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Let $K$ be an algebraic number field of degree $n$. Let $\mathcal{O}_K$ be the ring of algebraic integers of $K$. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $n$. Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$. Let $I$ be a non-zero ideal of $R$. If $I + \mathfrak{f} = R$, we call $I$ regular. For the properties of regular ideals, see this. We denote by $N(I)$ the number of elements of the finite ring $R/I$.

If $I$ and $J$ are regular ideals of $R$, we have $N(IJ) = N(I)N(J)$ as proved in my answer to this question. I wonder if the formula still holds when one of them is not regular. And I came up with the following proposition.

Proposition Let $R$ be an order of an an algebraic number field. Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$. Then $N(IJ) = N(I)N(J)$.

Outline of my proof I generalized the result of this and proved $R/J$ is $R$-isomorophic to $I/IJ$. See my answer below for the detail.

My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.

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    $\begingroup$ When the vast majority of your post is a justification for why it shouldn't be closed, perhaps it's better for a meta thread. $\endgroup$ – user61527 Nov 9 '13 at 2:17
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    $\begingroup$ The vast majority of your question seems to be trying to justify its existence by citing SE policies. Isn't that the sort of thing that should go with a meta post, as you've done a number of times? $\endgroup$ – user61527 Nov 9 '13 at 2:28
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    $\begingroup$ @T.Bongers It's not admonition. I wrote the reason in the remark why I refer to the SE blog. Anyway, not everybody knows such a meta thread. $\endgroup$ – Makoto Kato Nov 9 '13 at 2:58
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    $\begingroup$ This question in its present form requires significant revision to bring it into the normal usage of the site. For that reason, I am voting to put it "on hold". $\endgroup$ – Carl Mummert Nov 12 '13 at 21:45
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    $\begingroup$ Here are my personal suggestions, in no particular order. (1) Don't use boilerplate. (2) Make the question engaging to read. (3) Decide whether you are asking people to check your proof, or provide other proofs. Don't ask several different things in the same question. (5) Don't tell people how to answer the question (the "please provide full proofs" sentence). (6) Tell where you encountered the question. (7) Explain why you think the question is interesting. (8) Don't ask separate questions that are just slight variations of each other. I cannot control how other people vote, of course. $\endgroup$ – Carl Mummert Nov 12 '13 at 23:42
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Lemma Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$. Then there exists non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = A$.

Proof: By this question, $I$ is uniquely decomposed as a product of regular prime ideals. Let $I = P_1^{e_1}...P_n^{e_n}$ be the prime decomposition, where $P_1,\cdots,P_n$ are distinct regular prime ideals and $e_i \ge 1$ for all $i$. Let $Q_1, ..., Q_m$ be all the distinct prime ideals which contain $J$, but not contain $I$. For each $i$ choose $\alpha_i\in\ P_i^{e_i}\setminus P_i^{e_i+1}$. For each $j$ choose $\beta_j\in\ A\setminus Q_j$. Note that $P_1,\cdots,P_n$ and $Q_1, ..., Q_m$ are maximal ideals. Hence, by the Chinese Remainder Theorem, there exists $\alpha\in A$ such that $\alpha\equiv\alpha_i\pmod{P_i^{e_i+1}}$ for all $i$ and $\alpha\equiv\beta_j\pmod{Q_j}$ for all $j$. Since $\alpha \in P_i^{e_i}$ for all $i$, $\alpha \in I$. Since $I$ is invertible by this question, there exists an ideal $M$ such that $\alpha R = IM$. Clearly $M + J = A$

Proof of the proposition It suffices to prove that $R/J$ is isomorphic to $I/IJ$. By the lemma, there exist non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = R$. Then $IM + IJ = I$. Let $\psi\colon R \rightarrow I/IJ$ be the $R$-homomorphism sending $x$ to $\alpha x$ (mod $IJ$). Since $\alpha R + IJ = I$, $\psi$ is surjective. It remains to prove that $\psi$ is injective. Suppose $\alpha x \in IJ$ for $x \in R$. It suffices to prove that $x \in J$. Since $\alpha R = IM$, $IMx \subset IJ$. Since $I$ is invertible, $Mx \subset J$. Since $M + J = R$, there exist $\mu \in M$ and $\beta \in J$ such that $\mu + \beta = 1$. Since $\mu x + \beta x = x$, $x \in J$ as desired.

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  • $\begingroup$ I noticed that someone serially upvoted for my questions including this one. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. $\endgroup$ – Makoto Kato Nov 27 '13 at 7:09

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