ANT week 4,5 AG point of view

Classic ResultUnique Factorization of Ideals in Dedekind Domain.Define Ideal Group Via Grothendieck Group, Weil DivisorProjective Module, line bundle, fractional ideal, Picard Group.Projective ModulePicard GroupFractional ideal, Cartier Divisor, line bundle and the exact sequence

 

Classic Result

Let $I,J$$I,J$ be two ideals in $R$$R$. If $I\supseteq J$$I\supseteq J$, we use the notation $I|J$$I|J$, Notice that $a|b⟺(a)|(b)$$a|b\iff(a)|(b)$ in an integral domain.

Then we have

$$\boxed{{\displaystyle P\text{ is prime ideal}⟺P|IJ⟹P|I\vee P|J}}$$

Proof.

Assume that $P$$P$ is prime ideal and $P|IJ$$P|IJ$, but $P\nmid I\wedge P\nmid J$$P\nmid I\wedge P\nmid J$, then $\mathrm{\exists }x\in I,y\in J$$\exists x\in I,y\in J$ such that $x,y\notin P$$x,y\notin P$. Since $P$$P$ is a prime ideal, $R-P$$R-P$ is a ''Closed Monoid''. i,e, $xy\in R-P⟺x\in R-P\wedge y\in R-P$$xy\in R-P\iff x\in R-P\wedge y\in R-P$. But $xy\in IJ\subseteq P$$xy\in IJ\subseteq P$, that is a contradiction.

Conversely, assume $P|IJ⟹P|I\vee P|J$$P|IJ\implies P|I\vee {P|J}$.

Then

$$xy\in P⟹(x)(y)\subseteq P⟺P|(x)(y)⟹P|(x)\vee P|(y)⟺x\in P\vee y\in P◻$$

The duality between prime ideal and closed monoid appears at Monoidal Category as well. Usually prime ideals corresponds to some bad property. See my ongoing paper: Click here.

Definition. For an integral domain $R$$R$ that is not a field, we say $R$$R$ is Dedekind Domain if $R$$R$ is an integrally closed, Noetherian domain with Krull dimension one.

Remark. Every PID is Dedekind domain since PID is UFD hence integrally closed, PID is Noetherian and krull dimension one.

Remark. For a Dedekind domain $R$$R$, you should image that $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$ as a smooth curves. We will see that the ideal group is just the Weil Divisor Group of $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$.

Lemma 1. Let $R$$R$ be a Noetherian integral domain, for every ideal $I\subseteq R$$I\subseteq R$, there exists finite many prime ideal ${\mathfrak{p}}_1,...,{\mathfrak{p}}_n$$\mathfrak p_1,...,\mathfrak p_n$ such that $I|{\mathfrak{p}}_1{\mathfrak{p}}_2...{\mathfrak{p}}_n$$I|\mathfrak p_1\mathfrak p_2...\mathfrak p_n$

Proof. Let $S$$S$ be the set of all the ideal do not satisfies this property, we need to prove $S=\mathrm{\varnothing }$$S=\emptyset$. Since $R$$R$ is a Noetherian ring, then $S$$S$ has a maximal element $M.$$M.$ $M$$M$ is not a prime ideal since $M\in S$$M\in S$. Hence there exists $r,s\in R-M$$r,s\in R-M$ such that $rs\in M$$rs\in M$. Hence there exists some prime ideal ${\mathfrak{p}}_1,...,{\mathfrak{p}}_n,{\mathfrak{q}}_1,...,{\mathfrak{q}}_m$$\mathfrak p_1,...,\mathfrak p_n,\mathfrak q_1,...,\mathfrak q_m$ such that $M+(r)|{\mathfrak{p}}_1...{\mathfrak{p}}_n,M+(s)|{\mathfrak{q}}_1,...,{\mathfrak{q}}_m$$M+(r)|\mathfrak p_1...\mathfrak p_n,M+(s)|\mathfrak q_1,...,\mathfrak q_m$. Hence we have

$$M|(M+(r))(M+(s))|{\mathfrak{p}}_1...{\mathfrak{p}}_n{\mathfrak{q}}_1,...,{\mathfrak{q}}_m$$

That is a contradiction with $M\in S$$M\in S$. $◻$$\square$

Lemma 2. Let $I\subseteq R$$I\subseteq R$ be a proper ideal in Dedekind domain $R$$R$, then there exists a $k\in K-R$$k\in K-R$ such that $kI\subseteq R$$kI\subseteq R$. Here $K=\mathrm{Frac}(R)$$K= \mathrm{Frac}(R)$.

