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.)

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Yes

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 the unit group of $U(D(R))$ is $n$-divisible. Here, if we take $R = \mathbb{Z}$ as you asked, the Dirichlet ring answer is divisible.

Nothat for all $n \geq 2$, $U(D(\mathbb{Z}))$ is not $n$-divisible and that even the unit group Dirichlet series $\zeta(s)$ is not divisiblean $n$th power in $D(\mathbb{Z})$.

[Now, for some reason, as can be seen just by writing out I switch back to the operation. 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.]

$\mathbb{Q}$ (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.

1

Yes, your question is equivalent to asking whether the unit group of the Dirichlet ring is divisible.

No, the unit group is not divisible, as can be seen just by writing out the operation. 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 $n$.