Section 23 – Factorization of polynomialsover a field

Instructor: Yifan Yang

Spring 2007

Notation

Throughout this section, the letter F will always denote a field.

Division algorithm for F [x ]

Theorem (23.1, Division algorithm for F [x ])Let F be a field. Let f (x) and g(x) be polynomials in F [x ].Suppose that g(x) is not the zero polynomial. Then there existunique polynomials q(x) and r(x) such that

1. f (x) = g(x)q(x) + r(x),

2. r(x) = 0 or deg r(x) < deg g(x).

Proof.• Assume that g(x) = bmxm + · · ·+ b0, where m ≥ 0 and

bm 6= 0. Consider the setS = {f (x)− g(x)s(x) : s(x) ∈ F [x ]}.

• If 0 ∈ S, then ∃ s(x) ∈ F [x ] such that f (x) = g(x)s(x).Then we can take q(x) = s(x), r(x) = 0, and we are done.

Proof of Theorem 23.1, continued

• Assume that 0 6∈ S. Let r(x) be an element of minimaldegree in S. We have f (x) = g(x)q(x) + r(x) for someq(x) ∈ F [x ]. We need to show that deg r(x) < deg g(x).

• Suppose that r(x) = ckxk + · · ·+ c0 with k ≥ m andck 6= 0. Consider f (x)− g(x)(q(x) + ck/bmxk−m).

• We have

f (x)−g(x)q(x)− (ck/bm)xk−mg(x) = r(x)− (ck/bm)xk−mg(x)

= (ckxk + · · ·+ c0)− (ck/bm)(bmxk + · · · ),

whose degree is less than r(x). This contradicts to theassumption that r(x) is of minimal degree in S. Thus,deg r(x) must be less than deg g(x).

Proof of Theorem 23.1, continued

We now show that q(x) and r(x) are unique.

• Suppose that

f (x) = g(x)q1(x) + r1(x)

f (x) = g(x)q2(x) + r2(x),

where ri(x) either are the zero polynomial or satisfydeg ri(x) < deg g(x).

• Then we have r1(x)− r2(x) = g(x)(q2(x)− q1(x)).

• Now r1(x)− r2(x) is either zero or a polynomial of degree< deg g(x). However, if the right-hand side is not zero,then the degree is at least deg g(x).

• Thus, the only possibility is that r1(x) = r2(x), andq1(x) = q2(x). This completes the proof. �

Example

Example. Let F = Z5, f (x) = x4 + x2 + 3x + 2, andg(x) = x2 + 2x + 3. Let us find the polynomials q(x) and r(x).

Solution.x2 + 3x + 2

x2 + 2x + 3 x4 +x2 +3x +1

x4 + 2x3 + 3x2

3x3 + 3x2 + 3x3x3 + x2 + 4x

2x2 + 4x + 12x2 + 4x + 1

0

Thus, we find q(x) = x2 + 3x + 2 and r(x) = 0.

Remark

Note that the condition that F is a field is crucial. For example,assume that F = Z instead. Let f (x) = x + 1 and g(x) = 2x .Then it is impossible to find q(x), r(x) ∈ Z[x ] such thatf (x) = g(x)q(x) + r(x) with r(x) = 0 or deg r(x) < deg g(x).

Zeros of f (x) ∈ F [x ]

Corollary (23.3)An element a ∈ F is a zero of f (x) ∈ F [x ] if and only if x − a isa factor of f (x) in F [x ].

Proof.• Suppose that f (a) = 0. By Theorem 23.1, we have

f (x) = (x − a)g(x) + r(x) for some q(x) ∈ F [x ], wherer(x) = 0 or deg r(x) < deg(x − a) = 1, i.e., or r(x) = c is aconstant polynomial.

• Then we have 0 = f (a) = 0g(a) + c. That is, r(x) is thezero polynomial. This proves that f (a) = 0 ⇒ (x − a)|f (x).

• The proof of the converse statement is easy. �

Zeros of f (x) ∈ F [x ]

Corollary (23.5)A nonzero polynomial f (x) ∈ F [x ] of degree n can have at mostn zeros in F .

