# Advanced Number Theory Note #12: Separability property of Gauss sums associated with Dirichlet characters

In a previous note I studied a multiplicative property of quadratic Gauss sums. (I intend to use this result in a future note which will explore in detail a proof of the law of quadratic reciprocity using Cauchy’s residue theorem from Complex Analysis). Quadratic Gauss sums are a special case of Gauss sums associated with Dirichlet characters, and in the present note I want to explore some important properties of the latter, particularly a separability property which plays a role in other areas I want to look into in future notes. As usual I will try to illustrate key abstract ideas with concrete examples.

For any Dirichlet character $\chi$ mod $k$, the Gauss sum associated with $\chi$ is

$G(n, \chi) = \sum_{m=1}^k \chi(m)e^{2\pi imn/k}$

For the principal character mod $k$, usually denoted $\chi_1$, we have $\chi_1(m) = 1$ if $gcd(m, k) = 1$ and $\chi_1(m) = 0$ otherwise, and in this case the Gauss sum reduces to Ramanujan’s sum:

$G(n, \chi_1) = c_k(n) = \sum_{\substack{m=1 \\ gcd(m, k)=1}}^k e^{2\pi imn/k}$

Therefore the Gauss sums $G(n, \chi)$ can be viewed as a generalisation of Ramanujan’s sums.

The following is the key separability property of Gauss sums which is of interest here:

Theorem 1. If $\chi$ is any Dirichlet character mod $k$, then

$G(n, \chi) = \overline{\chi}(n)G(1, \chi)$      whenever     $gcd(n, k) = 1$

Remark: Note that if $\chi$ is the principal character $\chi_1$ mod $k$, then this statement reduces to $c_k(n) = \mu(k)$ if $gcd(n, k) = 1$. I mentioned in Advanced Number Theory Note #10 that the Möbius function as a sum of primitive $k$th roots of unity is an interesting special case of Ramanujan’s sums and Gauss sums, and this connection between the three is made beautifully clear here.

Proof of Theorem 1: Let $gcd(n, k) = 1$ and suppose $r$ runs through a complete residue system mod $k$. Then the numbers $nr$ also run through a complete residue system mod $k$ and since

$|\chi(n)|^2 = \chi(n)\overline{\chi}(n) = 1$

we have

$\chi(r) = \chi(n)\overline{\chi}(n)\chi(r) = \overline{\chi}(n)\chi(nr)$

Using this observation in the sum defining $G(n, \chi)$ we get

$G(n, \chi) = \sum_{r\hspace{1 mm} mod \hspace{1 mm}k}\chi(r)e^{2\pi inr/k}$

$= \overline{\chi}(n) \sum_{r\hspace{1 mm} mod \hspace{1 mm}k}\chi(nr)e^{2\pi inr/k}$

$= \overline{\chi}(n) \sum_{m\hspace{1 mm} mod \hspace{1 mm}k}\chi(m)e^{2\pi im/k}$

$= \overline{\chi}(n) G(1, \chi)$

which proves the theorem. QED

Gauss sums $G(n, \chi)$ which satisfy

$G(n, \chi) = \overline{\chi}(n)G(1, \chi)$               (1)

are said to be separable.

Example: As an example, let $\chi$ be the (only) nonprincipal character mod 4. Then we have $\chi(1) = 1$, $\chi(3) = -1$, $\chi(2) = \chi(4) = 0$, so

$G(n, \chi) = \chi(1)e^{2\pi in/4} + \chi(3)e^{2\pi i3n/4}$

$= i^n - (-i)^n = i^n(1 - (-1)^n)$

Therefore

$G(n, \chi) = 2i$    if    $n \equiv 1$ (mod 4)

$G(n, \chi) = -2i$    if    $n \equiv 3$ (mod 4)

$G(n, \chi) = 0$    if    $n \equiv 0$ or $2$ (mod 4)

But

$G(1, \chi) = \chi(1)e^{2\pi i/4} + \chi(3)e^{2\pi i3/4} = 2i$

so it is obvious that $G(n, \chi) = \overline{\chi}(n)G(1, \chi)$ in each case. END

Theorem 1 says that $G(n, \chi)$ is separable whenever $n$ is relatively prime to the modulus $k$. The following theorem applies in the case of integers $n$ which are not relatively prime to $k$:

Theorem 2. If $\chi$ is a character mod $k$, the Gauss sum $G(n, \chi)$ is separable for every $n$ if and only if

$G(n, \chi) = 0$       whenever    $gcd(n, k) > 1$

Remark: In the context of the example above where $\chi$ is the nonprincipal character mod 4, observe that we indeed do have $G(n, \chi) = 0$ if $gcd(n, 4) > 1$.

Proof of Theorem 2: By Theorem 1, separability always holds if $gcd(n, k) = 1$, but in the case $gcd(n, k) > 1$ we have $\overline{\chi}(n) = 0$ (by definition of Dirichlet characters) so equation (1) above holds if and only if $G(n, \chi) = 0$. QED

The next theorem gives a useful consequence of separability:

Theorem 3. If $G(n, \chi)$ is separable for every $n$, then

$|G(1, \chi)|^2 = k$                       (2)

Remark: Again in the context of the example above where $\chi$ is the nonprincipal character mod 4, observe that we indeed do have $|G(1, \chi)|^2 = 4$.