Proof. Let $0\ne a\in I$$0\ne a\in I$, let $r$$r$ be the minimal natural number such that $(a)|{\mathfrak{p}}_1...{\mathfrak{p}}_r$$(a)|\mathfrak p_1...\mathfrak p_r$. Then we have $I|{\mathfrak{p}}_1...{\mathfrak{p}}_r$$I|\mathfrak p_1...\mathfrak p_r$ and there exists a maximal ideal $\mathfrak{m}$$\mathfrak m$ such that $\mathfrak{m}|I|{\mathfrak{p}}_1...{\mathfrak{p}}_r$$\mathfrak m|I|\mathfrak p_1...\mathfrak p_r$. Since $\mathfrak{m}$$\mathfrak m$ is prime ideal, there exists a number $i$$i$ such that $\mathfrak{m}|{\mathfrak{p}}_i$$\mathfrak{m}|\mathfrak{p}_i$. But since in Dedekind domain, prime ideal is maximal ideal as well, we have ${\mathfrak{m}\mathfrak{=}\mathfrak{p}}_i$$\mathfrak{m=p}_i$. Since we pick the minimal $r$$r$, there exists $b\in {\mathfrak{p}}_2...{\mathfrak{p}}_r$$b\in\mathfrak p_2...\mathfrak p_r$ such that $b\notin (a)$$b\notin(a)$. Now let $k={\displaystyle \frac{b}{a}},bI\subseteq b\mathfrak{m}=b{\mathfrak{p}}_1\subseteq {\mathfrak{p}}_1...{\mathfrak{p}}_n\subseteq aR⟹kI\subseteq R$$k=\dfrac{b}{a},bI\subseteq b\mathfrak m=b\mathfrak p_1\subseteq\mathfrak p_1...\mathfrak p_n\subseteq aR\implies kI\subseteq R$. $◻$$\square$

This lemma will help us to define the fractional ideal.

Lemma 3. Let $I\subseteq R$$I\subseteq R$ be an ideal in Dedekind domain, then there exits another ideal $J$$J$ such that $IJ$$IJ$ is a principal ideal.

Proof. Let $0\ne a\in I$$0\ne a\in I$, let $J=(a):I:=\{r\in R:rI\subseteq (a)\}$$J=(a):I:=\{r\in R:rI\subseteq(a)\}$. Easy to see that $(a):I$$(a):I$ is a sub $R$$R$-module of $R$$R$ hence an ideal.

Easy to see that $(a)|IJ⟹A=\frac{1}{a}IJ\subseteq R$$(a)|IJ\implies A=\frac{1}{a}IJ\subseteq R$ is an ideal. We need to show that $A=R$$A=R$ hence $IJ=(a)$$IJ=(a)$.

Now assume that $A\ne R$$A\ne R$, by lemma 3 we know that there exists $k\in K=\mathrm{Frac}(R)$$k\in K= \mathrm{Frac}(R)$ such that $kA\subseteq R$$kA\subseteq R$. Since $a\in I$$a\in I$, we have $IJ|(a)J⟹A=\frac{1}{a}IJ|J$$IJ|(a)J\implies A=\frac{1}{a}IJ|J$. We have $R|kA|kJ$$R|kA|kJ$. For all $\beta \in J,k\beta \in R$$\beta\in J,k\beta\in R$, and $k\beta I\subseteq kIJ=kaA\subseteq (a)$$k\beta I\subseteq kIJ=kaA\subseteq (a)$. By the definition of $J$$J$ we know that $k\beta \in J$$k\beta\in J$, hence $kJ\subseteq J$$k J\subseteq J$.

