Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me.

Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers.


  1. In my lecture we have defined the norm $N_{K/\mathbb{Q}}(x)$ for an element $x \in K$ in the usual way. In a following lecture we "extended" this to the norm of an ideal $\mathfrak{a}$ of $\mathcal{O}_K$ by setting $$ \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|$$ and showed that for the principal ideal $(a)$ with $a \in \mathcal{O}_K$ we always have $\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|$.

  2. Later, we introduced the notion of a fractional ideal in $K$ being a non-zero, finitely generated $\mathcal{O}_K$-submodule of $K$. And we showed that a non-zero $\mathcal{O}_K$-submodule $\mathfrak{a}$ of $K$ is a fractional ideal iff there is a $c \in \mathcal{O}_K \setminus \{0\} $ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (thus $c\mathfrak{a}$ is an ideal of $\mathcal{O}_K$).

  3. We now extended the norm $ \mathfrak{N}$ by defining for each fractional ideal $\mathfrak{a}$ of $K$ the norm $$ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, $$ $c$ being an element of $\mathcal{O}_K \setminus \{0\}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (exists according to 2.).


How can I show that this latter definition is independent of the chosen $c$?


I changed the definition in (3) from $ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} $ to $\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} $ and added the absolute value everywhere else, too. I think that was a typo in my lecture notes.

Thoughts (Is it better to add them as comments? I was not sure.)

  1. Let $c, c' \in \mathcal{O}_K \setminus \{0\}$ such that $c \mathfrak{a} \subseteq \mathcal{O}_K $ and $c' \mathfrak{a} \subseteq \mathcal{O}_K$. At the moment I am contemplating if and in which way $c$ and $c'$ are related. I was wondering whether one always has $c'=bc$ (or vice versa) for an $b \in \mathcal{O}_K$. Then we had $$ \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}$$ and it remained to show that $[\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}]$ (maybe because $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).
  • $\begingroup$ Just added a first thought. But I am not sure whether this leads in the right direction. $\endgroup$ – puck29 Aug 19 '15 at 16:56
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    $\begingroup$ The direction is fine (I think). And I think it will lead to an argument that works, whenever you can find a $b\in\mathcal{O}_K$. But you are not guaranteed to always find such a $b$. But if $c,c'$ are any two multipliers such that $c\mathfrak{a}, c'\mathfrak{a}\subseteq\mathcal{O}_K$, can you, perhaps compare both of them to what you would get with $cc'$? $\endgroup$ – Jyrki Lahtonen Aug 19 '15 at 19:43
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    $\begingroup$ And to continue in your chosen direction: Observe that $$bc\mathfrak{a}\subseteq c\mathfrak{a}\subseteq{\mathcal O}_K.$$ Therefore theory of finitely generated abelian groups will give you $$ [\mathcal{O}_K:bc\mathfrak{a}]=[\mathcal {O}_K:c\mathfrak{a}]\cdot [c\mathfrak{a}:bc\mathfrak{a}].$$ Exercise: Show that $$c\mathfrak{a}/bc\mathfrak{a}\cong \mathcal{O}_K/b\mathcal{O}_K.$$ as abelian groups. $\endgroup$ – Jyrki Lahtonen Aug 19 '15 at 19:44
  • $\begingroup$ Just wanted to say thank you for the hints :-). I will try to work on them and then report back. $\endgroup$ – puck29 Aug 19 '15 at 21:21
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    $\begingroup$ I would like to invite you to write a summary of all this as an answer. That way our site hygiene is improved by the tiny amount of getting this question out of the unanswered queue. Also, hopefully, you will get pointers from people who actually know this stuff. $\endgroup$ – Jyrki Lahtonen Aug 20 '15 at 13:46

Thanks to the hints in the comments, I think I now can summarize an answer myself. But I still need help on how to prove the isomorphy in point (2) below. Please still feel free to point out mistakes or to post alternative (easier?) answers.

Let $c,c' \in \mathcal{O}_K \setminus \{ 0 \}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ and $c'\mathfrak{a} \subseteq \mathcal{O}_K$.

  1. We have the chain $ c'c\mathfrak{a} \subseteq c\mathfrak{a} \subseteq \mathcal{O}_K $ of $\mathcal{O}_K$-modules. Because of the isomorphy $$ \frac{\mathcal{O}_K / c'c\mathfrak{a} }{ c\mathfrak{a} / c'c\mathfrak{a}} \cong \mathcal{O}_K / c\mathfrak{a} $$ of quotient modules it follows that $[\mathcal{O}_K : c'c\mathfrak{a}] = [\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]$ (note: all indeces involved are indeed finite – something I do not prove here).

  2. We have the isomorphy $ c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1}$ (why?) of $\mathcal{O}_K$-modules and therefore $$ c\mathfrak{a} / c'c\mathfrak{a} = c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1} = \mathcal{O}_K / (c') · \mathcal{O}_K = \mathcal{O}_K / (c'). $$ This implies $[c\mathfrak{a} : c'c\mathfrak{a}] = [\mathcal{O}_K:(c')]$.

  3. By (1) and (2) we now get $$ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · \overbrace{[\mathcal{O}_K:(c')]}^{=\left| N_{K/\mathbb{Q}}(c') \right|}}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|}$$

  4. Since we analogously (just switch the roles of $c$ and $c'$ in (1-3)) can show $ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} = \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|}$ we get $$ \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|} $$ and therefore the norm $\mathfrak{N}(\mathfrak{a}) $ of a fractional ideal $\mathfrak{a}$ in $K$ is indeed well-defined.


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