TITLE: Why the roots of unity are the analogs of constants ?
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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).