# Conductor of the Hecke character- power residue symbol

The power residue symbol is the multiplicative character $$\bigl(\frac a \cdot\bigr): \mathcal I_\mathfrak m\to \mu_n$$ that satistfies $$\bigl(\frac a {\mathfrak p}\bigr)\equiv a^{\frac{N\mathfrak p-1}{n}} \mod {\mathfrak p}$$ for every prime ideal $$\mathfrak p\in\mathcal I_\mathcal m$$, where $$N\mathfrak p\in\mathbb N$$ is the norm of $$\mathfrak p$$.

I am curious about the conductor of the above Hecke character. By Artin's reciprocity, I understand that it exists, but I am not sure how to determine it.

This MO post states that above Hecke character's conductor divides a power of $$\mathfrak m$$. This fact follows from class field theory, from example, from [1, VIII.5.5, page 241].

I studied the following reference that post had suggested, but I cannot figure out why the conductor divides a power of $$\mathfrak{m}$$. Can anyone help me with this?

• What does $I_m$ stand for, please? Commented Jul 30 at 7:03
• $\mathcal I_\mathfrak m$ be the set of fractional ideals of $K$ coprime with $\mathfrak m$ Commented Jul 30 at 7:58
• @GerryMyerson And $\mathfrak{m}$ is the ideal generated by $a$ and $n$. Commented Jul 30 at 10:57

Let us recall the notations of the referenced MO post. Let $$n$$ be a positive integer, let $$\mu_n\subset\mathbb C$$ be the set of $$n$$-th roots of unity, let $$K$$ be a number field containing $$\mu_n$$, let $$R$$ be the ring of integers of $$K$$, let $$a\in R\setminus\{0\}$$, let $$\mathfrak m\subset R$$ be the ideal generated by $$a$$ and $$n$$, let $$\mathcal I_\mathfrak m$$ be the set of fractional ideals of $$K$$ coprime with $$\mathfrak m$$.
Now let us consider the Kummer extension $$L=K(\sqrt[n]{a})$$. Then the power residue symbol in the post belongs to the Galois character $$\mathrm{Gal}(L/K)\to\mu_n$$ given by $$\sigma\mapsto\sigma(a)/a$$. In particular, the conductor of the power residue symbol divides the discriminant of the extension $$L/K$$. However, the extension $$L/K$$ is unramified outside the prime divisors of $$an$$, whence its discriminant divides a power of $$an$$. Equivalently, the discriminant of $$L/K$$ divides a power of $$\mathfrak{m}n$$. The factor $$n$$ cannot be saved. For example, in the case of $$K=\mathbb{Q}$$, $$n=2$$, $$a=-1$$, we are dealing with the Jacobi symbol, whose conductor is $$4$$, not $$1$$.
• Isn't the conductor of the residue symbol simply the conductor of $L/K$ since a prime splits completely in $L/K$ if and only if the residue symbol is trivial? Commented Aug 11 at 8:29
• @FranzLemmermeyer That is correct. At any rate, I talked about the discriminant of $L/K$, because it is easier to compute (I guess). Commented Aug 11 at 16:13