Let $R$ be the ring of Dirichlet series with integer coefficients. I'd often wondered about whether $R$ was a UFD; [this post](https://mathoverflow.net/questions/5522/dirichlet-series-with-integer-coefficients-as-a-ufd) 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](http://books.google.com/books?id=ZpDt86aI15kC&lpg=PP1&pg=PA318#v=onepage&q=&f=false) 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
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$\mu$ 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 $\frac{a}{b}\in\mathbb{Q}$.



Fractional powers of Dirichlet series?
For your first question: let $R$ be any commutative ring, and let $D(R)$ be the ring of formal Dirichlet series over $R$, i.e., the set of all functions $f: \mathbb{Z}^+ \rightarrow R$ under pointwise addition and convolution product. 


Then the unit group of $R$ is precisely the set of formal Dirichlet series $f$ such that 
$f(1)$ is a unit in $R$.


As for your second question, it is indeed equivalent to asking whether $U(D(R))$ is $n$-divisible. Here, if we take $R = \mathbb{Z}$ as you asked, the answer is that for all $n \geq 2$, $U(D(\mathbb{Z}))$ is **not** $n$-divisible and that even the Dirichlet series $\zeta(s)$ is not an $n$th power in $D(\mathbb{Z})$.


[Now, for some reason, I switch back to the classical notation, i.e., I replace the arithmetical function $f$ by its "Dirichlet generating series" $\sum\_{n=1}^{\infty} \frac{f(n)}{n^s}$. It would have been simpler not to do this, but too late.]


Let
$f(s) = \sum\_{n=1}^{\infty} \frac{a\_n}{n^s}$ be any formal Dirichlet series, and suppose 
that $g(s) = \sum\_{n=1}^{\infty} \frac{b\_n}{n^s}$ be a formal Dirichlet series such that 
$g^2 = f$. Thus


$a\_1 + \frac{a\_2}{2^s} + \ldots = (b\_1 + \frac{b\_2}{2^s} + ... )(b\_1 + \frac{b\_2}{2^s} + \ldots)$


$= b\_1^2 + \frac{2 b\_1 b\_2}{2^s} + \frac{2 b\_1 b\_3}{3^s} + \frac{2b\_1 b\_4 + b\_2^2}{4^s} + 
\ldots$


(This multiplication is formal, i.e., it is true by definition.) 


Thus $b\_1 = \pm \sqrt{a\_1}$. Suppose we take the plus sign, for simplicity. Then for all primes $p$,


$a\_p = 2 b\_1 b\_p$, so


$b\_p = \frac{a\_p}{2 \sqrt{a\_1} }$, 


so we need $2 \sqrt{a\_1}$ to divide $a\_p$, so at least we need $a\_p$ to be even for all primes $p$. Further conditions will come from the composite terms. 


These same considerations show that if we replaced the coefficient ring $\mathbb{Z}$ by 
$\mathbb{Q}$ (or any coefficient field of characteristic $0$), then any formal Dirichlet series with $a\_1 = 1$ is $n$-divisible for all positive integers $n$. In particular, you can write $\zeta(s)^{\frac{1}{n}}$ as a Dirichlet series with $\mathbb{Q}$-coefficients just by applying the above procedure and successively solving for the coefficients. Whether there is a nice formula for these coefficients is a question for a better combinatorialist than I to answer. 


**EDIT**: Based on your comments below, I now understand that you are looking for a characterization of $U(D(\mathbb{Z}))$ as an astract abelian group. I believe it is isomorphic to 
$\{ \pm 1 \} \times \prod\_{i=1}^{\infty} \mathbb{Z}$. (Or, more transparently, to 
the product of $\{ \pm 1\}$ with the product of infinitely many copies of $\prod\_{i=1}^{\infty} \mathbb{Z}$, one for each prime number. But as abstract groups it amounts to the same thing.)


