Let $K=\mathbb{Q(\sqrt{d})}$ and $\mathbb{a}$ be a nonzero ideal in the ring of integers $O_K$ . Why does the equation hold : $\mathbb{a}\mathbb{\overline a}=N(\mathbb{a}O_K)$ ? One can prove this by taking a prime ideal $\mathbb{a}=\mathbb{p}$ but this does not bring me further .
Product of ideals in a quadratic field
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pka
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QUOTE="Minami, post: 470057, member: 75851"]
Let \(\displaystyle K=\mathbb{Q(\sqrt{d})}\) and \(\displaystyle \mathbb{a}\) be a nonzero ideal in the ring of integers \(\displaystyle O_K\) . Why does the equation hold : \(\displaystyle \mathbb{a}\mathbb{\overline a}=N(\mathbb{a}O_K)\) ? One can prove this by taking a prime ideal \(\displaystyle \mathbb{a}=\mathbb{p}\) but this does not bring me further .[/QUOTE]
Fixed LaTeX
Let \(\displaystyle K=\mathbb{Q(\sqrt{d})}\) and \(\displaystyle \mathbb{a}\) be a nonzero ideal in the ring of integers \(\displaystyle O_K\) . Why does the equation hold : \(\displaystyle \mathbb{a}\mathbb{\overline a}=N(\mathbb{a}O_K)\) ? One can prove this by taking a prime ideal \(\displaystyle \mathbb{a}=\mathbb{p}\) but this does not bring me further .[/QUOTE]
Fixed LaTeX