CYBERNETICS AND CONTROL THEORY

R. R. VARSHAMOV

ON A THEOREM FROM THE THEORY OF REDUCIBILITY OF POLYNOMIALS

(Presented by Academician V. S. Kulebakin, 5 X 1963)

The theory of reducibility of polynomials modulo a prime, which is of considerable interest in itself, occupies a special place in the modern theory of linear coding \((^{1})\). One of its important and at the same time most difficult problems is the problem of synthesizing irreducible polynomials of a given degree. The present note is devoted to this direction. All polynomials considered below are assumed to have coefficients in the Galois field \(GF(q)\). Denote by \(L^p f(x)\) the expression

\[ F(x)=p(x)^{-1}\sum_{v=0}^{n}\sum_{u=0}^{m} a_v b_u x^{u q^v}, \tag{1} \]

where

\[ f(x)=\sum_{v=0}^{n} a_v x^v,\quad p(x)=\sum_{u=0}^{m} b_u x^u \quad (p(x)\ne \mathrm{const}). \]

Lemma. The polynomial \(L^p r(x)\) divides \(L^p f(x)\) without remainder, where \(r(x)\) is any divisor of \(f(x)\).*

Proof. Substituting into (1) \(f(x)=r(x)h(x)\), where

\[ r(x)=\sum_{v=0}^{R} r_v x^v \quad \text{and} \quad h(x)=\sum_{j=0}^{H} h_j x^j, \]

we obtain

\[ L^p f(x)=p(x)^{-1}\sum_{j=0}^{H}\sum_{v=0}^{R}\sum_{u=0}^{m} h_j r_v b_u x^{u q^{v+j}}= \]

\[ =\sum_{j=1}^{H} h_j p(x)^{-1}\left(\sum_{u=0}^{m} b_u x^u\right)^{q^j} \left(L^p r(x)\right)^{q^j} =\sum_{j=0}^{H} h_j\left(L^p x^j\right)\left(L^p r(x)\right)^{q^j}, \tag{2} \]

i.e. \(L^p r(x)\mid L^p f(x)\). The lemma is proved.

Main theorem. Let

\[ f(x)=\sum_{v=0}^{n} a_v x^v \quad (a_0\ne 0); \]

\(s_1(x),\ldots,s_k(x)\) be the set of all proper divisors of \(f(x)\); \(K_f[L^p f(x)]\) be the least common multiple of \(L^p s_1(x),\ldots,L^p s_k(x)\); \(N\) be the exponent of the polynomial \(f(x)\),** and \(g(x)\ne \mathrm{const}\) an irreducible factor of (1) not dividing \(K_f[L^p f(x)]\).

Then \(c\), the degree of the polynomial \(g(x)\), is a multiple of \(N\), i.e. \((c,N)=N\).

Proof. Suppose the contrary, i.e. \((c,N)\ne N\) \((c\ge 1)\). Then, obviously, the expression \(s(x)=(x^c-1,f(x))\) is a proper divisor of \(f(x)\), and, by the lemma,

\[ L^p s(x)=\Lambda_1(x)L^p(x^c-1)+\Lambda_2(x)F(x), \tag{3} \]

since

\[ s(x)=\lambda_1(x)(x^c-1)+\lambda_2(x)f(x). \]

By definition,

\[ p(x)L^p(x^c-1)=\sum_{u=0}^{m} b_u\left(x^{u q^c}-x^u\right), \]

which shows that

* In fact, a stronger assertion holds:

\[ \left(L^p f_1(x),\, L^p f_2(x)\right)=L^p\left(f_1(x),\, f_2(x)\right). \]

** That is, the least natural number satisfying \(x^N\equiv 1 \pmod{f(x)}\).

\(x^{c-1}-1/p(x)L^p(x^c-1)L^p(x^c-1)\) and, simultaneously, \(g(x)/p(x)L^p(x_c-1)\), since the degree of the irreducible polynomial \(g(x)\) is equal to \(c\). Let us note, however, that \((g(x),p(x))=\text{const}\) in view of \((p(x)L^p f(x))'=a_0p'(x)\) \((a_0\ne 0)\). Therefore it is clear that \(g(x)/L^p(x^c-1)\), and also, by (3), \(g(x)/L^p s(x)\). This contradicts the condition of the theorem. Thus our assumption is false; consequently, \((c,N)=N\). The theorem is proved.

The main theorem has an application in the constructive theory of the synthesis of irreducible polynomials of a given degree modulo a prime. We shall set out some separate results obtained with its aid.

Theorem 1. Let

\[ f(x)=\sum_{\nu=0}^{n} a_\nu x^\nu \quad (a_0\ne 0), \qquad p(x)=x. \]

Then \(c\), the degree of the irreducible polynomial \(g(x)\) satisfying the conditions of the main theorem, coincides with the exponent of \(f(x)\), i.e. \(c=N\).

Proof. Since \(f(x)/x^N-1\), it follows, according to the lemma, that \(F(x)/L^x(x^N-1)\) and \(g(x)/x^{q^N-1}-1\), since \(L^x(x^N-1)=x^{q^N-1}-1\). The polynomial \(g(x)\) is irreducible; therefore \(c/N\), and this, together with the main theorem, gives \(c=N\). The theorem is proved.

As a consequence of Theorem 1 we obtain some previously known results of Yore, Carlitz, Gleason, and Marsh (see, for example, \((^2)\)) for the cases when the polynomial \(f(x)\) is irreducible or when \(f(x)\) is a primitive polynomial. Thus, for example, in the field \(GF(3)\) the expression \(L^x(x^2+x+2)=x^8+x^2+2\) is irreducible, since \(x^2+x+2\) is a primitive polynomial. In addition, for \(q=2\)* we have:

Theorem 2. Let

\[ f(x)=\prod_{i=1}^{\sigma} f_i(x), \]

where \(f_i(x)\) \((i=1,\ldots,\sigma)\) are primitive polynomials with pairwise relatively prime degrees \(n_i\), respectively,

\[ |\alpha|=\sum_{i=1}^{\sigma}\alpha_i, \]

where \(\alpha_i\) are integers \((0\le \alpha_i\le 1)\), and

\[ f^{(\alpha)}(x)=\prod_{i=1}^{\sigma} f_i^{\alpha_i}(x). \]

Then the polynomial

\[ F_1(x)=\prod_{j=0}^{[\sigma/2]} \left( \prod_{|\alpha|=\sigma-2j-1} L^x f^{(\alpha)}(x) \right)^{-1} \left( \prod_{|\alpha|=\sigma-2j} L^x f^{(\alpha)}(x) \right) \]

is irreducible.

Proof. It is easy to show that the polynomial \(F_1(x)\), being a divisor of \(L^x f(x)\), is relatively prime to \(K_f[L^x f(x)]\). Moreover, a simple calculation establishes that its degree is

\[ M_{n_1,\ldots,n_\sigma}=\prod_{i=1}^{\sigma}(2^{n_i}-1). \]

As is known, the exponent of the polynomial \(f(x)\) also coincides with \(M_{n_1,\ldots,n_\sigma}\). Hence, according to Theorem 1, it follows that the polynomial \(F_1(x)\) is irreducible. The theorem is proved.

Theorem 2 makes it comparatively easy to find irreducible polynomials whose degrees are representable in the form of a product of pairwise relatively prime Mersenne numbers. Thus, for example, in the case when \(f(x)=(x^2+x+1)(x^3+x^2+1)=x^5+x+1\), we shall have the irreducible polynomial of degree 21

\[ F_1'(x)=\frac{L^x(x^5+x+1)}{L^x(x^2+x+1)L^x(x^3+x^2+1)}= \]

\[ = x^{21}+x^{19}+x^{18}+x^{15}+x^{14}+x^{11}+x^8+x^7+x^5+x^4+1. \]

* The case most interesting in practical applications. In what follows, we have \(q=2\).

In an analogous way, in the case \(f(x)=(x^2+x+1)(x^5+x^2+1)\) we obtain the irreducible polynomial of degree 93

\[ F_1(x)=(x^7+1)^3(x^4+x^2+x+1)(x^{68}+x^{37}+x^{12}+x^9+x^6+x^4)+1. \]

Theorem 3. Let

\[ f(x)=\sum_{v=0}^{n} a_v x^v,\qquad \omega(x)=\sum_{u=0}^{\Omega}\omega_u x^u, \]

\[ \sigma(\omega)=f(1)+xL^x\omega(x),\qquad \Lambda(x)=(x+f(1))\omega(x)f(x),\qquad \Lambda(0)\ne 0, \]

\[ H_{\omega f}=K_f[L^{\sigma(\omega)}f(x)]K_\omega[L^{\sigma(\omega)}f(x)] \]

and let \(g(x)\ne \mathrm{const}\) be an irreducible divisor of \(L^{\sigma(\omega)}f(x)\) which does not divide \(H_{\omega f}\). Then the degree of the polynomial \(g(x)\) coincides with the exponent of \(\Lambda(x)\).

Proof. We have

\[ \sigma(\omega)L^{\sigma(\omega)}f(x) =\sum_{v=0}^{n}\sum_{u=0}^{\Omega}a_v\omega_u(x^u q)^v+f(1)\sum_{v=0}^{n}a_v= \]

\[ =xL^x\left(\sum_{v=0}^{n}\sum_{u=0}^{\Omega}a_v\omega_u x^{u+v}\right)+f(1)\sum_{v=0}^{n}a_v =xL^x\omega(x)f(x)+f(1). \]

However, according to (2),

\[ xL^x\omega(x)f(x)+f(1)=(L^x(x+f(1))\omega(x)f(x))(L^x\omega(x)f(x))^{-1}, \]

therefore

\[ \sigma(\omega)(L^x\omega(x)f(x))L^{\sigma(\omega)}f(x)=L^x\Lambda(x). \tag{4} \]

It follows from this (since the expression \(L^x\Lambda(x)\) has no multiple roots, in view of \((xL^x\Lambda(x))'=\Lambda(0)\ne 0\)) that

\[ (\sigma(\omega)L^x\omega(x)f(x),\,L^{\sigma(\omega)}f(x))=1. \tag{5} \]

We now show that \(g(x)\nmid K_{\Lambda}[L^x\Lambda(x)]\). Indeed, in the case
\(g(x)/K_{\Lambda}[L^x\Lambda(x)]\) we would have \(g(x)/L^x\Lambda^1(\lambda)\), where \(\Lambda^1(x)\) is a proper divisor of \(\Lambda(x)\), and, by virtue of (5), \(x+f(1)/\Lambda^1(x)\). Let
\[ \Lambda^1(x)=(x+f(1))\omega^1(x)f^1(x)\quad (\omega^1/\omega,\ f^1/f,\ \omega^1 f^1\ne \omega f). \]
According to (4),
\[ \sigma(\omega^1)(L^x\omega^1(x)f^1(x))L^{\sigma(\omega^1)}f^1(x)=L^x\Lambda^1(x), \]
and therefore \(g(x)/\sigma(\omega^1)(L^x\omega^1(x)f^1(x))L^{\sigma(\omega^1)}f^1(x)\). But, by the lemma,
\[ L^x\omega^1(x)f^1(x)/L^x\omega(x)f(x), \]
and in view of the obvious identity
\[ \sigma(\omega)L^x\omega(x)\equiv L^x\omega(x)(x+f(1)), \]
also
\[ \sigma(\omega^1)L^x\omega^1(x)/\sigma(\omega)L^x\omega(x). \]
Moreover, by virtue of (5), \(g(x)\nmid L^x\omega(x)f(x)\) and
\(g(x)\nmid \sigma(\omega)L^x\omega(x)\). Therefore it is clear that
\(g(x)/L^{\sigma(\omega^1)}f^1(x)\), or, what is the same,
\(g(x)/H_{\omega f}\), and this contradicts the condition of the theorem; consequently,
\[ g(x)\nmid K_{\Lambda}[L^x\Lambda(x)]. \]
Thus we are convinced that the polynomial \(g(x)\), being a divisor of \(L^x\Lambda(x)\), is relatively prime to \(K_{\Lambda}[L^x\Lambda(x)]\). Hence, according to Theorem 1, it follows that its degree coincides with the exponent of the polynomial \(\Lambda(x)\). The theorem is proved.

Relying on Theorem 3, one can prove the following fact.

Theorem 4. Let \(\delta\) denote one of the numbers 0 or 1; let \(\omega(x)=x^{1+\delta}+1\), and let \(f_i(x)\) \((i=1,\ldots,\sigma)\) be primitive polynomials with pairwise relatively prime degrees \(n_i\), respectively.

Then the polynomial

\[ F_2(x)=\prod_{j=0}^{\left[\frac{\sigma}{2}\right]} \left( \prod_{|\alpha|=\sigma-2j-1} L^{\sigma(\omega)}f^{(\alpha)}(x) \right)^{-1} \left( \prod_{|\alpha|=\sigma-2j} L^{\sigma(\omega)}f^{(\alpha)}(x) \right) \tag{6} \]

is irreducible.

Proof. We first show that \(F_2(x) \nmid K_\Lambda[L^x\Lambda(x)]\). Indeed, suppose the contrary, i.e. \(F_2(x) \mid K_\Lambda[L^x\Lambda(x)]\). Then, obviously, \(F_2(x) \mid L^x\Lambda^1(x)\), where \(\Lambda^1(x)\) is a proper divisor of \(\Lambda(x)\), and, by virtue of (5), and also according to the condition of the theorem \((\omega(x)=x^{1+\delta}+1)\), \(\Lambda^1(x)=(x+1)\omega(x)f^1(x)\) \((f^1\ne f)\). Hence, by (4), (5), and since \(F_2(x)\mid L^{\sigma(\omega)}f(x)\), it follows that \(F_2(x)\mid L^{\sigma(\omega)}f^1(x)\). This contradicts the condition of the theorem, since from formula (6) it is seen that \(F_2(x)\nmid K_f[L^{\sigma(\omega)}f(x)]\).

Thus, our assumption is false; consequently, \(F_2(x)\nmid K_\Lambda[L^x\Lambda(x)]\). We now note that the degree of the polynomial \(F_2(x)\) \((=2^{1+\delta}M_{n_1,\ldots,n_\sigma})\) coincides with the exponent of the polynomial \(\Lambda(x)\). Therefore, by Theorem 3, the polynomial \(F_2(x)\) is irreducible. The theorem is proved.

Theorem 4 extends the possibilities for constructing irreducible polynomials. With its help, for example, one can find irreducible polynomials whose degrees are represented in the form \(2^{1+\delta}M_{n_1,\ldots,n_\sigma}\). Thus, for example, in the case \(\delta=0\) and \(f(x)=x^5+x+1\), we obtain the irreducible polynomial of degree 42

\[ F_2(x)= \frac{L^{\sigma(x+1)}(x^5+x+1)} {L^{\sigma(x+1)}(x^2+x+1)\,L^{\sigma(x+1)}(x^3+x^2+1)} = \]

\[ = x^{42}+x^{41}+x^{40}+x^{35}+x^{32}+x^{30}+x^{29}+x^{28}+x^{27}+ \]

\[ + x^{26}+x^{25}+x^{23}+x^{18}+x^{15}+x^{14}+x^{13}+x^8+x^6+1. \]

Similarly, for \(\delta=1\) and \(f(x)=x^3+x+1\), we shall have the irreducible polynomial of degree 28

\[ F_2(x)=L^{\sigma(x^2+1)}(x^3+x+1)= \]

\[ =(x^{17}+x^2)(x^{11}+x^8+x^7+x^5+x^3+x^2+x+1)+1. \]

Institute of Automation
and Telemechanics

Received
4 X 1963

CITED LITERATURE

1. W. W. Peterson, Error-Correcting Codes, N. Y.—London, 1961.
2. N. Zierler, J. Soc. Ind. Appl. Math., 7, No. 1, 31 (1959).