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Let $K$ be an algebraic number field of degree $n$. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $n$. The ideal theory of $R$ is useful at least when $K$ is a quadratic number field, because it is related to the theory of binary quadratic forms. Let $I$ be a fractional ideal of $R$, i.e. $I$ is a non-zero $R$-submodule of $K$ and there exists non-zero $\alpha \in K$ such that $\alpha I \subset R$. The norm $N(I)$ of $I$ may be defined as follows. There exist $\alpha \in R$ and an ideal $J$ of $R$ such that $I = (1/\alpha)J$. We would like to define $N(I)$ as $N(J)/N(\alpha R)$, where $N(J)$ is defined as $|R/J|$. Of course we need to prove that this is well-defined.

My question Let $I$ be a fractional ideal of an order $R$. Are the following statements correct? If yes, how do we prove them?

  1. $N(I)$ is well-defined.

  2. Let $\gamma$ be non-zero element of $K$. Then $N(\gamma I) = |N(\gamma)|N(I)$.

  3. Let $\alpha_1, \cdots, \alpha_n$ be $\mathbb{Z}$-basis of $I$. Let $\theta_1, \cdots, \theta_n$ be $\mathbb{Z}$-basis of $R$. Suppose $\alpha_i = \sum_j a_{ij} \theta_j$ for $i = 1,\cdots,n$. Then $N(I) = |$det $(a_{ij})|$.

  4. Let $I, J$ be fractional ideals of $R$ such that $J \subset I$. Then $|I/J| = N(J)/N(I)$.

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This is not really a problem about orders or fractional ideals, but about lattices. Let $V$ be a finite-dimensional ${\mathbf Q}$-vector space (such as a number field) and set $n = \dim_{\mathbf Q}(V)$. A lattice in $V$ is a finite free ${\mathbf Z}$-module in $V$ of rank $n$. If $V$ is a number field $K$, examples of lattices in $V$ include any order $R$ in $K$ and any $R$-fractional ideal.

When $L$ and $L'$ are lattices in $V$, check their sum $L+L' = \{x + y : x \in L, y \in L'\}$ is a lattice. If $L' \subset L$, the usual index $[L:L'] = |L/L'|$ is finite. We want to define an index $[L:L']$ even if $L'$ is not contained in $L$. Here's how we can do it. For any two lattices $L$ and $L'$ in $V$, define the index $[L:L']$ to be the positive rational number $$ \frac{[M:L']}{[M:L]}, $$ where $M$ is any lattice in $V$ containing $L$ and $L'$, and the numerator and denominator here are the usual notion of index (because $L$ and $L'$ are contained in $M$).

Exercises.

1) Check this is independent of the choice of $M$ and thus is well-defined. (Hint: use multiplicativity of the usual notion of index and the fact that any lattice containing $L$ and $L'$ must contain $L+L'$.)

2) Check this equals $|L/L'|$ if $L' \subset L$.

3) Check for any three lattices $L, L', L''$ in $V$ that $[L:L''] = [L:L'][L':L'']$.

4) For any lattices $L$ and $L'$ in $V$, and any ${\mathbf Q}$-linear automorphism $\varphi \colon V \rightarrow V$, check $[L:L'] = [\varphi(L):\varphi(L')]$.

5) For any lattice $L$ in $V$ and ${\mathbf Q}$-linear automorphism $\varphi \colon V \rightarrow V$, check $\varphi(L)$ is a lattice in $V$ and $[L:\varphi(L)] = |\det \varphi|$.

6) For any lattices $L$ and $L'$ in $V$, show there is a ${\mathbf Q}$-linear automorphism $\varphi \colon V \rightarrow V$ such that $\varphi(L) = L'$, and for any such $\varphi$ we have $[L:L'] = |\det \varphi|$. This provides a different way of defining the index $[L:L']$.

Using $V = K$ and considering the lattices $R$, $I$, and $J$, and using as $\varphi \colon K \rightarrow K$ suitable multiplication maps $\varphi(x) = \alpha{x}$, you can recover the properties you want. Define ${\rm N}(I) = [R:I]$, even if $I$ is not contained in $R$.

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  • $\begingroup$ A great answer. I will post detailed proofs of these facts. $\endgroup$ – Makoto Kato Dec 13 '13 at 18:18
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    $\begingroup$ I don't think you need to post proofs. Work on it yourself as a series of exercises without having to feel the need to show the solution to the world. Others who are interested and did not know this material can do it themselves too. $\endgroup$ – KCd Dec 14 '13 at 5:24
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After proving a few lemmas, we prove the assertions stated in KCd's answer. We fix a $\mathbb{Q}$-vector space $V$ of finite dimension $n$.

