# The product of all the conjugates of an ideal is a principal ideal generated by the norm.

Let $K\mid \mathbb{Q}$ be a Galois extension with Galois group $\text{Gal}(K\mid \mathbb{Q})=\{\sigma_1,...,\sigma_n\}$ and $\mathfrak{a}\subset \mathcal{O}_K$ an ideal.

It is true that $\prod_{i=1}^n\sigma_i(\mathfrak{a})=(N(\mathfrak{a}))$?

I know that this is true for quadratic extensions but I'm interested in the general case because it gives a pretty easy way to compute the norm of an ideal once you know their generators.

Yes. It clearly suffices to check only for prime ideals, as both the left hand side and the right hand side are multiplicative on ideals. Therefore, wlog assume $\mathfrak a = \mathfrak p_1$ prime, and let $p$ be the rational prime below $\mathfrak p_1$. $p$ factors as

$$p \mathcal O_K = (\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_g)^e$$

where the $\mathfrak p_i$ run over the distinct "conjugates" of the prime ideal $\mathfrak p$. (This follows, since the Galois group acts transitively on the set of prime ideals lying over a fixed prime.) Letting $f$ be the shared inertia degree, by definition, $(N(\mathfrak p_1)) = (|\mathcal O_K / \mathfrak p_1|) = (p^f)$. On the other hand, we have:

$$\prod_i \sigma_i(\mathfrak p_1) = (\prod_i \mathfrak p_i)^{ef} = (\mathfrak p_1 \mathfrak p_2 \ldots \mathfrak p_g)^{ef} = (p \mathcal O_K)^f = (p)^f = (p^f)$$

and the result follows.