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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.