5. polynomial rings Let R be a commutative ring. A polynomial of degree n in an indeterminate (or variable) x with coefficients in R is an expression of the form f = f(x)=a + a x + + a xn, 0 1 ··· n where a0, ,an R and an =0.Wesaythata0, ,an are the coefficients of f and n ··· ∈ ̸ ··· that anx is the highest degree term of f. A polynomial is determined by its coeffiecients. m i n i Two polynomials f = i=0 aix and g = i=1 bix are equal if m = n and ai = bi for i =0, 1, ,n. ! ! Degree··· of a polynomial is a non-negative integer. The polynomials of degree zero are just the elements of R, these are called constant polynomials, or simply constants. Let R[x] denote the set of all polynomials in the variable x with coefficients in R. The addition and 2 multiplication of polynomials are defined in the usual manner: If f(x)=a0 + a1x + a2x + a x3 + and g(x)=b + b x + b x2 + b x3 + are two elements of R[x], then their sum is 3 ··· 0 1 2 3 ··· f(x)+g(x)=(a + b )+(a + b )x +(a + b )x2 +(a + b )x3 + 0 0 1 1 2 2 3 3 ··· and their product is defined by f(x) g(x)=a b +(a b + a b )x +(a b + a b + a b )x2 +(a b + a b + a b + a b )x3 + · 0 0 0 1 1 0 0 2 1 1 2 0 0 3 1 2 2 1 3 0 ··· Let 1 denote the constant polynomial 1 and 0 denote the constant polynomial zero. Then (R[x], +, , 0, 1) is a commutative ring. The ring axioms are easily verified. · 5.1. Lemma. Let R be a domain. Then deg(fg)=deg(f)+deg(g).

m j Proof. Suppose deg(f)=m and deg(g)=n.Thenf and g has the form f = j=1 ajx and n k g = k=0 bkx where am =0andbn = 0. From the definition of polynomial multiplication,! ̸ ̸ m+n one notices! that the highest degree term of fg is ambnx . Since R is a domain, ambn =0, so fg has degree m + n. ̸ ! 5.2. Corollary. Let R be a domain. (a) Then R[x] is a domain. (b) The units in R[x] are precisely the constant polynomials that are also units in R. Proof. (a) Let f,g R[x]suchthatfg =0.thendeg(f)+deg(g) = deg(0) = 0 (since the zero polynomial∈ is a constant, so it has degree zero). Since degree is a nonzero integer, it follows that deg(f)=0anddeg(g)=0,sof and g are constants, i.e, f and g are just elements of R. Now, since R is a domain, it follows that f =0org =0. (b) Let u be an unit of R[x]. Then there exists v R[x]suchthatuv =1,sodeg(u)+ deg(v)=deg(1)=0,sodeg(u)=deg(v) = 0, that is,∈u and v are constants, that is, u and v are elements of R.Nowuv =1saysthatu is an unit of R. !

