Roots of unity in algebraic K-theory

For any commutative ring $R$, the tensor product of (finitely generated, projective) $R$-modules equips the algebraic K-theory $K(R) = K_0(R)$ with the structure of a commutative ring with unit.

For $n>2$, I'm wondering what assumptions one might need to impose upon $R$ to ensure that $K(R)$ contains primitive $n^{\rm th}$ roots of unity (i.e., a cyclic subgroup of $K(R)^{\times}$ of order $n$). I'd even be happy with a good family of examples.

flag	asked Jun 7 at 12:20 Craig Westerland 1,563●5●12

Since $K(R)^{\times}$ contains the Picard group of $R$, it is enough that the latter contains a cyclic subgroup of order $n$. There are lots of examples of this; for example, when $R$ is a regular integral finitely generated $\mathbb C$-algebra of dimension 1, whose function field is not purely transcendental over $\mathbb C$. – Angelo Jun 7 at 13:46

Right, of course! Similarly you get the same result from a positive proportion of the rings of integers of quadratic imaginary number fields, according to the Cohen-Lenstra heuristics. – Craig Westerland Jun 7 at 13:57

(replaced earlier comment) Do you really want a cyclic subgroup of units? I bet you really want a map from $\mathbb Z[\zeta_n]$, but I doubt you'll get that. – Ben Wieland Jun 7 at 18:02

@Ben Wieland: I guess either would make me happy, but your suggestion would make me more happy. – Craig Westerland Jun 8 at 9:08

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