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Hello,

Joel Dogde, in a comment on his question "Roots of unity in different completions of a number field", says the following, about the analogy between number fields and function fields :

Number of roots of unity in number fields is something like the size of the constant field for function fields.

Could anyone explain that ?

Thanks.

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    $\begingroup$ A comment: counterexamples to Malle's conjecture on counting extensions of a global field with given Galois group were given in the function field case and the number field case. In the function field case, it is believed that these counterexamples are all due to the presence of constant field extensions, whereas in the number field case the analogous problem seems to be with cyclotomic subfields. $\endgroup$
    – Pete L. Clark
    Commented Jun 14, 2010 at 14:54
  • $\begingroup$ What is Malle conjecture ? $\endgroup$
    – user2330
    Commented Jun 14, 2010 at 14:59
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    $\begingroup$ @ Mister Clark: Aren't you on a MO Summer vacation...? $\endgroup$
    – Max Lonysa Muller
    Commented Jun 14, 2010 at 15:06
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    $\begingroup$ @MM: Yes, but some of the resorts have nice email access. $\endgroup$
    – Pete L. Clark
    Commented Jun 14, 2010 at 19:56

2 Answers 2

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One answer: the roots of unity in $K$ are precisely the elements of $K$ which have absolute value $1$ for every absolute value on $K$; the elements of the constant field have this property for function fields.

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    $\begingroup$ Right. And, even more basically, if $K$ is a global field with constant field $\mathbb{F}_q$, then the roots of unity in $K$ are precisely the nonzero elements of $\mathbb{F}_q$. (There are also some differences: any constant field extension is everywhere unramified, whereas adjoining sufficiently many roots of unity to a number field gives a ramified extension.) $\endgroup$
    – Pete L. Clark
    Commented Jun 14, 2010 at 14:51
  • $\begingroup$ @Pete: Doesn't adjoining roots of unity to a number field that weren't there before always result in a ramified extension? $\endgroup$
    – Dror Speiser
    Commented Jun 14, 2010 at 21:00
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    $\begingroup$ Not always: Let $K= \mathbb{Q}(\sqrt{-5})$ and $L=K(i)$. Then $L$ is unrammified over $K$. $\endgroup$
    – David E Speyer
    Commented Jun 14, 2010 at 22:52
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The number of roots of unity in a local field (of order prime to the characteristic) plus 1 is the cardinality of the residue class field (i.e. constant field in the function field case). This follows because the $q$-th roots for $q$ prime to the characteristic are distinct in the residue class field, because $\prod_i (1 - \zeta_q^i) = q$. In addition, each element $\bar{x}$ of the residue class field must satisfy $\bar{x}^r= 1$ for some $r$ prime to the characteristic, and $\bar{x}$ can be lifted to the local field by Hensel's lemma to an $r$-th root of unity. The plus one was added to account for zero (thanks to KConrad for pointing this out).

An unramified extension of local fields is obtained by adjoining a root of unity (prime to the characteristic of the residue class field). In the case when the local fields are completions of function fields, the additional roots of unity correspond precisely to increasing the field of constants (because constants are roots of unity, as Pete Clark as mentioned).

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    $\begingroup$ Akhil, I think it's more natural to refer to the order of a root of unity rather than its degree (that is more intrinsic to the root of unity) and also you forgot about 0 being in the residue field: you meant to count the cardinality of the nonzero elements in the residue field, e.g., Q_p has residue field of size p and the number of roots of unity (with order prime to p) is p-1 rather than p. $\endgroup$
    – KConrad
    Commented Jun 14, 2010 at 15:59

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