# Polynomial Rings. (Last Updated: December 8, 2017)

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1 Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters Polynomials Throughout this section, R is a commutative ring with identity Definition. An expression of the form n f(x) = a i x i where a i R is called a polynomial over R in indeterminate x. denoted R[x]. i=0 The set of such polynomials is The elements a i are the coefficients of f. The degree of f is the largest m such that 0 a m if such an m exists. We write deg(f) = m and say a m is the leading coefficient. Otherwise f = 0 and we set deg(f) =. A nonzero polynomial with leading coefficient 1 is called monic. Let p(x), q(x) R[x] be nonzero polynomials with degrees n and m, respectively. Write p(x) = a 0 + a 1 x + + a n x n q(x) = b 0 + b 1 x + + b m x m. The polynomials p(x) and q(x) are equal (p(x) = q(x)) if and only if n = m and a i = b i for all i. We can define two binary operations, addition and multiplication, on R[x]. Suppose n m and set b i = 0 for i > m. Then p(x) + q(x) = (a 0 + b 0 ) + (a 1 + b 1 )x + + (a n + b n )x n. This is similar if m > n. Now p(x)q(x) = c 0 + c 1 x + + c m+n x m+n, where i c i = a 0 b i + a 1 b i a i 1 b 1 + a i b 0. k=0 Theorem 1. Let R be a commutative ring with identity. Then R[x] is a commutative ring with identity.

2 Proof. Above we showed that addition and multiplication are binary operations. It is easy to check that (R[x], +) is an abelian group where the zero polynomial is the (additive) identity. The multiplicative identity is the constant polynomial 1. Associativity of multiplication and the distributive property are easy (albeit annoying) proofs. The details are left as an exercise. Example. Suppose p(x) = 3 + 2x 3 and q(x) = 2 x 2 + 4x 4 are polynomials in Z[x]. Note that deg(p(x)) = 3 and deg(q(x)) = 4. Then p(x) + q(x) = (3 + 2) + (0 + 0)x + (0 1)x 2 + (2 + 0)x 3 + (0 + 4)x 4 = 5 x 2 + 2x 3 + 4x 4 and p(x)q(x) = (3 + 2x 3 )(2 x 2 + 4x 4 ) = 6 3x 2 + 4x x 4 2x 5 + 8x 7. We ll now discuss a variety of properties that a ring may or may not possess. Example. Let p(x) = 3+3x 3 and q(x) = 4+4x 2 +4x 4 be polynomials in Z 12 [x]. Then p(x)+q(x) = 7 + 4x 2 + 3x 3 + 4x 4 and p(x)q(x) = 0. Proposition 2. If R be an integral domain, then R[x] is an integral domain. Proof. Let p(x), q(x) R[x] be nonzero polynomials with degrees n and m, respectively. Write p(x) = a 0 + a 1 x + + a n x n q(x) = b 0 + b 1 x + + b m x m. Then the leading term of p(x)q(x) is a n b m x n+m. By hypothesis, a n, b m 0 and because R is an integral domain, a n b m 0 so p(x)q(x) 0. Remark. Note that what we actually proved in the last proposition was that for an integral domain R, deg(p(x), q(x)) = deg(p(x)) + deg(q(x)), for any polynomials p(x), q(x) R[x]. This justifies why we set deg(0) =. So far we ve discussed polynomials in one variable, but it is relatively straightforward, albeit very tedius, to define polynomials in two or more variables. By the above theorem, if R is a commutative ring with identity then so is R[x]. If y is another indeterminate, then it makes sense to define (R[x])[y]. One could then show that this ring is isomorphic to (R[y])[x]. Both of these rings will be identified with the ring R[x, y] and call this the ring of polynomials in two indeterminates x and y with coefficients in R. Similarly (or inductively), one can then define the ring of polynomials in n indeterminates with coefficients in R, denoted R[x 1,..., x n ].

3 Let S be a commutative ring with identity and R a subring of S containing 1 (so that R is also a commutative ring with identity). Let α S. If p(x) = a 0 + a 1 x + + a n x n, then we define p(α) to be p(α) = a 0 + a 1 α + + a n α n S. Theorem 3. Let S be a commutative ring with identity and R a subring of S containing 1. For α S, the map φ α : R[x] S p(x) p(α) where p(x) = a 0 + a 1 x + + a n x n is a ring homomorphism. Proof. Exercise. Definition. The map in the previous theorem is called the evaluation homomorphism at α. We say α R is a root (or zero) of p(x) R[x] if φ α (p(x)) = Divisibility We will now prove a version of the division algorithm for polynomials. This will be applied to determine when polynomials are irreducible over certain rings. Theorem 4. Let F be a field and f(x), g(x) F [x] with g(x) 0. polynomials f(x), r(x) F (x) such that Then there exist unique f(x) = g(x)q(x) + r(x), where deg r(x) < deg g(x). Proof. For simplicity throughout we will write f = f(x), g = g(x), etc. Set deg f = n and deg g = m. If n < m, then let q = 0 and r = f. Note that this may happen if f = 0. Now suppose m n. Write f = a 0 + a 1 x + + a n x n g = b 0 + b 1 x + + b m x m. Because b m 0 and F is a field, b 1 m exists. If n = 0, then m = 0 so f = c 0 and g = b 0. Set q = a 0 b 1 0 and r = 0. We proceed inductively. That is, assume the division algorithm holds when deg f < n. Note that f a n b 1 m x n m g has degree strictly less than n. Hence, there exists q 0, r such that f a n b 1 m x n m g = q 0 g + r with deg r < deg g. This implies that f = (q 0 + a n b 1 m x n m )g + r.

