Dirichlet Characters L Functions and Primes in Progressions -- Examples

For $q=3$, the unit group is $\{1,2\}\cong C_2$, so there are two characters with values on the classes $(0,1,2)$ given by ${\chi }_0=(0,1,1)$ and the nonprincipal character $(0,1,-1)$. For $q=4$, the unit group is also $C_2$, giving ${\chi }_0=(0,1,0,1)$ and ${\chi }_4=(0,1,0,-1)$.

For $q=5$, the unit group is cyclic of order $4$, generated by $2$, so the four characters are determined by $\chi (2)\in \{1,-1,i,-i\}$. Writing values on the classes $0,1,2,3,4$ gives ${\chi }_0=(0,1,1,1,1)$, ${\chi }_1=(0,1,i,-i,-1)$, ${\chi }_2=(0,1,-1,-1,1)$, and ${\chi }_3=(0,1,-i,i,-1)$. Each table is zero exactly off the units and multiplicative on the unit classes, so these are exactly the Dirichlet characters for the three moduli.

The unit groups $(\mathbb{Z}\mathrm{/}8\mathbb{Z}{)}^{\times }=\{1,3,5,7\}$ and $(\mathbb{Z}\mathrm{/}12\mathbb{Z}{)}^{\times }=\{1,5,7,11\}$ are both isomorphic to $C_2\times C_2$. Hence each has four homomorphisms to $\{\pm 1\}$. For modulus $8$, taking signs independently on the generators $3$ and $5$ produces the four characters with values $(0,1,0,\pm 1,0,\pm 1,0,\pm 1)$ on the classes $0,\dots ,7$, subject to $\chi (7)=\chi (3)\chi (5)$.

The same construction for modulus $12$ uses the generators $5$ and $7$, with $\chi (11)=\chi (5)\chi (7)$. Thus all four characters have zeroes on the nonunits and values $(0,1,0,0,0,\pm 1,0,\pm 1,0,0,0,\pm 1)$ on $0,\dots ,11$. Because every homomorphism from $C_2\times C_2$ to $\{\pm 1\}$ is determined by the chosen signs on a basis, these are all the Dirichlet characters modulo $8$ and $12$.

Using the four rows $(1,1,1,1)$, $(1,i,-i,-1)$, $(1,-1,-1,1)$, and $(1,-i,i,-1)$ on the unit classes $1,2,3,4$, the row inner products are $4$ on the diagonal and $0$ off the diagonal. Likewise the column sums ${\sum }_{\chi }\chi (a)\overline{\chi (b)}$ are $4$ when $a=b$ and $0$ otherwise. This is the theorem Orthogonality relations for Dirichlet characters modulo q in the concrete case $q=5$.

Taking $a=2$, the weighted average $\frac{1}{4}{\sum }_{\chi }\overline{\chi (2)}\chi (n)$ is $1$ at the class $2$ and $0$ at the other residue classes, exactly as A residue-class indicator from character sums predicts.

For $q=4$, the theorem gives $L(s,{\chi }_0)=\zeta (s)(1-2^{-s})$. For $q=6$, it gives $L(s,{\chi }_0)=\zeta (s)(1-2^{-s})(1-3^{-s})$. In each case the omitted Euler factors are exactly those at the bad primes dividing the modulus.

Evaluating the finite factor at $s=1$ gives the residues $1-1\mathrm{/}2=1\mathrm{/}2=\phi (4)\mathrm{/}4$ and $(1-1\mathrm{/}2)(1-1\mathrm{/}3)=1\mathrm{/}3=\phi (6)\mathrm{/}6$, matching the general residue formula.

The unit group modulo $4$ is $\{1,3\}$, with $3^2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$. Its unique nontrivial character sends $1$ to $1$ and $3$ to $-1$, and extension by zero sends the even classes to $0$. This is the character ${\chi }_4$ listed in Dirichlet character tables modulo 3, 4, and 5, so ${\chi }_4(2m)=0$, ${\chi }_4(4m+1)=1$, and ${\chi }_4(4m+3)=-1$. The convergence theorem then identifies $L(1,{\chi }_4)={\sum }_{n\ge 1}\frac{{\chi }_4(n)}{n}={\sum }_{m\ge 0}(\frac{1}{4m+1}-\frac{1}{4m+3})=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \mathrm{.}$

The last series is exactly the Gregory-Leibniz series, so The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+... gives $L(1,{\chi }_4)=\pi \mathrm{/}4$.

The reduced residue classes modulo $3$ are $1$ and $2$, so Primes in one reduced residue class have Dirichlet density 1 over phi(q) gives ${\sum }_{p\equiv 1(3)}{p}^{-s}=\frac{1}{2}\log \frac{1}{s-1}+O(1)$ and the same formula for the class $2$.

Thus each reduced class carries half of the logarithmic divergence among the primes. The small primes $7,13,19,31,\dots$ illustrate the statement, but the proof is the theorem from step 1.1, not the finite list.

Take $q=6$ and $a=3$. Every integer congruent to $3$ modulo $6$ is $6m+3=3(2m+1)$, so it is divisible by $3$.

Hence the only prime in that residue class is $3$ itself. In particular, there are not infinitely many such primes. So the reduced-residue hypothesis in Dirichlet's theorem on primes in arithmetic progressions is indispensable.

Let $P=\{p\text{ prime}:p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)\}$. By Primes in one reduced residue class have Dirichlet density 1 over phi(q), $P$ has relative Dirichlet density $1\mathrm{/}2$ among the primes.

As a subset of the integers, however, its counting function is at most the prime-counting function $\pi (x)$. By Chebyshev bounds for the prime-counting function, $\pi (x)\mathrm{/}x\le c_2\mathrm{/}\log x$ for all sufficiently large $x$, and this upper bound tends to $0$. Hence $P(x)\mathrm{/}x\to 0$, so $P$ has natural density $0$ among the positive integers despite its positive Dirichlet density among the primes.