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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.

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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.

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