§3P. Polynomial Rings

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 08 March 2011

Polynomial rings

Let $R$ be a commutative ring. Then $R [x]$ is a commutative ring.

		Proof.
	To show: If $f (x), g (x) \in R [x]$ then $f (x) + g (x) = g (x) + f (x)$ . If $f (x), g (x), h (x) \in R [x]$ then $(f (x) + g (x)) + h (x) = f (x) + (g (x) + h (x))$ . There exists $0 \in R [x]$ such that if $f (x) \in R [x]$ then $0 + f (x) = f (x) + 0 = f (x)$ . If $f (x) \in R [x]$ then there exists $- f (x) \in R [x]$ such that $f (x) + (- f (x)) = 0$ . If $f (x), g (x), h (x) \in R [x]$ then $(f (x) g (x)) = f (x) (g (x) h (x))$ . There exists $1 \in R [x]$ such that $1 \cdot f (x) = f (x) = f (x) \cdot 1$ for all $f (x) \in R [x]$ . If $f (x), g (x), h (x) \in R [x]$ then $f (x) (g (x) + h (x)) = f (x) g (x) + f (x) h (x)$ and $(g (x) + h (x)) f (x) = g (x) f(x)+ h (x) f (x)$ . If $f (x), g (x) \in R [x]$ then $f (x) g (x) = g (x) h (x)$ . Let $f (x), g (x) \in R [x]$ such that $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots$ . Then $\begin{array}{rcl} f (x) + g (x) & = & (r_{0} + s_{0}) + (r_{1} + s_{1}) x + (r_{2} + s_{2}) x^{2} + \dots & and \\ g (x) + f (x) & = & (s_{0} + r_{0}) + (s_{1} + r_{1}) x + (s_{2} + r_{2}) x^{2} + \dots \end{array}$ Since addition in $R$ is a commutative operation, $r_{i} + s_{i} = s_{i} + r_{i} for all i .$ So $f (x) + g (x) = g (x) + f (x)$ . Let $f (x), g (x), h (x) \in R [x]$ and $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ , $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots$ , and $h (x) = t_{0} + t_{1} x + t_{2} x^{2} + \dots$ . Then $\begin{array}{rcl} (f (x) + g (x)) + h (x) & = & ((r_{0} + s_{0}) + t_{0}) + ((r_{1} + s_{1}) + t_{1}) x + ((r_{2} + s_{2}) + t_{2}) x^{2} + \dots & and \\ f (x) + (g (x) + h (x)) & = & (r_{0} + (s_{0} + t_{0})) + (r_{1} + (s_{1} + t_{1})) x + (r_{2} + (s_{2} + t_{2})) x^{2} + \dots . \end{array}$ Since addition in $R$ is an associative operation, $(r_{i} + s_{i}) + t_{i} = r_{i} + (s_{i} + t_{i}) for all i .$ So $(f (x) + g (x)) + h (x) = f (x) + (g (x) + h (x))$ . Let $0$ denote the zero polynomial $0 = 0 + 0 x + 0 x^{2} + \dots .$ Let $f (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots .$ Then $0 + f (x) = (0 + r_{0}) + (0 + r_{1}) x + (0 + r_{2}) x_{2} + \dots .$ Since $0 + r_{i} = r_{i}$ for all $i$ , $0 + f (x) = f (x) .$ Since addition of polynomials is commutative by a), $0 + f (x) = f (x) + 0 = f (x) .$ Let $f (x) \in R [x]$ such that $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ . Then let $- f (x) = - r_{0} + (- r_{1}) x + (- r_{2}) x^{2} + \dots$ . Since $- r_{i} \in R$ for all $i$ , $- f (x) \in R [x]$ . Then $f (x) + (- f (x)) = (r_{0} - r_{0}) + (r_{1} - r_{1}) x + (r_{2} - r_{2}) x^{2} + \dots .$ So $f (x) + (- f (x)) = 0 + 0 x + 0 x^{2} + \dots = 0 .$ Since addition of polynomials is commutative by a), $f (x) + (- f (x)) = - f (x) + f (x) = 0$ . Let $f (x), g (x), h (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots,$ $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots,$ and $h (x) = t_{0} + t_{1} x + t_{2} x^{2} + \dots .$ Then $\begin{array}{c} f (x) g (x) = c_{0} + c_{1} x + c_{2} x^{2} + \dots, where c_{k} = \sum_{i + j = k} r_{i} s_{j}, and \\ (f (x) g (x)) h (x) = d_{0} + d_{1} x + d_{2} x^{2} + \dots where d_{n} = \sum_{k + l = n} c_{k} t_{l} . \end{array}$ So $d_{n} = \sum_{k + l = n} (\sum_{i + j = k} r_{i} s_{j}) t_{l} = \sum_{i + j + l = n} r_{i} s_{j} t_{l},$ by the distributive law in $R$ . Also $\begin{array}{c} g (x) h (x) = e_{0} + e_{1} x + e_{2} x^{2} + \dots, where e_{q} = \sum_{a + b = q} s_{a} t_{b}, and \\ (f (x)) (g (x) h (x)) = d_{0}^{'} + d_{1}^{'} x + d_{2}^{'} x^{2} + \dots, where d_{n}^{'} = \sum_{p + q = n} r_{p} e_{q} . \end{array}$ Then $d_{n}^{'} = \sum_{p + q = n} r_{p} (\sum_{a + b = q} s_{a} t_{b}) = \sum_{p + a + b = n} r_{p} s_{a} t_{b},$ by the distributive law in $R$ . So $d_{n} d_{n}^{'}$ . So $(f (x) g (x)) (h (x)) = (f (x)) (g (x) h (x))$ . Let $f (x), g (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots .$ Then $\begin{array}{c} f (x) g (x) = c_{0} + c_{1} x + c_{2} x^{2} + \dots, where c_{m} = \sum_{i + j = k} r_{i} s_{j}, and \\ g (x) f (x) = c_{0}^{'} + c_{1}^{'} x + c_{2}^{'} x^{2} + \dots, where c_{k}^{'} = \sum_{i + j = k} s_{j} r_{i} . \end{array}$ Since $R$ is a commutative ring, $c_{k} = \sum_{i + j = k} r_{i} r_{j} = \sum_{i + j = k} s_{j} r_{i} = c_{k}^{'}, for all k .$ So $f (x) g (x) = g (x) f (x)$ . Let $1 \in R [x]$ be the polynomial given by $1 = 1 + 0 x + 0 x^{2} + \dots .$ Let $f (x) \in R [x]$ and $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots .$ Then $\begin{array}{c} 1 \cdot f (x) = c_{0} + c_{1} x + c_{2} x^{2} + \dots where c_{k} = \sum_{i + j = k} a_{i} r_{j}, and \\ a_{1} = 1 and a_{i} = 0 for all i > 0 . \end{array}$ So $c_{k} = a_{0} r_{k} + 0 + 0 + \dots + 0 = r_{k}$ So $1 \cdot f (x) = f (x) .$ Since multiplication in $R [x]$ is commutative by h), $1 \cdot f (x) = f (x) \cdot 1 = f (x) .$ Let $f (x), g (x), h (x) \in R [x]$ and suppose $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ , $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots$ , and $h (x) = t_{0} + t_{1} x + t_{2} x^{2} + \dots .$ Then $f (x) (g (x) + h (x)) = c_{0} + c_{1} x + c_{2} x^{2} + \dots where c_{k} = \sum_{i + j = k} r_{i} (s_{j} + t_{j}) .$ Also $f (x) g (x) + f (x) h (x) = c_{0}^{'} + c_{1}^{'} x + c_{2}^{'} x^{2} + \dots$ where $c_{k}^{'} = \sum_{m + m = k} r_{m} s_{n} + \sum_{m + n = k} r_{m} t_{n} = \sum_{m + n = k} (r_{m} s_{n} + r_{m} t_{n}) .$ So $c_{k} = c_{k}^{'}$ for all $k$ . Thus $f (x) (g (x) + h (x)) = f (x) g (x) + f (x) h (x) .$ Sine multiplication in $R [x]$ is commutative by h), $\begin{array}{rcl} (g (x) + h (x)) f (x) & = & f (x) (g (x) + h (x)) \\ = & f (x) g (x) + f (x) h (x) \\ = & g (x) f (x) + h (x) f (x) . \end{array}$ So $R [x]$ is a commutative ring. $□$