Since $R$$R$ is Noetherian ring, $J$$J$ is finite generated by $x_1,...,x_m$$x_1,...,x_m$. By $kJ\subseteq J$$kJ\subseteq J$ we have

$$\begin{array}{c}k\left(\begin{array}{c}{x}_1\\ ...\\ x_m\end{array}\right)=M\left(\begin{array}{c}{x}_1\\ ...\\ x_m\end{array}\right)\end{array}$$

Hence $\det (M-kI)=0$$\det(M-kI)=0$, hence $k$$k$ is the root of the monic polynomial $\det (M-xI)$$\det(M-xI)$. Since $R$$R$ is integrally closed, we have $k\in R$$k\in R$.

That is a contradiction with $k\in K-R$$k\in K-R$. Hence $A$$A$ has to equal $R$$R$. $◻$$\square$

Lemma 4. Let $A,B,C$$A,B,C$ be some ideals in a Dedekind Domain $R$$R$, then $AB=AC⟹B=C$$AB=AC\implies B=C$.

Proof. By Lemma 3 we could find $J$$J$ such that $JA=(a)$$JA=(a)$, then we have $(JA)B=(JA)C,aB=aC⟹B=C$$(JA)B=(JA)C,aB=aC\implies B=C$. $◻$$\square$

Lemma 5. Let $A,B$$A,B$ be two ideals in a Dedekind Domain $R$$R$, then

$$\boxed{{\displaystyle A|B⟺\mathrm{\exists }C,B=AC}}$$

Proof.

$\Leftarrow$$\Leftarrow$ is obvious. For $\Rightarrow$$\Rightarrow$, take $J$$J$ such that $AJ=(a)$$AJ=(a)$, by $A|B$$A|B$ we know that $AJ=(a)|BJ$$AJ=(a)|BJ$, then $C=\frac{1}{a}BJ$$C=\frac{1}{a}BJ$ is an ideal. $B=AC◻$$B=AC\quad\square$

Unique Factorization of Ideals in Dedekind Domain.

Proposition.

Let $R$$R$ be a Dedekind Domain, then every proper ideal could be unique factor to product of finitely many prime ideal..

Proof.

$\boxed{{\displaystyle \text{Existence}:}}$$\boxed{\text{Existence}:}$

Let $S$$S$ be the set of ideals that do not have this property. If $S\ne \mathrm{\varnothing }$$S\ne\emptyset$, then it contains a maximal element $M$$M$, then there exists a prime ideal(maximal ideal) $P,P|M$$P,P|M$. By Lemma 5 we have $M=PI$$M=PI$ for some ideal $I$$I$. Hence $I|M$$I|M$ and $I\ne M$$I\ne M$. Therefore $I={\mathfrak{p}}_1...{\mathfrak{p}}_k$$I=\mathfrak p_1...\mathfrak p_k$. Hence $M$$M$ is product of prime ideal, that is a contradiction.

$\boxed{{\displaystyle \text{Uniqueness}:}}$$\boxed{\text{Uniqueness}:}$

It follows from prime ideals in maximal ideal and cancel law of the ideal monoid directly. $◻$$\square$

Define Ideal Group Via Grothendieck Group, Weil Divisor

Definition. Let $I(R)$$I(R)$ be the monoid of nonzero ideal for a Dedekind Domain. Then we define the ideal group of $R$$R$ by the Grothendieck Group of $I(R)$$I(R)$, which is a free monoid. We use the notation $K_0(I(R))$$K_0(I(R))$. Or, we could define the ideal group via Weil Divisor. The free abelian group generated by all the irreducible closed subset of $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$ with codimension $1$$1$. In Dedekind Domain, this will be the free abelian group generated by no zero prime ideal.

We will construct the inverse of each ideal in $R$$R$ via fractional ideal(projective module) in next section.

Projective Module, line bundle, fractional ideal, Picard Group.

Projective Module

We admit the following fact without proof.

Kaplansky's theorem on projective modules. Let $(A,\mathfrak{m})$$(A,\mathfrak m)$ be a local ring and $P$$P$ be a projective module over $A$$A$, then $P$$P$ is free.

If $P$$P$ is finite generated, then $P\cong A^n,n={\dim }_{A/\mathfrak{m}}(P{\otimes }_{A}A/\mathfrak{p})$$P\cong A^n,n=\dim_{A/\mathfrak m}(P\otimes_A A/\mathfrak p)$.

Definition.

