# Math Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

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1 Math Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will show that F [x] has many properties we associate with Z: a division algorithm, prime (irreducible) polynomials, and unique factorization into irreducible polynomials. We will also write f instead of f(x). The degree of f is deg(f) = k where f = k i=0 a kx k and a k Division Algorithm: For each f, g F [x] such that f 0 and g 0 there exists r, q F [x] such that f = gq + r and deg(r) < deg(g). Proof: we will first show that this is true when deg(f) < deg(g). In this case we have f = g 0 + f, so can take q = 0 and r = f. Now we ll show that q and r can be found when deg(f) deg(g) by induction on n = deg(f) deg(g). The base case is n = 0. In this case f = f 0 + a k x k and g = g 0 + x k where f 0 and g 0 are smaller degree polynomials and a k 0, 0. Note that f a k Therefore we have f = g a k + (f a k g), and we set q = a k g has degree < k. and r = f a k g. Now assume that we can find q and r when deg(f) deg(g) = n. Consider now deg(f) deg(g) = n+1. Then deg(f) deg(gx) = n, and by the induction hypothesis we can find q 1, r 1 such that f = (gx)q 1 + r 1. If deg(r 1 ) < deg(g) then we are done, setting q = q 1 x and r = r 1. But if deg(r) = deg(g) then by the base base we can find q 2 and r 2 with r = gq 2 + r 2 with deg(r 2 ) < deg(g). Hence we have f = (gx)q 1 + gq 2 + r 2 = g(xq 1 + q 2 ) + r 2, and we can set q = xq 1 + q 2 and r = r The polynomials q and r in the Division Algorithm are unique. Proof: Let f and g be given and assume f = gq 1 + r 1 and f = gq 2 + r 2 where deg(r 1 ) < deg(g) and deg(r 2 ) < deg(g). Then we have gq 1 + r 1 = gq 2 + r 2, g(q 1 q 2 ) = (r 2 r 1 ). If neither expression is zero, then the lefthand side has degree deg(g) and the righthand side has degree < deg(g). Contradiction. Therefore r 2 r 1 = 0, which forces g(q 1 q 2 ) = 0, which forces q 1 q 2 = 0 since F [x] is an integral domain. Hence q 1 = q 2 and r 1 = r A non-constant polynomial f F [x] is irreducible if and only if f = ab implies deg(a) = deg(f) or deg(b) = deg(f). A non-zero polynomial f divides a non-zero polynomial g if and only if f = gh for some h F [x]. Two 1

2 polynomials f and g are said to be relatively prime if and only if d f and f g implies deg(d) = f and g are relatively prime polynomials if and only if there exist r, s F [x] such that rf + sg = 1. Proof: Assume f and g are relatively prime in F [x]. Let S = {hf +kg : h, k F [x]}. Then there is at least one non-zero polynomial in S. Let d be the smallest degree non-zero polynomial in S. We claim that d f and d g, hence deg(d) = 0. To see this, write f = dq + r where deg(r) < deg(d). Note that we can write d = hf +kg for some h, k F [x]. So we have f = (hf +kg)q+r, r = (1 h)f + ( kq)g S. To avoid a contradiction, we must have r = 0. Hence f = dq and d f. Similarly, d g. Therefore deg(d) = 0. Conversely, suppose that f and g are relatively prime. Then rf + sg = 1 is possible. If d f and d g then we can write f = df 0 and g = dg 0, hence 1 = rdf 0 + sdg 0 = d(rf 0 + sg 0 ), which can only be possible if both d and rf 0 + sg 0 both have degree Theorem: Let f and g be relatively prime polynomials in F [x]. If f gh in F [x] then f h. Proof: We can write rf + sg = 1. Now suppose f gh. Then gh = fk for some polynomial k. Therefore h = h(rf + sg) = hrf + hsg = hrf + sfk = f(hr + sk), therefore f h. 6. Corollary: if p is irreducible and p rs in F [x] then p r or p s. Proof: Suppose p does not divide r. Then p and r are relatively prime: Let d p and d r. We will show that d must have degree 0. If not, then it must have degree equal to the degree of p, therefore p = dα where α F. Since d r, we have r = dr 0 = α 1 pr 0 = p(α 1 r 0 ), which contradicts our hypothesis that p does not divide r. Hence p and r are relatively prime. By the previous theorem, we must have p s. 7. Every non-constant polynomial f can be factored into a product of irreducible polynomials. Proof: by induction on deg(f) = n. If n = 1, then f = ab implies deg(a) = 1 = deg(f) or deg(b) = 1 = deg(f), so f is irreducible, and f is its own product of irreducible polynomials. Now assume that all polynomials of degree 1, 2,..., n can be factored into a product of irreducible polynomials. Consider f with degree n + 1. If f 2

