ANT week 6. Sth around Ideal Norm

The partial order of ideal in Dedekind DomainIdeal NormIdeal Norm for $\mathcal O_K$ $\mathcal O_K$ as lattice in $\mathbb R^n$ via Minkowski embeddingEuler Function on Dedekind DomainArithmetic function ring on the ideal monoid.

The partial order of ideal in Dedekind Domain

Recall Lemma 5 in last section, in a Dedekind Domain $R$$R$, we have

$$A|B⟺B=AC$$

Let $I(R)$$I(R)$ be the set of ideals. We already know that, it is a partial order set under $|$$|$, and it is a monoid under $\cdot$$\cdot$.

Lemma 5 tells us that the order and the operation are congruent.

Indeed, let $P$$P$ be the set of prime ideals in $R$$R$, then we have the embedding of poset.

$$I(R)\to \prod _{\mathfrak{p}\in P}\mathbb{Z},I=\prod _{\mathfrak{p}\in P}{\mathfrak{p}}^{e_{\mathfrak{p}}}$$

Here $\prod _{\mathfrak{p}\in P}\mathbb{Z}$$\prod_{\mathfrak p\in P}\mathbb Z$ has the product order.

Ideal Norm

Definition. Let $A$$A$ be a ring and $I\subseteq A$$I\subseteq A$ be an ideal, we define $N(I)=|A/I|$$N(I)=|A/I|$ to be the absolute norm of the ideal.

Example. For $⟨m⟩\subseteq \mathbb{Z}$$\langle m\rangle\subseteq\mathbb Z$, then $N(⟨m⟩)=|\mathbb{Z}/m\mathbb{Z}|=m$$N(\langle m\rangle)=|\mathbb Z/m\mathbb Z|=m$.

In general, for $I,J$$I,J$ such that $I+J=R$$I+J=R$, then $IJ=I\cap J$$IJ=I\cap J$ since $I+J=R⟹\mathrm{\exists }a\in I,b\in J,a+b=1$$I+J=R\implies\exists a\in I,b\in J,a+b=1$

For any $x\in I\cap J,x=x(a+b)=xa+xb\in IJ$$x\in I\cap J,x=x(a+b)=xa+xb\in IJ$. Hence we have $N(IJ)=|A/IJ|=|A/I\cap J|=|A/I\times A/J|=N(I)N(J).$$N(IJ)=|A/IJ|=|A/I\cap J|=|A/I\times A/J|=N(I)N(J).$

Proposition. Let $A$$A$ be a Dedekind Domain, then for any ideal $I,J\subseteq A,N(IJ)=N(I)N(J)$$I,J\subseteq A,N(IJ)=N(I)N(J)$.

Proof. If $N(I)=\mathrm{\infty }$$N(I)=\infty$, then

$$\boxed{{\displaystyle IJ\subseteq I⟹A/IJ\twoheadrightarrow A/I⟹N(IJ)=\mathrm{\infty }}}$$

Now let us deal with finite case.

Consider the following short exact sequence for finite set

$$0\stackrel{}{\to }A\stackrel{\alpha }{\to }B\stackrel{\beta }{\to }C\stackrel{}{\to }0$$

Then we have $|B|=|A|\cdot |C|$$|B|=|A|\cdot |C|$. Since $C\cong B/A$$C\cong B/A$.

Then consider

$$0\stackrel{}{\to }{\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}\stackrel{}{\to }R/{\mathfrak{p}}^{k+1}\stackrel{}{\to }R/{\mathfrak{p}}^k\stackrel{}{\to }0$$

Notice that ${\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}\cong A/\mathfrak{p}$$\mathfrak p^k/\mathfrak p^{k+1}\cong A/\mathfrak p$ since $\mathrm{Ann}({\mathfrak{p}}^k/{\mathfrak{p}}^{k+1})=\{a\in A:a{\mathfrak{p}}^k\subseteq {\mathfrak{p}}^{k+1}\}=({\mathfrak{p}}^{k+1}:{\mathfrak{p}}^k)=\mathfrak{p}$$\mathrm{Ann}(\mathfrak p^k/\mathfrak p^{k+1})=\{a\in A:a\mathfrak p^k\subseteq\mathfrak p^{k+1}\}=(\mathfrak p^{k+1}:\mathfrak p^{k})=\mathfrak p$.