Let $A$$A$ be a ring, and let $X$$X$ be a finitely generated $A$$A$-module. Define the rank function

$${\mathrm{rk}}_X:\mathrm{Spec}A⟶\mathbb{N},\mathfrak{p}⟼{\dim }_{\kappa (\mathfrak{p})}(X{\otimes }_A\kappa (\mathfrak{p})),$$

where $\kappa (\mathfrak{p})=A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$$\kappa(\mathfrak{p})=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$.

Remark. Given $\mathbb{N}$$\mathbb N$ discrete topology, if $X$$X$ is projective module, then ${\mathrm{rk}}_X$$\mathrm{rk}_X$ is continuous. Hence when $X$$X$ is projective and $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ is connected, we have ${\mathrm{rk}}_X$$\mathrm{rk}_X$ is constant.

The following theorem suggests us view projective module over a Noetherian ring as a vector bundle.

Proposition. For $\mathfrak{p}\in \mathrm{Spec}(A),X_{\mathfrak{p}}\cong A_{\mathfrak{p}}^n$$\mathfrak p\in\mathrm{Spec}(A),X_\mathfrak p\cong A_{\mathfrak p}^n$ for some $n\in \mathbb{N}$$n\in\mathbb N$.

Theorem. Let $A$$A$ be a Noetherian ring and $M$$M$ is a finite generated $A$$A$-module then the following are equivalent:

$†.$$\dagger.$ $M$$M$ is projective.

$†.$$\dagger.$ $M$$M$ is locally free

$†.$$\dagger.$ $M_P$$M_P$ is free $A_P$$A_P$-module for every $P$$P$ in $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$.

Picard Group

Definition. A line bundle over a ring $A$$A$ is a finite generated projective $A$$A$-module such that the rank function ${\mathrm{rk}}_M$$\mathrm{rk}_M$ is constant with value $1$$1$. We call $A$$A$ itself the trivial line bundle.

We would like to put an abelian group structure on the set of isomorphism classes of line bundle. The product is tensor product and the inverse is the dual object.

Remark. In general, we could define the Picard Group for a symmetric monoidal category.

Definition. For a finite generated projective module over $A$$A$, we define its dual by

$$\boxed{{\displaystyle X^{\ast }:={\mathrm{Hom}}_A(X,A)}}$$

Lemma. Let $X$$X$ and $Y$$Y$ be finite generated projective $A$$A$-modules,

$(1):X{\otimes }_{A}Y$$(1):X\otimes_AY$ is a finite generated projective $A$$A$-module and ${\mathrm{rk}}_{X{\otimes }_{A}Y}={\mathrm{rk}}_X\cdot {\mathrm{rk}}_Y$$\mathrm{rk}_{X\otimes_AY}=\mathrm{rk}_X\cdot \mathrm{rk}_Y$ point wise.

$(2):X^{\ast }$$(2): X^*$ is a finite generated $A$$A$-module and ${\mathrm{rk}}_{X^{\ast }}={\mathrm{rk}}_X$$\mathrm{rk}_{X^*}=\mathrm{rk}_X$

$(3):$$(3):$ If $X$$X$ is a line bundle then $X^{\ast }{\otimes }_{A}X\cong A$$X^*\otimes_A X\cong A$.

Proof.

1. Since $X$$X$ is finite generated projective, there exists a $Z$$Z$ such that $X\oplus Z\cong A^n$$X\oplus Z\cong A^n$.

Then $(X{\otimes }_{A}Y)\oplus (Z{\otimes }_{A}Y)\cong (X\oplus Z){\otimes }_{A}Y\cong A^n{\otimes }_{A}Y\cong Y^n$$(X\otimes_A Y)\oplus(Z\otimes_AY)\cong (X\oplus Z)\otimes_A Y\cong A^n\otimes_A Y\cong Y^n$. Hence $X{\otimes }_{A}Y$$X\otimes_A Y$ is a direct summand of a finite generated projective module, hence it is finite generated projective module.

