Norm of generator of prime ideal

Question

Let $K=\mathbb{Q}(\zeta_n)$ be a cyclotomic field. For a prime ideal $\mathfrak{p}$ in the ring of integers $\mathcal{O}_K$, we can write $\mathfrak{p} = (p,f_i(\zeta_n))$, where $\mathfrak{p}\cap\mathbb{Z}=p$ and and $f_i(x)$ is an irreducible polynomial in the factorization of $\Phi_n(x)\bmod p$, where $\Phi_n(x)$ is the $n$th cyclotomic polynomial. My question is: suppose $f_j(x)$, $i\ne j$, is another polynomial in the factorization of $\Phi_n(x)\bmod p$. Is it true that $N_{K/\mathbb{Q}}(f_i(\zeta_n)) = N_{K/\mathbb{Q}}(f_j(\zeta_n))$?

Answer 1

When one says that $\mathfrak{p} = (p,f_i(x))$ for some factor $f_i(x)$ of $\Phi_n(x) \bmod p$, this doesn't quite make sense, and reflects an an abuse of notation. First of all, $\mathfrak{p}$ is an ideal inside a ring of characteristic zero, so it only makes sense to say that $f_i(x) \in \mathfrak{p}$ if $f_i(x)$ is an element of $\mathbf{Z}[x]$ rather than $\mathbf{F}_p[x]$. So the more precise statement is that $\mathfrak{p} = (p,f_i(x))$ where $f_i(x)$ is any polynomial in $\mathbf{Z}[x]$ whose mod-$p$ reduction is a factor of $\Phi_n(x) \bmod p$. The point is that any two choices of lift $f_i(x)$ and $g_i(x)$ will differ by a multiple of $p$, and so the ideals $(p,f_i(x))$ and $(p,g_i(x))$ will be the same.

But in light of this, the original question doesn't make much sense, because the quantity $N_{K/\mathbf{Q}}(f_i(\zeta))$ even for a single $i$ is not really well defined and depends on how you choose $f_i(x)$.

As an example, for $n = 4$, you have $x^2 + 1 = (x-2)(x-3) \bmod 5$ and thus (for example) $\mathfrak{p} = (5,i-2)$ is prime, and $N(i-2)= 5$ where $f_1(x) = x - 2$. But you could just as well have taken $f_1(x) = x+3$ and then $\mathfrak{p} = (5,i+3)$ and $N(i+3) = 10$. So the two things you are asking to be equal are themselves not well-defined.