§3P. Polynomial Rings

## Polynomial rings

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

 Proof. To show: If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then $f\left(x\right)+g\left(x\right)=g\left(x\right)+f\left(x\right)$. If $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ then $\left(f\left(x\right)+g\left(x\right)\right)+h\left(x\right)=f\left(x\right)+\left(g\left(x\right)+h\left(x\right)\right)$. There exists $0\in R\left[x\right]$ such that if $f\left(x\right)\in R\left[x\right]$ then $0+f\left(x\right)=f\left(x\right)+0=f\left(x\right)$. If $f\left(x\right)\in R\left[x\right]$ then there exists $-f\left(x\right)\in R\left[x\right]$ such that $f\left(x\right)+\left(-f\left(x\right)\right)=0$. If $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ then $\left(f\left(x\right)g\left(x\right)\right)=f\left(x\right)\left(g\left(x\right)h\left(x\right)\right)$. There exists $1\in R\left[x\right]$ such that $1\cdot f\left(x\right)=f\left(x\right)=f\left(x\right)\cdot 1$ for all $f\left(x\right)\in R\left[x\right]$. If $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ then $f\left(x\right)\left(g\left(x\right)+h\left(x\right)\right)=f\left(x\right)g\left(x\right)+f\left(x\right)h\left(x\right)$ and $\left(g\left(x\right)+h\left(x\right)\right)f\left(x\right)=g\left(x\right)f\left(x\right)+h\left(x\right)f\left(x\right)$. If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then $f\left(x\right)g\left(x\right)=g\left(x\right)h\left(x\right)$. Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ such that $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots$. Then $fx+ gx = r0+s0+ r1+s1x+ r2+s2 x2+⋯ and gx+ fx = s0+r0+ s1+r1x+ s2+r2 x2+⋯$ Since addition in $R$ is a commutative operation, So $f\left(x\right)+g\left(x\right)=g\left(x\right)+f\left(x\right)$. Let $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ and $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$, $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots$, and $h\left(x\right)={t}_{0}+{t}_{1}x+{t}_{2}{x}^{2}+\cdots$. Then $fx+ gx + hx = r0+s0+ t0 + r1+s1+ t1 x+ r2+s2+ t2 x2+⋯ and fx+ gx+ hx = r0+ s0+t0 + r1+ s1+t1 x+ r2+ s2+t2 x2+⋯.$ Since addition in $R$ is an associative operation, So $\left(f\left(x\right)+g\left(x\right)\right)+h\left(x\right)=f\left(x\right)+\left(g\left(x\right)+h\left(x\right)\right)$. Let $0$ denote the zero polynomial $0= 0+0x+0x2+⋯.$ Let $f\left(x\right)\in R\left[x\right]$ and let $fx= r0+r1x +r2x2+⋯.$ Then $0+fx= 0+r0+ 0+r1x+ 0+r2x2 +⋯.$ Since $0+{r}_{i}={r}_{i}$ for all $i$, $0+fx =fx.$ Since addition of polynomials is commutative by a), $0+f\left(x\right)=f\left(x\right)+0=f\left(x\right).