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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.

Preludium

  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.).

Question

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

Edit

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 $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).
puck29
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  • Just added a first thought. But I am not sure whether this leads in the right direction. – puck29 Aug 19 '15 at 16:56
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    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 $cc'$? – Jyrki Lahtonen Aug 19 '15 at 19:43
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    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. – Jyrki Lahtonen Aug 19 '15 at 19:44
  • Just wanted to say thank you for the hints :-). I will try to work on them and then report back. – puck29 Aug 19 '15 at 21:21
  • I thought, I nearly got it. But I just realized that my idea to show $\mathcal{O}_K / b\mathcal{O}_K \cong c\mathfrak{a} / bc\mathfrak{a}$ does not work out. I tried to use the fact that both $\mathcal{O}_K$ and $c\mathfrak{a}$ have the same(!) rank $n:=[K:\mathbb{Q}]$ as $\mathbb{Z}$-modules, since we had a theorem that implies this. I thus wanted to construct a surjective homomorphism $$ \mathcal{O}_K \longrightarrow c\mathfrak{a} \longrightarrow c\mathfrak{a} / bc\mathfrak{a} $$ and show that its kernel is just $b\mathcal{O}_K$. – puck29 Aug 20 '15 at 01:46
  • By the way, I think I understand the other parts now, including your suggestion to work with $cc'\mathfrak{a}$. – puck29 Aug 20 '15 at 01:54
  • Hmm. I'm no longer sure that my suggestion(to show that $c\mathfrak{a}/bc\mathfrak{a}\cong \mathcal{O}_K/b\mathcal{O}_K$) works as simply as I thought (that they would "trivially" be isomorphic as $\mathcal{O}_K$-modules). If you have covered the uniqueness of factorization of ideals into prime ideals then it follows from that: multiplying any ideal of $\mathcal{O}_K$ by $b$ simply tags on the prime ideals that make $(b)$. Sorry... – Jyrki Lahtonen Aug 20 '15 at 07:21
  • Thanks for your time @JyrkiLahtonen :-) – much appreciated. Yes, we have covered the whole topic of (unique) ideal factorization in Dedekind rings in the lecture. I also shortly thought of just writing $$ c\mathfrak{a} / bc\mathfrak{a} = c\mathfrak{a} / (b) · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (b) · c\mathfrak{a} · (c\mathfrak{a})^{-1} = \mathcal{O}_K / (b) · \mathcal{O}_K = \mathcal{O}_K / (b) $$ but I am too shy ;-) to do so since I don't know how to justify the isomorphy (can I somehow?). But maybe you meant something else. – puck29 Aug 20 '15 at 13:27
  • Oh, YESS! That's the way to do it! No reason to be shy :-) – Jyrki Lahtonen Aug 20 '15 at 13:44
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    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. – Jyrki Lahtonen Aug 20 '15 at 13:46
  • @JyrkiLahtonen: Yes, I can try to summarize this as an answer. Just learned that it is okay to answer your own questions. But I am still pretty confused why exactly the isomorphy holds. Let $A,B,C$ be submodules of an $R$-module $K$ and $B \subseteq A$. Can we always just multiply the "numerator" and "denominator" of the quotient module like $A/B \cong A · C / B · C$? I know the usual theorems like $\frac{K/B}{A/B} \cong K/A$ but at the moment do not know how to prove something like the former – especially because it involves this multiplication from "outside" the "module world". – puck29 Aug 20 '15 at 14:17
  • I changed the definition of the norm of a fractional ideal by introducing absolute values. I think this was a typo in my lecture notes (and otherwise my proof does not work ;-). Is it true that the norm of a fractional ideal is always a positive rational number? – puck29 Aug 21 '15 at 08:00
  • Sorry @JyrkiLahtonen. I realized only now that it is possible to upvote comments as well. – puck29 Aug 21 '15 at 15:16
  • Related: https://math.stackexchange.com/questions/604050/ – Watson Nov 25 '18 at 17:31

1 Answers1

1

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.

puck29
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