Proof.• We will prove by induction on the degree of f (x).

• When f (x) has degree 0, i.e., when f (x) = c for somec 6= 0 ∈ F , it is clear that f (x) has no zeros.

• Now assume that the statement holds for all polynomials ofdegree n ≤ k . We will prove that the statement also holdsfor polynomials f (x) of degree k + 1.

Proof of Corollary 23.5, continued

• If f (x) has no zeros in F , we are done. Otherwise, assumethat a ∈ F is a zero of f (x).

• By Corollary 23.3, f (x) = (x − a)g(x) for some polynomialg(x) of degree k .

• Now if b ∈ F is a zero of f (x), then (b − a)g(b) = 0.

• Since F has no zero divisors, this implies that b − a = 0 org(b) = 0, i.e., b = a or b is a zero of g(x).

• By the induction hypothesis, the polynomial g(x) has atmost k zeros. It follows that f (x) has at most k + 1 zeros.

• By the principle of mathematical induction, we concludethat the statement holds for all polynomials. �

The multiplicative group F×

Corollary (23.6)Let F be a field. If G is a finite subgroup of the multiplicativegroup F× of all nonzero elements in F . Then G is cyclic. Inparticular, if F is a finite field, then F× is cyclic.

Example

1. In Z5 we have 22 = 4, 23 = 3, and 24 = 1. Thus, Z×5 = 〈2〉is cyclic.

2. We have |Z×7 | = 6. Thus, the multiplicative order of anelement in Z×7 is 1, 2, 3, or 6. Now 32 = 2 6= 1 and33 = 6 6= 1. Therefore, the multiplicative order of 3 is 6,and Z×7 = 〈3〉 is cyclic.

Proof of Corollary 23.6

• Recall that by the fundamental theorem for finitelygenerated abelian groups (Theorem 11.12), the finiteabelian group G is isomorphic to Zp

e11× · · · × Zp

ekk

for some

primes pi and integers ei ≥ 1.

• Since the maximal order of an element in Zpe11× · · · × Zp

ekk

is lcm(pe11 , . . . , pek

k ), the group G is cyclic if and only if pi

are all distinct.

• Observe also that the order of every element in G is adivisor of m = lcm(pe1

1 , . . . , pekk ). In other words, every

element of G is a zero of xm − 1.

• If G is not cyclic, then m is strictly less than |G|.• Now the polynomial xm − 1 has degree m, but has at least|G| > m zeros. This contradicts to Corollary 23.5. �

In-class exercises

Find q(x) and r(x) for the following pairs of polynomials in thespecified rings of polynomials.

1. F = Z2, f (x) = x5 + x3 + x + 1, g(x) = x3 + x2 + 1.

2. F = Z3, f (x) = x5 + x3 + x + 1, g(x) = x3 + x2 + 1.

3. F = Z5, f (x) = x5 + x3 + x + 1, g(x) = x3 + x2 + 1.

Find a generator for each of the following groups.

1. Z×13.

2. Z×17.

Irreducible polynomials

DefinitionA nonconstant polynomial f (x) ∈ F [x ] is irreducible over F or isan irreducible polynomial in F [x ] if f (x) cannot be expressed asa product g(x)h(x) of two polynomials in F [x ] with0 < deg g(x), deg h(x) < deg f simultaneously.If a polynomial f (x) ∈ F [x ] is nonconstant and is not irreducibleover F , then it is reducible over F .

Example

1. x2 + 1 is irreducible over R.

2. x2 + 1 is reducible over C since x2 + 1 = (x − i)(x + i) in C.

Remarks

• Observe that if f (x) ∈ F [x ] is irreducible over F , then inf (x) = g(x)h(x) in F [x ], one of g(x) and h(x) has degreeequal to deg f (x), and the other has degree 0. Since thenonzero constant polynomials in F [x ] are precisely theunits in F [x ], the definition of irreducible polynomials canalso be given as “f (x) is irreducible over F if in anyfactorization f (x) = g(x)h(x) in F [x ], one of g(x) and h(x)is a unit”.