$ Let $f\left(x\right)\in R\left[x\right]$ such that $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$. Then let $-f\left(x\right)=-{r}_{0}+\left(-{r}_{1}\right)x+\left(-{r}_{2}\right){x}^{2}+\cdots$. Since $-{r}_{i}\in R$ for all $i$, $-f\left(x\right)\in R\left[x\right]$. Then $fx+ -fx= r0-r0+ r1-r1x+ r2-r2x2 +⋯.$ So $fx+ -fx= 0+0x+0x2+⋯=0.$ Since addition of polynomials is commutative by a), $f\left(x\right)+\left(-f\left(x\right)\right)=-f\left(x\right)+f\left(x\right)=0$. Let $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots ,$ $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots ,$ and $h\left(x\right)={t}_{0}+{t}_{1}x+{t}_{2}{x}^{2}+\cdots .$ Then $fxgx =c0+c1x+ c2x2+⋯, where ck= ∑ i+j=k risj, and fx gx hx= d0+d1x+ d2x2+⋯ where dn= ∑ k+l=n cktl.$ So $dn= ∑ k+l=n ∑ i+j=k risj tl= ∑ i+j+l=n risjtl,$ by the distributive law in $R$. Also $gxhx =e0+e1x+ e2x2+⋯, where eq= ∑ a+b=q satb, and fx gx hx = d0′+d1′x + d2′ x2+⋯, where dn′ = ∑ p+q=n rpeq.$ Then $dn′ = ∑ p+q=n rp ∑ a+b=q satb = ∑ p+a+b=n rpsatb,$ by the distributive law in $R$. So ${d}_{n}{d}_{n}^{\prime }$. So $\left(f\left(x\right)g\left(x\right)\right)\left(h\left(x\right)\right)=\left(f\left(x\right)\right)\left(g\left(x\right)h\left(x\right)\right)$. Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots .$ Then $fxgx =c0+ c1x+ c2x2+⋯, where cm= ∑ i+j=k risj, and gxfx = c0′ + c1′ x+ c2′ x2+⋯, where ck′ = ∑ i+j=k sjri.$ Since $R$ is a commutative ring, So $f\left(x\right)g\left(x\right)=g\left(x\right)f\left(x\right)$. Let $1\in R\left[x\right]$ be the polynomial given by $1= 1+0x+0x2+⋯.$ Let $f\left(x\right)\in R\left[x\right]$ and $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots .$ Then So ${c}_{k}={a}_{0}{r}_{k}+0+0+\cdots +0={r}_{k}$ So $1\cdot f\left(x\right)=f\left(x\right).$ Since multiplication in $R\left[x\right]$ is commutative by h), $1\cdot f\left(x\right)=f\left(x\right)\cdot 1=f\left(x\right).$ Let $f\left(x\right),g\left(x\right),h\left(x\right)\in R\left[x\right]$ and suppose $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$, $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots$, and $h\left(x\right)={t}_{0}+{t}_{1}x+{t}_{2}{x}^{2}+\cdots .$ Then $fx gx+ hx = c0+ c1x+ c2x2+⋯ where ck= ∑ i+j=k ri sj+tj.$ Also $f\left(x\right)g\left(x\right)+f\left(x\right)h\left(x\right)={c}_{0}^{\prime }+{c}_{1}^{\prime }x+{c}_{2}^{\prime }{x}^{2}+\cdots$ where $ck′ = ∑ m+m=k rmsn+ ∑ m+n=k rmtn= ∑ m+n=k rmsn+ rmtn .$ So ${c}_{k}={c}_{k}^{\prime }$ for all $k$. Thus $fx gx+ hx = fx gx + fx hx.$ Sine multiplication in $R\left[x\right]$ is commutative by h), $gx+ hx fx = fx gx+ hx = fx gx + fx hx = gx fx + hx fx.$ So $R\left[x\right]$ is a commutative ring. $\square$