Let $R$ be an integral domain. Then $R [x]$ is an integral domain.

		Proof.
	To show: If $a (x), b (x) \in R [x]$ and $a (x) b (x) = 0$ then either $a (x) = 0$ or $b (x) = 0$ . Let $a (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ and let $b (x) = b_{0} + b_{1} x + b_{2} x^{2} + \dots .$ Let $c (x) = a (x) b (x) = c_{0} + c_{1} x + c_{2} x^{2} + \dots .$ Assume $a (x) \neq 0$ . Then $a_{l} \neq 0$ for some $l \geq 0$ . Let $k$ be the smallest $k \geq 0$ such that $a_{k} \neq 0$ . To show: $b (x) = 0$ . To show: $b_{N} = 0$ for all $N \geq 0$ . Proof by induction on $N$ . Base case: $N = 0$ . We know $c_{k} = 0$ since $c (x) = a (x) b (x) = 0 .$ So $\sum_{i + j = k} a_{i} b_{j} = 0 .$ Since $a_{i} = 0$ for all $i < k$ , $0 = \sum_{i + j = k} a_{i} b_{j} = a_{k} b_{0} .$ Since $R$ is an integral domain and $a_{k}, b_{0} \in R$ and $a_{k} \neq 0$ , we have $b_{0} = 0$ . Induction assumption: assume $b_{n} = 0$ for all $n < N$ . We know $c_{k + N} = 0$ since $c (x) = a (x) b (x) = 0$ . So $\sum_{i + j = k + N} a_{i} b_{j} = 0 .$ Since $a i = 0$ for all $i < k$ and $b_{n} = 0$ for all $n < N$ , we have $0 = \sum_{i + j = k + N} a_{i} b_{j} = a_{k} b_{N} .$ Since $R$ is an integral domain, and $a_{k}, b_{N} \in R$ and $a_{k} \neq 0$ , we have $b_{N} = 0$ . So $b_{N} = 0$ for all $N \geq 0$ . So $b (x) = 0$ . So $R [x]$ is an integral domain. $□$

