Extension of $N_{\mathbb{Q}[\sqrt{d}]/\mathbb{Q}}(\mathfrak{p})=\mathfrak{p}\overline{\mathfrak{p}}$ to general Algebraic Number Fields

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?

• 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. Commented Dec 5, 2021 at 21:38
• How are Galois Conjugate Ideals defined? Commented Dec 5, 2021 at 21:40
• Apply an automorphism to all the elements at once. Commented Dec 5, 2021 at 21:42

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.