Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 23 / 2 / c / Solution

Let $q_0$ be the conductor of a Dirichlet character and ${\chi }_0$ the inducing primitive even character. Removing the Euler factors absent from $L(s,\chi )$ gives the imprimitive Dirichlet L-function Euler correctionThe primitive functional equation therefore givesEquivalently, replace the final primitive function by $L(1-s,\overline{\chi })/P_{\overline{\chi }}(1-s)$, interpreted as a meromorphic identity with removable values handled by continuation. It is the conductor of a Dirichlet character $q_0$, rather than the possibly inflated modulus $q$, that enters the gamma factor and root number. The complex conjugation bar in the original PDF is lost in the converted TeX.

The zeros are those of $L(s,{\chi }_0)$ together with the zeros of the finite Euler product, and multiplicities add. Since $|{\chi }_0(p)|=1$ at each extra prime, an extra factor vanishes at the imaginary points determined byEach extra prime creates infinitely many such points. Its nonzero-imaginary points are not zeros of the primitive function: the functional equation and nonvanishing of Dirichlet L-functions on the line one exclude them. Thus the zero sets are identical precisely when every prime dividing $q$ already divides $q_0$, making $P=1$. Increasing prime-power exponents alone can make a character imprimitive without changing its L-function. If $q_0=1$, the primitive function is zeta; the same Euler correction applies, with its pole at one retained.