Notation Let $x_1,\cdots,x_m$ be a finite sequence of elements of $V$. We denote by $[x_1,\cdots,x_m]$ the $\mathbb{Z}$-submodule of $V$ generated by $x_1,\cdots,x_m$.

Lemma 1 Let $L$ be a finite free $\mathbb{Z}$-module of rank $n$. Let $M$ be a $\mathbb{Z}$-submodule of $L$. Suppose there exists an integer $d \gt 0$ such that $dL \subset M$, Then $M$ is a finite free $\mathbb{Z}$-module of rank $n$.

Proof: We use induction on $n$. If $n = 1$, the assertion is clear. Suppose $n \gt 1$. Let $\theta_1,\cdots,\theta_n$ be a free $\mathbb{Z}$-basis of $L$. Let $p_n\colon L \rightarrow \mathbb{Z}$ be the map defined by $p_n(x) = x_n$, where $x = x_1\theta_1 + \cdots + x_n\theta_n$. Clearly $p_n$ is a $\mathbb{Z}$-homomorphism. If $p_n(M) = 0$, then $M \subset [\theta_1,\cdots,\theta_{n-1}]$. Since $d\theta_n \in M, d\theta_n \in [\theta_1,\cdots,\theta_{n-1}]$. This is a contradiction. Hence $p_n(M) \ne 0$. Then there exists an integer $a_n \gt 0$ such that $p_n(M) = a_n\mathbb{Z}$. Hence there exists $\omega_n \in M$ such that $p_n(\omega_n) = a_n$. Then $M = M \cap [\theta_1,\cdots,\theta_{n-1}] + [\omega_n]$. This is a direct sum. Since $d[\theta_1,\cdots,\theta_{n-1}] \subset M \cap [\theta_1,\cdots,\theta_{n-1}]$, by the induction hypothesis, we are done. QED

Lemma 2 Let $L$ be a lattice of $V$. Let $M$ be a finitely generated $\mathbb{Z}$-submodule of $V$. Then there exists an integer $d \gt 0$ such that $dM \subset L$.

Proof: Let $\theta_1,\cdots,\theta_n$ be a $\mathbb{Z}$-basis of $L$. Suppose $M = [\alpha_1,\cdots,\alpha_m]$. Then $\alpha_i = \sum_j a_{ij} \theta_j$ for $i = 1,\cdots,m$, where $a_{ij} \in \mathbb{Q}$. There exists an integer $d \gt 0$ such that $da_{ij} \in \mathbb{Z}$ for all $i, j$. Then $d\alpha_i \in L$ for all $i$. Hence $dM \subset L$. QED

Corollary Let $L, M$ be lattices of $V$. Then $L \cap M$ is a lattice of $V$.

Proof: By the lemma, there exists an integer $d \gt 0$ such that $dL \subset M$. Since $dL \subset L \cap M \subset L$, the assertion follows from Lemma 1. QED

Lemma 3 Let $L$ be a subset of $V$. The following assertions are equivalent.

  1. $L$ is a lattice of $V$.

  2. $L$ is a finitely generated $\mathbb{Z}$-submodule of $V$ and it contains a $\mathbb{Q}$-basis of $V$.

Proof: $1. \Rightarrow 2.$ Clear.

$2. \Rightarrow 1.$ Let $\theta_1,\cdots,\theta_n$ be a $\mathbb{Q}$-basis of $V$. Let $M = [\theta_1,\cdots,\theta_n]$. By Lemma 2, there exist integers $c \gt 0, d \gt 0$ such that $cM \subset L, dL \subset M$. Then $dcM \subset dL \subset M$. By Lemma 1, $dL$ is a lattice of $V$. Hence $L$ is also a lattice of $V$. QED

Corollary Let $L, M$ be lattices of $V$. Then $L + M$ is a lattice of $V$.

Proposition 1 Let $L$ and $L'$ be lattices of $V$. By the corollarys of Lemma 2 and Lemma 3, $L\cap L'$ and $L + L'$ are lattices of $V$. Let $M$ be a lattice of $V$ containing $L$ and $L'$. Let $N$ be a lattice of $V$ contained in $L$ and $L'$. Then $$\frac{[M:L']}{[M:L]} = \frac{[L+L':L']}{[L+L':L]} = \frac{[L:L\cap L']}{[L':L\cap L']} = \frac{[L:N]}{[L':N]}$$.

