Let $L(s)$ be an L-function given by an Euler product \[ L(s)= \prod_{p\not\in S} \prod_{j=1}^r \left (1- \frac{\alpha_j}{p^s}\right)^{-1} \times \prod_{p \in S} L_p(s), \] where $S$ is a finite set of primes. The symmetric $n$-th power of $L(s)$ is an L-function given by an Euler product \[ L(s,\mathrm{sym}^n) = \prod_{p\not\in S}\prod_{ { \mathrm{degree}-n} \atop \mathrm{monomials}\,m} \left( 1- \frac{m(\alpha_1,\dotsc,\alpha_r)}{p^s} \right)^{-1} \times \prod_{p\in S} L_p(s,\mathrm{sym}^n). \] The Euler factors at the primes $p\in S$ (the "bad" primes) are computed via a more complicated recipe which involves a non-trivial amount of information about the underlying object. The degree of an Euler factor at one of the bad primes will be smaller than the degree of the Euler factors outside the set $S$.

**Authors:**

**Knowl status:**

- Review status: reviewed
- Last edited by Alex J. Best on 2018-12-13 12:51:16

**Referred to by:**

**History:**(expand/hide all)

- 2018-12-13 12:51:16 by Alex J. Best (Reviewed)