Section 23 – of over a field

Instructor: Yifan Yang

Spring 2007 Notation

Throughout this section, the letter F will always denote a field. Division 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 . Then there exist unique 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. m • Assume that g(x) = bmx + ··· + b0, where m ≥ 0 and bm 6= 0. Consider the set S = {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 minimal degree in S. We have f (x) = g(x)q(x) + r(x) for some q(x) ∈ F[x]. We need to show that deg r(x) < deg g(x). k • Suppose that r(x) = ck x + ··· + c0 with k ≥ m and k−m ck 6= 0. Consider f (x) − g(x)(q(x) + ck /bmx ). • We have

k−m k−m f (x)−g(x)q(x) − (ck /bm)x g(x) = r(x) − (ck /bm)x g(x) k k = (ck x + ··· + c0) − (ck /bm)(bmx + ··· ),

whose degree is less than r(x). This contradicts to the assumption 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 satisfy deg 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), and q1(x) = q2(x). This completes the proof.  Example

4 2 Example. Let F = Z5, f (x) = x + x + 3x + 2, and g(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 + 3x 3x3 + x2 + 4x 2x2 + 4x + 1 2x2 + 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 that f (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 is a 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], where r(x) = 0 or deg r(x) < deg(x − a) = 1, i.e., or r(x) = c is a constant polynomial. • Then we have 0 = f (a) = 0g(a) + c. That is, r(x) is the zero 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 most n 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 some c 6= 0 ∈ F, it is clear that f (x) has no zeros. • Now assume that the statement holds for all polynomials of degree n ≤ k. We will prove that the statement also holds for 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, assume that a ∈ F is a zero of f (x). • By Corollary 23.3, f (x) = (x − a)g(x) for some polynomial g(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 or g(b) = 0, i.e., b = a or b is a zero of g(x). • By the induction hypothesis, the polynomial g(x) has at most k zeros. It follows that f (x) has at most k + 1 zeros. • By the principle of mathematical induction, we conclude that 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 multiplicative group F × of all nonzero elements in F. Then G is cyclic. In particular, if F is a finite field, then F × is cyclic. Example

2 3 4 × 1. In Z5 we have 2 = 4, 2 = 3, and 2 = 1. Thus, Z5 = h2i is cyclic. × 2. We have |Z7 | = 6. Thus, the multiplicative order of an × 2 element in Z7 is 1, 2, 3, or 6. Now 3 = 2 6= 1 and 33 = 6 6= 1. Therefore, the multiplicative order of 3 is 6, × and Z7 = h3i is cyclic. Proof of Corollary 23.6

• Recall that by the fundamental theorem for finitely generated abelian groups (Theorem 11.12), the finite

abelian group G is isomorphic to Z e1 × · · · × Z ek for some p1 pk primes pi and ei ≥ 1.

• Since the maximal order of an element in Z e1 × · · · × Z ek p1 pk e1 ek is lcm(p1 ,..., pk ), the group G is cyclic if and only if pi are all distinct. • Observe also that the order of every element in G is a e1 ek divisor of m = lcm(p1 ,..., pk ). 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 the specified rings of polynomials. 5 3 3 2 1. F = Z2, f (x) = x + x + x + 1, g(x) = x + x + 1. 5 3 3 2 2. F = Z3, f (x) = x + x + x + 1, g(x) = x + x + 1. 5 3 3 2 3. F = Z5, f (x) = x + x + x + 1, g(x) = x + x + 1. Find a generator for each of the following groups. × 1. Z13. × 2. Z17. Irreducible polynomials

Definition A nonconstant polynomial f (x) ∈ F[x] is irreducible over F or is an irreducible polynomial in F[x] if f (x) cannot be expressed as a product g(x)h(x) of two polynomials in F[x] with 0 < deg g(x), deg h(x) < deg f simultaneously. If a polynomial f (x) ∈ F[x] is nonconstant and is not irreducible over 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 in f (x) = g(x)h(x) in F[x], one of g(x) and h(x) has degree equal to deg f (x), and the other has degree 0. Since the nonzero constant polynomials in F[x] are precisely the units in F[x], the definition of irreducible polynomials can also be given as “f (x) is irreducible over F if in any factorization f (x) = g(x)h(x) in F[x], one of g(x) and h(x) is a ”. • Note that whether a polynomial f (x) is irreducible or not depends on which field we are talking about. For example, 2 x + 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) is reducible 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), where 0 < 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

Remark When the degree of f (x) is 4, it is possible that f (x) = g(x)h(x), where each of g and h is of degree 2. To determine whether f (x) is irreducible over F, we also need to consider this possibility. Examples

Example. Determine whether f (x) = x3 + x + 1 is irreducible over 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 of these is equal to 0.