Hence we could view ${\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}$$\mathfrak p^k/\mathfrak p^{k+1}$ as a $A/\mathfrak{p}$$A/\mathfrak p$ vector space. If there is a subspace $V$$V$ such that $0\subset V\subset {\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}$$0\subset V\subset\mathfrak p^k/\mathfrak p^{k+1}$

Then consider ${\pi }^{-1}(V)$$\pi^{-1}(V)$, where $\pi :A\to A/{\mathfrak{p}}^{k+1}$$\pi: A\to A/\mathfrak p^{k+1}$. We know that ${\pi }^{-1}(V)$$\pi^{-1}(V)$ is an $A$$A$ submodule of $A$$A$, hence an ideal. But there is no ideal between ${\mathfrak{p}}^{k+1}\subset {\mathfrak{p}}^k$$\mathfrak p^{k+1}\subset\mathfrak p^{k}$. Hence $V=0\vee V={\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}$$V=0\vee V=\mathfrak p^{k}/\mathfrak p^{k+1}$. Therefore, we have ${\dim }_{A/\mathfrak{p}}({\mathfrak{p}}^k/{\mathfrak{p}}^{k+1})=1$$\dim_{A/\mathfrak p}(\mathfrak p^{k}/\mathfrak {p}^{k+1})=1$, hence ${\mathfrak{p}}^k/{\mathfrak{p}}^{k+1}\cong A/\mathfrak{p}$$\mathfrak p^{k}/\mathfrak {p}^{k+1}\cong A/\mathfrak p$.

This implies that

$$\boxed{{\displaystyle N({\mathfrak{p}}^{k+1})=N(\mathfrak{p})N({\mathfrak{p}}^k)}}$$

By induction we have

$$\boxed{{\displaystyle N({\mathfrak{p}}^e)}}=N(\mathfrak{p}{)}^e$$

Then for any ideal $I,J$$I,J$, we have

$$N(IJ)=N({\mathfrak{p}}_{i_1}^{e_{i_1}}...{\mathfrak{p}}_{i_n}^{e_{i_n}}{\mathfrak{p}}_{j_1}^{e_{j_1}}...{\mathfrak{p}}_{j_m}^{e_{j_m}})=N(I)N(J)$$

even if they are not coprime.

Hence we could extend the norm map to a group homomorphism from ideal group of $A$$A$ to ${\mathbb{Q}}_{+}^{\ast }$$\mathbb Q_+^*$ by the universal property of Grothendieck Group. If the ideal norm is finite.

Ideal Norm for ${\mathcal{O}}_K$$\mathcal O_K$

${\mathcal{O}}_K$$\mathcal O_K$ as lattice in ${\mathbb{R}}^n$$\mathbb R^n$ via Minkowski embedding

The ideal norm for ${\mathcal{O}}_K$$\mathcal O_K$ is always finite, and it has some connections between ideal norm and $\mathrm{Disc}(x_1,...,x_n)$$\mathrm{Disc}(x_1,...,x_n)$.

Lemma. Let $I\subseteq {\mathcal{O}}_K$$I\subseteq \mathcal O_K$ be an ideal, then $I$$I$ is a free $\mathbb{Z}$$\mathbb Z$-module with $\mathrm{rank}$$\mathrm{rank}$ $n=[K:\mathbb{Q}]$$n=[K:\mathbb Q]$.

