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Let $d \in \mathbb{Z} \setminus \{0, 1\}$ square free, $K = \mathbb{Q}(\sqrt{d})$, $\mathcal{O}_K$ the corresponding ring of integers, $\mathfrak{a} \subseteq K$ a fractional ideal $n, n' \in \mathbb{N}$ with $n\mathfrak{a}, n'\mathfrak{a} \subseteq \mathcal{O}_K$.

I'm currently trying to prove that the definition of the idealnorm for fractional ideals is well defined. Inside this proof I need the following result: $[n\mathfrak{a} : nn'\mathfrak{a}] = [\mathcal{O}_K : (n')]$ or more general $n\mathfrak{a} / nn'\mathfrak{a} \cong \mathcal{O}_K / (n')$ as abelian groups.

I don't know how I could approach this step. Any help is welcome!

Duplicate claim: My proof looks similar to the one posted here, but the linked post doesn't prove the question I asked here.

Example
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1 Answers1

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I think both $[n\mathfrak a : nn' \mathfrak a]$ and $[\mathcal O_K : (n')]$ are equal to $(n')^d$, where $d$ is the degree of the number field $K$ over $\mathbb Q$.

For example, if $x_1, \dots, x_d$ is an integral basis for $\mathcal O_K$, then $n'x_1, \dots, n'x_d$ is an integral basis for $(n')$. Thus $\mathcal O_K / (n') \cong \mathbb Z_{n'}^{\oplus d}$ as an abelian group.

Similarly, if $y_1, \dots, y_d$ is an integral basis for the ideal $n\mathfrak a \subset \mathcal O_K$, then $n'y_1, \dots, n'y_d$ is an integral basis for $nn'\mathfrak a$. So $\mathcal n\mathfrak a / nn'\mathfrak a \cong \mathbb Z_{n'}^{\oplus d}$ as an abelian group too.

Kenny Wong
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  • Could you explain the notation $\mathbb Z_{n'}^{\oplus d}$, please? I'm not familiar with it. – Example Jul 05 '18 at 06:59
  • @Quadrat $\mathbb Z_{n'}$ is the cyclic group or order $n'$. $\mathbb Z_{n'}^{\oplus d}$ is the direct sum of $d$ copies of the cyclic group of order $n'$. – Kenny Wong Jul 05 '18 at 07:42
  • Thank you, sadly now I don't understand how you came to the conclusion of the isomorphism. – Example Jul 05 '18 at 08:12
  • @Quadrat Well, suppose $G$ is an abelian group freely generated by $x_1, \dots, x_d$. So every element $g \in G$ can be written as $g = a_1 x_1 + \dots + a_d x_d$ for a unique choice of $(a_1, \dots, a_d)$. Now suppose $H$ is the subgroup generated by $nx_1, \dots, nx_d$. So the elements in $ H$ are those elements $b_1 x_1 + \dots + b_d x_d$ in $G$ where $b_1, \dots, b_d$ are all divisible by $n$. Then what is the quotient $G / H$? – Kenny Wong Jul 05 '18 at 09:27
  • I got it. The quotient is $G/H \cong \oplus_{k=1}^{d} \mathbb{Z}/n\mathbb{Z}$. – Example Jul 05 '18 at 10:19