Let $F$ be a field. The ring $F [x]$ is a Euclidean domain with size function given by $\begin{matrix} deg : & F [x] & \to & ℕ \\ f (x) & \mapsto & deg (f (x)) . \end{matrix}$

		Proof.
	To show: If $a (x), b (x) \in F [x]$ and $a (x) \neq 0$ then there exist $q (x), r (x) \in F [x]$ such that $b (x) = a (x) q (x) + r (x)$ where either $r (x) = 0$ or $deg (r (x)) < deg (a (x)) .$ Assume $a (x), b (x) \in F [x]$ and $a (x) \neq 0$ . Case 1: $b (x) = 0$ . Then $b (x) = a (x) \cdot 0 + 0 .$ So $q (x) = 0$ and $r (x) = 0$ satisfies the condition. Case 2: $deg (b (x)) < deg (a (x))$ . Then, since $b (x) = a (x) \cdot 0 + b (x) and deg (b (x)) < deg (a (x)),$ $q (x) = 0$ and $r (x) = b (x)$ satisfies the condition. Case 3: $deg (b (x)) \geq deg (a (x))$ . Let $a (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{s} x^{s}$ and $b (x) = b_{0} + b_{1} x + b_{2} x^{2} + \dots + b_{t} x^{t}$ , where $a_{s}, b_{t} \in F [x]$ , $a_{s} \neq 0$ , $b_{t} \neq 0$ . Proof by induction on $deg (b (x))$ . Base case: $deg (b (x)) = 0$ . Then $deg (a (x)) = 0$ since $deg (a (x)) \leq deg (b (x))$ . So $b (x) = b_{0} \in F [x]$ and $a (x) = a_{0} \in F [x]$ . So $b (x) = (\frac{b_{0}}{a_{0}}) \cdot a (x) + 0$ . So $q (x) = b_{0} a_{0}^{- 1}$ and $r (x) = 0$ satisfies the condition. Induction assumption: Assume that if $b_{1} (x) \in F [x]$ and $deg (b_{1} (x)) < t$ then there exist $q_{1} (x), r_{1} (x) \in F [x]$ such that $b_{1} (x) = q_{1} (x) a (x) + r_{1} (x)$ where either $r_{1} (x) = 0$ or $deg (r_{1} (x)) < deg (a (x))$ . Assume that $deg (b (x)) = t$ . Let $b_{1} (x) = b (x) - (b_{t} a_{s}^{- 1} x^{t - s}) a (x)$ . Then $deg (b_{1} (x)) < deg (b (x))$ . Note that the coefficient of $x^{t}$ in $b_{1} (x)$ is $- b_{t} + b_{t} = 0$ . So $deg (b_{1} (x)) < t = deg (b (x))$ . Thus, by the induction assumption, there exist $q_{1} (x), r_{1} (x) \in F [x]$ such that $b_{1} (x) = q_{1} (x) + r_{1} (x)$ where either $r_{1} (x) = 0$ or $deg (r_{1} (x)) < deg (a (x))$ . Then $\begin{array}{rcl} b (x) & = & b_{1} (x) - (b_{t} a_{s}^{- 1} x^{s - t}) a (x) \\ = & q_{1} (x) a (x) + r_{1} (x) - (b_{t} a_{s}^{- 1} x^{s - t}) a (x) \\ = & (q_{1} (x) - b_{t} a_{s}^{- 1} x^{s - t}) a (x) + r_{1} (x) . \end{array}$ So, if $q (x) = q_{1} (x) - b_{t} a_{s}^{- 1} x^{s - t}$ and $r (x) = r_{1} (x)$ then $b (x) = q (x) a (x) + r (x)$ and either $r (x) = 0$ or $deg (r (x)) < deg (a (x))$ So $F [x]$ with the size function given by $deg$ is a Euclidean domain. $□$

Let $R, S$ be commutative rings and $ϕ : R \mapsto S$ be a ring homomorphism. Then the map $\begin{matrix} ψ : & R [x] & \to & S [x] \\ r_{0} + r_{1} x + r_{2} x^{2} + \dots & \mapsto & φ (r_{0}) + φ (r_{1}) x + φ (r_{2}) x^{2} + \dots \end{matrix}$ is a ring homomorphism.