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

 Proof. To show: If $a\left(x\right),b\left(x\right)\in R\left[x\right]$ and $a\left(x\right)b\left(x\right)=0$ then either $a\left(x\right)=0$ or $b\left(x\right)=0$. Let $a\left(x\right)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots$ and let $b\left(x\right)={b}_{0}+{b}_{1}x+{b}_{2}{x}^{2}+\cdots .$ Let $c\left(x\right)=a\left(x\right)b\left(x\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots .$ Assume $a\left(x\right)\ne 0$. Then ${a}_{l}\ne 0$ for some $l\ge 0$. Let $k$ be the smallest $k\ge 0$ such that ${a}_{k}\ne 0$. To show: $b\left(x\right)=0$. To show: ${b}_{N}=0$ for all $N\ge 0$. Proof by induction on $N$. Base case: $N=0$. We know ${c}_{k}=0$ since $c\left(x\right)=a\left(x\right)b\left(x\right)=0.$ So $∑ i+j=k aibj=0.$ Since ${a}_{i}=0$ for all $i, $0= ∑ i+j=k aibj= akb0.$ Since $R$ is an integral domain and ${a}_{k},{b}_{0}\in R$ and ${a}_{k}\ne 0$, we have ${b}_{0}=0$. Induction assumption: assume ${b}_{n}=0$ for all $n. We know ${c}_{k+N}=0$ since $c\left(x\right)=a\left(x\right)b\left(x\right)=0$. So $∑ i+j=k+N ai bj=0.$ Since $ai=0$ for all $i and ${b}_{n}=0$ for all $n, we have $0= ∑ i+j=k+N ai bj =akbN.$ Since $R$ is an integral domain, and ${a}_{k},{b}_{N}\in R$ and ${a}_{k}\ne 0$, we have ${b}_{N}=0$. So ${b}_{N}=0$ for all $N\ge 0$. So $b\left(x\right)=0$. So $R\left[x\right]$ is an integral domain. $\square$

