Namespaces
Variants
Actions

Conductor of an Abelian extension

From Encyclopedia of Mathematics
Revision as of 21:39, 22 November 2014 by Richard Pinch (talk | contribs) (specify context global fields)
Jump to: navigation, search

2020 Mathematics Subject Classification: Primary: 11R [MSN][ZBL]

Let $L/K$ be an Abelian extension of global fields and let $N_{L/K} C_L$ be the corresponding subgroup of the idèle class group $C_K$ (cf. Class field theory). The conductor of an Abelian extension is the greatest common divisor of all positive divisors $n$ such that $L$ is contained in the ray class field $K^n$ (cf Modulus in algebraic number theory).

For an Abelian extension of local fields $L/K$ the conductor of $L/K$ is $\mathfrak{p}_K^n$, where $\mathfrak{p}_K$ is the maximal ideal of (the ring of integers $A_K$ of) $K$ and $n$ is the smallest integer such that $N_{L/K} L^* \subset U_K^n = \{ x \in A_K : x \equiv 1 \pmod \mathfrak{p}_K^n \}$, $U_K^0 = U_k = A_K^*$. (Thus, an Abelian extension is unramified if and only if its conductor is $A_K$.) The link between the local and global notion of a conductor of an Abelian extension is given by the theorem that the conductor $\mathfrak{f}$ of an Abelian extension $L/K$ of number fields is equal to $\prod_{\mathfrak{p}} \mathfrak{f}_{\mathfrak{p}}$, where $\mathfrak{f}_{\mathfrak{p}}$ is the conductor of the corresponding local extension $L_{\mathfrak{p}} / K_{\mathfrak{p}}$. Here for the infinite primes, $\mathfrak{f}_{\mathfrak{p}} = \mathfrak{p}$ or $1$ according to whether $L_{\mathfrak{p}} \neq K_{\mathfrak{p}}$ or $L_{\mathfrak{p}} = K_{\mathfrak{p}}$.

The conductor ramification theorem of class field theory says that if $\mathfrak{f}$ is the conductor of a class field $L/K$, then $\mathfrak{f}$ is not divisible by any prime divisor which is unramified for $L/K$ and $\mathfrak{f}$ is divisible by any prime divisor that does ramify for $L/K$ (cf Ramification theory of valued fields).

If $L/K$ is the cyclic extension of a local field $K$ with finite or algebraically closed residue field defined by a character $\chi$ of degree 1 of $\mathrm{Gal}(K^{\mathrm{s}}/K)$, then the conductor of $L/K$ is equal to $\mathfrak{p}_K^{\mathfrak{f}(\chi)}$, where $\mathfrak{f}(\chi)$ is the Artin conductor of the character $\chi$ (cf. Conductor of a character). Here $K^{\mathrm{s}}$ is the separable algebraic closure of $K$. There is no such interpretation known for characters of higher degree.

References

[a1] J.-P. Serre, "Local fields" , Springer (1979) (Translated from French)
[a2] J. Neukirch, "Class field theory" , Springer (1986) pp. Chapt. 4, Sect. 8
How to Cite This Entry:
Conductor of an Abelian extension. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Conductor_of_an_Abelian_extension&oldid=42926