[Math] Norm of a fractional ideal of an order of an algebraic number field

algebraic-number-theory

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 $|I/J| = N(J)/N(I)$.

Best Answer

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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