Proof: Note that $M \supset L + L'$. Hence $$[M:L'] = [M:L + L'][L+L':L']$$ $$[M:L] = [M:L + L'][L+L':L]$$ Hence $$\frac{[M:L']}{[M:L]} = \frac{[L+L':L']}{[L+L':L]}$$

Note that $L \cap L' \supset N$. Hence $$[L:N] = [L:L \cap L'][L\cap L':N]$$ $$[L':N] = [L':L \cap L'][L\cap L':N]$$ Hence $$\frac{[L:N]}{[L':N]} = \frac{[L:L\cap L']}{[L':L\cap L']}$$.

Note that $[L:L\cap L'] = [L+L':L']$ and $[L':L\cap L'] = [L+L':L]$. Hence $$\frac{[L+L':L']}{[L+L':L]} = \frac{[L:L\cap L']}{[L':L\cap L']}$$ QED

Definition 1 Let $L$ and $L'$ be lattices of $V$. Let $M$ be a lattice of $V$ containing $L$ and $L'$. Let $N$ be a lattice of $V$ contained in $L$ and $L'$. We define $$[L:L'] = \frac{[M:L']}{[M:L]}$$ or $$[L:L'] = \frac{[L:N]}{[L':N]}$$ By Proposition 1, this is well-defined.

Proposition 2 Let $L$ and $L'$ be lattices of $V$ such that $L \supset L'$. Then $[L:L'] = |L/L'|$.

Proof: Since $L+L' = L$, $$[L:L'] = \frac{[L:L']}{[L:L]} = |L/L'|$$ QED

Proposition 3 Let $L, L', L''$ be lattices of $V$. Then $[L:L''] = [L:L'][L':L'']$.

Proof: Let $M = L+L'+L''$. Then $$[L:L''] = \frac{[M:L'']}{[M:L]}$$ $$[L:L'] = \frac{[M:L']}{[M:L]}$$ $$[L':L''] = \frac{[M:L'']}{[M:L']}$$ Hence $[L:L''] = [L:L'][L':L'']$. QED

Lemma 4 Let $L$ and $L'$ be lattices of $V$ such that $L \supset L'$. Let $\psi \colon V \rightarrow V$ be $\mathbb Q$-linear automorphism. Then $[L:L'] = [\psi(L):\psi(L')]$.

Proof: $\psi$ induces a surjective $\mathbb{Z}$-linear map $\phi\colon L \rightarrow \psi(L)/\psi(L')$. The kernel of $\phi$ is $L'$. Hence $\phi$ induces an $\mathbb{Z}$-linear isomorphism $L/L' \cong \phi(L)/\phi(L')$. QED

Proposition 4 For any lattices $L$ and $L'$ in $V$, and any $\mathbb Q$-linear automorphism $\psi \colon V \rightarrow V$, $[L:L'] = [\psi(L):\psi(L')]$.

Proof: Let $M$ be a lattice of $V$ containing $L$ and $L'$. Then $$[L:L'] = \frac{[M:L']}{[M:L]}$$ By Lemma 4, $$\frac{[M:L']}{[M:L]} = \frac{[\psi(M):\psi(L')]}{[\psi(M):\psi(L)]} = [\psi(L):\psi(L')]$$ QED

Lemma 5 Let $L$ be a lattice of $V$. Let $\psi \colon V \rightarrow V$ be a $\mathbb Q$-linear automorphism. Suppose $\psi(L) \subset L$. Then $[L: \psi(L)] = |\det \psi|$.

Proof: Let $L' = \psi(L)$. Let $\theta_1,\cdots,\theta_n$ be a free $\mathbb{Z}$-basis of $L$. Suppose $\psi(\theta_i) = \sum_j a_{ij}\theta_j$ for $i = 1, \cdots, n$, where $a_{ij} \in \mathbb{Q}$. Then $\det \psi = \det (a_{ij})$. On the other hand, by Lemma 2, there exists an integer $d \gt 0$ such that $dL \subset L'$. Hence we can apply Lemma 1. By its proof, there exists a free $\mathbb{Z}$-basis $\omega_1,\cdots,\omega_n$ of $L'$ of the form $\omega_i = b_{i1}\theta_1 + \cdots + b_{ii}\theta_i$ for $i = 1, \cdots, n$, where $b_{ii} \gt 0$. Then $[L: L'] = b_{11}\cdots b_{nn} = \det (b_{ij})$(see the proof of Lemma 3 of my answer to this question). Since $|\det (b_{ij})| = |\det \psi|$(see the proof of the proposition of my answer to this question), we are done. QED

Proposition 5 For any lattice $L$ in $V$ and $\mathbb Q$-linear automorphism $\psi \colon V \rightarrow V$, $[L:\psi(L)] = |\det \psi|$.