• Note that whether a polynomial f (x) is irreducible or notdepends on which field we are talking about. For example,x2 + 1 is irreducible over R, but is reducible over C. Also,x2 − 2 is irreducible over Q, but reducible over R or Q(

√2).

Polynomials of degree 2 or 3

Theorem (23.10)Suppose that f (x) ∈ F [x ] is of degree 2 or 3. Then f (x) isreducible over F if and only if f (x) has a zero in F .

Proof.Assume that f (x) is reducible, say, f (x) = g(x)h(x), where0 < deg g(x), deg h(x) < deg f (x) = 2 or 3. Then one of g(x)and h(x) is of degree 1 taking the form x − a for some a ∈ F .Thus f (x) has a zero a in F .The converse statement is trivial. This proves the theorem.

Remark

RemarkWhen the degree of f (x) is 4, it is possible thatf (x) = g(x)h(x), where each of g and h is of degree 2. Todetermine whether f (x) is irreducible over F , we also need toconsider this possibility.

Examples

Example. Determine whether f (x) = x3 + x + 1 is irreducibleover Z5.

Solution.• Since f (x) = x3 + x + 1 is of degree 3, it is reducible over

Z5 if and only if it has a zero in Z5.

• Now we have f (0) = 1, f (1) = 3, f (2) = 8 + 2 + 1 = 1,f (3) = 27 + 3 + 1 = 1, and f (4) = 64 + 4 + 1 = 4. None ofthese is equal to 0.

• Thus, f (x) is irreducible over Z5.

Examples

Example. Determine whether f (x) = x4 + x2 + x + 1 isirreducible over Z5.Solution.• We have f (0) = 1, f (1) = 4, f (2) = 3, f (3) = 4, and

f (4) = 2. Thus, f (x) has no linear factors.

• Now assume that f (x) = (x2 + ax + b)(x2 + cx + d). Thenwe have x4 + x2 + x + 1 =x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd .

• Comparing the constant terms, we get(b, d) = (1, 1), (2, 3), (3, 2), or (4, 4).

• From a + c = 0, we get a = −c. Then (ad + bc) = 1 givesa(d − b) = 1.

• It follows that (b, d , a) = (2, 3, 1), or (3, 2, 4).

• But ac + b + d = 4 6= 1. Thus, f (x) is irreducible over Z5.

Gauss’s lemma

Theorem (23.11, Gauss’s lemma)Let f (x) ∈ Z[x ]. Then f (x) factors into a product of twopolynomials of lower degrees r and s in Q[x ] if and only if it hassuch a factorization with polynomials of the same degrees rand s in Z[x ].

Proof.See the supplemental material.

Corollary 23.12If f (x) = xn + an−1xn−1 + · · ·+ a0 is in Z[x ] with a0 6= 0, and iff (x) has a zero in Q, then the zero is in Z and satisfies m|a0.

Example

Example. Factor f (x) = x4 + 10x3 + 13x2 − 32x + 12 ∈ Z[x ]into a product of irreducible polynomials over Q.

Solution.• We first check if it has a linear factor. By Corollary 23.12,

such a factor is of the form x −m for some integer m|12.

• By a direct computation, we find ±1,±2,±3,±4,±6,±12are not zeros of f (x). Thus, f (x) has no linear factors.

• Assume that f (x) = (x2 + ax + b)(x2 + cx + d). Then byGauss’s lemma, (b, d) = ±(1, 12),±(2, 6), or ±(3, 4).

Example, continued

Example. Factor f (x) = x4 + 10x3 + 13x2 − 32x + 12 ∈ Z[x ]into a product of irreducible polynomials over Q.

Solution.• Assume that (b, d) = (1, 12). We have

x4 + 10x3 + 13x2 − 32x + 12 =x4 + (a + c)x3 + (ac + 13)x2 + (12a + c)x + 12.

• Comparing the coefficients, we get a + c = 10, ac = 0, and12a + c = −32. However, no integers can satisfy theseequations.

• Likewise, the choices (b, d) = (−1,−12) and (2, 6) do notwork either.