4 Setting q = q 0 + a n b 1 m x n m completes the existence part of the proof. For uniqueness, assume there exists q, q, r, r F [x] such that f = qg + r = q g + r with deg r, deg r < deg g. Then g(q q ) = r r. If q q, then because g 0 we have deg(r r) = deg(g(q q )) deg g. This is a contradiction since both r and r have degrees less than deg g. r = r. Thus, q = q and so Corollary 5. Let F be a field. An element α F is a root of p(x) F [x] if and only if (x α) divides (is a factor of) p(x). Proof. By the division algorithm, p(x) = (x α)q(x) + r(x) for some q(x), r(x) F [x] with deg r(x) < deg(x α) = 1. Applying the evaluation homomorphism we get p(α) = (α α)q(α) + r(α) = r(α). If x α divides p(x), then r(x) = 0 so p(α) = 0 and α is a root of p(x). Conversely, if α is a root, then r(α) = 0. But deg r < 1 and since r(x) cannot be a nonzero constant, then r(x) = 0. Corollary 6. Let F be a field. A nonzero polynomial p(x) F [x] of degree n can have at most n distinct roots. Proof. First note that a polynomial of degree 0 has no roots. Let p(x) F [x] have degree n. If n = 1, then p(x) has 1 root. We proceed by induction on deg p(x). Suppose all polynomials of degree m, m 1, have at most m distinct roots. Let deg p(x) = m + 1 and let α be a root of p(x). By the previous corollary, p(x) = (x α)q(x) for some q(x) with deg q(x) = m. Thus, m has at most m distinct roots and so p(x) has at most m + 1 distinct roots. The next proof is very similar to the corresponding result for Z. Corollary 7. Let F be a field. Every ideal in F [x] is principal. Proof. Let I be a nonzero ideal in F [x]. Set S = {deg p(x) : p(x) I, p(x) 0}. Since I 0, then S and S N. Thus, by the Well-Ordering Principal, S has a least element, d. Let g(x) I be a polynomial of degree d. We claim I = g(x). It is clear that g(x) I and so it is left only to prove the reverse inclusion. Let f(x) I. If deg f(x) = d, then either f(x) = ag(x) for some a F or else deg(f(x) g(x)) < d, a contradiction since f(x) g(x) I. Now assume deg f(x) > d. By the division algorithm, f(x) = g(x)q(x) + r(x) for some q(x), r(x) I with deg r(x) < deg g(x). But then r(x) = f(x) g(x)q(x) I. If deg r(x) 0, then this contradicts the minimality of d. Thus, r(x) = 0 and f(x) g(x)

5 Disclaimer/Warning: The above result does not hold for F [x, y]. In particular, the ideal x, y is not principal. 3. Irreducible polynomials Definition. Let F be a field. A nonconstant polynomial f(x) F [x] is irreducible over F if f(x) cannot be written as a product of two polynomials g(x), h(x) F [x] with deg g(x), deg h(x) < deg f(x). Example. (1) x 2 2 is irreducible over Q. (2) x is irreducible over R. (3) p(x) = x 3 + x is irreducible over Z 3. To see this, just note that p(0) = 2, p(1) = 1, and p(2) = 2, so p(x) does not have a root in Z 3. Lemma 8. Let p(x) Q[x]. Then p(x) = r s (a 0 + a 1 x + + a n x n ), where r, s, a 0,..., a n Z, the a i relatively prime, and r, s relatively prime. Proof. Write, a i = b i /c i with b i, c i Z. Then p(x) = 1 c 0 c n (d 0 + d 1 x + + d n x n ), where the d i are integers. Set d = gcd{d 0,..., d n }. Set a i = d i d 1 and let r/s = lowest terms. The result follows. d c 0 c n written in Theorem 9 (Gauss Lemma). If a non-constant monic polynomial p(x) Z[x] is irreducible over Z, then it is irreducible over Q. Proof. We will prove the contrapositive. Let p(x) Z[x] be a non-constant polynomial and suppose it factors over Q. We will show that it factors over Z. Write p(x) = α(x)β(x) for some monic polynomials α(x), β(x) Q[x] with deg α(x), deg β(x) < deg p(x). By the previous lemma, α(x) = c 1 d 1 (a 0 + a 1 x + + a m x m ) = c 1 d 1 α 1 (x) β(x) = c 2 d 2 (b 0 + b 1 x + + n n x n ) = c 2 d 2 β 1 (x), where the a i are relatively prime and the b i are relatively prime. Then Set c/d = c 1c 2 d 1 d 2 cases. p(x) = c 1c 2 d 1 d 2 α 1 (x)β 1 (x). expressed in lowest terms. Thus, dp(x) = cα 1 (x)β 1 (x). We now consider several