Proof. We already know that $I\subseteq {\mathcal{O}}_K\cong {\mathbb{Z}}^n$$I\subseteq\mathcal O_K\cong\mathbb Z^n$, hence $I$$I$ is a free $\mathbb{Z}$$\mathbb Z$ module with $\mathrm{rank}(I)\le n$$\mathrm{rank}(I)\le n$. Now we only need to prove that

the rank is $n$$n$. Let ${\omega }_1,...,{\omega }_n$$\omega_1,...,\omega_n$ be a $\mathbb{Z}$$\mathbb Z$ basis of ${\mathcal{O}}_K$$\mathcal O_K$ and $\alpha \in I$$\alpha\in I$. Let $t=N_{K/\mathbb{Q}}(\alpha )\in \mathbb{Z}$$t=N_{K/\mathbb Q}(\alpha)\in\mathbb Z$. Then take $t/\alpha$$t/\alpha$, which is product of some conjugate of $\alpha$$\alpha$, hence $t/\alpha \in {\mathcal{O}}_K$$t/\alpha\in\mathcal O_K$. Hence $t=(t/\alpha )\cdot \alpha \in I$$t=(t/\alpha)\cdot \alpha\in I$. Then we have

$$\boxed{{\displaystyle t{\omega }_1,...,t{\omega }_n\in I\text{ and linearly independent}}}◻$$

Definition. Minkowski embedding

Let $K$$K$ be a number field, with $[K:\mathbb{Q}]=n=r+2s$$[K:\mathbb Q]=n=r+2s$. Where $r$$r$ is the number of real embedding and $s$$s$ is the number of pair of complex embedding. Let ${\mathrm{Hom}}_{\mathbb{Q}}(K,\mathbb{C})=\{{\sigma }_1,...,{\sigma }_r,{\tau }_1,{\bar{\tau }}_1,...,{\tau }_s,{\bar{\tau }}_s\}$$\mathrm{Hom}_{\mathbb Q}(K,\mathbb C)=\{\sigma_1,...,\sigma_r,\tau_1,\overline\tau_1,...,\tau_s,\overline\tau_s\}$.

Take the primitive element $\theta$$\theta$ such that $K=\mathbb{Q}[\theta ]$$K=\mathbb Q[\theta]$, we have $K\cong \mathbb{Q}[x]/(f(x))$$K\cong\mathbb Q[x]/(f(x))$.

$$\boxed{{\displaystyle K\hookrightarrow K{\otimes }_{\mathbb{Q}}\mathbb{R}\cong \mathbb{R}[x]/(f(x))\cong {\mathbb{R}}^r\times {\mathbb{C}}^s}}$$

Which is

$$\boxed{{\displaystyle \mathrm{\Phi }:=k⟼({\sigma }_1(k),...,{\sigma }_r(k),{\tau }_1(k),...,{\tau }_s(k))}}$$

We call this map Minkowski embedding.

We define the Inner product on ${\mathbb{R}}^r\times {\mathbb{C}}^s$$\mathbb{R}^r\times\mathbb C^s$ by

$$\boxed{{\displaystyle ⟨x,y⟩=\sum _{i=1}^r{x}_i{y}_i+\sum _{j=1}^s({\bar{z}}_j{w}_j+z_j\overline{w_j})}}$$

Where $x=(x_1,...,x_r,z_1,...,z_s),y=(y_1,...,y_r,w_1,...,w_s)$$x=(x_1,...,x_r,z_1,...,z_s),y=(y_1,...,y_r,w_1,...,w_s)$.

Then we have ${\mathrm{Tr}}_{K/\mathbb{Q}}(xy)=⟨\mathrm{\Phi }(x),\mathrm{\Phi }(y)⟩$$\mathrm{Tr}_{K/\mathbb Q}(xy)=\langle \Phi(x),\Phi(y)\rangle$.

