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S Apr 4, 2020 at 0:02 history bounty ended CommunityBot
S Apr 4, 2020 at 0:02 history notice removed CommunityBot
Apr 3, 2020 at 4:26 comment added GH from MO In the definition on p.182 using the Kronecker symbol, $\mathfrak{p}$ and $\mathfrak{c}$ are meant to be ideals of $\mathcal{O}_k$, and their norms are meant to be $[\mathcal{O}_k:\mathfrak{p}]$ and $[\mathcal{O}_k:\mathfrak{c}]$. So I think you need to consider the product $\mathfrak{c}_k\mathcal{O_k}$, which is a fractional ideal of $\mathcal{O_k}$: write it as a quotient of two ideals of $\mathcal{O}_k$, and then apply the definition on p.182 using the Kronecker symbol.
Mar 28, 2020 at 17:20 answer added Franz Lemmermeyer timeline score: 1
Mar 27, 2020 at 9:21 comment added Franz Lemmermeyer ${\mathbb Z}-modules$ in ${\mathcal O}_k$ of course - sorry. I'll go through the stuff carefully until tomorrow.
S Mar 26, 2020 at 22:52 history bounty started Shimrod
S Mar 26, 2020 at 22:52 history notice added Shimrod Authoritative reference needed
Mar 26, 2020 at 21:12 comment added Shimrod @FranzLemmermeyer How they can be $\mathcal O_K$-modules?
Mar 26, 2020 at 21:08 comment added Franz Lemmermeyer You're right. But they are ${\mathcal O}_k$-modules, and these have norm $f$. Siegel's "Ringideal" is defined as in Hecke's lecture "Analysis und Zahlentheorie" (edited by Roquette), p. 125ff.
Mar 26, 2020 at 20:46 comment added Shimrod @FranzLemmermeyer Please see Update 2 above.
Mar 26, 2020 at 20:39 history edited Shimrod CC BY-SA 4.0
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Mar 26, 2020 at 20:11 comment added Shimrod @FranzLemmermeyer That is not true. We actually have $\mathcal O =\lbrace \alpha \in K\colon \alpha \mathfrak c_k\subset c_k\rbrace$.
Mar 26, 2020 at 20:01 comment added Franz Lemmermeyer These are ideals in ${\mathcal O}_k$, and of course we are talking about the ordinary norms of these ordinary ideals. You seem to want to compute the norm of ideals in a ring to which they do not belong.
Mar 26, 2020 at 19:28 comment added Shimrod @FranzLemmermeyer But $\mathfrak c_K \not \subset \mathcal O$, so what do you mean by the residue class ring in this case? One cannot use $k-\omega$ to reduce elements of $\mathcal O$, because $k-\omega \not \in \mathcal O$. I have updated the question to give a more detailed computation of the norm.
Mar 26, 2020 at 19:26 history edited Shimrod CC BY-SA 4.0
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Mar 26, 2020 at 14:29 comment added Franz Lemmermeyer Anyway, using the definition of the norm of an ideal as the cardinality of its residue class ring, use the element $k - \omega$ to reduce any algebraic integer to an ordinary integer; using the element $f$ of the ideal we find that the residue class ring is simply ${\mathbb Z}/f{\mathbb Z}$, so the norm if the ideal is $f$.
Mar 26, 2020 at 14:21 comment added Franz Lemmermeyer I don't understand what you are doing - what references are you using?
Mar 26, 2020 at 13:11 comment added Shimrod @FranzLemmermeyer Then what is wrong with my computation of the norm (see above)?
Mar 26, 2020 at 13:10 history edited Shimrod CC BY-SA 4.0
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Mar 26, 2020 at 7:48 comment added Franz Lemmermeyer The norm of an ideal with basis $[m, a+n\omega]$ is $mn$.
Mar 25, 2020 at 17:03 comment added Shimrod @FranzLemmermeyer We have $\mathfrak c_k\subset \mathcal O_K$, but $\mathfrak c_k\not\subset \mathcal O $. Could you please show how you calculated the norm?
Mar 25, 2020 at 16:56 comment added Franz Lemmermeyer Your ideals are integral ideals, and their norms seem to be $f$, not $1$.
Mar 24, 2020 at 20:45 history edited Shimrod CC BY-SA 4.0
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Mar 24, 2020 at 17:19 history asked Shimrod CC BY-SA 4.0