Let $F$ be a field. The ring $F\left[x\right]$ is a Euclidean domain with size function given by $deg: F[x] → ℕ f(x) ↦ deg(f(x)).$

 Proof. To show: If $a\left(x\right),b\left(x\right)\in F\left[x\right]$ and $a\left(x\right)\ne 0$ then there exist $q\left(x\right),r\left(x\right)\in F\left[x\right]$ such that $bx= axqx+ rx$ where either $r\left(x\right)=0$ or $deg\left(r\left(x\right)\right) Assume $a\left(x\right),b\left(x\right)\in F\left[x\right]$ and $a\left(x\right)\ne 0$. Case 1: $b\left(x\right)=0$. Then $b\left(x\right)=a\left(x\right)\cdot 0+0.$ So $q\left(x\right)=0$ and $r\left(x\right)=0$ satisfies the condition. Case 2: $deg\left(b\left(x\right)\right). Then, since $bx= ax⋅0+ bx and degbx< degax,$ $q\left(x\right)=0$ and $r\left(x\right)=b\left(x\right)$ satisfies the condition. Case 3: $deg\left(b\left(x\right)\right)\ge deg\left(a\left(x\right)\right)$. Let $a\left(x\right)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\cdots +{a}_{s}{x}^{s}$ and $b\left(x\right)={b}_{0}+{b}_{1}x+{b}_{2}{x}^{2}+\cdots +{b}_{t}{x}^{t}$, where ${a}_{s},{b}_{t}\in F\left[x\right]$, ${a}_{s}\ne 0$, ${b}_{t}\ne 0$. Proof by induction on $deg\left(b\left(x\right)\right)$. Base case: $deg\left(b\left(x\right)\right)=0$. Then $deg\left(a\left(x\right)\right)=0$ since $deg\left(a\left(x\right)\right)\le deg\left(b\left(x\right)\right)$. So $b\left(x\right)={b}_{0}\in F\left[x\right]$ and $a\left(x\right)={a}_{0}\in F\left[x\right]$. So $b\left(x\right)=\left(\frac{{b}_{0}}{{a}_{0}}\right)\cdot a\left(x\right)+0$. So $q\left(x\right)={b}_{0}{a}_{0}^{-1}$ and $r\left(x\right)=0$ satisfies the condition. Induction assumption: Assume that if ${b}_{1}\left(x\right)\in F\left[x\right]$ and $deg\left({b}_{1}\left(x\right)\right) then there exist ${q}_{1}\left(x\right),{r}_{1}\left(x\right)\in F\left[x\right]$ such that $b1x= q1x ax+ r1x$ where either ${r}_{1}\left(x\right)=0$ or $deg\left({r}_{1}\left(x\right)\right). Assume that $deg\left(b\left(x\right)\right)=t$. Let ${b}_{1}\left(x\right)=b\left(x\right)-\left({b}_{t}{a}_{s}^{-1}{x}^{t-s}\right)a\left(x\right)$. Then $deg\left({b}_{1}\left(x\right)\right). Note that the coefficient of ${x}^{t}$ in ${b}_{1}\left(x\right)$ is $-{b}_{t}+{b}_{t}=0$. So $deg\left({b}_{1}\left(x\right)\right). Thus, by the induction assumption, there exist ${q}_{1}\left(x\right),{r}_{1}\left(x\right)\in F\left[x\right]$ such that $b1x= q1x+ r1x$ where either ${r}_{1}\left(x\right)=0$ or $deg\left({r}_{1}\left(x\right)\right). Then $bx = b1x- bt as-1 xs-t ax = q1x ax+ r1x - bt as-1 xs-t ax = q1x- bt as-1 xs-t ax +r1x.$ So, if $q\left(x\right)={q}_{1}\left(x\right)-{b}_{t}{a}_{s}^{-1}{x}^{s-t}$ and $r\left(x\right)={r}_{1}\left(x\right)$ then $bx= qxax +rx$ and either $r\left(x\right)=0$ or $deg\left(r\left(x\right)\right) So $F\left[x\right]$ with the size function given by $deg$ is a Euclidean domain. $\square$

