V. V. GLAZKOV

ON A CLASS OF FINITE HOMOMORPHISMS

(Presented by Academician I. M. Vinogradov on 7 IV 1964)

Let \(h(n)\) be a finite homomorphism of the natural sequence, i.e. a completely multiplicative function of a natural argument whose values form a finite set of complex numbers, at least one of which is different from zero; let \(m\) be the number of nonzero values of the function \(h(n)\);

\[ S(x)=\sum_{n\le x} h(n) \]

is the summatory function of the finite homomorphism \(h(n)\); \(p, q\), with or without subscripts, are prime numbers; \(\chi_0(n,k)\) is the principal Dirichlet character of modulus \(k\); \(\varphi(n)\) is Euler’s function; \(\zeta(s)\) is the Riemann zeta-function; \(s=\sigma+it\).

A finite homomorphism will be called a generalized character if its summatory function satisfies the condition \(S(x)=\alpha x+O(1)\) with some constant \(\alpha\), generally speaking complex; a generalized character is called principal if \(\alpha\ne 0\), and nonprincipal if \(\alpha=0\). The so-called problem of characters, not solved up to now, is as follows: do there exist (apart from one trivial example) generalized characters distinct from Dirichlet characters? (see \((^{1,2})\)). In the case of principal characters the answer to this question is given by the following theorem.

Theorem. Every principal generalized character is a principal Dirichlet character.

Basic lemma. If \(h(n)\) is not a Dirichlet character and the series

\[ \sum_{h(p)\ne 1}\frac1p \]

converges, then \(h(n)\) cannot be a principal generalized character.

Lemma 1. The product of a principal generalized character by a principal Dirichlet character \(\chi_0(n,k)\) is again a principal generalized character.

The proof is obtained without difficulty by induction on the number of primes entering into \(k\) (cf. \((^1)\)).

Lemma 2. Let \(h(n)\) not be a Dirichlet character and let \(h(p)\ne 1\) for only a finite set of primes \(p\). Then \(h(n)\) cannot be a generalized character.

Proof. Let \(h(p)\ne 1\) for the primes \(p_1,p_2,\ldots,p_r\). Since \(h(n)\) is not a Dirichlet character, for some \(i\le r\) we have \(h(p_i)\ne 1,0\). We shall assume that \(h(p_1)=\lambda\ne 1,0\). Suppose now that \(S(x)=\alpha x+O(1)\). Putting \(h_1(n)=h(n)\chi_0(n,p_2,p_3\ldots p_r)\), we shall have, by Lemma 1,

\[ S_1(x)=\sum_{n\le x} h_1(n)=\alpha_1 x+O(1) \tag{1} \]

with some constant \(\alpha_1\). Choose a prime \(p\ne p_i\) \((i=1,2,\ldots,r)\) and consider the functions

\[ f(n)= \begin{cases} \chi_0(n,p_2\ldots p_r), & \text{if } p\nmid n,\\ (1-p)\chi_0\!\left(\dfrac{n}{p},p_2\ldots p_r\right), & \text{if } p\mid n; \end{cases} \]

\[ \widetilde f(n)= \begin{cases} f(n), & \text{if } p_1\nmid n,\\ \lambda^\beta f\!\left(\dfrac{n}{p_1^\beta}\right), & \text{if } p_1^\beta\parallel n. \end{cases} \]

Applying to the pair of functions \(f(n)\), \(\widetilde f(n)\) literally the same arguments as in the proof of Theorem 1 of paper \((^3)\), we obtain the unboundedness of the summatory function \(\sum_{n\le x}\widetilde f(n)\). On the other hand, evidently,

\[ \widetilde f(n)=h_1(n),\quad \text{if } p\nmid n,\qquad \widetilde f(np)=(1-p)h_1(n). \]

Hence

\[ \sum_{n\le x}\widetilde f(n) = \sum_{\substack{n\le x\\ p\nmid n}}\widetilde f(n) + \sum_{n\le x/p}\widetilde f(np) = \sum_{n\le x} h_1(n) - \sum_{n\le x/p} h_1(np) + \sum_{n\le x/p}(1-p)h_1(n) = S_1(x)-S_1\!\left(\frac{x}{p}\right) + (1-p)S_1\!\left(\frac{x}{p}\right) =O(1) \]

by virtue of relation (1). The contradiction obtained proves the lemma.