Proof: By Lemma 2, there exists an integer $d \gt 0$ such that $d\psi(L) \subset L$. By Proposition 3 and Proposition 4, $[L:d\psi(L)] = [L:dL][dL:d\psi(L)] = d^n[dL:d\psi(L)] = d^n[L:\psi(L)]$. Since $\det d\psi = d^n \det \psi$, replacing $\psi$ by $d\psi(L)$, we may suppose that $\psi(L) \subset L$. Then the assertion follows from Lemma 5. QED

Proposition 6 For any lattices $L$ and $L'$ in $V$, there is a $\mathbb Q$-linear automorphism $\psi \colon V \rightarrow V$ such that $\psi(L) = L'$.

Proof: Let $\theta_1,\cdots,\theta_n$ be a free $\mathbb{Z}$-basis of $L$. Let $\theta'_1,\cdots,\theta'_n$ be a free $\mathbb{Z}$-basis of $L'$. There is a unique $\mathbb Q$-linear automorphism $\psi \colon V \rightarrow V$ such that $\psi(\theta_i) = \theta'_i$ for $i = 1, \cdots, n$. Clearly $\psi(L) = L'$. QED

Now we prove the assertions of the question. Let $K$ be an algebraic number field of degree $n$. Let $R$ be an order of $K$.

Lemma 6 Let $I$ be a fractional ideal of an order $R$. Then $I$ is a lattice of $K$.

Proof: We may suppose that $I \subset R$. Since $R$ is a latice of $K$, $I$ is a finitely generated $\mathbb{Z}$-submodule of $R$. Since $I \ne 0$, there exists a non-zero element $\alpha \in I$. Then $\alpha R \subset I$. Since $\alpha R$ is a lattice of $K$, $I$ is a lattice by Lemma 3. QED

Definition 2 Let $I$ be a fractional ideal of an order $R$. We define $N(I) = [R:I]$. The right hand side is defined by Definition 1.

Proposition 7 Let $I$ be a fractional ideal of an order $R$. Let $\gamma$ be non-zero element of $K$. Then $N(\gamma I) = |N(\gamma)|N(I)$.

Proof: By Proposition 3, $N(\gamma I) = [R:\gamma I] = [R:\gamma R][\gamma R: \gamma I]$. Let $\psi\colon K \rightarrow K$ be the map defined by $\psi(x) = \gamma x$. $\psi$ is a $\mathbb{Q}$-linear automorphism of $K$ and $\det \psi = N(\gamma)$. By Propositio 5, $[R:\gamma R] = |N(\gamma)|$. By Propositio 4, $[\gamma R: \gamma I] = [R:I] = N(I)$. Hence $N(\gamma I) = |N(\gamma)|N(I)$. QED

Proposition 8 Let $I$ be a fractional ideal of an order $R$. Let $\alpha_1, \cdots, \alpha_n$ be $\mathbb{Z}$-basis of $I$. Let $\theta_1, \cdots, \theta_n$ be $\mathbb{Z}$-basis of $R$. Suppose $\alpha_i = \sum_j a_{ij} \theta_j$ for $i = 1,\cdots,n$. Then $N(I) = |$det $(a_{ij})|$.

Proof: There exists a unique $\mathbb{Q}$-linear automorphism $\psi\colon K \rightarrow K$ such that $\psi(\theta_i) = \alpha_i$ for $i = 1,\cdots,n$. Since $I = \psi(R)$, the assertion follows from Proposition 5. QED

Proposition 9 Let $I, J$ be fractional ideals of $R$ such that $J \subset I$. Then $|I/J| = N(J)/N(I)$.

Proof: By Proposition 3, $[R:J] = [R:I][I:J]$. Hence $|I/J| = N(J)/N(I)$. QED

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The following proofs are based on a different idea from KCd's answer.

Let $K$ be an algebraic number field of degree $n$. Let $\{\sigma_1\cdots,\sigma_n\}$ be the set of $\mathbb{Q}$-algebra homomorphisms from $K$ into $\mathbb{C}$. We suppose $\sigma_1 =1$. Let $R$ be an order of $K$.