n 5.3. Definition (evaluation at a point). Let f(x)=a0 + a1x + + anx R[x]bea ···n ∈ polynomial and r R. Then we let f(r)=a0 + a1r + + anr .Wesaythatf(r) is the value of the polynomial∈ f(x)atr.Wesaythata ···R is a root of the polynomial f if f(a) = 0. This way each polynomial f(x) determines∈ a function f : R R that takes r R to f(r). Notice that if f = g,thenf(r)=g(r) for all r R. Define→ a function ev∈: R[x] R given by ev (f)=f(r). One verifies that ev is a ring∈ homomorphism from r → r r R[x]toR; it is called the evaluation homomorphism. We also say that evr(f)=f(r) is the element of R obtained by evaluating f at r. 13 For the rest of this section, we let F be a field and we consider polynomial ring in one variable over the field F .Ifp Z is a prime number, then Z/pZ is a field with p elements. ∈ We write Fp = Z/pZ. 5.4. Theorem (Euclidean algorithm). Let f(x),g(x) F [x] with g(x) =0.Thenthere exists unique polynomials q(x) and r(x) such that f(x)=∈ g(x)q(x)+r(x)̸ and deg(r(x)) < deg(q(x)). Sketch of proof. Follows from the Euclidean algorithm for division of polynomials. See the book for details. ! 5.5.ExampleExamples of polynomial division in Q[X]: (2x2 x +1)(x3 1 x + 1 )+(x + 1 )=2x5 x4 + 3 x2 +1 − − 2 2 2 − 2 2 3 1 1 Divide the right hand side by 2x x +1andcheckthat(x 2 x + 2 ) is the quotient and 1 − − (x + 2 ) is the remainder. 5.6. Corollary. F [x] is a PID. sketch of proof. Given an ideal I find the polynomial of least degree in I and show that it generates I. ! 5.7. Corollary. Let F be a field. A polynomial f F [x] has a root a F if and only if (x a) divides f in F [x]. ∈ ∈ − Proof. If (x a) divides f in F [x], then f(x)=(x a)g(x)forsomeg F [x], so f(a)= (a a)g(a)=0,so− a is a root of f. Conversely, suppose− a is a root of∈f. By Euclidean algorithm,− we can divide f by (x a) and write f(x)=(x a)g(x)+r where g,r F [x] and deg(r) < deg(x a)=1,sodeg(− r) = 0, that is r F −is a constant. Evaluating∈ both sides at a we obtain 0− = f(a)=(a a)g(a)+r = r,sor∈=0,hencef(x)=(x a)g(x). ! − − 5.8.Reducingcoefficients modulo an ideal: Let φ : R S be a ring homomorphism. → Then verify that φ determines a ring homomorphism Φ : R[x] S[x] given by Φ(a0 + a1x + n n → +anx )=φ(a0)+φ(a1)x+ +φ(an)x . In particular, if I is an ideal in R,thenwehave ···the natural quotient homomorphism··· π : R R/I and hence we have a ring homomorphsim R[x] R/I[x] obtained by taking a polynomial→ and reducing its coefficients modulo I. This → n homomorphism takes f = a0 + a1x + + anx R[x]tof mod I,wheref mod I is defined ··· ∈ n to be f mod I =(a0 mod I)+(a1 mod I)x+ +(an mod I)x .Ifa is a root of f in R,then reducing coefficients modulo I,wefindthat···a mod I is a root of f mod I. In particular, if f has a root in R,thenf mod I has a root in R/I. Recall that an element f is a PID R is called irreducible, if f = uv with u, v R implies that either u or v is an unit. An element is called reducible if it is not irreducible.∈ The units in F [x] are just the nonzero constants (since F is a field). Given any f F [x]anda constant u F 0 , we can always write f = u(u−1f); this is called a trivial factorization.∈ So a non-trivial∈ \{ factorization} of f F [x] is a factorization f = gh where both g and h has degree at least 1. The irredicible polynomials∈ in F [x] are the polynomials that “cannot be non-trivially factored”. 5.9.Exercise:(a) Show that f(x)=x3 3x + 3 is irreducible in Q[x]. 3 − (b) Show that f(x)=x 3x + 3 is reducible in F5[x] − 14 sketch of proof. Suppose f is reducible. If f = gh is a nontrivial factorization of f,then deg(g)+deg(h) = 3, so either g has degree 1 and h has degree 2 or vice versa (since g and h are not constants). So f has a factor of the form (ax+b)wherea, b Q and a = 0. It follows that b/a is a root of f,sof has a root in Q. Write the root of f∈in the form̸ m/n where m, n − Z with gcd(m, n)=1.Thenf(m/n) = 0 implies m3 3mn2 + n3 =0.Son divides 3mn2∈ n3 = m3 but gcd(m, n)=1,son = 1. It follows that f−has an integer root. Now since f Z−[x], we can reduce coefficients modulo 2 and obtain a polynomial f mod 2 F [x]. ∈ ∈ 2 Note that f mod 2 does not have a solution in F2, since f(0) 1mod2andf(1) 1mod2. It follows that f does not have an integer root, which is a contradiction.≡ So f is≡ irreducible in Q[x]. On the other hand, note that f(2) 0 mod 5. so 2 is a solution of f in F5. Divide f by (x 2) in F [x], to find the factorization≡ f(x) (x +1)2(x 2) mod 5. ! − 5 ≡ −

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