Definition. Ideal norm in ${\mathcal{O}}_K$$\mathcal O_K$, $[K:\mathbb{Q}]=n$$[K:\mathbb Q]=n$

Let $I\subseteq {\mathcal{O}}_K$$I\subseteq\mathcal O_K$ be an ideal. Let ${\omega }_1,...,{\omega }_k$$\omega_1,...,\omega_k$ be the basis of ${\mathcal{O}}_K$$\mathcal O_K$ and ${\alpha }_1,...,{\alpha }_n$$\alpha_1,...,\alpha_n$ be the basis of $I$$I$.

Then there exists a matrix $T$$T$ such that $T({\omega }_i)={\alpha }_i$$T(\omega_i)=\alpha_i$.

This is invariant under the change of basis since we have $\det :GL(R)\to U(R)$$\det:GL(R)\to U(R)$, and $U(\mathbb{Z})=\pm 1$$U(\mathbb Z)=\pm 1$.

We claim that $|\det T|=N(I)$$|\det T|=N(I)$.

By the structure theorem of finitely generated module over PID, we could find a basis of $I$$I$ such that ${\alpha }_i=d_i{\omega }_i$$\alpha_i=d_i\omega_i$.

Hence we have a linear map:

$$T=\mathrm{diag}(d_1,...,d_n):\mathbb{Z}{\omega }_1\oplus ...\oplus \mathbb{Z}{\omega }_n\to \mathbb{Z}{d}_1{\omega }_1\oplus ...\oplus \mathbb{Z}{d}_n{\omega }_n$$

Easy to see that $\det \left(T\right)=d_1...d_n$$\det(T)=d_1...d_n$, Also we have

$${\mathcal{O}}_K/I\cong (\mathbb{Z}{\omega }_1\oplus ...\oplus \mathbb{Z}{\omega }_n)/(\mathbb{Z}{d}_1{\omega }_1\oplus ...\oplus \mathbb{Z}{d}_n{\omega }_n)\cong \mathbb{Z}/d_1\mathbb{Z}\oplus ...\oplus \mathbb{Z}/d_n\mathbb{Z}$$

Thus

$$|\det \left(T\right)|=|{\mathcal{O}}_K/I|$$

Then you can see the geometric meaning of the ideal norm. $◻$$\square$

Corollary. The ideal norm for ${\mathcal{O}}_K$$\mathcal O_K$ is finite.

Corollary. ${\mathcal{O}}_K$$\mathcal O_K$ is Dedekind Domain.

Proof. We already know that ${\mathcal{O}}_K$$\mathcal O_K$ is integrally closed and Noetherian. Now we need to prove the Krull dimension is 1.

Since $N(\mathfrak{p})=|A/\mathfrak{p}|$$N(\mathfrak p)=|A/\mathfrak p|$ is finite and every finite integral domain is field. $◻$$\square$

Remark. Indeed, let $A$$A$ be an Artinian integral domain, then $A$$A$ is a field. Let $0\ne a\in A$$0\ne a\in A$, and consider the following descending chain

$$\boxed{{\displaystyle (a)\supseteq (a^2)\supseteq (a^3)...}}$$

Assume this chain stable at $(a^n)$$(a^n)$, then $(a^n)=(a^{n+1})$$(a^n)=(a^{n+1})$, there exists a $b$$b$ such that $b{a}^{n+1}=a^n⟹ba=1$$ba^{n+1}=a^n\implies ba=1$. $◻$$\square$

Let $I$$I$ be an ideal, and ${\alpha }_1,...,{\alpha }_n$$\alpha_1,...,\alpha_n$ be a $\mathbb{Z}$$\mathbb Z$ basis of $I$$I$.

We have

$$\mathrm{Disc}({\alpha }_1,...,{\alpha }_n)=\det {\left(T\right)}^2\mathrm{Disc}(K)$$

But we know that $\det \left(T\right)=N(I)$$\det(T)=N(I)$. Hence $\mathrm{Disc}({\alpha }_1,...{\alpha }_n)=N(I{)}^2\mathrm{Disc}(K)$$\mathrm{Disc}(\alpha_1,...\alpha_n)=N(I)^2\mathrm{Disc}(K)$.