		Proof.
	To show: If $f (x), g (x) \in R [x]$ then $ψ (f (x) + g (x)) = ψ (f (x)) + ψ (g (x))$ . If $f (x), g (x) \in R [x]$ then $ψ (f (x) g (x)) = ψ (f (x)) ψ (g (x))$ . $ψ (1_{R}) = 1_{S}$ where $1_{R}$ and $1_{S}$ are the identities in $R$ and $S$ . Let $f (x), g (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = r_{0}^{'} + r_{1}^{'} x + r_{2}^{'} x^{2} + \dots$ . Then $\begin{array}{rcl} ψ (f (x) + g (x)) & = & ψ ((r_{0} + r_{0}^{'}) + (r_{1} + r_{1}^{'}) x + (r_{2} + r_{2}^{'}) x^{2} + \dots) \\ = & φ (r_{0} + r_{0}^{'}) + φ (r_{1} + r_{1}^{'}) x + φ (r_{2} + r_{2}^{'}) x^{2} + \dots \end{array}$ Since $φ$ is a homomorphism, $\begin{array}{rcl} φ (f (x) + g (x)) & = & (φ (r_{0}) + φ (r_{0}^{'})) + (φ r_{1} + φ r_{1}^{'}) x + (φ (r_{2}) + φ (r_{2}^{'})) x^{2} + \dots \\ = & (φ (r_{0}) + φ (r_{1}) x + φ (r_{2}) x^{2} + \dots) + (φ (r_{0}^{'}) + φ (r_{1}^{'}) x + φ (r_{2}^{'}) x^{2} + \dots) \\ = & ψ (f (x)) + ψ (g (x)) . \end{array}$ Let $f (x), g (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = r_{0}^{'} + r_{1}^{'} x + r_{2}^{'} x^{2} + \dots$ . Then $ψ (f (x) g (x)) = ψ (c_{0} + c_{1} x + c_{2} x^{2} + \dots), where c_{k} = \sum_{i + j = k} r_{i} r_{j}^{'} .$ So $ψ (f (x) g (x)) = φ (c_{0}) + φ (c_{1}) x + φ (c_{2}) x^{2} + \dots .$ Since $φ$ is a homomorphism, $φ (c_{k}) = φ (\sum_{i + j = k} r_{i} r_{j}^{'}) = \sum_{i + j = k} φ (r_{i} r_{j}^{'}) = \sum_{i + j = k} φ (r_{i}) φ (r_{j}^{'}) .$ So $ψ (f (x) g (x)) = d_{0} + d_{1} x + d_{2} x^{2} + \dots, where d_{k} = \sum_{i + j = k} φ (r_{i}) φ (r_{j}^{'}) .$ So, by the distributive law in $S$ , $\begin{array}{rcl} ψ (f (x) g (x)) & = & (φ (r_{0}) + φ (r_{1}) x + φ (r_{2}) x^{2} + \dots) (φ (r_{0}^{'}) + φ (r_{1}^{'}) x + φ (r_{2}^{'}) x^{2} + \dots) \\ = & φ (f (x)) φ (g (x)) . \end{array}$ Let $1_{R}$ be the identity in $R$ . $\begin{array}{rcl} ψ (1_{R}) & = & ψ (1_{R} + 0_{R} x + 0_{R} x^{2} + \dots) \\ = & φ (1_{R}) + φ (0_{R}) x + φ (0_{R}) x^{2} + \dots \end{array}$ Since $φ$ is a homomorphism, $φ (1_{R}) = 1_{S}$ and $φ (0_{R}) = 0_{S}$ . So $ψ (1_{R}) = 1_{S} + 0_{S} x + 0_{S} x^{2} + \dots .$ So $ψ$ is a homomorphism. $□$

Let $R$ be a commutative ring and let $α \in R$ . Then the evaluation homomorphism ${ev}_{α} : R [x] \to R$ is a ring homomorphism.