Let $R,S$ be commutative rings and $\varphi :R↦S$ be a ring homomorphism. Then the map $ψ: R[x] → S[x] r0+r1x+r2x2 +⋯ ↦ φ(r0)+φ(r1)x+φ(r2)x2 +⋯$ is a ring homomorphism.

 Proof. To show: If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then $\psi \left(f\left(x\right)+g\left(x\right)\right)=\psi \left(f\left(x\right)\right)+\psi \left(g\left(x\right)\right)$. If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then $\psi \left(f\left(x\right)g\left(x\right)\right)=\psi \left(f\left(x\right)\right)\psi \left(g\left(x\right)\right)$. $\psi \left({1}_{R}\right)={1}_{S}$ where ${1}_{R}$ and ${1}_{S}$ are the identities in $R$ and $S$. Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={r}_{0}^{\prime }+{r}_{1}^{\prime }x+{r}_{2}^{\prime }{x}^{2}+\cdots$. Then $ψ fx+gx = ψ r0+ r0′ + r1+ r1′ x+ r2+ r2′ x2+⋯ = φ r0+ r0′ +φ r1+ r1′ x+φ r2+ r2′ x2+⋯$ Since $\phi$ is a homomorphism, $φ fx+gx = φr0+ φ r0′ + φr1+φ r1′ x+ φr2+ φ r2′ x2+⋯ = φr0+ φr1x+ φr2x2+⋯ + φ r0′ +φ r1′ x+φ r2′ x2+⋯ = ψ fx + ψ gx .$ Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={r}_{0}^{\prime }+{r}_{1}^{\prime }x+{r}_{2}^{\prime }{x}^{2}+\cdots$. Then $ψ fxgx = ψ c0+ c1x+ c2x2+⋯ , where ck= ∑ i+j=k ri rj′ .$ So $\psi \left(f\left(x\right)g\left(x\right)\right)=\phi \left({c}_{0}\right)+\phi \left({c}_{1}\right)x+\phi \left({c}_{2}\right){x}^{2}+\cdots .$ Since $\phi$ is a homomorphism, $φck= φ ∑ i+j=k ri rj′ = ∑ i+j=k φ ri rj′ = ∑ i+j=k φri φ rj′ .$ So $ψ fxgx = d0+d1x+ d2x2+⋯, where dk= ∑ i+j=k φri φ rj′ .$ So, by the distributive law in $S$, $ψ fxgx = φr0+ φr1x+ φr2x2+⋯ φ r0′ + φ r1′ x+ φ r2′ x2+⋯ = φfx φgx.$ Let ${1}_{R}$ be the identity in $R$. $ψ1R = ψ 1R+0Rx +0Rx2+⋯ = φ1R+ φ0Rx+ φ0Rx2+⋯$ Since $\phi$ is a homomorphism, $\phi \left({1}_{R}\right)={1}_{S}$ and $\phi \left({0}_{R}\right)={0}_{S}$. So $\psi \left({1}_{R}\right)={1}_{S}+{0}_{S}x+{0}_{S}{x}^{2}+\cdots .$ So $\psi$ is a homomorphism. $\square$

