Aurifeuillean factorization with number fields

Question

Basically the question is if number fields can be used in Aurifeuillean factorization.

Probably this is easy and the answer is "no".

Let $f,g \in \mathbb{Z}[x], a \in \mathbb{N}$. Let $f(x)$ and $f(g(x))$ be irreducible over $\mathbb{Q}[x]$.

The goal is to find nontrivial factor of $f(g(a))$ using number fields.

Suppose over some number field $K$ $f(g(x))$ is reducible and factors $$f(g(x)) = f_1(x) \cdot f_2(x) \cdots f_n(x)$$

The norms of $f_i(a)$ will be related to the factorization of $f(g(a))$, usually giving the trivial factor.

Is it possible for some $f,g,K$, the norm of $f_i(a)$ to give nontrivial factor of $f(g(a))$?

I suspect the answer is "no".

Answer 1

I don't see the relevance of the extra polynomial $g$ in this question.

Anyway the answer is indeed no because if you let $\sigma_1,\ldots,\sigma_d$ be all different embeddings of $K \to \bar{\mathbb Q}$, then the norm of $f_i(a)$ is $\prod_j \sigma_j(f_i(a))$. However $f(g(x))$ is irreducible over $\mathbb Q$ so it has to divide $\prod_j \sigma_j(f_i(x)) \in \mathbb Q(x)$. In particular the norm of $f_i(a)$ will be a multiple of $f(g(a))$.