Lemma 3. The principal generalized character is different from zero at all primes, with the possible exception of only finitely many of them.

Proof. Suppose \(h(p_i)=0\) for an infinite sequence of primes \(p_1,p_2,\ldots\), and at the same time \(S(x)=ax+O(1)\), where \(\alpha\ne 0\). Then find a natural \(y\) from the system of congruences \(y+i\equiv 0\pmod {p_i}\), \(i=1,2,\ldots,t\), where \(t\) is any natural number. Evidently, \(h(y+i)=0\), and therefore \(S(y+t)=S(y)\), whence \(\alpha t=O(1)\), which is impossible for \(\alpha\ne 0\) and arbitrary \(t\).

Proof of the main lemma will be carried out by contradiction. Suppose \(h(p_i)\ne 1\) on a set of primes \(p_1,p_2,\ldots\) such that the series \(\sum_i 1/p_i\) converges, and \(S(x)=\alpha x+O(1)\), where \(\alpha\ne 0\). Obviously, using Lemmas 2, 3, and 1, we see that one may assume the following additional conditions to be satisfied: 1) the sequence \(p_1,p_2,\ldots\) is infinite; 2) \(h(p_i)=0\) only for a finite number of primes \(p_i\); denote their product by \(k\); 3) for each of the \(m\) nonzero values of \(h(n)\), the set of primes for which \(h(p)\) is equal to this value is either empty or infinite. Assuming all the listed conditions to be satisfied, put

\[ m=q_1^{\delta_1}q_2^{\delta_2}\cdots q_u^{\delta_u} \]

—the canonical factorization of the number \(m\); let \(\xi_1,\xi_2,\ldots,\xi_u\) be fixed primitive roots of unity of orders \(q_1^{\delta_1}, q_2^{\delta_2}, \ldots, q_u^{\delta_u}\), respectively. It is clear that any nonzero value \(h(n)\) is represented in a unique way in the form

\[ h(n)=\prod_{i=1}^{u}\xi_i^{\varepsilon_i(n)}, \]

where \(0\le \varepsilon_i(n)<q_i^{\delta_i}\). We shall suppose the primitive roots \(\xi_1,\ldots,\xi_u\) chosen so that from the sequence \(p_1,p_2,\ldots\) one can extract \(u\) disjoint infinite subsequences \(p_{i1},p_{i2},\ldots\) \((i=1,2,\ldots,u)\) with the condition: \(\varepsilon_i(p_{ij})=1\) for \(i=1,2,\ldots,u\) and all \(j\). Such a choice is always possible.

Next, let \(\eta=\xi_i^{\varepsilon}\), where \(\varepsilon\) is equal to 0 or 1 and is chosen so that \(\alpha\ne \eta\varphi(k)/k\); \(a\) is any natural number exceeding \(2m/k\). Consider all natural

\[ n\le ak. \]

If \((n,k)=1\), then

\[ h(n)=\prod_{i=1}^{u}\xi_i^{\varepsilon_i(n)}, \]

whereas if \((n,k)>1\), then \(h(n)=0\). Choose a natural \(\nu\) so large that between \(ak\) and \(p_\nu\), in each of the \(u\) subsequences \(p_{i1},p_{i2},\ldots\), there are no fewer than \(a\varphi(k)\) primes and

\[ \sum_{i=\nu+1}^{\infty}\frac{1}{p_i}<\frac{1}{2ak}. \tag{2} \]

Take from each subsequence \(a\varphi(k)\) primes lying between \(ak\) and \(p_\nu\), and put them in one-to-one correspondence with the natural numbers \(n\le ak\),

coprime to \(k\); the number from the \(i\)-th subsequence \(p_{i1}, p_{i2}, \ldots\) corresponding to the given \(n\) will be denoted by \(\bar p_{ni}\). As already stated, \(ak < \bar p_{ni} < p\), \(\varepsilon_i(\bar p_{ni}) = 1\).