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

 Proof. To show: If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then ${ev}_{\alpha }\left(f\left(x\right)+g\left(x\right)\right)={ev}_{\alpha }\left(f\left(x\right)\right)+{ev}_{\alpha }\left(g\left(x\right)\right)$. If $f\left(x\right),g\left(x\right)\in R\left[x\right]$ then ${ev}_{\alpha }\left(f\left(x\right)g\left(x\right)\right)={ev}_{\alpha }\left(f\left(x\right)\right){ev}_{\alpha }\left(f\left(x\right)\right).$ ${ev}_{\alpha }\left({1}_{R}\right)={1}_{R}$ where ${1}_{R}$ is the identity in $R$. Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots .$ Then $evα fx+gx = evα r0+s0+ r1+s1x+ r2+s2x2 +⋯ = r0+s0+ r1+s1α+ r2+s2 α2+⋯.$ By the distributive law in $R$, $evα fx+gx = r0+s0+ r1α+s1α+ r2α2+ s2α2+⋯ = r0+ r1α+ r2α2+⋯ + s0+ s1α+ r2α2+⋯ = evα fx+ evα gx.$ Let $f\left(x\right),g\left(x\right)\in R\left[x\right]$ and let $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots .$ Then $evα fxgx = evα c0+c1x+ c2x2+⋯ , where ck= ∑ i+j=k rirj.$ So ${ev}_{\alpha }\left(f\left(x\right)g\left(x\right)\right)={c}_{0}+{c}_{1}\alpha +{c}_{2}{\alpha }^{2}+\cdots$. Now compute ${ev}_{\alpha }\left(f\left(x\right)\right){ev}_{\alpha }\left(g\left(x\right)\right).$ $evα fx evα gx= r0+r1α+ r2α2+⋯ s0+s1α+ s2α2+⋯ .$ By the distributive law in $R$, $evα fx evα gx = r0s0+ r1s0α+ r0s1α+ r0s2α2+ r1s1α2+ r2s0α2+⋯ = r0s0+ r1s0+ r0s1 α+ r0s2+ r1s1+ r2s0 α2+⋯ = c0+ c1α+ c2α2+⋯ where$ $ck= ∑ i+j=k risj.$ So $evα fxgx = evα fx evα gx.$ Let ${1}_{R}$ be the identity in $R$ and ${0}_{R}$ be the zero in $R$. Then $evα1R= evα 1R+0Rx+ 0Rx2+⋯ = 1R+0Rα+ 0Rα2+⋯ =1R.$ So ${ev}_{\alpha }$ is a ring homomorphism. $\square$

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

 Proof. Assume $f\left(x\right)={r}_{0}+{r}_{1}x+{r}_{2}{x}^{2}+\cdots$ and $g\left(x\right)={s}_{0}+{s}_{1}x+{s}_{2}{x}^{2}+\cdots$ are primitive polynomials in $R\left[x\right]$. Proof by contradiction. Assume $f\left(x\right)g\left(x\right)$ is not primitive. Then there exists an irreducible element $p\in R$ that divides all the coefficients of $f\left(x\right)g\left(x\right).$ Since $f\left(x\right)$ is primitive there must be at least one coefficient of $f\left(x\right)$ which is not divisible by $p$. Since $g\left(x\right)$ is primitive there must be at least one coefficient of $g\left(x\right)$ 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 $f\left(x\right)g\left(x\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots .$ Then, since $p$ divides ${r}_{i}$ for all $i, and $p$ divides ${s}_{j}$ for all $j, $cm+n = rmsn+ rm-1sn+1+ rm+1sn-1+ ⋯+ r0sm+n+ rm+ns0 = rmsn+pc,$ where $c$ is some element of $R$. Since ${c}_{m+n}$ is divisible by $p$ it follows that ${r}_{m}{s}_{n}={c}_{m+n}-pc$ is divisible by $p$. Suppose that $d\in R$ such that ${r}_{m}{s}_{n}=pd.$ Let $rm= a1⋯ak, sn= b1⋯bl, and d= d1⋯dq,$ 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 $a1⋯ak b1⋯bl= pd1⋯dq.$ By the uniqueness of factorizations, So either ${p}_{i}=up$ or ${q}_{j}=up$ for some unit $u\in R$. Then, either $a=up{a}_{1}\cdots {a}_{i-1}{a}_{i+1}\cdots {a}_{k}$, or $b=up{d}_{1}\cdots {d}_{j-1}{d}_{j+1}\cdots {d}_{q}$. Thus, either ${r}_{m}$ or ${s}_{n}$ is divisible by ${a}_{i}$ or ${d}_{j}$. Contradiction. So $f\left(x\right)g\left(x\right)$ is a primitive polynomial in $R\left[x\right]$. $\square$

