Norm of a fractional ideal of an order of an algebraic number field
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 $R$.
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.
$L$ is a lattice of $V$.
$L$ 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
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