• When (b, d) = (−2,−6), we find a + c = 10, ac − 8 = 13,−6a− 2c = −32. A solution is (a, c) = (3, 7). Thus,f (x) = (x2 + 3x − 2)(x2 + 7x − 6).

Reduction modulo m

Lemma (Exercise 37, reduction modulo m)Let m > 1 be a positive integer. Define φm : Z[x ] → Zm[x ] by

φm(anxn + · · ·+ a0) = anxn + · · ·+ a0,

where ai denotes the residue class modulo m containing ai .Then φm is a ring homomorphism, called reduction modulo m.

Example

1. The reduction of x2 + 4x + 7 modulo 2 is x2 + 1.

2. The reduction of x3 + 3x + 9 modulo 3 is x3.

Reduction modulo p

Idea.• Let f (x) ∈ Z[x ]. Consider the reduction f (x) modulo p,

where p is a prime.

• If f (x) = g(x)h(x) is reducible over Q, then the reductionmodulo p gives f (x) = g(x)h(x). (Exercise 37.)

• Thus, if f (x) is irreducible over Zp, then f (x) is irreducibleover Q. However, note that if f (x) is reducible over Zp, it isstill possible that f (x) is irreducible over Q. For example,f (x) = x2 + 1 and p = 2 with f (x) = (x + 1)2.

Example. The polynomial x3 + 13x + 81 is irreducible over Qbecause its reduction modulo 2 is x3 + x + 1, which isirreducible over Z2.

Eisenstein criterion

Theorem (23.15, Eisenstein criterion)Let p be a prime. Suppose that f (x) = anxn + · · ·+ a0 is in Z[x ],and an 6≡ 0 mod p, ai ≡ 0 mod p for all i < n, but a0 6≡ 0mod p2. Then f (x) is irreducible over Q.

ExampleThe polynomial x2 − 2 is irreducible over Q by the Eisensteincriterion. This shows that

√2 is an irrational number.

Proof of Theorem 23.15

• Let f (x) denote the reduction of f (x) modulo p. Byassumption, we have f (x) = dxn for some d 6= 0 ∈ Zp.

• Now if f (x) = g(x)h(x) for some g(x), h(x) ∈ Z[x ], theng(x)h(x) = dxn.

• We claim that this implies that g(x) = bxk , h(x) = cxm forsome nonnegative integers k , m, and some a, b ∈ Zp

satisfying k + m = n and ab = d mod p.

• Assume that the claim is true for the moment. Theng(x) = bkxk + · · ·+ b0 and cmxm + · · ·+ c0, wherebi , cj ≡ 0 mod p for all i < k and all j < m.

• If k and m are both > 0, then the constant term a0 = b0c0

of f (x) is divisible by p2, contradicting to a0 6≡ 0 mod p2.

• Therefore, one of g(x) and h(x) must be a constantpolynomial. That is, f (x) is irreducible over Q. �

Proof of the claim

• Write g(x) = bkxk + · · ·+ b0 and h(x) = cmxm + · · ·+ c0.

• Let r be the smallest integer such that br 6≡ 0 mod p, ands be the smallest integer such that cs 6≡ 0 mod p.

• Now the `th coefficient a` of f (x) is equal to∑`

i=0 bic`−j .

• For ` = r + s, we have

a` = (b0c` + · · ·+ br−1cs+1) + br cs + (br+1cs−1 + · · ·+ b`c0)

≡ br cs 6≡ 0 mod p.

• Since a` ≡ 0 mod p for all ` < n, we conclude thatr + s = n, which in turn implies r = k and s = m. �

Remark

RemarkIn the proof above, the assumption that p is a prime is cruciallyused in the step br cs 6≡ 0 mod p. Without the assumption thatp is a prime, the claim is not true at all. For example, in Z4[x ]we have x2 = (x + 2)(x + 2).

Cyclotomic polynomial

Corollary 23.17Let p be a prime. Then the polynomial

Φp(x) = xp−1 + xp−2 + · · ·+ 1 =xp − 1x − 1

is irreducible over Q.

