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Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as locally they are isomorphic(Localization of $I$ is a principal ideal as projective module over local ring is free) hence they are isomorphic because $R/J$ is Artin.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?

Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as locally they are isomorphic  (Localization of $I$ is a principal ideal as projective module over local ring is free) hence they are isomorphic because $R/J$ is Artin.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?

Edit: As examples in the answer below about the first part of the question shows, in general we cannot hope $N(I^n)=kq^n, \forall n>0$. However, does there exist $n_0 \in \Bbb N$ independent on $I$ such that $N(I^n)=k_Iq_I^n, \forall n>n_0, I \subseteq O$ ?

Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as there is a natural morphism from $I \otimes R/J$ to $R/J$ whose localization at every prime is an isomorphism (Localization of $I$ is a principal ideal as projective module over local ring is free  ) hence itself is an isomorphism.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?

Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as locally they are isomorphic(Localization of $I$ is a principal ideal as projective module over local ring is free) hence they are isomorphic because $R/J$ is Artin.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?

Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as there is a natural morphism from $I \otimes R/J$ to $R/J$ whose localization at every prime is an isomorphism (Localization of $I$ is a principal ideal as projective module over local ring is free ) hence itself is an isomorphism.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. So essentially we can analysis the problem locally.

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?

Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.

For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.

First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.

Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as there is a natural morphism from $I \otimes R/J$ to $R/J$ whose localization at every prime is an isomorphism (Localization of $I$ is a principal ideal as projective module over local ring is free ) hence itself is an isomorphism.

By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)

For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings.

Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way?