Now we compute the rank. For $\mathfrak{p}\in \mathrm{Spec}(A),X_{\mathfrak{p}}\cong A_{\mathfrak{p}}^n$$\mathfrak p\in\mathrm{Spec}(A),X_\mathfrak p\cong A_{\mathfrak p}^n$ and $Y_{\mathfrak{p}}\cong A_{\mathfrak{p}}^m$$Y_{\mathfrak{p}}\cong A_{\mathfrak p}^m$, then $(X{\otimes }_{A}Y{)}_{\mathfrak{p}}\cong X_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}{Y}_{\mathfrak{p}}\cong A_{\mathfrak{p}}^{mn}$$(X\otimes_A Y)_{\mathfrak p}\cong X_{\mathfrak p}\otimes_{A_{\mathfrak p}} Y_{\mathfrak p}\cong A_{\mathfrak p}^{mn}$.

2. Again, we have $X\oplus Z\cong A^n$$X\oplus Z\cong A^n$, hence ${\mathrm{Hom}}_A(X,A)\oplus {\mathrm{Hom}}_A(Z,A)\cong {\mathrm{Hom}}_A(X\oplus Z,A)\cong (A^n{)}^{\ast }\cong A^n$$\mathrm{Hom}_A(X,A)\oplus\mathrm{Hom}_A(Z,A)\cong\mathrm{Hom}_A(X\oplus Z,A)\cong (A^n)^*\cong A^n$. Hence $X^{\ast }$$X^*$ is a finite generated projective module. Also we have $(X^{\ast }{)}_{\mathfrak{p}}\cong (X_{\mathfrak{p}}{)}^{\ast }$$(X^*)_{\mathfrak p}\cong(X_{\mathfrak p})^*$.

3. If ${\mathrm{rk}}_X=1$$\mathrm{rk}_X=1$ Define $\theta :X^{\ast }\otimes X\to A$$\theta:X^*\otimes X\to A$ by $f\otimes x⟼f(x)$$f\otimes x\longmapsto f(x)$, we only need to check it is an isomorphism at each stalk.(Form the Sheaf theory point of view)

   $${\theta }_{\mathfrak{p}}:X_{\mathfrak{p}}^{\ast }{\otimes }_{\mathfrak{p}}{X}_{\mathfrak{p}}\to A_{\mathfrak{p}}$$

   Since we have $X_{\mathfrak{p}}$$X_{\mathfrak p}$ is free, hence $X_{\mathfrak{p}}\cong A_{\mathfrak{p}}$$X_\mathfrak p\cong A_{\mathfrak p}$. Thus ${\theta }_{\mathfrak{p}}$$\theta_\mathfrak p$ is an isomorphism. $◻$$\square$

   Definition. Define the Picard Group of a ring $A$$A$ be the isomorphism classes of line bundle with $\otimes$$\otimes$, denote as $(\mathrm{Pic}(A),\otimes )$$(\mathrm{Pic}(A),\otimes)$, for each line bundle, the inverse is just $X^{\ast }$$X^*$. Picard Group is a functor from $\mathrm{CRing}$$\mathrm{CRing}$ to $\mathrm{Ab}$$\mathrm{Ab}$. For a ring homomorphism $f:A\to B$$f:A\to B$, we have $\mathrm{Pic}(f):X⟼X{\otimes }_{A}B$$\mathrm{Pic}(f):X\longmapsto X\otimes_A B$.

Fractional ideal, Cartier Divisor, line bundle and the exact sequence

Definition. A fractional ideal $I$$I$ of $A$$A$ is a nonzero submodule of $\mathrm{Frac}(A)$$\mathrm{Frac}(A)$ that is contained in a cyclic $A$$A$-submodule of $\mathrm{Frac}(A)$$\mathrm{Frac}(A)$.

Let us denote $\mathrm{Frac}(A)$$\mathrm{Frac}(A)$ as $K$$K$. Then a fractional ideal is just a submodule like $I\subseteq (f/g)A$$I\subseteq(f/g)A$.

Hence for any fractional ideal we have $gI\subseteq fA\subset A$$gI\subseteq fA\subset A$, hence $gI$$gI$ is a submodule of $A$$A$, which is an ideal of $A$$A$.

Every finite generated $A$$A$ submodule of $K$$K$ are fractional ideal. For any two fractional ideal $I,J$$I,J$, $IJ$$IJ$ is still a fractional ideal.

