Let $\zeta(s)$ denote the Riemann zeta function. Is the set $\{
\zeta(s-j)\, \colon\, j\in\mathbb{Z}\}$, or even $\{\zeta(s-z)\,
\colon\, z\in\mathbb{C}\}$, algebraically independent over
$\mathbb{C}$? If not, then expanding a polynomial equation satisfied
by these functions into a Dirichlet series and taking the coefficient
of $n^{-s}$ would yield an "unlikely" number-theoretic identity.



Algebraic independence of shifts of the Riemann zeta function
$\zeta(s - z)$ has an Euler product $\prod\_p \frac{1}{1 - p^{z-s}}$, and so a monomial $\prod\_i \zeta(s - z\_i)$ (with the $z\_i$ not necessarily distinct) has an Euler product


$$\prod\_i \zeta(s - z\_i) = \prod\_p \prod\_i \frac{1}{1 - p^{z\_i - s}}.$$


We want to show that these monomials are linearly independent. Now here's an observation: it's quite hard for Dirichlet series with Euler products to be linearly dependent. This is because any linear dependence must, by examining only the coefficients of $\frac{1}{p^{ks}}$ for each prime separately, be a linear dependence for every Euler factor separately, but also must be a linear dependence for all of the Euler factors multiplied together, and even for any subset of the Euler factors multiplied together.


In fact we can prove the following, passing from Dirichlet series to coefficients. If $S$ is a set of primes, write $\mathbb{N}\_S$ for the set of positive integers divisible only by the primes in $S$, and write $\mathbb{N}\_{-S}$ for the set of positive integers divisible only by the primes not in $S$.



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> **Lemma:** Let $f\_0, \dots f\_k : \mathbb{N} \to \mathbb{C}$ be multiplicative arithmetic functions which are
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> * *essentially nonzero* in the sense that for any finite set of primes $S$, $f\_i(n) \neq 0$ for some $n \in \mathbb{N}\_{-S}$, and
> * *essentially distinct* in the sense that for any finite set of primes $S$, if $f\_i(n) = f\_j(n)$ for all $n \in \mathbb{N}\_{-S}$ then $i = j$.
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> 
> Then the functions $f\_i$ are *essentially linearly independent* in the sense that for any finite set of primes $S$ they are linearly independent over $\mathbb{C}$ when restricted to $\mathbb{N}\_{-S}$.
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*Proof.* This ends up being a slight variant of the standard proof of [linear independence of characters](https://stacks.math.columbia.edu/tag/0CKK) (which would apply directly if "multiplicative" were replaced by "completely multiplicative"). We induct on $k$. When $k = 0$ the result follows from the assumption that the $f\_i$ are essentially nonzero. For general $k$, let $S$ be a finite set of primes and suppose by contradiction that we have a nontrivial linear dependence, which WLOG we take to be of the form


$$f\_0(n) = \sum\_{i=1}^k c\_i f\_i(n), n \in \mathbb{N}\_{-S}.$$


Since $f\_0$ is essentially nonzero this requires that at least one of the $c\_i$ also be nonzero. Now, if $m, n \in \mathbb{N}\_{-S}$ are positive integers such that $\gcd(m, n) = 1$, then on the one hand


$$f\_0(mn) = \sum\_{i=1}^k c\_i f\_i(mn) = \sum\_{i=1}^k c\_i f\_i(m) f\_i(n)$$


and on the other hand


$$f\_0(mn) = f\_0(m) f\_0(n) = f\_0(m) \sum\_{i=1}^k c\_i f\_i(n) = \sum\_{i=1}^k c\_i f\_0(m) f\_i(n).$$


Subtracting gives


$$\sum\_{i=1}^k c\_i (f\_0(m) - f\_i(m)) f\_i(n) = 0.$$


If $T$ is any finite set of primes, letting $m$ be any element of $\mathbb{N}\_T \cap \mathbb{N}\_{-S}$ (divisible by the primes in $T$ but not the primes in $S$) and letting $n$ range over $\mathbb{N}\_{-(S \cup T)}$ gives, by the inductive hypothesis, that for each value of $m$ the above is a linear dependence of the $f\_i$ which must be trivial, hence the coefficients $c\_i (f\_0(m) - f\_i(m))$ must vanish for all $m \in \mathbb{N}\_T \cap \mathbb{N}\_{-S}$. (This bit of the argument is why we need the freedom to ignore finitely many primes.)


Since one of the $c\_i$, say $c\_j$, is nonzero, it follows that $f\_0(m) = f\_j(m)$ for all $m \in \mathbb{N}\_T \cap \mathbb{N}\_{-S}$, but since this is true independent of the choice of $T$, we in fact have $f\_0(m) = f\_j(m)$ for all $m \in \mathbb{N}\_{-S}$, which contradicts essential distinctness. $\Box$


Now it suffices to check that the monomials $\prod\_i \zeta(s - z\_i)$ are essentially nonzero and essentially distinct. Essential distinctness is a bit less straightforward than I thought, since deleting finitely many factors from the Euler product of $\zeta(s - z\_i)$ produces zeroes at $s = z\_i$ which may cancel some of the poles from other factors. But it doesn't affect the order of the pole at $s = z\_i + 1$, which is further to the right, so we can still consider the rightmost $z\_i$'s and the corresponding poles. We get that if two monomials are essentially equal then the rightmost $z\_i$'s which occur in each must match (with matching multiplicities) so we can factor these out and inductively conclude that all of the $z\_i$ must match.


We should also get algebraic independence for a broader class of Dirichlet series (anything for which it's clear that we can still show essential distinctness), e.g. shifts of Dirichlet L-functions.