Corollary. Ideal norm and norm of element.

Let $0\ne a\in {\mathcal{O}}_K$$0\ne a\in\mathcal O_K$, then we claim that $|N_{K/\mathbb{Q}}(a)|=N(⟨a⟩)$$|N_{K/\mathbb Q}(a)|=N(\langle a\rangle)$.

Proof. Let ${\omega }_1,...,{\omega }_n$$\omega_1,...,\omega_n$ be the basis of ${\mathcal{O}}_K$$\mathcal O_K$, then $a{\omega }_1,...,a{\omega }_n$$a\omega_1,...,a\omega_n$ is the basis of $⟨a⟩$$\langle a\rangle$. Then we have

$$\mathrm{Disc}(a{\omega }_1,...,a{\omega }_n)=N(⟨a⟩{)}^2\mathrm{Disc}(K)$$

But $\mathrm{Disc}(a{\omega }_1,...,a{\omega }_n)=\det {({\sigma }_i(a{\omega }_j))}^2=(\prod _{i=1}^n{\sigma }_i(a){)}^2\det ({\sigma }_i({\omega }_j))=N_{K/\mathbb{Q}}(a{)}^2\mathrm{Disc}(K)$$\mathrm{Disc}(a\omega_1,...,a\omega_n)=\det(\sigma_i(a\omega_j))^2=(\prod_{i=1}^n\sigma_i(a))^2\det(\sigma_i(\omega_j))=N_{K/\mathbb Q}(a)^2\mathrm{Disc}(K)$. $◻$$\square$

Application. Consider $\mathbb{Z}[i]$$\mathbb Z[i]$, how do you compute $\mathbb{Z}[i]/(3+i)$$\mathbb Z[i]/(3+i)$? Well, we have $|\mathbb{Z}[i]/(3+i)|=3^2+1^2=10$$|\mathbb Z[i]/(3+i)|=3^2+1^2=10$. But the only abelian group with order $10$$10$ is $\mathbb{Z}/10\mathbb{Z}$$\mathbb Z/10\mathbb Z$, and there is only one ring structure on $\mathbb{Z}/10\mathbb{Z}$$\mathbb Z/10\mathbb Z$.

Theorem. Let $I$$I$ be a nonzero ideal in ${\mathcal{O}}_K$$\mathcal O_K$.

$(1)$$(1)$ If $N(I)$$N(I)$ is prime, then $I$$I$ is prime. $(2)$$(2)$ $N(I)\in I$$N(I)\in I$. $(3)$$(3)$ If $I$$I$ is a prime ideal, then $I\cap \mathbb{Z}=(p)$$I\cap\mathbb Z=(p)$ for some $(p)$$(p)$ and $N(I)=p^n$$N(I)=p^n$ for some $n\in \mathbb{N},n\le [K:\mathbb{Q}]$$n\in\mathbb N,n\le[K:\mathbb Q]$.

Proof. We prove (3) first. This follows directly from the definition of $\mathrm{Spec}:\mathrm{CRing}\to \mathbf{Top}$$\mathrm{Spec}:\mathrm{CRing}\to\mathbf{Top}$. We have $I\cap \mathbb{Z}=i^{-1}(I)\in \mathrm{Spec}\mathbb{Z}$$I\cap\mathbb Z=i^{-1}(I)\in\mathrm{Spec}\mathbb Z$, where $i$$i$ is the universal map from $\mathbb{Z}$$\mathbb Z$ to ${\mathcal{O}}_K$$\mathcal O_K$. Since $I$$I$ is prime, hence a maximal ideal in ${\mathcal{O}}_K$$\mathcal O_K$, we have that ${\mathcal{O}}_K/I$$\mathcal O_K/I$ is a finite field, hence ${\mathcal{O}}_K/I\cong {\mathbb{F}}_{p^m}$$\mathcal O_K/I\cong\mathbb F_{p^m}$. Thus $N(I)=p^m$$N(I)=p^m$. To see that $m\le [K:\mathbb{Q}]=n$$m\le[K:\mathbb Q]=n$, we have the following maps of $\mathbb{Z}$$\mathbb Z$-module homomorphism.