Consider the obviously solvable system of \(a\varphi(k)+1\) congruences

\[ x \equiv 0\left(\bmod \prod_{j=1}^{\nu}{}' p_j^{ak}\right), \tag{A} \]

\[ x+n \equiv \prod_{i=1}^{u} \bar p_{ni}^{\,mq_i^{-\delta_i}\beta_i(n)} \left(\bmod \prod_{i=1}^{u}\bar p_{ni}^{ak}\right),\quad n\le ak,\quad (n,k)=1, \]

where \(\prod'\) denotes the product over primes distinct from all \(\bar p_{ni}\), and \(q_i^{\beta_i(n)}\) is the least natural number satisfying the congruence

\[ \frac{m}{q_i^{\delta_i}}x+\varepsilon_i(n)\equiv \omega_i\pmod{q_i^{\delta_i}};\quad \omega_1=\varepsilon,\quad \omega_2=\omega_3=\cdots=\omega_u=0. \]

If \(b\) is the least natural number satisfying the system (A), then it is satisfied by all numbers of the form \(b+Qy\), where \(Q=\prod_{j=1}^{\nu}p_j^{ak}\).

It is now not hard to observe that the numbers

\[ b+1+Qy,\ b+2+Qy,\ldots,\ b+ak+Qy \tag{3} \]

for any natural \(y\) have the form

\[ b+n+Qy=p_1^{\gamma_{1n}}p_2^{\gamma_{2n}}\cdots p_\nu^{\gamma_{\nu n}}\cdot b_n(y)=a_n b_n(y), \tag{4} \]

where \((b_n(y),Q)=1\), \(a_n\) does not depend on \(y\), and \(h(a_n)=\eta\chi_0(n,k)\).

We shall show that there exists a natural \(y\) for which all the numbers (3) are coprime to \(p_{\nu+1}, p_{\nu+2}, \ldots\). Let \(N(z)\) denote the number of natural \(y\), not exceeding \(z\), for which at least one of the numbers (3) is divisible by one of \(p_{\nu+1}, p_{\nu+2}, \ldots\). Since \(p_{\nu+i}>ak\), among the numbers (3) no more than one can be divisible by a given \(p_{\nu+i}\). Hence, when \(y\) runs through a complete system of residues modulo \(p_{\nu+i}\), among the sets of the form (3) there will be exactly \(ak\) such sets in which one of the numbers is divisible by \(p_{\nu+i}\). It follows that, for \(z\ge 2\), we obtain:

\[ N(z)\le ak\sum_{p_\nu<p_i\le 2Qz}\left(\left[\frac{z}{p_i}\right]+1\right) <akz\sum_{i=\nu+1}^{\infty}\frac1{p_i}+ak\pi(2Qz). \]

The first term on the right-hand side is less than \(z/2\) by virtue of inequality (2), and the second will be less than \(z/3\) if we choose \(z=e^{12akQ}\). Hence we obtain \(N(z)<\frac56 z\le z-2\). Consequently, for some \(y\le e^{12akQ}\) all the numbers (3), and therefore all the numbers \(b_n(y)\) in the representation (4), are coprime to \(p_{\nu+1}, p_{\nu+2}, \ldots\). But \((b_n(y),Q)=1\), therefore \((b_n(y),p_i)=1\) also for \(i=1,2,\ldots\), whence \(h(b_n(y))=1\), i.e. \(h(b+n+Qy)=\eta\chi_0(n,k)\) for \(n=1,2,\ldots,ak\). It follows that for the summatory function we obtain

\[ S(b+ak+Qy)-S(b+Qy)=\sum_{n=1}^{ak}h(b+n+Qy)=\eta a\varphi(k). \]

On the other hand, this difference is \(aak+O(1)\), since we assumed that \(S(x)=ax+O(1)\). Hence \(\eta a\varphi(k)=aak+O(1)\). But this is impossible by the choice of \(\eta\) and the arbitrariness of \(a\). The lemma is proved.

