# When is the product of two Dirichlet $L$-functions also a Dirichlet $L$-function?

This question came up while trying to learn about the Riemann zeta function - as is the guilty pleasure of many of us.

A Dirichlet $L$-function is notated as $$L\{z,\chi\}(z) = \sum_{n=0}^{\infty}\frac{\chi(n)}{n^z}$$ Where it is usually supposed that $z$ is complex and $\chi$ is a multiplicative function: $\chi(a)\chi(b)=\chi(ab)$

One can further show that $$L\{z,\chi\}(z)=\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)}{p^z}\Big)^{-1}$$ Where $\prod_{p\in\mathbb{P}}$ means take a product over the prime numbers.

Next I tried multiplying two $L$-functions as follows: $$L\{z,\chi\}(z)\cdot L\{\omega,\phi\}(\omega)$$ $$||$$ $$\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)}{p^z}\Big)^{-1}\Big(1-\frac{\phi(p)}{p^\omega}\Big)^{-1}$$ $$||$$ $$\prod_{p\in\mathbb{P}}\Big(1-\frac{\chi(p)p^\omega + \phi(p)p^z - \chi(p)\phi(p)}{p^{z+\omega}}\Big)^{-1}$$ So let's choose a new function for this product that we might clean it up a little $$M(a)=\chi(a)a^\omega + \phi(a)a^z - \chi(a)\phi(a)$$ Now we may say $$L\{z,\chi\}(z)\cdot L\{\omega,\phi\}(\omega) =\prod_{p\in\mathbb{P}}\Big(1-\frac{M(p)}{p^{z+\omega}}\Big)^{-1} =L\{z+\omega,M\}(z+\omega)$$ And we (almost) have another Dirichlet $L$-function. The catch is that $M$ has to be multiplicative.

I noticed that $$M(1)=\chi(1)1^\omega + \phi(1)1^z - \chi(1)\phi(1)=(1)(1)+(1)(1)-(1)(1)=1$$ So it seems reasonable that there exists some choice of $\chi$ and $\phi$ that make $M$ multiplicative as well. It would be best if this choice $\chi$ and $\phi$ could be done for all $z,\omega \in \mathbb{C}$ but I suspect that $z$ and $\omega$ will have to be fixed to some particular values.

So finally

For what $\chi, \phi, z,$ and $\omega$ is $M(a)=\chi(a)a^\omega + \phi(a)a^z - \chi(a)\phi(a)$ a multiplicative function.

• Have you tried restricting to $z=\omega$? – Somos Aug 30 '17 at 21:48
• Now if $\chi_1,\chi_2$ are Dirichlet characters then so is $\psi(n) = \chi_1(n) \chi_2(n)$ – reuns Aug 31 '17 at 21:35

First off, you are using highly nonstandard notation. In number theory, a Dirichlet $L$-function is written as $L(s,\chi)$ (or occasionally $L(\chi,s)$), where $s$ is a complex variable and $\chi$ is a Dirichlet character, not $L\{z,\chi\}(z)$. (Why is $\{z,\chi\}$ in brackets? Why does $z$ appear twice?) This has the Euler product $L(s,\chi) = \prod_p \frac{1}{1 - \chi(p) p^{-s}},$ where the product is over all primes $p$. (There is no need to write $p \in \mathbb{P}$; it is always implicitly understood that the product is over all primes.)
In any case, if $\chi_1, \chi_2$ are Dirichlet characters, then $L(s_1,\chi_1) L(s_2,\chi_2)$ is never equal to some Dirichlet $L$-function $L(s_3,\chi_3)$ for some Dirichlet character $\chi_3$ and complex variable $s_3$. This is basically because $L(s_1,\chi_1) L(s_2,\chi_2) = \sum_{n = 1}^{\infty} \sum_{ab = n} \frac{\chi_1(a) \chi_2(b)}{a^{s_1} b^{s_2}},$ whereas $L(s_3,\chi_3) = \sum_{n = 1}^{\infty} \frac{\chi_3(n)}{n^{s_3}},$ and then one can easily check that the arithmetic function $\frac{\chi_3(n)}{n^{s_3}}$ is completely multiplicative (i.e. $\frac{\chi_3(m)}{m^{s_3}} \frac{\chi_3(n)}{n^{s_3}} = \frac{\chi_3(mn)}{(mn)^{s_3}}$), but that $\sum_{ab = n} \frac{\chi_1(a) \chi_2(b)}{a^{s_1} b^{s_2}}$ is not (take $n$ to be a prime power).