• Thus, f (x) is irreducible over Z5. Examples

Example. Determine whether f (x) = x4 + x2 + x + 1 is irreducible 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). Then we 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 gives a(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 two polynomials of lower degrees r and s in Q[x] if and only if it has such a factorization with polynomials of the same degrees r and s in Z[x]. Proof. See the supplemental material. Corollary 23.12 n n−1 If f (x) = x + an−1x + ··· + a0 is in Z[x] with a0 6= 0, and if f (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 m|12. • By a direct computation, we find ±1, ±2, ±3, ±4, ±6, ±12 are not zeros of f (x). Thus, f (x) has no linear factors. • Assume that f (x) = (x2 + ax + b)(x2 + cx + d). Then by Gauss’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, and 12a + c = −32. However, no integers can satisfy these equations. • Likewise, the choices (b, d) = (−1, −12) and (2, 6) do not work 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

n n φm(anx + ··· + a0) = anx + ··· + a0,

where ai denotes the residue class modulo m containing ai . Then φm is a 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 reduction modulo p gives f (x) = g(x)h(x). (Exercise 37.)

• Thus, if f (x) is irreducible over Zp, then f (x) is irreducible over Q. However, note that if f (x) is reducible over Zp, it is still 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 Q because its reduction modulo 2 is x3 + x + 1, which is irreducible over Z2. Eisenstein criterion

Theorem (23.15, Eisenstein criterion) n Let p be a prime. Suppose that f (x) = anx + ··· + a0 is in Z[x], and an 6≡ 0 mod p, ai ≡ 0 mod p for all i < n, but a0 6≡ 0 mod p2. Then f (x) is irreducible over Q. Example 2 The polynomial x − 2 is irreducible√ over Q by the Eisenstein criterion. This shows that 2 is an irrational number. Proof of Theorem 23.15

• Let f (x) denote the reduction of f (x) modulo p. By n assumption, we have f (x) = dx for some d 6= 0 ∈ Zp. • Now if f (x) = g(x)h(x) for some g(x), h(x) ∈ Z[x], then g(x)h(x) = dxn. • We claim that this implies that g(x) = bxk , h(x) = cxm for some 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. Then k m g(x) = bk x + ··· + b0 and cmx + ··· + c0, where bi , 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 2 2 of f (x) is divisible by p , contradicting to a0 6≡ 0 mod p . • Therefore, one of g(x) and h(x) must be a constant polynomial. That is, f (x) is irreducible over Q.  Proof of the claim

k m • Write g(x) = bk x + ··· + b0 and h(x) = cmx + ··· + c0.

• Let r be the smallest integer such that br 6≡ 0 mod p, and s be the smallest integer such that cs 6≡ 0 mod p. P` • Now the `th coefficient a` of f (x) is equal to i=0 bi c`−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 that r + s = n, which in turn implies r = k and s = m.  Remark

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

Corollary 23.17 Let p be a prime. Then the polynomial

xp − 1 Φ (x) = xp−1 + xp−2 + ··· + 1 = p x − 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’s Last Theorem. He proved that X n + Y n = Z n has no non-trivial solutions if 3 ≤ n ≤ 100, with possible exceptions n = 37, 59, 67. Proof of Corollary 23.17

• In general, a polynomial f (x) ∈ F[x] is irreducible over F if and only if f (x + 1) is irreducible over F. Thus, it suffices to prove that Φp(x + 1) is irreducible over Q. p p p p−1 • We have (x + 1) − 1 = x + 1 x + ··· + px. Thus, p Φ (x + 1) = xp−1 + xp−2 + ··· + p. p 1

p • The coefficients 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 over Q. In-class exercises

Factors the following polynomials into products of irreducible polynomials in the given fields. 3 1. x + x + 1 over Z3. 4 2. x + 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. Remark In other words, irreducible polynomials in F[x] are in many ways similar to primes in Z. Unique factorization in F[x]

Theorem (23.20) Every nonconstant polynomial f (x) ∈ F[x] can be factored in F[x] into a product of irreducible polynomials. The factorization is unique, except for order and for units. Remarks • The phrase “except for order” means that we consider the f (x) = p1(x)p2(x) and f (x) = p2(x)p1(x) as the same one. • The phrase “except for units” means that the factorizations −1 p1(x)p2(x) and (cp1(x))(c p2(x)) are considered as the same, where c 6= 0 ∈ F. (Note that the nonzero elements in F are the units in F[x].) Proof of Theorem 23.20

We first prove by induction that every nonconstant polynomial f (x) is a product of irreducible polynomials over F. • Polynomials of degree 1 are clearly irreducible polynomials. • Suppose that every polynomial of degree ≤ n is a product of irreducible polynomials. • Now let f (x) be a polynomial of degree n + 1. • If f (x) is irreducible, we are done. Otherwise, assume that f (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 of irreducible polynomials, and thus so is f (x) = g(x)h(x). • We conclude that every nonconstant polynomial is a product of irreducible polynomials. Proof of uniqueness

• Assume that f (x) = p1(x) ... pr (x) and f (x) = q1(x) ... qr (x) are two factorizations of f (x) into a product 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) = ui pi (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.