3 is irreducible, we are done. But if f is not, then f = ab where deg(a) < deg(f) and deg(b) < deg(f). This implies that neither a nor b is a constant polynomial. By the induction hypothesis, both a and b can be factored into products of irreducible polynomials. Multiplying them together, so can f. 8. If f = p 1 p 2 p k and f = q 1 q 2 q j are two factorizations of f into irreducible polynomials then k = j and the irreducible factors can be paired into associates (two polynomials are associates if one is a non-zero constant multiple of the other). Proof: By induction on k 1. For the base case, suppose p 1 = f = q 1 q 2 q j. Since p 1 is irreducible, j = 1 and we have p 1 = q 1. Now assume the theorem is true for k n. Consider k = n + 1. Write p 1 p n+1 = f = q 1 q j. Then p n+1 q 1 q j, so we must have p n+1 q i for some i. Since q i is irreducible, we must have deg(p n+1 ) = deg(q i ). This implies q i = α i p n+1 for some α i F. Hence p n+1 and q i are associates. So now we have p 1 p n = (α i q 1 ) q i q j. The polynomial α i q 1 is irreducible. Using the induction hypothesis, we must have j 1 = n and we can pair off the remaining irreducible factors into associates. Note that if α i q 1 is an associate of p j then so is q 1. So we have j = n + 1 and all irreducibles paired off into associates. 9. We have proved the main theorems. We are ready for some applications. The first application is the Factor Theorem: if f F [x], a F, then f(a) = 0 if and only if x a divides f. Proof: Assume f(a) = 0. Using the division algorithm, write f(x) = (x a)q + r where degr < 1. Then r is a constant polynomial. Applying the homomorphism φ a to both sides, we get f(a) = (a a)q(a) + r(a), i.e. 0 = r. Therefore f(x) = (x a)q and x a divides f. Conversely, suppose x a divides f. Then f = (x a)q. Applying φ a, we get f(a) = (a a)q(a) = Corollary: If f has n distinct roots then deg(f) n. Proof: by induction on n. For n = 1 we have f = (x a)q where f(a) = 0, therefore deg(f) 1. Now assume if f has n distinct roots then deg(f) has degree n. Consider f F [x] with roots a 1,... a n+1. We can write f = (x a n+1 )g for some polynomial g. Since g has roots a 1,..., a n, the induction hypothesis implies that g has degree n, hence f has degree n Corollary: if F is a field and G F is a finite group under multiplication then G is cyclic. 3

4 Proof: Write G = {a 1,..., a k }. Let m be the least common multiple of o(a 1 ), o(a 2 ),..., o(a k ). Then a m i = 1 for all i, so x m 1 has k distinct roots in F [x]. Hence m k. On the other hand, a k i = 1 for all i, so o(a i ) k for all i, hence k is a common multiple of o(a 1 ), o(a 2 ),..., o(a k ), hence m k. Therefore m = k. Now if G = Z d1 Z dr, then the order of (1, 1,..., 1) is lcm(d 1,..., d r ) = m = k = G, therefore G = Z k and G is cyclic. 12. The next few results are methods of proving that a polynomial is irreducible. 13. If f F [x] has degree 2 or 3 then f is reducible if and only if f has a root in F. Proof: If f is reducible then f = hk for some h, k F [x], and one of these factors has the form ax + b where a 0. Therefore f( b ) = 0. Conversely, a if f(r) = 0 then f = (x r)k for some k F [x]. 14. Example: f(x) = x 2 6 is irreducible in Q[x]. Otherwise, it would have a root in Q, which clearly it does not. See also Example 23.8, page Theorem: if f Z[x] and f = bc with b, c Q[x] then f = BC with B, C Z[x] and deg(b) = deg(b) and deg(c) = deg(c). (Proof omitted.) 16. Corollary: if f = x n + lower terms Z[x] has a root r Z then it has a root m Z and m f(0). Proof: Write f = (x r)c. Then f = (jx + k)c for some j, k Z and C Z[x]. Since the leading term of f has coefficient 1, this forces j = ±1. Without loss of generality f = (x + k)c. Therefore f( k) = Eisenstein s Criterion: Let f = a 0 + a 1 x + + a n x n be a non-constant polynomial with integer coefficients. If there is prime number p such that p divides a 0 through a n 1, p a n, and p 2 a 0, then f is irreducible in Q[x]. Proof. Suppose not. Then we can write f = bc for some pair of polynomials b and c with integer coefficients, each of degree between 1 and n 1. Write b = b 0 + b 1 x + + b r x r and c = c 0 + c 1 x + + c n r x n r. Then a 0 = b 0 c 0, therefore p b 0 c 0, therefore p divides b 0 or c 0 but not both of them. Without loss of generality we will say that p b 0 and p c 0. We will now prove that p for all k by induction on k, 0 k r. This will yield a contradiction because a n = b r c n r and p b r implies p a n, contrary to hypothesis. Base Case: p b 0. We already know this. 4

5 Induction Hypothesis: p divides b 0 through for some k < r. We will now show that p +1. We have a k+1 = +1 c 0 + c c b 0 c k+1. Since k < r and r < n, we must have k + 1 < n. By hypothesis we know that p a k+1. On the other hand, by the induction hypothesis we know that p divides b 0 through. Therefore p divides a k+1 ( c 1 + +b 0 c k+1 ) = +1 c 0. Since p +1 c 0 and p c 0, p +1. This completes the induction proof. Hence p b r, which contradicts p a n. Therefore f must be irreducible. 18. Example: f = x 5 + 5x 27 is irreducible in Q[x] using p = 5. Homework for Section 23, due??? (only the starred problems will be graded): Hints: 4, 8, 10, 12, 14, 19, 20, 30, 34, Use long division. Write answer with coefficients in the range 0, 1, 2,..., Look for roots. 12. Look for roots. Prove that your irreducible factors are actually irreducible. 14. Look for roots in Q, R, C. 30. Look for polynomials in Z 3 [x] with no root in Z First argue that a p = a in Z p (Euler s Theorem application when a 0, trivial if a = 0), then factor x p + a p using high-school algebra formula. 36. Apply the division algorithm to the polynomials f(x) and x α. 5

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