6 Case 1: (d = 1) Then ca m b n = 1 because p(x) is monic. Hence, c = ±1. If c = 1, then a m = b n = ±1. If c = 1, then p(x) = α 1 (x)β 1 (x). Moreover, a m = b n = 1, then α 1 (x) and β 1 (x) are monic polynomials, deg α(x) = deg α 1 (x), and deg β(x) = deg β 1 (x), so this proves the claim. The case a m = b n = 1 is similar with p(x) = ( α 1 (x))( β 1 (x)). The case c = 1 is also similar and left as an exercise. Case 2: (d 1) Since gcd(c, d) = 1, there exists a prime p such that p d and p c. The coefficients of α 1 (x) are relatively prime and so there exists a coefficient a i of α 1 (x) such that p a i. Similarly, there exists a coefficient b j of β 1 (x) such that p b j. Let α 1 (x) and β 1 (x) be the images of α 1(x) and β 1 (x) in Z p [x]. Since p d, α 1 (x)β 1 (x) = 0. But this is impossible since neither α 1 (x) or β 1 (x) are the zero polynomial and Z p [x] is an integral domain. Therefore, d = 1. Corollary 10. Let p(x) = x n + a n 1 x n a 0 be a polynomial with coefficients in Z and a 0 0. If p(x) has a zero in Q, then p(x) also has a zero in Z. Furthermore, α divides a 0. Proof. Let p(x) have a zero a Q. Then p(x) has a linear factor x a Q[x]. By Gauss Lemma, p(x) has a factorization with a linear factor in Z[x]. Hence, for some α ZZ, p(x) = (x α)(x n 1 + a 0 /α). Thus, a 0 /α Z and so α a 0. Example. Let p(x) = x 4 2x 3 + x + 1. We will show that p(x) is irreducible over Q. Case 1: Suppose p(x) has a linear factor in Q[x], so p(x) = (x a)q(x) for some q(x) Q[x]. Then p(x) has a zero in Z (by Gauss Lemma) and α 1 (by the corollary), so α = ±1. But p(1) = 1 and p( 1) = 3, so p(x) has no linear factors. Case 2: Suppose p(x) is the product of two (irreducible) quadratic factors. By Gauss s Lemma, p(x) factors over Z[x] and so p(x) = (x 2 + ax + b)(x 2 + cx + d) = x 4 + (a + c)x 3 + (b + d + ac)x 2 + (ad + bc)x + bd, = x 4 2x 3 + x + 1 where a, b, c, d Z. Thus, bd = 1 so b = d = ±1. Then 1 = ad + bc = (a + c)b = 2b. This implies 1 = ±2, a contradiction. Hence, p(x) is irreducible. Theorem 11 (Eisenstein s Criterion). Let p be a prime and suppose that f(x) = a n x n + + a 0 Z[x]. If p a i for i = 0, 1,..., n 1, but p a n and p 2 a 0, then f(x) is irreducible over Q.

7 Proof. If suffices by Gauss Lemma to prove that f(x) does not factor over Z[x]. Let f(x) = (b r x r + + b 0 )(c s x s + + c 0 ) be such a factorization with b r, c s 0 and r, s < n. Since p 2 a 0 = b 0 c 0, either b 0 or c 0 is not divisible by p. Suppose p b 0 and p c 0 (the other case is similar). Since p a n and a n = b r c s, neither b r nor c s is divisible by p. Let m be the smallest value of k such that p c k. Then a m = b 0 c m + b 1 c m b m c 0 is not divisible by p since all but one of the terms (b 0 c m ) is divisible by p. Therefore, m = n since a i is divisible by p for m < n. Hence, f(x) cannot be factored and is therefore irreducible. Example. Let f(x) = 16x 5 9x 4 +3x 2 +6x 21 and set p = 3. Then clearly p divides all coefficients except for that of x 5 and p Thus, f(x) is irreducible over Q by Eisenstein s Criterion.

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