		Proof.
	To show: If $f (x), g (x) \in R [x]$ then ${ev}_{α} (f (x) + g (x)) = {ev}_{α} (f (x)) + {ev}_{α} (g (x))$ . If $f (x), g (x) \in R [x]$ then ${ev}_{α} (f (x) g (x)) = {ev}_{α} (f (x)) {ev}_{α} (f (x)) .$ ${ev}_{α} (1_{R}) = 1_{R}$ where $1_{R}$ is the identity in $R$ . Let $f (x), g (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots .$ Then $\begin{array}{rcl} {ev}_{α} (f (x) + g (x)) & = & {ev}_{α} ((r_{0} + s_{0}) + (r_{1} + s_{1}) x + (r_{2} + s_{2}) x^{2} + \dots) \\ = & (r_{0} + s_{0}) + (r_{1} + s_{1}) α + (r_{2} + s_{2}) α^{2} + \dots . \end{array}$ By the distributive law in $R$ , $\begin{array}{rcl} {ev}_{α} (f (x) + g (x)) & = & r_{0} + s_{0} + r_{1} α + s_{1} α + r_{2} α^{2} + s_{2} α^{2} + \dots \\ = & (r_{0} + r_{1} α + r_{2} α^{2} + \dots) + (s_{0} + s_{1} α + r_{2} α^{2} + \dots) \\ = & {ev}_{α} (f (x)) + {ev}_{α} (g (x)) . \end{array}$ Let $f (x), g (x) \in R [x]$ and let $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots .$ Then ${ev}_{α} (f (x) g (x)) = {ev}_{α} (c_{0} + c_{1} x + c_{2} x^{2} + \dots), where c_{k} = \sum_{i + j = k} r_{i} r_{j} .$ So ${ev}_{α} (f (x) g (x)) = c_{0} + c_{1} α + c_{2} α^{2} + \dots$ . Now compute ${ev}_{α} (f (x)) {ev}_{α} (g (x)) .$ ${ev}_{α} (f (x)) {ev}_{α} (g (x)) = (r_{0} + r_{1} α + r_{2} α^{2} + \dots) (s_{0} + s_{1} α + s_{2} α^{2} + \dots) .$ By the distributive law in $R$ , $\begin{array}{rcl} {ev}_{α} (f (x)) {ev}_{α} (g (x)) & = & r_{0} s_{0} + r_{1} s_{0} α + r_{0} s_{1} α + r_{0} s_{2} α^{2} + r_{1} s_{1} α^{2} + r_{2} s_{0} α^{2} + \dots \\ = & r_{0} s_{0} + (r_{1} s_{0} + r_{0} s_{1}) α + (r_{0} s_{2} + r_{1} s_{1} + r_{2} s_{0}) α^{2} + \dots \\ = & c_{0} + c_{1} α + c_{2} α^{2} + \dots where \end{array}$ $c_{k} = \sum_{i + j = k} r_{i} s_{j} .$ So ${ev}_{α} (f (x g (x))) = {ev}_{α} (f (x)) {ev}_{α} (g (x)) .$ Let $1_{R}$ be the identity in $R$ and $0_{R}$ be the zero in $R$ . Then ${ev}_{α} (1_{R}) = {ev}_{α} (1_{R} + 0_{R} x + 0_{R} x^{2} + \dots) = 1_{R} + 0_{R} α + 0_{R} α^{2} + \dots = 1_{R} .$ So ${ev}_{α}$ is a ring homomorphism. $□$

(Gauss' Lemma) Let $R$ be a unique factorization domain. Let $f (x), g (x) \in R [x]$ be primitive polynomials. Then $f (x) g (x)$ is a primitive polynomial.

		Proof.
	Assume $f (x) = r_{0} + r_{1} x + r_{2} x^{2} + \dots$ and $g (x) = s_{0} + s_{1} x + s_{2} x^{2} + \dots$ are primitive polynomials in $R [x]$ . Proof by contradiction. Assume $f (x) g (x)$ is not primitive. Then there exists an irreducible element $p \in R$ that divides all the coefficients of $f (x) g (x) .$ Since $f (x)$ is primitive there must be at least one coefficient of $f (x)$ which is not divisible by $p$ . Since $g (x)$ is primitive there must be at least one coefficient of $g (x)$ which is not divisible by $p$ . Let $m$ be the smallest $m$ such that $r_{m}$ is not divisible by $p$ . Let $n$ be the smallest $n$ such that $s_{n}$ is not divisible by $p$ . Suppose that $f (x) g (x) = c_{0} + c_{1} x + c_{2} x^{2} + \dots .$ Then, since $p$ divides $r_{i}$ for all $i < m$ , and $p$ divides $s_{j}$ for all $j < n$ , $\begin{array}{rcl} c_{m + n} & = & r_{m} s_{n} + r_{m - 1} s_{n + 1} + r_{m + 1} s_{n - 1} + \dots + r_{0} s_{m + n} + r_{m + n} s_{0} \\ = & r_{m} s_{n} + p c, \end{array}$ where $c$ is some element of $R$ . Since $c_{m + n}$ is divisible by $p$ it follows that $r_{m} s_{n} = c_{m + n} - p c$ is divisible by $p$ . Suppose that $d \in R$ such that $r_{m} s_{n} = p d .$ Let $r_{m} = a_{1} \dots a_{k}, s_{n} = b_{1} \dots b_{l}, and d = d_{1} \dots d_{q},$ be factorizations of $r_{m}, s_{n}$ and $d$ into irreducible elements $a_{1}, \dots, a_{k}, b_{1}, \dots, b_{l}, d_{1}, \dots, d_{q} \in R .$ Then $a_{1} \dots a_{k} b_{1} \dots b_{l} = p d_{1} \dots d_{q} .$ By the uniqueness of factorizations, $\begin{matrix} either & p is an associate to a_{i} for some 1 \leq i \leq k . \\ or & p is an associate to b_{j} for some 1 \leq j \leq l . \end{matrix}$ So either $p_{i} = u p$ or $q_{j} = u p$ for some unit $u \in R$ . Then, either $a = u p a_{1} \dots a_{i - 1} a_{i + 1} \dots a_{k}$ , or $b = u p d_{1} \dots d_{j - 1} d_{j + 1} \dots d_{q}$ . Thus, either $r_{m}$ or $s_{n}$ is divisible by $a_{i}$ or $d_{j}$ . Contradiction. So $f (x) g (x)$ is a primitive polynomial in $R [x]$ . $□$