$${\mathcal{O}}_K\twoheadrightarrow {\mathcal{O}}_K/p{\mathcal{O}}_K\cong {\mathcal{O}}_K{\otimes }_{\mathbb{Z}}\mathbb{Z}/p\mathbb{Z}\cong {\mathbb{Z}}^n{\otimes }_{\mathbb{Z}}\mathbb{Z}/p\mathbb{Z}\cong {\mathbb{F}}_p^n\twoheadrightarrow {\mathcal{O}}_K/I\cong {\mathbb{F}}_{p^m}$$

Now let us prove $(2)$$(2)$. We prove sth more general first.

Let $I\subset R$$I\subset R$ be an ideal and view $R/I$$R/I$ as an $R$$R$-module, then we cliam that ${\mathrm{Ann}}_R(R/I)=I$$\mathrm{Ann}_R(R/I)=I$. It follows from

$$\boxed{{\displaystyle r\in {\mathrm{Ann}}_R(R/I)⟺r\cdot 1\in I}}$$

Now since $|{\mathcal{O}}_K/I|=N(I)$$|\mathcal O_K/I|=N(I)$, Lagrange’s theorem tells us that $N(I)\in {\mathrm{Ann}}_{{\mathcal{O}}_K}({\mathcal{O}}_k/I)=I$$N(I)\in\mathrm{Ann}_{\mathcal O_K} (\mathcal O_k/I)=I$.

For $(1)$$(1)$, let us prove something much more general.

Definition. Let $\mathcal{C}$$\mathcal C$ be a category and $f:X\to Y$$f:X\to Y$ be a morphism that is not an isomorphism. We say $f$$f$ is irreducible if

$$\boxed{{\displaystyle f=g\circ h⟹g\text{ is invertible or }h\text{ is invertible}}}$$

Theorem. Let $F:\mathcal{C}\to \mathcal{D}$$F:\mathcal C\to\mathcal D$ be a functor that reflects isomorphisms, i.e., $F(f)$$F(f)$ is invertible implies $f$$f$ is invertible, then $F$$F$ reflects irreducibility, i.e., $F(f)$$F(f)$ is irreducible implies $f$$f$ is irreducible.

Proof. Let $F(f)$$F(f)$ be irreducible, and assume that $f$$f$ is not irreducible, so $f=g\circ h$$f=g\circ h$ such that both $g$$g$ and $h$$h$ are not invertible. Then $F(f)=F(g\circ h)=F(g)\circ F(h)$$F(f)=F(g\circ h)=F(g)\circ F(h)$. But since $F$$F$ reflects isomorphisms and every functor preserves isomorphisms, hence $F(g)$$F(g)$ and $F(h)$$F(h)$ are not isomorphisms, which contradicts the fact that $F(f)$$F(f)$ is irreducible. $◻$$\square$

To apply this theorem to (1), we need to associate a monoid with a category, hence we define the following functor.

Definition. Let $B:\mathbf{Mon}\to \mathbf{Cat}$$B:\mathbf{Mon}\to\mathbf{Cat}$ be a functor from the category of monoids to the category of small categories as follows:

For a monoid $M$$M$, define $B(M)$$B(M)$ to be the category with $\mathrm{Ob}(B(M)):=\{\ast \}$$\mathrm{Ob}(B(M)):=\{*\}$, $\mathrm{Mor}(B(M)):=M$$\mathrm{Mor}(B(M)):=M$. The composition of morphisms is given by the binary operation in the monoid. It is easy to see that a monoid homomorphism will induce a functor automatically.