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

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

 Proof. Let $f\left(x\right)=\frac{{a}_{0}}{{b}_{0}}+\frac{{a}_{1}}{{b}_{1}}x+\cdots +\frac{{a}_{k}}{{b}_{k}}{x}^{k}\in F\left[x\right]$. Then $f\left(x\right)=\frac{1}{{b}_{0}{b}_{1}\cdots {b}_{k}}\left({c}_{0}+{c}_{1}x+\cdots {c}_{k}{x}^{k}\right)$ where ${c}_{i}={a}_{i}{b}_{1}\cdots {b}_{i-1}{b}_{i+1}\cdots {b}_{k}$. Let $d=gcd\left({c}_{0}{c}_{1}\cdots {c}_{k}\right)$. Then $fx= d b0⋯bk c0′ + c1′ x+⋯+ ck′ xk$ where ${c}_{i}^{\prime }=\frac{{c}_{i}}{d}$. Note that ${c}_{i}^{\prime }\in R$ since $d$ divides ${c}_{i}$. Furthermore ${c}_{0}^{\prime }+{c}_{1}^{\prime }x+\cdots +{c}_{k}^{\prime }{x}^{k}=g\left(x\right)$ is primitive since $gcd\left({c}_{0}^{\prime },{c}_{1}^{\prime },\dots ,{c}_{k}^{\prime }\right)=1$. So $fx= cgx$ where $c=\frac{d}{{b}_{0}{b}_{1}\dots {b}_{k}}\in F$ and $g\left(x\right)={c}_{0}^{\prime }+{c}_{1}^{\prime }x+\cdots +{c}_{k}^{\prime }{x}^{k}\in R\left[x\right]$ is a primitive polynomial. Suppose $f\left(x\right)=cg\left(x\right)$ and $f\left(x\right)=CG\left(x\right)$ where $c,C\in F$ and $g\left(x\right),G\left(x\right)\in R\left[x\right]$ are primitive polynomials. Let $g\left(x\right)={a}_{0}+{a}_{1}x+\cdots +{a}_{k}{x}^{k}$ and let $G\left(x\right)={b}_{0}+{b}_{1}x+\cdots +{b}_{k}{x}^{k}$. Suppose $c=\frac{a}{b}$ and $C=\frac{A}{B}$ where $a,b,A,B\in R$. Since $f\left(x\right)=\frac{a}{b}g\left(x\right)=\frac{A}{B}G\left(x\right)$, $aBgx= bAGx.$ So $aB{a}_{i}=bA{b}_{i}$ for all $1\le i\le k$. Since $g\left(x\right)$ is primitive, $gcd\left(aB{a}_{0},aB{a}_{1},\dots ,aB{a}_{k}\right)=aB$. Since $G\left(x\right)$ is primitive, $gcd\left(bA{b}_{0},bA{b}_{1},\dots ,bA{b}_{k}\right)=bA$. Thus by Proposition 1.6 of §2.T, So $\frac{a}{b}=u\left(\frac{A}{B}\right)$. So $c=uC$ where $u\in R$ is a unit. So $CG\left(x\right)=cg\left(x\right)=uCg\left(x\right)$. By the cancelation law, $G\left(x\right)=ug\left(x\right)$. So $c=uC$ and $G\left(x\right)=ug\left(x\right)$. So $c$ and $g\left(x\right)$ are unique up to multiplication by a unit. $⇒$ Assume that $f\left(x\right)$ is irreducible in $F\left[x\right]$. Proof by contradiction. Assume $g\left(x\right)$ is not irreducible in $R\left[x\right]$. Then there are ${g}_{1}\left(x\right)$ and ${g}_{2}\left(x\right)$ in $R\left[x\right]$ such that $g\left(x\right)={g}_{1}\left(x\right){g}_{2}\left(x\right)$. So $f\left(x\right)=cg\left(x\right)=c{g}_{1}\left(x\right){g}_{1}\left(x\right)$. Since $R\left[x\right]\subseteq F\left[x\right]$ then ${g}_{1}\left(x\right),{g}_{2}\left(x\right)\in F\left[x\right]$. Contradiction. So $g\left(x\right)$ is irreducible in $R\left[x\right]$. ⇐ Assume $g\left(x\right)$ is irreducible in $R\left[x\right]$. Proof by contradiction. Assume $f\left(x\right)$ is not irreducible in $F\left[x\right]$. Then there are ${f}_{1}\left(x\right)$ and ${f}_{2}\left(x\right)$ in $F\left[x\right]$ such that $f\left(x\right)={f}_{1}\left(x\right){f}_{2}\left(x\right)$. So by a), there exist ${c}_{1},{c}_{2}\in F$ and primitive polynomials ${g}_{1}\left(x\right),{g}_{2}\left(x\right)\in R\left[x\right]$ such that $f1x= c1g1x and f2x= c2g2x.$ Let $c={c}_{1}{c}_{2}$. Then $f\left(x\right)={c}_{1}{c}_{2}{g}_{1}\left(x\right){g}_{2}\left(x\right)$. By Gauss' lemma ${g}_{1}\left(x\right){g}_{2}\left(x\right)$ is a primitive polynomial in $R\left[x\right]$. So, by b), $g\left(x\right)=u{g}_{1}\left(x\right){g}_{2}\left(x\right)$, where $u\in R$. So $g\left(x\right)$ is not irreducible in $R\left[x\right]$. Contradiction. So $f\left(x\right)$ is irreducible in $F\left[x\right]$. $\square$