Proof of the theorem. Consider the function defined, for \(\sigma>1\), by the Dirichlet \(L\)-series for \(h(n)\), and apply Abel’s formula to this series. We obtain

\[ L(s)=\sum_{n=1}^{\infty}\frac{h(n)}{n^s} = -s\int_{1}^{\infty}\frac{S(x)\,dx}{x^{s+1}} = \alpha\,\frac{s}{s-1}+O\!\left(\frac{|s|}{\sigma}\right), \tag{5} \]

since \(S(x)=\alpha x+O(1)\). From equality (5) we conclude that \(L(s)\) is regular in the entire half-plane \(\sigma>0\), except for a pole of the first order at the point \(s=1\). Hence the function \(g(s)=\zeta(s)/L(s)\) is regular in some neighborhood of the point \(s=1\). For \(\sigma>1\), expansion into an Euler product gives

\[ g(s)=\prod_p\left(1-\frac{h(p)}{p^s}\right) \left(1-\frac{1}{p^s}\right)^{-1} = \prod_{h(p)\ne 1} \left(1-\frac{h(p)}{p^s}\right) \left(1-\frac{1}{p^s}\right) = \]

\[ = \prod_{h(p)\ne 1} \left(1+\frac{1-h(p)}{p^s-1}\right). \]

Let us note that, for \(\sigma>1\) and \(h(p)\ne 1\),

\[ \left|1+\frac{1-h(p)}{p^\sigma-1}\right|^2 \ge 1+\frac{1-\operatorname{Re} h(p)}{p^\sigma} \ge 1+\frac{c}{p^\sigma}, \]

where the constant \(c>0\) depends only on \(m\). Therefore, for \(\sigma>1\),

\[ |g(\sigma)|^2 = \prod_{h(p)\ne 1} \left|1+\frac{1-h(p)}{p^\sigma-1}\right|^2 \ge \prod_{h(p)\ne 1} \left(1+\frac{c}{p^\sigma}\right) > c\sum_{h(p)\ne 1}\frac{1}{p^\sigma}. \]

If the last series diverged at \(\sigma=1\), then its sum would tend to infinity as \(\sigma\to 1+0\). But then, as \(\sigma\to 1+0\), \(|g(\sigma)|\to\infty\), which contradicts the continuity of \(g(s)\) at the point \(s=1\). Hence the series

\[ \sum_{h(p)\ne 1}\frac{1}{p} \]

converges, and our principal generalized character \(h(n)\) is a Dirichlet character by virtue of the main lemma. The theorem is proved.

Remark 1. The condition of generality of the character, i.e. \(S(x)=\alpha x+O(1)\), cannot be weakened. It can be shown that, for any function \(F(x)\) increasing arbitrarily slowly to infinity, there exists a finite homomorphism, distinct from a Dirichlet character, whose summatory function is \(\alpha x+\Omega(F(x))\) with \(\alpha\ne 0\).

Remark 2. The main lemma remains valid also in the case of nonprincipal characters in the following form: if \(h(n)\) is not a Dirichlet character, then for some nonprincipal Dirichlet character \(\chi(n)\) the series

\[ \sum_{(hp)\ne \chi(p)}\frac{1}{p} \]

converges, whereas the summatory function of the finite homomorphism \(h(n)\) is unbounded. The corresponding \(\Omega\)-estimates can be obtained. This is much stronger than the previous results on “spoiled” characters \((^3,^4)\).

Saratov State University
named after N. G. Chernyshevsky

Received
2 IV 1964

REFERENCES

1. N. G. Chudakov, K. A. Rodosskii, DAN, 73, No. 6, 1137 (1950).
2. N. G. Chudakov, UMN, 8, issue 3, 149 (1953).
3. B. S. Bronshtein, Scientific Notes of Moscow Univ., issue 165, mathematics, 7, 212 (1954).
4. N. G. Chudakov, B. M. Bredikhin, Ukrainian Mathematical Journal, 8, No. 4, 347 (1956).