Proof of (1) Now let $M=I({\mathcal{O}}_K)$$M=I(\mathcal O_K)$ be the monoid of ideals, consider $B(N):B(M)\to B(\mathbb{N})$$B(N):B(M)\to B(\mathbb N)$, we know that $N(I)=1⟺I={\mathcal{O}}_K$$N(I)=1\iff I=\mathcal O_K$, and the unique isomorphism in $B(\mathbb{N})$$B(\mathbb N)$ is 1. Hence we have that $N$$N$ reflects isomorphisms, thus reflects irreducibility. Thus if $N(I)$$N(I)$ is prime, then $I$$I$ is prime. $◻$$\square$

Remark. Another example of a functor that reflects isomorphisms and hence reflects irreducibility is given by the degree of field extension. Here we define

$$\mathrm{deg}:\mathbf{Field}\to B(\mathbb{N})$$

by $\mathrm{deg}(L/F)=[L:F]$$\deg(L/F)=[L:F]$. It is easy to see that $\mathrm{deg}(L/F)=1⟺L\cong F$$\deg(L/F)=1\iff L\cong F$. Hence if we have $\mathrm{deg}(L/F)=p$$\deg(L/F)=p$, then there is no nontrivial field extension between $F$$F$ and $L$$L$.

Remark. You should view $B(M)$$B(M)$ as the syntax category of $M$$M$ action, and category of $M$$M$-$\mathrm{Set}$$\mathrm{Set}$, i.e. ${\mathrm{Set}}^{B(M)}$$\mathrm{Set}^{B(M)}$ as the category of semantics of $B(M)$$B(M)$ in $\mathbf{Set}$$\mathbf{Set}$.

Euler Function on Dedekind Domain

Let $I(A)$$I(A)$ be the ideal monoid of a Dedekind Domain $A$$A$. We define the Euler function on $I(A)$$I(A)$ by

$$\phi (I)=|U(A/I)|$$

Obviously when $I+J=A$$I+J=A$, we have $\phi (IJ)=\phi (I)\phi (J)$$\varphi(IJ)=\varphi(I)\varphi(J)$ by CRT and unit functor is right adjoint of group algebra.

Now let us study the value of $\phi (\mathfrak{p})$$\varphi(\mathfrak p)$ .

Lemma. $\phi ({\mathfrak{p}}^e)=N(\mathfrak{p}{)}^e-N(\mathfrak{p}{)}^{e-1}$$\varphi(\mathfrak p^{e})=N(\mathfrak p)^{e}-N(\mathfrak p)^{e-1}$.

Proof. Since $\mathfrak{p}$$\mathfrak p$ is maximal ideal, hence $A/{\mathfrak{p}}^e$$A/\mathfrak p^{e}$ is local ring. $\phi ({\mathfrak{p}}^e)=N({\mathfrak{p}}^e)-|\mathfrak{p}/{\mathfrak{p}}^e|$$\varphi(\mathfrak p^{e})=N(\mathfrak p^e)-|\mathfrak p/\mathfrak p^e|$.

We have the following short exact sequence.

$$0\stackrel{}{\to }\mathfrak{p}/{\mathfrak{p}}^e\stackrel{}{\to }R/{\mathfrak{p}}^e\stackrel{}{\to }R/\mathfrak{p}\stackrel{}{\to }0$$

$$N({\mathfrak{p}}^e)=N(\mathfrak{p}{)}^e=|\mathfrak{p}/{\mathfrak{p}}^e|\cdot N(\mathfrak{p})⟹|\mathfrak{p}/{\mathfrak{p}}^e|=N(\mathfrak{p}{)}^{e-1}◻$$

Arithmetic function ring on the ideal monoid.

Indeed, we should study Arithmetic function ring on the ideal monoid for a Dedekind Domain.

I think the theory will look very similar to the traditional case (click here). But I do not have time to write it.