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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 $\frac{a}{b}\in\mathbb{Q}$.

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The explicit isomorphism sends p_i^s to x_i. –  Qiaochu Yuan Jan 14 '10 at 11:17
Rather, 1/p_i^s to x_i. –  Qiaochu Yuan Jan 14 '10 at 19:24

3 Answers 3

up vote 3 down vote accepted

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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Ah, thanks! I suppose that makes sense that there aren't roots - Dirichlet series with values in $\mathbb{Z}$ is a bit restrictive. However, in regards to the group of units - I understood that characterization of units, but couldn't figure out how to turn it into a statement like $U(D(\mathbb{Z}))\simeq$ $G$ (some abelian group, which I would guess is some combination of copies of $\mathbb{Z}$) –  Zev Chonoles Jan 14 '10 at 14:19
Regarding the unit group, Z[[x]] is generated by -1 and 1 + x^n, n \ge 1. But I'm not sure this generalizes to more than one variable, since there you're taking tensor products instead of direct products. –  Qiaochu Yuan Jan 14 '10 at 19:31
Sorry, I meant U(Z[[x]]). I suppose it's reasonable to suspect that Z[[x_1, x_2, ...]] is generated by 1 + (monomials). –  Qiaochu Yuan Jan 14 '10 at 19:33
@Qiaochu: the unit group of $\mathbb{Z}[[x]]$ is uncountable, whereas your suggested set of generators is countable. –  Pete L. Clark Jan 14 '10 at 22:23
Sorry, not "generators" in the sense of finite Z-linear combinations, but infinite ones (which converge in the x-adic topology). –  Qiaochu Yuan Jan 16 '10 at 7:17

Alright, let me try to justify the assertions I made in the comments yesterday. The formal argument is a bit tiresome to write out, but the basic idea of the proof goes like this: suppose we are given $f \in \mathbb{Z}[[x_1, x_2, ... ]]$ with constant term $1$. Let the linear terms of $f$ be $\sum_{i \ge 1} a_i x_i$. Then $f_1(x) = \frac{f(x)}{\prod_{i \ge 1} (1 + x_i)^{a_i}}$ has no linear terms. Let the quadratic terms of $f_1$ be $\sum_{i \le j} a_{ij} x_i x_j$. Then $f_2(x) = \frac{f_1(x)}{\prod_{i \le j} (1 + x_i x_j)^{a_{ij}}}$ has no linear or quadratic terms. Rinse and repeat. Conclusion: $U(\mathbb{Z}[[x_1, x_2, ...]])$ is topologically generated by $\pm 1$ and $1 + \text{monomials}$ in the appropriate topology (which I am not totally sure how to define, but am firmly convinced exists). In particular I agree with Pete about the structure of the unit group.

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I just want to point out that there is a nice formula for the coefficients of complex powers of Dirichlet series with completely multiplicative coefficients. You only need to use the binomial series, as follows. If: $$F(s)=\prod_p\left(1-f(p)p^{-s}\right)^{-1}$$ then formally we can define $$F^{z}(s)=\prod_p\left(1-f(p)p^{-s}\right)^{-z},$$ where obviously the usual rules for defining the complex logarithm apply. Multiplying this out gives $$F^{z}(s)=\sum_{n=1}^{\infty}\frac{f(n)\lambda(n)\prod_{i=1}^{k}{{-z}\choose{\alpha_i}}}{n^s},$$ where $n=\prod_{i=1}^{k}p_i^{\alpha_i}$ is the canonical factorization of $n$.

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