1]. Then $I\supseteq J$ exactly when $a_{\mathfrak{p}}\le b_{\mathfrak{p}}$ for every $\mathfrak{p}$, and that coordinatewise inequality is exactly the condition that $J=I{\prod }_{\mathfrak{p}}{\mathfrak{p}}^{b_{\mathfrak{p}}-a_{\mathfrak{p}}}$.

Therefore $I\supseteq J$ if and only if $I$ divides $J$.

Form the integral ideal $J:=(a)I^{-1}$. By [L1], only finitely many prime ideals $\mathfrak{p}$ satisfy $v_{\mathfrak{p}}(a)>v_{\mathfrak{p}}(I)$; call this finite set $S$. For each $\mathfrak{p}\in S$, choose $b_{\mathfrak{p}}\in I$ with $v_{\mathfrak{p}}(b_{\mathfrak{p}})=v_{\mathfrak{p}}(I)$. By [L3], choose elements $u_{\mathfrak{p}}\in R$ such that $u_{\mathfrak{p}}\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{p})$ and $u_{\mathfrak{p}}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{q})$ for every distinct $\mathfrak{q}\in S$. Put $b:={\sum }_{\mathfrak{p}\in S}{u}_{\mathfrak{p}}{b}_{\mathfrak{p}}$, and if $S=\mathrm{\varnothing }$ put $b=0$. Then $b\in I$.

For $\mathfrak{p}\in S$, the term $u_{\mathfrak{p}}{b}_{\mathfrak{p}}$ is a unit multiple of $b_{\mathfrak{p}}$ in $R_{\mathfrak{p}}$, while every other summand lies in $\mathfrak{p}{I}_{\mathfrak{p}}$. Hence $v_{\mathfrak{p}}(b)=v_{\mathfrak{p}}(I)$. For $\mathfrak{q}\mathrm{\notin }S$, one has $v_{\mathfrak{q}}(a)=v_{\mathfrak{q}}(I)$ by definition of $S$. Therefore $(a,b)$ and $I$ have the same prime valuations.

By [L1], ideals with the same prime valuations are equal. Hence $(a,b)=I$.

If $(a)$ and $(b)$ are principal fractional ideals with $a,b\in K^{\times }$, then $(a)(b)=(ab)$ and $(a{)}^{-1}=(a^{-1})$. Therefore the principal fractional ideals form a subgroup.

If $I^{\mathrm{\prime }}=(a)I$ and $J^{\mathrm{\prime }}=(b)J$, then $I^{\mathrm{\prime }}{J}^{\mathrm{\prime }}=(ab)(IJ)$. So changing representatives by principal multiples changes the product by another principal factor. Hence the class of $IJ$ depends only on the classes of $I$ and $J$.

The map $\mathrm{cl}$ is well defined and surjective: by [L1], every divisor is a finite sum and therefore determines a unique fractional ideal $\prod {\mathfrak{p}}^{n_{\mathfrak{p}}}$, and every ideal class has such a representative.

If $x\in K^{\times }$, then [L1] applied to the principal fractional ideal $(x)$ gives

Therefore $\mathrm{cl}(\mathrm{div}(x))=[(x)]=0$ in $\mathrm{Cl}(R)$. So $\mathrm{im}(\mathrm{div})\subseteq \ker (\mathrm{cl})$.

Conversely, if $D=\sum n_{\mathfrak{p}}[\mathfrak{p}]$ satisfies $\mathrm{cl}(D)=0$, then the ideal $\prod {\mathfrak{p}}^{n_{\mathfrak{p}}}$ is principal, say equal to $(x)$ with $x\in K^{\times }$. Uniqueness in [L1] then forces $D=\mathrm{div}(x)$. Thus $\ker (\mathrm{cl})=\mathrm{im}(\mathrm{div})$.

If $x\in R^{\times }$, then $(x)=R$, so all valuations of $(x)$ are zero and $\mathrm{div}(x)=0$. Conversely, if $\mathrm{div}(x)=0$, then [L1] gives $(x)=R$, which is exactly the statement that $x$ is a unit of $R$. Hence $\ker (\mathrm{div})=R^{\times }$.

Steps 1.1, 1.2, 1.3, and 2.1 prove exactness of the displayed sequence.

If $R$ is a PID, then every nonzero fractional ideal is principal, so the quotient in [F2] has exactly one class. Thus $\mathrm{Cl}(R)$ is trivial.

Conversely, assume $\mathrm{Cl}(R)$ is trivial. Then every nonzero prime ideal has trivial class and is therefore principal. By [L1], every nonzero integral ideal is a finite product of principal prime ideals, hence principal. Therefore [F1] makes $R$ a PID.

This proves the equivalence.