Definition 3 For any sequence $\alpha_1,\cdots, \alpha_n \in K$, we denote $\det (\sigma_i(\alpha_j))$ by $\Delta(\alpha_1,\cdots, \alpha_n)$.

Lemma 7 Let $L$ be a lattice of $K$, i.e. a free $\mathbb{Z}$-submodule of $K$ of rank $n$. Let $\alpha_1,\cdots, \alpha_n$ be a free $\mathbb{Z}$-basis of $L$. Then $\Delta(\alpha_1,\cdots, \alpha_n)^2$ is a non-zero rational number and independent of the choice of a $\mathbb{Z}$-basis of $L$.

Proof: Note that $(\sigma_k(\alpha_i))^t(\sigma_k(\alpha_j)) = (\sum_k (\alpha_i\alpha_j)) = (\text{Tr}(\alpha_i\alpha_j))$, where $A^t$ denotes the transpose of a matrix $A$. Hence $\Delta(\alpha_1,\cdots, \alpha_n)^2 = \det (\text{Tr}(\alpha_i\alpha_j)) \in \mathbb{Q}$. Let $\beta_1,\cdots, \beta_n$ be another free $\mathbb{Z}$-basis of $L$. There exists $(p_{ij}) \in \text{GL}_n(\mathbb{Z})$ such that $\alpha_i = \sum_j p_{ij}\beta_j$ for all $i$. Then $\Delta(\alpha_1,\cdots, \alpha_n) = \det (p_{ij}) \Delta(\beta_1,\cdots, \beta_n)$. Since $\det (p_{ij}) = \pm 1$, $\Delta(\alpha_1,\cdots, \alpha_n)^2 = \Delta(\beta_1,\cdots, \beta_n)^2$

It remains to prove that $\Delta(\alpha_1,\cdots, \alpha_n)^2 \ne 0$. There exists $\theta \in K$ such that $K = \mathbb{Q}(\theta)$. Since $1, \theta,\cdots,\theta^{n-1}$ is a $\mathbb{Q}$-basis of $K$, There exists $(a_{ij})\in \text{GL}_n(\mathbb{Q})$ such that $\alpha_i = \sum_j a_{ij}\theta^{j-1}$ for all $i$. Then $\Delta(\alpha_1,\cdots, \alpha_n) = \det (a_{ij}) \Delta(1, \theta,\cdots,\theta^{n-1})$. Since $\Delta(1, \theta,\cdots,\theta^{n-1}) = \prod_{j\gt i}(\sigma_j(\theta) - \sigma_i(\theta)) \ne 0$(for example, see this article), $\Delta(\alpha_1,\cdots, \alpha_n) \ne 0$. QED

Defnition 4 Let $L$ be a lattice of $K$. Let $\alpha_1,\cdots, \alpha_n$ be a free $\mathbb{Z}$-basis of $L$. We denote $\Delta(\alpha_1,\cdots, \alpha_n)^2$ by $d(L)$. This is well-defined by Lemma 7.

Lemma 8 Let $I \ne 0$ be an ideal of $R$. Note that $I$ is a lattice of $K$ by Lemma 6 of my previous answer to the question. Then $d(I) = N(I)^2d(R)$, where $N(I) = |R/I|$.

Proof: Let $\alpha_1,\cdots, \alpha_n$ be a free $\mathbb{Z}$-basis of $I$. Let $\theta_1,\cdots, \theta_n$ be a free $\mathbb{Z}$-basis of $R$. Suppose $\alpha_i = \sum_j a_{ij}\theta_j$ for all $i$. Then $d(I) = (\det (a_{ij}))^2 d(R)$. Since $N(I) = |\det (a_{ij})|$(see the proof of Lemma 5 of my previous answer to the question), we are done. QED

Lemma 9 Let $I \ne 0$ be an ideal of $R$. Let $\gamma \ne 0$ be an element of $R$. Then $N(\gamma I) = |N(\gamma)|N(I)$.