Let $R$ be a unique factorization domain. Let $F$ be the field of fractions of $R$ and let $f (x) \in F [x]$ . Then

There exists an element $c \in F$ and a primitive polynomial $g (x) \in R [x]$ such that $f (x) = c g (x) .$
The factors $c$ and $g (x)$ are unique up to multiplication by a unit.
$f (x)$ is irreducible in $F [x]$ if and only if $g (x)$ is irreducible in $R [x]$ .

		Proof.
	Let $f (x) = \frac{a_{0}}{b_{0}} + \frac{a_{1}}{b_{1}} x + \dots + \frac{a_{k}}{b_{k}} x^{k} \in F [x]$ . Then $f (x) = \frac{1}{b_{0} b_{1} \dots b_{k}} (c_{0} + c_{1} x + \dots c_{k} x^{k})$ where $c_{i} = a_{i} b_{1} \dots b_{i - 1} b_{i + 1} \dots b_{k}$ . Let $d = gcd (c_{0} c_{1} \dots c_{k})$ . Then $f (x) = \frac{d}{b_{0} \dots b_{k}} (c_{0}^{'} + c_{1}^{'} x + \dots + c_{k}^{'} x^{k})$ where $c_{i}^{'} = \frac{c_{i}}{d}$ . Note that $c_{i}^{'} \in R$ since $d$ divides $c_{i}$ . Furthermore $c_{0}^{'} + c_{1}^{'} x + \dots + c_{k}^{'} x^{k} = g (x)$ is primitive since $gcd (c_{0}^{'}, c_{1}^{'}, \dots, c_{k}^{'}) = 1$ . So $f (x) = c g (x)$ where $c = \frac{d}{b_{0} b_{1} \dots b_{k}} \in F$ and $g (x) = c_{0}^{'} + c_{1}^{'} x + \dots + c_{k}^{'} x^{k} \in R [x]$ is a primitive polynomial. Suppose $f (x) = c g (x)$ and $f (x) = C G (x)$ where $c, C \in F$ and $g (x), G (x) \in R [x]$ are primitive polynomials. Let $g (x) = a_{0} + a_{1} x + \dots + a_{k} x^{k}$ and let $G (x) = b_{0} + b_{1} x + \dots + b_{k} x^{k}$ . Suppose $c = \frac{a}{b}$ and $C = \frac{A}{B}$ where $a, b, A, B \in R$ . Since $f (x) = \frac{a}{b} g (x) = \frac{A}{B} G (x)$ , $a B g (x) = b A G (x) .$ So $a B a_{i} = b A b_{i}$ for all $1 \leq i \leq k$ . Since $g (x)$ is primitive, $gcd (a B a_{0}, a B a_{1}, \dots, a B a_{k}) = a B$ . Since $G (x)$ is primitive, $gcd (b A b_{0}, b A b_{1}, \dots, b A b_{k}) = b A$ . Thus by Proposition 1.6 of §2.T, $a B = u b A for some unit u \in R .$ So $\frac{a}{b} = u (\frac{A}{B})$ . So $c = u C$ where $u \in R$ is a unit. So $C G (x) = c g (x) = u C g (x)$ . By the cancelation law, $G (x) = u g (x)$ . So $c = u C$ and $G (x) = u g (x)$ . So $c$ and $g (x)$ are unique up to multiplication by a unit. $\Rightarrow$ Assume that $f (x)$ is irreducible in $F [x]$ . Proof by contradiction. Assume $g (x)$ is not irreducible in $R [x]$ . Then there are $g_{1} (x)$ and $g_{2} (x)$ in $R [x]$ such that $g (x) = g_{1} (x) g_{2} (x)$ . So $f (x) = c g (x) = c g_{1} (x) g_{1} (x)$ . Since $R [x] \subseteq F [x]$ then $g_{1} (x), g_{2} (x) \in F [x]$ . Contradiction. So $g (x)$ is irreducible in $R [x]$ . ⇐ Assume $g (x)$ is irreducible in $R [x]$ . Proof by contradiction. Assume $f (x)$ is not irreducible in $F [x]$ . Then there are $f_{1} (x)$ and $f_{2} (x)$ in $F [x]$ such that $f (x) = f_{1} (x) f_{2} (x)$ . So by a), there exist $c_{1}, c_{2} \in F$ and primitive polynomials $g_{1} (x), g_{2} (x) \in R [x]$ such that $f_{1} (x) = c_{1} g_{1} (x) and f_{2} (x) = c_{2} g_{2} (x) .$ Let $c = c_{1} c_{2}$ . Then $f (x) = c_{1} c_{2} g_{1} (x) g_{2} (x)$ . By Gauss' lemma $g_{1} (x) g_{2} (x)$ is a primitive polynomial in $R [x]$ . So, by b), $g (x) = u g_{1} (x) g_{2} (x)$ , where $u \in R$ . So $g (x)$ is not irreducible in $R [x]$ . Contradiction. So $f (x)$ is irreducible in $F [x]$ . $□$