Remark

• Φp(x) is called the pth cyclotomic polynomial.

• The zeros of Φp(x) are precisely the pth roots of unity,except for 1.

• It plays the central role in Kummer’s approach to Fermat’sLast Theorem. He proved that X n + Y n = Z n has nonon-trivial solutions if 3 ≤ n ≤ 100, with possibleexceptions n = 37, 59, 67.

Proof of Corollary 23.17

• In general, a polynomial f (x) ∈ F [x ] is irreducible over F ifand only if f (x + 1) is irreducible over F . Thus, it suffices toprove that Φp(x + 1) is irreducible over Q.

• We have (x + 1)p − 1 = xp +(p

1

)xp−1 + · · ·+ px . Thus,

Φp(x + 1) = xp−1 +

(p1

)xp−2 + · · ·+ p.

• The binomial coefficients(p

k

)are all multiples of p when

0 < k < p.

• But the constant term is p, which is not divisible by p2.

• Thus, by the Eisenstein criterion, Φp(x) is irreducible overQ.

In-class exercises

Factors the following polynomials into products of irreduciblepolynomials in the given fields.

1. x3 + x + 1 over Z3.

2. x4 + x + 1 over Z5.

3. x4 − 3x3 − 3x2 + 11x − 6 over Q.

4. x4 + 3x3 + 6x2 − 3x + 12 over Q.

Unique factorization in F [x ]

Theorem (23.18)Let p(x) be an irreducible polynomial in F [x ]. If p(x)|r(x)s(x)for r(x), s(x) ∈ F [x ], then p(x)|r(x) or p(x)|s(x)| in F [x ].

Proof.Will be proved in Theorem 27.27.

RemarkIn other words, irreducible polynomials in F [x ] are in manyways similar to primes in Z.

Unique factorization in F [x ]

Theorem (23.20)Every nonconstant polynomial f (x) ∈ F [x ] can be factored inF [x ] into a product of irreducible polynomials. The factorizationis unique, except for order and for units.

Remarks

• The phrase “except for order” means that we consider thefactorizations f (x) = p1(x)p2(x) and f (x) = p2(x)p1(x) asthe same one.

• The phrase “except for units” means that the factorizationsp1(x)p2(x) and (cp1(x))(c−1p2(x)) are considered as thesame, where c 6= 0 ∈ F . (Note that the nonzero elementsin F are the units in F [x ].)

Proof of Theorem 23.20

We first prove by induction that every nonconstant polynomialf (x) is a product of irreducible polynomials over F .

• Polynomials of degree 1 are clearly irreduciblepolynomials.

• Suppose that every polynomial of degree ≤ n is a productof irreducible polynomials.

• Now let f (x) be a polynomial of degree n + 1.

• If f (x) is irreducible, we are done. Otherwise, assume thatf (x) = g(x)h(x), where deg g(x), h(x) ≤ deg f (x)− 1 = n.

• By the induction hypothesis, g(x) and h(x) are products ofirreducible polynomials, and thus so is f (x) = g(x)h(x).

• We conclude that every nonconstant polynomial is aproduct of irreducible polynomials.

Proof of uniqueness

• Assume that f (x) = p1(x) . . . pr (x) andf (x) = q1(x) . . . qr (x) are two factorizations of f (x) into aproduct of irreducible polynomials.

• By Theorem 23.18, p1(x) divides one of qi(x).

• By rearranging the indices, we assume that p1(x)|q1(x),i.e., q1(x) = p1(x)r1(x) for some r1(x) ∈ F [x ].

• Since q1(x) is irreducible, r1(x) must be a unit, that is,r1(x) = u1 is a constant polynomial in F [x ].

• Then p1(x) . . . pr (x) = (u1p1(x))q2(x) . . . qs(x).

• Canceling p1(x), we get p2(x) . . . pr (x) = u1q2(x) . . . qs(x).

• Continuing this way, we get qi(x) = uipi(x) for i = 1, 2 . . .,for some units ui ∈ F , and we have r = s.

Homework

Problems 4, 10, 14, 16, 20, 30, 34, 36, 37 of Section 23.