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In a quadratic number field $\mathbb{Q}[\sqrt{d}]$, we have that the norm is given by $N_{\mathbb{Q}[\sqrt{d}]/\mathbb{Q}}(\mathfrak{p})=\mathfrak{p}\overline{\mathfrak{p}}$.

For general algebraic number fields, the norm is defined to be $N_{K/\mathbb{Q}}(\mathfrak{p})=|\mathcal{O}_K/\mathfrak{p}|$. Thus when $K$ is a quadratic number field, we get $\mathfrak{p}\overline{\mathfrak{p}}=|\mathcal{O}_K/\mathfrak{p}|$.

Is there such an identity in general? For example, since we can write $\alpha\in K$ of degree $n$ with integers $a_j$ and $a$ as $\alpha_0=\sum a_j \sqrt[n]{a^j}$, we can consider its 'conjugates' $\alpha_k=\sum a_j e^{2\pi ij/n}\sqrt[n]{a^j}$, and note that $\prod_{k=0}^{n-1} \alpha_k\in\mathbb{Q}$.

Question: Can we show that $|\mathcal{O}_K/(\alpha_0)|=\prod_{k=0}^{n-1} \alpha_k$ for an algebraic number field of dimension $n$ and principal prime ideal $(\alpha_0)$?

If not, is there a similar formula for this? How about for non principal prime ideals?

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    $\begingroup$ In general it's a product of all the Galois conjugates (you can't always write algebraic numbers in terms of n-th roots). For ideals you either take the ideal generated by all the norms or take the product of all Galois conjugate ideals (maybe those are the same). Unfortunately I don't have my ANT textbook with me right now to provide a citation. $\endgroup$ Dec 5, 2021 at 21:38
  • $\begingroup$ How are Galois Conjugate Ideals defined? $\endgroup$
    – Tejas Rao
    Dec 5, 2021 at 21:40
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    $\begingroup$ Apply an automorphism to all the elements at once. $\endgroup$ Dec 5, 2021 at 21:42

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This is going to be an incomplete answer since I can't remember all the details.

Let $L/K$ be an extension of number fields. Let $\alpha \in L$ and define the $K$-linear operator $m_\alpha : L \to L$. Then the norm of $\alpha$ is $\det(m_\alpha)$ (again, as a $K$-linear not $L$-linear map).

For "generic" $\alpha$, the characteristic polynomial of $m_\alpha$ equals the minimal polynomial of $m_\alpha$ which always equals the minimal polynomial of $\alpha$ (over $K$). Thus $\det(m_\alpha)$ is the constant term of this minimal polynomial. That constant term is equal to the product of the roots of the minimal polynomial. So if $L$ is the splitting field of $\alpha$ (a Galois extension), then

$$ \det(m_\alpha) = \prod_{\sigma \in G(L/K)} \sigma(\alpha).$$

If $L/K$ is Galois and the splitting field of $\alpha$ is a proper subfield of $L$ then the values $\sigma(\alpha)$ will repeat for different automorphisms. But crucially, they repeat the same number of times for each distinct root of the minimal polynomial.

If the roots of the minimal polynomial are $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$, then

$$\det(m_\alpha) = \prod_{\sigma \in G(L/K)} \sigma(\alpha) = \left( \prod_{i = 1}^n \sigma_i(\alpha) \right)^{[L:K(\alpha)]}.$$

There is a second way to see this. If the splitting field is a proper subfield of $L$ then the minimal polynomial of $m_\alpha$ is not equal to the characteristic polynomial (it's a factor). In this case, write down the rational canonical form of $m_i$ as the block diagonal matrix $\operatorname{diag}(C_1, C_2, \dots, C_k)$ and let $q_i$ be the minimal polynomial of $C_i$. As usual, $q_1 \mid q_2 \mid \dots \mid q_k$ and $q_k$ is the minimal polynomial of $\alpha$. On the other hand, $q_1(m_\alpha) = m_{q_1(\alpha)}$ is a non-invertible operator. The only way that multiplication by $q_1(\alpha)$ can be non-invertible is if $q_1(\alpha) = 0$. Hence, $q_1 = q_2 = \dots = q_k$ since $q_1, \dots, q_k$ all divide the minimal polynomial $q_k$ and $q_i(\alpha) = 0$ for each $i$. It follows that the characteristic polynomial of $m_\alpha$, which is $q_1 \cdot q_2 \cdot \dots q_k$ is a power of the minimal polynomial. Then by looking at the dimensions, that power has to be $[L:K(\alpha)]$.


Ok, so that's the norm of a single element. And so we have defined $N_{L/K} : L \to K$. When $K = \mathbf{Q}$, this norm is a rational number. If $\alpha$ is integral, then $N_{L/\mathbf{Q}}(\alpha) \in \mathbf{Z}$ and the absolute value of that integer (because it could be negative) is equal to $|\mathcal{O}_{L}/\alpha\mathcal{O}_L|$.

And the way to connect the determinant/product of roots formula to this finite ring is to use the fact that $\det(m_\alpha)$ is a measure of relative area. Specifically, if $\Lambda$ is the fundamental parallelepiped in $\mathcal{O}_K$ then $m_\alpha(\Lambda)$ has a relative volume of $|N_{L/\mathbf{Q}}(\alpha)|$ in the lattice $\mathcal{O}_L$. This relative volume is then used to show that there are exactly that many elements of $\mathcal{O}_L/\alpha\mathcal{O}_L$.

I forget what parts of this can be done relative to $K$ and which should be done relative to $\mathbf{Q}$. One thing is that $N_{L/K}(\alpha) \in \mathcal{O}_K$ so taking the "absolute value" really means the absolute value of $N_{K/\mathbf{Q}}(N_{L/K}(\alpha)) = N_{L/\mathbf{Q}}(\alpha) \in \mathbf{Z}$. Of course, $|\mathcal{O}_L/\alpha\mathcal{O}_L|$ is always a positive integer so that part cannot be done relative to $K$.

If $I$ is any ideal of $\mathcal{O}_L$, we can define the norm $N_L(I) = |\mathcal{O}_L/I|$ which, for principal ideals, is $|N_{L/\mathbf{Q}}(\alpha)|$. For non-principal ideals, I am fairly certain that there is some comparison between $N_L(I)$ and the ideal $(N_{L/\mathbf{Q}}(\alpha) : \alpha \in I) \subseteq \mathbf{Z}$ but I'm not 100% certain how this all works.

You should be able to find details and precise statements of anything I was vague on in any algebraic number theory textbook.

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