Let $R$ be a unique factorization domain. Then $R [x]$ is a unique factorization domain.

		Proof.
	Assume $g (x) \in R [x]$ and let $g (x) = a_{0} + a_{1} x + + \dots + a_{k} x^{k}$ . To show: $g (x)$ has a unique factorization into irreducible factors in $R [x]$ . The factorization of $g (x)$ is unique up to multiplication by units in $R [x]$ and rearrangement of factors. By Theorem 1.3 and Theorems 1.1 of §2T and 1.3 of §2T, $F [x]$ is a UFD, so $g (x)$ has a factorization in $F [x]$ , $g (x) = f_{1} (x) f_{2} (x) \dots f_{r} (x), where f_{i} (x) \in F [x] are irreducible in F [x] .$ Then, by Proposition 1.7 a), there exist elements $c_{1}, \dots, c_{r} \in F$ and primitive polynomials $g_{1} (x), \dots, g_{r} (x)$ such that $f_{i} (x) = c_{i} g_{i} (x), for each 1 \leq i \leq k .$ Since the factors $f_{i} (x)$ are irreducible in $F [x]$ , it follows from Proposition 1.7 c), that the polynomials $g_{i} (x)$ are irreducible in $R [x]$ . Since the $g_{i} (x)$ are primitive, by Gauss's lemma, the product $g_{g} (x) \dots g_{r} (x)$ is primitive. So $g (x) = c g_{1} (x) g_{2} (x) \dots g_{r} (x),$ where $c = c_{1} c_{2} \dots c_{r} \in F$ . We also know that $g (x) = gcd (a_{0} \dots a_{k}) g^{'} (x)$ where $g^{'} (x)$ is a primitive polynomial in $R [x]$ . Thus, by Proposition 1.7 b), $c = u gcd (a_{0} \dots a_{k})$ where $u \in R$ is a unit. It follows that $c \in R$ . Since $R$ is a UFD, $c$ has a factorization $c = d_{1} \dots d_{s},$ where the elements $d_{j}$ are irreducible elements in $R$ . So $g (x) = d_{1} \dots d_{s} \dots g_{1} (x) \dots g_{r} (x),$ is a factorization of $g (x)$ into irreducibles in $R [x]$ . Suppose that $g (x) = d_{1}^{'} d_{2}^{'} \dots d_{l}^{'} g_{1}^{'} \dots g_{m}^{'},$ is another factorization of $g (x)$ into irredicibles in $R [x]$ . By Proposition 1.7 c), each of the factors $g_{i}^{'} (x)$ is irreducible in $F [x]$ . So $g (x) = d_{1}^{'} d_{2}^{'} \dots d_{l}^{'} g_{1}^{'} \dots g_{m}^{'},$ and $g (x) = d_{1} \dots d_{s} g_{1} (x) \dots g_{r} (x)$ are both factorizations of $g (x)$ in $F [x]$ . By Theorem 1.3 and Theorems 1.1 of §2T and 1.3 of §2T, $F [x]$ is a UFD, so $r = m$ and there is a permutation $σ$ such that $g_{σ (i)}^{'} = α_{i} g_{i} (x)$ for some $α_{i}$ which are units in $F$ . Proposition 1.7 b), gives that each $α_{i}$ is a unit in $R$ . Let $u = α_{1} α_{2} \dots α_{r}$ . Then $\begin{array}{rcl} g (x) & = & d_{1} \dots d_{s} \cdot g_{1} (x) \dots g_{r} (x) \\ = & d_{1}^{'} d_{2}^{'} \dots d_{l}^{'} g_{1}^{'} (x) g_{2}^{'} (x) \dots g_{m}^{'} (x) \\ = & u d_{1}^{'} d_{2}^{'} \dots d_{l}^{'} g_{1} (x) g_{2} (x) \dots g_{m} (x) . \end{array}$ Theny Proposition 1.7 b) implies there is a unit $v \in R$ such that $d_{1} \dots d_{l} = v u d_{1}^{'} \dots d_{l}^{'} .$ Since $R [x]$ is a UFD, $s = l$ and there is a permutation $τ$ such that $d_{τ (i)} = u_{i} d_{i}^{'},$ wehere the $u_{i}$ are units in $R$ . So there is a rearrangement of the factors $d_{i}^{'}$ and $g_{j}^{'}$ such that, up to multipication by units in $R$ , they are the same as the factors $d_{i}$ and $g_{j} (x)$ . So the factorization of $g (x)$ in $R [x]$ is unique. So $R [x]$ is a UFD. $□$

