2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ What does $I_m$ stand for, please? $\endgroup$ Commented Jul 30 at 7:03
  • 1
    $\begingroup$ $\mathcal I_\mathfrak m$ be the set of fractional ideals of $K$ coprime with $\mathfrak m$ $\endgroup$ Commented Jul 30 at 7:58
  • 1
    $\begingroup$ @GerryMyerson And $\mathfrak{m}$ is the ideal generated by $a$ and $n$. $\endgroup$
    – GH from MO
    Commented Jul 30 at 10:57

1 Answer 1

2
$\begingroup$

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 $1$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Commented Aug 11 at 8:29
  • $\begingroup$ @FranzLemmermeyer That is correct. At any rate, I talked about the discriminant of $L/K$, because it is easier to compute (I guess). $\endgroup$
    – GH from MO
    Commented Aug 11 at 16:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .