FACTORING POLYNOMIALS OVER FINITE FIELDS

Example 9.31. Show that φ(x) = xp−1 + xp−2 + · · · + x + 1 is irreducible over Q for any prime p. This is called a cyclotomic polynomial and can be written

φ(x) = (xp − 1)/(x − 1).

Solution. We cannot apply Eisenstein’s criterion to φ(x) as it stands. However,

FACTORING POLYNOMIALS OVER FINITE FIELDS

The roots of a polynomial in Zp [x] can be found by trying all the p possible values.

Example 9.32. Does x4 + 4 Z7[x] have any roots in Z7?

Solution. We see from Table 9.3 that x4 + 4 is never zero and therefore has no roots in Z7.

Proposition 9.33. A polynomial in Z2[x] has a factor x + 1 if and only if it has an even number of nonzero coefficients.

Proof. Let p(x) = a0 + a1x + · · · + anxn Z2[x]. By the factor theorem, (x + 1) is a factor of p(x) if and only if p(1) = 0. (Remember that x − 1 = x + 1

TABLE 9.3. Values Modulo 7

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in Z2[x].) Now p(1) = a0 + a1 + · · · + an, which is zero in Z2 if and only if p(x) has an even number of nonzero coefficients.

Example 9.34. Find all the irreducible polynomials of degree less than or equal to 4 over Z2.

Solution. Degree 1 polynomials are irreducible; in Z2[x] we have x and x + 1.

Let p(x) = a0 + · · · + anxn Z2[x]. If p(x) has degree n, then an is nonzero, so an = 1. The only possible roots are 0 and 1. The element 0 is a root if and only if a0 = 0, and 1 is a root if and only if p(x) has an even number of nonzero terms. Hence the following are the polynomials of degrees 2, 3, and 4 in Z2[x] with no linear factors:

If a polynomial of degree 2 or 3 is reducible, it must have a linear factor; hence the polynomials of degree 2 and 3 above are irreducible. If a polynomial of degree 4 is reducible, it either has a linear factor or is the product of two irreducible quadratic factors. Now there is only one irreducible quadratic in Z2[x], and its square (x2 + x + 1)2 = x4 + x2 + 1 is reducible.

LINEAR CONGRUENCES AND THE CHINESE

REMAINDER THEOREM

The euclidean algorithm for integers can be used to solve linear congruences. We first find the conditions for a single congruence to have a solution and then show how to find all its solutions, if they exist. We then present the Chinese remainder theorem, which gives conditions under which many simultaneous congruences, with coprime moduli, have solutions. These solutions can again be found by using the euclidean algorithm.

First let us consider a linear congruence of the form

ax ≡ b mod n.

This has a solution if and only if the equation ax + ny = b

has integer solutions for x and y. The congruence is also equivalent to the equation [a][x] = [b] in Zn.

Theorem 9.35. The equation ax + ny = b has solutions for x, y Z if and only if gcd(a, n)|b.

Proof. Write d = gcd(a, n). If ax + ny = b has a solution, then d|b because d|a and d|n. Conversely, let d|b, say b = k · d. By Theorem 9.9, there exist s, t Z such that as + nt = d. Hence ask + ntk = k · d and x = sk, y = tk is a solution to ax + ny = b.

The euclidean algorithm gives a practical way to find the integers s and t in Theorem 9.35. These can then be used to find one solution to the equation.

Theorem 9.36. The congruence ax ≡ b mod n has a solution if and only if d|b, where d = gcd(a, n). Moreover, if this congruence does have at least one solution, the number of noncongruent solutions modulo n is d; that is, if [a][x] = [b] has a solution in Zn, then it has d different solutions in Zn.

Proof. The condition for the existence of a solution follows immediately from Theorem 9.35. Now suppose that x0 is a solution, so that ax0 ≡ b mod n. Let d = gcd(a, n) and a = da , n = dn . Then gcd(a , n ) = 1, so the following statements are all equivalent.

(i)x is a solution to the congruence ax ≡ b mod n.

(ii)x is a solution to the congruence a(x − x0) ≡ 0 mod n.

(iii)n|a(x − x0).

(iv)n |a (x − x0).

Reducing modulo 17, we have 7t ≡ 3 mod 17. Solving this by the euclidean algorithm, we have

We have shown that if x = 36 + 41t is a solution to both congruences, then

t= 15 + 17u, where u Z. That is,

x = 36 + 41t = 36 + 41(15 + 17u) = 651 + 697u

Example 9.40. Find the smallest positive integer that has remainders 4, 3, and 1 when divided by 5, 7, and 9, respectively.

Solution. We have to solve the three simultaneous congruences

x ≡ 4 mod 5, x ≡ 3 mod 7, and x ≡ 1 mod 9.

The first congruence implies that x = 4 + 5t, where t Z. Substituting into the second congruence, we have

4 + 5t ≡ 3 mod 7.

Hence 5t ≡ −1 mod 7. Now 5−1 = 3 in Z7, so t ≡ 3 · (−1) ≡ 4 mod 7. Therefore, t = 4 + 7u, where u Z, and any integer satisfying the first two congruences is of the form

x = 4 + 5t = 4 + 5(4 + 7u) = 24 + 35u.

Substituting this into the third congruence, we have 24 + 35u ≡ 1 mod 9 and −u ≡ −23 mod 9. Thus u ≡ 5 mod 9 and u = 5 + 9v for some v Z.

The Chinese remainder theorem was known to ancient Chinese astronomers, who used it to date events from observations of various periodic astronomical phenomena. It is used in this computer age as a tool for finding integer solutions to integer equations and for speeding up arithmetic operations in a computer.

Addition of two numbers in conventional representation has to be carried out sequentially on the digits in each position; the digits in the ith position have to

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be added before the digit to be carried over to the (i + 1)st position is known. One method of speeding up addition on a computer is to perform addition using residue representation, since this avoids delays due to carry digits.

Let m = m1m2 · · · mr , where the integers mi are coprime in pairs. The residue representation or modular representation of any number x in Zm is the r-tuple

(a1, a2, . . . , ar ), where x ≡ ai mod mi .

For example, every integer from 0 to 29 can be uniquely represented by its residues modulo 2, 3, and 5 in Table 9.4.

This residue representation corresponds exactly to the isomorphism

Z30 → Z2 × Z3 × Z5.

Since this is a ring isomorphism, addition and multiplication are performed simply by adding and multiplying each residue separately.

For example, to add 4 and 7 using residue representation, we have

(0, 1, 4) + (1, 1, 2) = (0 + 1, 1 + 1, 4 + 2) = (1, 2, 1).

Similarly, multiplying 4 and 7, we have

(0, 1, 4) · (1, 1, 2) = (0 · 1, 1 · 1, 4 · 2) = (0, 1, 3).

Fast adders can be designed using residue representation, because all the residues can be added simultaneously. Numbers can be converted easily into residue form; however, the reverse procedure of finding a number with a given residue representation requires the Chinese remainder theorem. See Knuth [19, Sec. 4.3.2] for further discussion of the use of residue representations in computers.

TABLE 9.4. Residue Representation of the Integers from 0 to 29