Let $R$ be the ring of Dirichlet series with integer coefficients. I'd often wondered about whether $R$ was a UFD; this post cleared that up, because it turns out that $R\simeq\mathbb{Z}[[x_1,x_2,\cdots]]$ (the $x_i$ correspond to primes, apparently, but I'm not sure what the explicit isomorphism is).

My first (slightly mundane) question is: what is the group of units $U(R)$? I know that $f\in R$ is a unit iff $f(0)$ is a unit (which in this case means $f(0)=\pm1$); similarly, $f\in\mathbb{Z}[[x_1,x_2,\cdots]]$ is a unit iff $f$'s constant term is $\pm1$, and part D of this link would seem to help a bit (using $\mathbb{Z}[[x_1,x_2,\cdots]]\simeq \mathbb{Z}[[x_2,\cdots]][[x_1]]$), but I couldn't get very far figuring out what $U(R)$ actually is.

Now, my main question: Can we take arbitrary $n$th roots (and hence, arbitrary fractional powers) of Dirichlet series which are units in $R$? I believe this is equivalent to asking whether the group of units is divisible, but I'm not sure.

## A motivating example / special case of my question

$\mu$ and 1 (where $\mu$ is the Mobius function, $1$ is the constant 1 function) are units in $R$. In fact, $\mu\cdot 1=\epsilon$ (where $\epsilon(0)=1$, $\epsilon(n)=0$ for $n>0$ is the identity). Expressing this using the actual series,

$\displaystyle\left(\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{1}{n^s}\right)=\sum_{n=1}^\infty \frac{\epsilon(n)}{n^s} = 1$

and hence $\displaystyle\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$. Indeed, we can find the Dirichlet series for $\zeta(s)^k$ for any $k\in \mathbb{Z}$ by looking at the corresponding element $1^k\in R$ (note that $1^{-k}=\mu^k$). However, I would like to know what Dirichlet sequence corresponds to $\displaystyle\zeta(s)^{\frac{a}{b}}$ for $\frac{a}{b}\in\mathbb{Q}$.