We use the notation $I_F(A)$$I_F(A)$ for the monoid of fractional ideal.

Definition. We call a fractional ideal invertible or Cartier Divisor if it admit an inverse in $I_F(A)$$I_F(A)$. We denote the group of Cartier Divisors as $\mathrm{Cart}(A)$$\mathrm{Cart}(A)$ .

If $f\in K^{\ast }$$f\in K^*$, then we call the invertible fractional ideal $Af$$Af$ with inverse $A{f}^{-1}$$Af^{-1}$, we call the divisor principal divisors of $A$$A$.

Now if $I,J\in \mathrm{Cart}(A)$$I,J\in\mathrm{Cart}(A)$ with $IJ=A$$IJ=A$. Then there $x_1,...,x_n$$x_1,...,x_n$ in $I$$I$ and $y_1,..,y_n\in J$$y_1,..,y_n\in J$ such that $\sum x_i{y}_i=1$$\sum x_i y_i=1$.

Define $f:I\to A^n$$f:I\to A^n$ by $f(a)=(a{y}_1,...,a{y}_n)$$f(a)=(ay_1,...,ay_n)$ and $g:A^n\to I$$g:A^n\to I$ by $g(e_i)=x_i$$g(e_i)=x_i$. Then $g\circ f(a)=a\sum x_i{y}_i=a$$g\circ f(a)=a\sum x_iy_i=a$. Hence $I$$I$ is a summand of $A^n$$A^n$, thus it is projective.

Now suppose that $I,I\in \mathrm{Cart}(A)$$I,I\in\mathrm{Cart}(A)$, we would like to show that $IJ\cong I{\otimes }_{A}J$$IJ\cong I\otimes_A J$. Since $I$$I$ is projective hence it is flat, we have

$$\boxed{{\displaystyle I{\otimes }_{A}J\to I{\otimes }_{A}K\cong K}}$$

is injective, send $x\otimes y\to xy\in K$$x\otimes y\to xy\in K$, hence the image is just $IJ$$IJ$.

By the way, we shows that every invertible fractional ideal is a finite generated projective module, we still need to show that $\mathrm{rk}$$\mathrm{rk}$ of a fractional ideal is constant 1. But we are working over integral domain, hence $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ is connected, thus the rank function defined by a projective module is constant since it is continuous. Also we have shown that $I{\otimes }_{A}K\cong K$$I\otimes_A K\cong K$, hence the invertible fractional ideal has rank 1, i.e. it is a line bundle.

Now we prove every line bundle is isomorphic to a fractional ideal.

Let $X$$X$ be a line bundle, then $X$$X$ is flat

$$0\to A\to K⟹0\to A{\otimes }_{A}X\cong X\to K{\otimes }_{A}X\cong K$$

Hence $X$$X$ is a isomorphic to a finine generated submodule of $K$$K$, hence it is isomorphic to a fractional ideal.

Let $I$$I$ be any fractional ideal isomorphic to a line bundle $X$$X$, then it is invertible.

Let $I\cong X,J\cong X^{\ast },IJ\cong X{\otimes }_A{X}^{\ast }\cong A$$I\cong X,J\cong X^*,IJ\cong X\otimes_AX^*\cong A$, hence $IJ=Af$$IJ=Af$, $I(J{f}^{-1})=A$$I(Jf^{-1})=A$, hence $I$$I$ is invertible.

Indeed, we get the following exact sequence:

$$1\to U(A)\to U(K)\to \mathrm{Cart}(A)\to \mathrm{Pic}(A)\to 0$$

It is very easy to see in Dedekind Domain, every fractional ideal is invertible. Since every fractional ideal could be written as $d^{-1}I$$d^{-1} I$.

By lemma 3 there exists an ideal $J$$J$ such that $IJ=(a)$$IJ=(a)$. Thus $d^{-1}I(d/a)J=A$$d^{-1}I (d/a)J=A$.

Then for an Dedekind Domain, $\mathrm{Cart}(A)$$\mathrm{Cart}(A)$ is isomorphic to the $\mathrm{Weil}(A)$$\mathrm{Weil}(A)$ is just the ideal group and $\mathrm{Pic}(A)$$\mathrm{Pic}(A)$ is just the ideal class group.