The compatiblity of the ideal norm of a principal ideal with the element norm

algebraic-number-theorycommutative-algebra

Let $\mathcal{O}_K$ be the ring of integers of $K$. Given prime number $p$ and let $\mathfrak{p} \subset \mathcal{O}_K$ such that $\mathfrak{p} \cap \mathbb{Z} = p \mathbb{Z}$. Set $f(\mathfrak{p}/p)$ be the degree of the field extension $\mathbb{F}_p \subset R/\mathfrak{p}$. One also says that $\mathfrak{p}$ is an ideal of the norm $p^{ f(\mathfrak{p}/p)}$.

We define $N_{K/\mathbb{Q}}: \mathcal{I}(\mathcal{O}_K) \to \mathbb{Q}^*$ the homomorphism that maps a prime $\mathfrak{p}$ of $\mathcal{O}_K$ to $N_{K/\mathbb{Q}}\mathfrak{p}=p^{f(\mathfrak{p}/p)}$. Show that the compatiblity
$$|N_{K/\mathbb{Q}}(x)|=N_{K/\mathbb{Q}}(x\mathcal{O}_K)$$
of the ideal norm of a principal ideal $x\mathcal{O}_K$ with the element norm $N_{K/\mathbb{Q}}(x)$ of $x$.

We know that $|N_{K/\mathbb{Q}}(x)|=$ #$\mathcal{O}_K/x\mathcal{O}_K$. How can i show that $N_{K/\mathbb{Q}}(x\mathcal{O}_K)$ is also equal to #$\mathcal{O}_K/x\mathcal{O}_K$?

Best Answer

We look at the factorization $x\mathcal{O}_K = \mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_r^{e_r}$. Note that $N_{K/\mathbb{Q}}(x\mathcal{O}_K) = p_1^{e_1 f_1}\cdots p_r^{e_r f_r}$ by multiplicativity (here the $p_i$ are generators of $\mathfrak{p}_i \cap \mathbb{Z}$ and the $f_i$ are the corresponding inertia degrees).

On the other hand, the Chinese Remainder Theorem states that $\mathcal{O}_K / x\mathcal{O}_K \cong \mathcal{O}_K/\mathfrak{p}_1^{e_1}\times\cdots\times\mathcal{O}_K/\mathfrak{p}_r^{e_r}$. Then $\left|\mathcal{O}_K/x\mathcal{O}_K\right| = \left|\mathcal{O}_K/\mathfrak{p}_1^{e_1}\right|\cdots\left|\mathcal{O}_K/\mathfrak{p}_r^{e_r}\right|$.

But $\left|\mathcal{O}_K/\mathfrak{p}_i^{e_i}\right| = \left|\mathcal{O}_K/\mathfrak{p}_i\right|^{e_i} = p_i^{e_i f_i}$ so the two expressions are the same. The first equality follows from the fact that $\mathcal{O}_K/\mathfrak{p}_i^{e_i}$ is a vector space over $\mathcal{O}_K/\mathfrak{p}_i$ of dimension $e_i$. The second equality follows from the fact that $|\mathcal{O}_K/\mathfrak{p}_i|$ is the size of the finite field $\mathcal{O}_K/\mathfrak{p}_i$ of degree $f_i$ and characteristic $p_i$.

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