Proof of Theorem 3: We have

$|G(1, \chi)|^2 = G(1, \chi)\overline{G(1, \chi)}$

$= G(1, \chi)\sum_{m=1}^k \overline{\chi}(m)e^{-2\pi im/k}$

$= \sum_{m=1}^k G(m, \chi)e^{-2\pi im/k}$

$= \sum_{m=1}^k \big(\sum_{r=1}^k \chi(r)e^{2\pi imr/k} \big) e^{-2\pi im/k}$

$= \sum_{r=1}^k \chi(r) \sum_{m=1}^k e^{2\pi im(r - 1)/k} = k\chi(1) = k$

since the last sum over $m$ is a geometric sum which vanishes unless $r = 1$. QED

By Theorem 2, separability of $G(n, \chi)$ is equivalent to the vanishing of $G(n, \chi)$ whenever $gcd(n, k) > 1$. It is natural to then ask: For which characters $\chi$ will $G(n, \chi)$ fail to vanish when $gcd(n, k) > 1$, and thus be nonseparable? The following (final) theorem addresses this question, and is one which leads to a whole subfield concerned with induced moduli and induced Dirichlet characters which I will explore in detail in a later note.

Theorem 4. Let $\chi$ be a Dirichlet character mod $k$ and assume that $G(n, \chi) \neq 0$ for some $n$ such that $gcd(n, k) > 1$. Then there exists a divisor $d$ of $k$, $d < k$, such that

$\chi(a) = 1$    whenever     $gcd(a, k) = 1$    and    $a \equiv 1$ (mod $d$)              (3)

Proof: For the given $n$, let $q = gcd(n, k)$ and let $d = k/q$. Then $d|k$ and since $q > 1$ we have $d < k$. Choose any $a$ satisfying $gcd(a, k) = 1$ and $a \equiv 1$ (mod $d$). Then we must have $\chi(a) = 1$ by the following argument.
Since $gcd(a, k) = 1$, in the sum defining $G(n, \chi)$ we can replace the index of summation $m$ by $am$ and we get

$G(n, \chi) = \sum_{m\hspace{1 mm} mod \hspace{1 mm}k}\chi(m)e^{2\pi inm/k}$

$= \sum_{m\hspace{1 mm} mod \hspace{1 mm}k}\chi(am)e^{2\pi inam/k}$

$= \chi(a) \sum_{m\hspace{1 mm} mod \hspace{1 mm}k}\chi(m)e^{2\pi inam/k}$

Since $a \equiv 1$ (mod $d$) and $d = k/q$ we can write

$a = 1 + (bk/q)$

for some integer $b$, and we then have

$\frac{anm}{k} = \frac{nm}{k} + \frac{bknm}{qk} = \frac{nm}{k} + \frac{bnm}{q} \equiv \frac{nm}{k}$ (mod 1)

since $q|n$. To say that $anm/k \equiv nm/k$ (mod 1) means simply that $anm/k$ and $nm/k$ differ by an integer, and hence we have

$e^{2\pi ianm/k} = e^{2\pi i(nm/k + v)} = e^{2\pi inm/k}$

so the sum for $G(n, \chi)$ becomes

$G(n, \chi) = \chi(a) \sum_{m\hspace{1 mm} mod \hspace{1 mm}k}\chi(m)e^{2\pi inm/k} = \chi(a)G(n, \chi)$

Since $G(n, \chi) \neq 0$, this implies $\chi(a) = 1$ as claimed, so the theorem is proved. QED

Example: The following concrete example illustrates several of the ideas above. Suppose $\chi$ is the nonprincipal character mod 6. The reduced residue classes mod 6 are $\{1, 5\}$ so there are two Dirichlet characters mod 6. One is the principal character taking the value 1 for both 1 and 5. The other is the nonprincipal character $\chi$ which must have the square roots of unity as its values, so $\chi(1) = 1$, $\chi(5) = -1$. We will show that $G(2, \chi)$ and $G(4, \chi)$ are not separable, and $|G(1, \chi)|^2 \neq 6$. We have

$G(2, \chi) = \chi(1)e^{2\pi i2/6} + \chi(5)e^{2\pi i5 \cdot 2/6}$

$= e^{2\pi i/3} - e^{4\pi i/3} = \sqrt{3}i \neq 0$

and

$G(4, \chi) = \chi(1)e^{2\pi i4/6} + \chi(5)e^{2\pi i5 \cdot 4/6}$

$= e^{4\pi i/3} - e^{2\pi i/3} = -\sqrt{3}i \neq 0$

Therefore we have $G(2, \chi) \neq 0$ even though $gcd(2, 6) = 2 > 1$, and $G(4, \chi) \neq 0$ even though $gcd(4, 6) = 2 > 1$. It follows from Theorem 2 that neither $G(2, \chi)$ nor $G(4, \chi)$ can be separable. Furthermore, we have

$G(1, \chi) = \chi(1)e^{2\pi i/6} + \chi(5)e^{2\pi i5/6}$

$= e^{\pi i/3} - e^{5\pi i/3} = \sqrt{3}i$

and so

$|G(1, \chi)|^2 = 3 \neq 6$

To illustrate Theorem 4, observe that the proper divisors of $k = 6$ are $1$, $2$, $3$. Now consider any $a$ satisfying $gcd(a, 6) = 1$ and $a \equiv 1$ (mod 3). Then we must necessarily have $a \equiv 1$ (mod 6), so $\chi(a) = \chi(1) = 1$. Thus, $d = 3$ in the context of Theorem 4.
But note that we cannot have $d = 2$. A counterexample is the case $a = 5$. We have $5 \equiv 1$ (mod 2) and $gcd(5, 6) = 1$, but $\chi(5) = -1 \neq 1$.
Similarly, we cannot have $d = 1$. A counterexample is again the case $a = 5$. We have $5 \equiv 1$ (mod 1) and $gcd(5, 6) = 1$, but $\chi(5) = -1 \neq 1$.
Thus we cannot have either $d = 1$ or $d = 2$. We can only have $d = 3$ in the context of Theorem 4 in this case. END