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

 Proof. Assume $g\left(x\right)\in R\left[x\right]$ and let $g\left(x\right)={a}_{0}+{a}_{1}x++\cdots +{a}_{k}{x}^{k}$. To show: $g\left(x\right)$ has a unique factorization into irreducible factors in $R\left[x\right]$. The factorization of $g\left(x\right)$ is unique up to multiplication by units in $R\left[x\right]$ and rearrangement of factors. By Theorem 1.3 and Theorems 1.1 of §2T and 1.3 of §2T, $F\left[x\right]$ is a UFD, so $g\left(x\right)$ has a factorization in $F\left[x\right]$, Then, by Proposition 1.7 a), there exist elements ${c}_{1},\dots ,{c}_{r}\in F$ and primitive polynomials ${g}_{1}\left(x\right),\dots ,{g}_{r}\left(x\right)$ such that Since the factors ${f}_{i}\left(x\right)$ are irreducible in $F\left[x\right]$, it follows from Proposition 1.7 c), that the polynomials ${g}_{i}\left(x\right)$ are irreducible in $R\left[x\right]$. Since the ${g}_{i}\left(x\right)$ are primitive, by Gauss's lemma, the product ${g}_{g}\left(x\right)\cdots {g}_{r}\left(x\right)$ is primitive. So $gx= cg1x g2x⋯ grx,$ where $c={c}_{1}{c}_{2}\cdots {c}_{r}\in F$. We also know that $g\left(x\right)=gcd\left({a}_{0}\cdots {a}_{k}\right){g}^{\prime }\left(x\right)$ where ${g}^{\prime }\left(x\right)$ is a primitive polynomial in $R\left[x\right]$. Thus, by Proposition 1.7 b), $c=ugcd\left({a}_{0}\cdots {a}_{k}\right)$ where $u\in R$ is a unit. It follows that $c\in R$. Since $R$ is a UFD, $c$ has a factorization $c=d1⋯ds,$ where the elements ${d}_{j}$ are irreducible elements in $R$. So $gx= d1⋯ds⋯ g1x⋯ grx,$ is a factorization of $g\left(x\right)$ into irreducibles in $R\left[x\right]$. Suppose that $g\left(x\right)={d}_{1}^{\prime }{d}_{2}^{\prime }\cdots {d}_{l}^{\prime }{g}_{1}^{\prime }\cdots {g}_{m}^{\prime },$ is another factorization of $g\left(x\right)$ into irredicibles in $R\left[x\right]$. By Proposition 1.7 c), each of the factors ${g}_{i}^{\prime }\left(x\right)$ is irreducible in $F\left[x\right]$. So $g\left(x\right)={d}_{1}^{\prime }{d}_{2}^{\prime }\cdots {d}_{l}^{\prime }{g}_{1}^{\prime }\cdots {g}_{m}^{\prime },$ and $g\left(x\right)={d}_{1}\cdots {d}_{s}{g}_{1}\left(x\right)\cdots {g}_{r}\left(x\right)$ are both factorizations of $g\left(x\right)$ in $F\left[x\right]$. By Theorem 1.3 and Theorems 1.1 of §2T and 1.3 of §2T, $F\left[x\right]$ is a UFD, so $r=m$ and there is a permutation $\sigma$ such that $gσi′ = αigix$ for some ${\alpha }_{i}$ which are units in $F$. Proposition 1.7 b), gives that each ${\alpha }_{i}$ is a unit in $R$. Let $u={\alpha }_{1}{\alpha }_{2}\cdots {\alpha }_{r}$. Then $gx = d1⋯ds⋅ g1x⋯ grx = d1′ d2′ ⋯ dl′ g1′ x g2′ x ⋯ gm′ x = u d1′ d2′ ⋯ dl′ g1x g2x⋯ gmx.$ Theny Proposition 1.7 b) implies there is a unit $v\in R$ such that $d1⋯dl= vu d1′ ⋯ dl′ .$ Since $R\left[x\right]$ is a UFD, $s=l$ and there is a permutation $\tau$ such that $dτi =ui di′ ,$ wehere the ${u}_{i}$ are units in $R$. So there is a rearrangement of the factors ${d}_{i}^{\prime }$ and ${g}_{j}^{\prime }$ such that, up to multipication by units in $R$, they are the same as the factors ${d}_{i}$ and ${g}_{j}\left(x\right)$. So the factorization of $g\left(x\right)$ in $R\left[x\right]$ is unique. So $R\left[x\right]$ is a UFD. $\square$

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

 Proof. $⇒$ Assume $f\left(x\right)={c}_{0}+{c}_{1}x+\cdots {c}_{k}{x}^{k}$ 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 ${c}_{0},{c}_{1},\dots ,{c}_{k}\in \left(p\right)$. So ${\pi }_{p}\left({c}_{0}\right)={\pi }_{p}\left({c}_{1}\right)=\cdots ={\pi }_{p}\left({c}_{k}\right)=0$. So ${\stackrel{ˆ}{\pi }}_{p}\left(f\left(x\right)\right)={\pi }_{p}\left({c}_{0}\right)+{\pi }_{p}\left({c}_{1}\right)x+\cdots +{\pi }_{p}\left({c}_{k}\right){x}^{k}=0$. $⇐$ Assume that $f\left(x\right)={c}_{0}+{c}_{1}x+\cdots +{c}_{k}{x}^{k}$ and that there exists an irreducible element $p\in R$ such that ${\stackrel{ˆ}{\pi }}_{p}\left(f\left(x\right)\right)=0$. Then ${\stackrel{ˆ}{\pi }}_{p}\left(f\left(x\right)\right)=0$. Then ${\pi }_{p}\left({c}_{0}\right)={\pi }_{p}\left({c}_{1}\right)=\cdots ={\pi }_{p}\left({c}_{k}\right)=0$. So ${c}_{0},{c}_{1},\dots ,{c}_{k}\in \left(p\right)$. So $p$ divides ${c}_{0}$, $p$ divides ${c}_{1}$,…, $p$ divides ${c}_{k}$. So $f\left(x\right)$ is not primitive. $\square$

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

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

## References

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