Let $R$ be a UFD. For each irreducible element $p \in R$ , let $π_{p} : R \to R / (p)$ be the canonical surjection (Part 1, Ex. 2.1.5). Let ${\hat{π}}_{p} : R [x] \to R [x] / (p) [x]$ be the corresponding homomorphism between polynomial rings (Prop 3.1.6). Let $f (x) \in R [x]$ . Then $f (x)$ is not primitive if and only if there exists an irreducible element $p \in R$ such that ${\hat{π}}_{p} (f (x)) = 0$ .

		Proof.
	$\Rightarrow$ Assume $f (x) = c_{0} + c_{1} x + \dots c_{k} x^{k}$ is not primitive. Then there exists $p \in R$ irreducible such that $p$ divides $c_{0}$ , $p$ divides $c_{1}$ ,…, $p$ divides $c_{k}$ . So $c_{0}, c_{1}, \dots, c_{k} \in (p)$ . So $π_{p} (c_{0}) = π_{p} (c_{1}) = \dots = π_{p} (c_{k}) = 0$ . So ${\hat{π}}_{p} (f (x)) = π_{p} (c_{0}) + π_{p} (c_{1}) x + \dots + π_{p} (c_{k}) x^{k} = 0$ . $\Leftarrow$ Assume that $f (x) = c_{0} + c_{1} x + \dots + c_{k} x^{k}$ and that there exists an irreducible element $p \in R$ such that ${\hat{π}}_{p} (f (x)) = 0$ . Then ${\hat{π}}_{p} (f (x)) = 0$ . Then $π_{p} (c_{0}) = π_{p} (c_{1}) = \dots = π_{p} (c_{k}) = 0$ . So $c_{0}, c_{1}, \dots, c_{k} \in (p)$ . So $p$ divides $c_{0}$ , $p$ divides $c_{1}$ ,…, $p$ divides $c_{k}$ . So $f (x)$ is not primitive. $□$

(Gauss' Lemma) Let $R$ be a UFD. Let $f (x), g (x) \in R [x]$ be primitive polynomials. Then $f (x) g (x)$ is a primitive polynomial.

		Proof.
	We shall prove the contrapositive: If $f (x) g (x)$ is not primitve then either $f (x)$ is not primitive or $g (x)$ is not primitive. Assume $f (x) g (x)$ is not primitve. Then by Lemma 1.9, there exists an irreducible elemnet $p \in R$ such that ${\hat{π}}_{p} (f (x) g (x)) = 0,$ where ${\hat{π}}_{p} : R [x] \to R [x] / (p) [x]$ is the homomorphism between polynomial rings induced by the canonical surjection $π_{p} : R \to R / (p) .$ Since ${\hat{π}}_{p}$ is a homomorphism, ${\hat{π}}_{p} (f (x) g (x)) = {\hat{π}}_{p} (f (x)) {\hat{π}}_{p} (g (x)) = 0 .$ Thus, by Lemma 1.4 of §2T, $(p)$ is a prime ideal. Thus by Proposition 1.4 of §1T, $R / (p)$ is an integral domain. So either ${\hat{π}}_{p} (f (x)) = 0 or {\hat{π}}_{p} (g (x)) = 0 .$ Thus, by Lemma 1.9, $\begin{matrix} either & f (x) is not primitive \\ or & g (x) is not primitive. \end{matrix}$ $□$

References

[CM] H. S. M. Coxeter and W. O. J. Moser, Generators and relations for discrete groups, Fourth edition. Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], 14. Springer-Verlag, Berlin-New York, 1980. MR0562913 (81a:20001)

[GW1] F. Goodman and H. Wenzl, The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334. MR1242201 (95c:16020)

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