Proof: Let $\alpha_1,\cdots, \alpha_n$ be a free $\mathbb{Z}$-basis of $I$. Then $\gamma\alpha_1,\cdots, \gamma\alpha_n$ is a free $\mathbb{Z}$-basis of $\gamma I$. $\Delta(\gamma\alpha_1,\cdots, \gamma\alpha_n) = N(\gamma)\Delta(\alpha_1,\cdots, \alpha_n)$. Hence $d(\gamma I) = N(\gamma)^2d(I)$. By Lemma 8, $d(I) = N(I)^2d(R)$. Hence $d(\gamma I) = N(\gamma)^2N(I)^2d(R)$. By Lemma 8, $d(\gamma I) = N(\gamma I)^2d(R)$. Hence $N(\gamma)^2N(I)^2d(R) = N(\gamma I)^2d(R)$. Since $d(R) \ne 0$ by Lemma 7, $N(\gamma I)^2 = N(\gamma)^2N(I)^2$. Hence $N(\gamma I) = |N(\gamma)|N(I)$. QED

Proposition 10 Let $I$ be a fractional ideal of $R$. There exist $\alpha \in R$ and an ideal $J$ of $R$ such that $I = (1/\alpha)J$. Let $N(I)$ be defined as $N(J)/N(\alpha R)$. Then $N(I)$ is well-defined.

Proof: Let $\beta \in R$ and $L$ be an ideal of $R$. Suppose $I = (1/\beta)L$. It suffices to prove that $N(J)/N(\alpha R) = N(L)/N(\beta R)$. Since $(1/\alpha)J = (1/\beta)L$, $\beta J = \alpha L$. By Lemma 9, $|N(\beta)|N(J) = |N(\alpha)|N(L)$. Hence $N(J)/|N(\alpha)| = N(L)/|N(\beta)|$. Since $N(\alpha R) = |N(\alpha)|$ and $N(\beta R) = |N(\beta)|$ by Lemma 9, we are done. QED

Proposition 11 Let $I$ be a fractional ideal of $R$. Let $\gamma$ be non-zero element of $K$. Then $N(\gamma I) = |N(\gamma)|N(I)$.

Proof: $\gamma$ can be written as $\gamma = \mu/\nu$, where $\mu, \nu \in R$. Hence $\gamma I = (\mu/\nu) I$. Hence $\nu \gamma I = \mu I$. On the other hand, there exists $\alpha \ne 0 \in R$ such that $\alpha I \in R$. Then $\alpha \nu \gamma I = \mu \alpha I \subset R$. Hence $N(\gamma I) = N(\mu \alpha I)/|N(\alpha\nu)| = |N(\mu\alpha)|N(I)/|N(\alpha\nu)|= |N(\gamma)|N(I).$

QED

Proposition 12 Let $I$ be a fractional ideal of $R$. Let $\alpha_1, \cdots, \alpha_n$ be a $\mathbb{Z}$-basis of $I$. Let $\theta_1, \cdots, \theta_n$ be a $\mathbb{Z}$-basis of $R$. Suppose $\alpha_i = \sum_j a_{ij} \theta_j$ for $i = 1,\cdots,n$. Then $N(I) = |\det (a_{ij})|$.

Proof: Since $I$ is a lattice of $R$, there exists a rational integer $d \gt 0$ such that $dI \subset R$(see the proof of Lemma 2 of my previous answer to the question). Then $d\alpha_1, \cdots, d\alpha_n$ is a $\mathbb{Z}$-basis of $dI$. Since $d\alpha_i = \sum_j da_{ij} \theta_j$ for $i = 1,\cdots,n$, $N(dI) = |\det (da_{ij})| = d^n |\det (a_{ij})|$. Since $N(dI) = d^nN(I)$ by Proposition 11, we are done. QED

Corollary Let $I$ be a fractional ideal of $R$. Then $d(I) = N(I)^2 d(R)$.

Proposition 13 Let $I, J$ be fractional ideals of $R$ such that $J \subset I$. Then $|I/J| = N(J)/N(I)$.

Proof: Let $\alpha_1, \cdots, \alpha_n$ be a $\mathbb{Z}$-basis of $I$. Let $\beta_1, \cdots, \beta_n$ be a $\mathbb{Z}$-basis of $J$. Suppose $\beta_i = \sum_j b_{ij} \alpha_j$ for $i = 1,\cdots,n$, where $b_{ij} \in \mathbb{Z}$. Then $|I/J| = |\det (b_{ij})|$. $d(J) = (\det (b_{ij}))^2 d(I) = |I/J|^2d(I)$. Since $d(J) = N(J)^2 d(R)$ and $d(I) = N(I)^2d(R)$ by the corollary of Proposition 12, we are done. QED

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    $\begingroup$ -1, for the same reason I have explained recently. $\endgroup$ – Carl Mummert Dec 18 '13 at 22:03
  • $\begingroup$ @CarlMummert You didn't explain why I shouldn't answer my own question. Anyway, why don't you leave me alone and mind your own business? You are wasting your time. You can't change my style. $\endgroup$ – Makoto Kato Dec 18 '13 at 23:10

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