{"id":413,"date":"2023-06-20T16:43:55","date_gmt":"2023-06-20T08:43:55","guid":{"rendered":"https:\/\/qwq.cafe\/?p=413"},"modified":"2023-06-20T16:57:03","modified_gmt":"2023-06-20T08:57:03","slug":"abstract-algebra-1-9-unique-factorization-domains","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=413","title":{"rendered":"Abstract Algebra \u2013 1.9 Unique Factorization Domains"},"content":{"rendered":"

Definition 1.9.1<\/strong> A domain $R$ is a unique factorization domain (UFD)<\/strong> or factorial ring<\/strong> if<\/p>\n

    \n
  1. every $r\\in R$, neither $0$ or a unit, is a product of irreducibles;<\/li>\n
  2. if $p_1\\cdots p_m=q_1\\cdots q_n$ where all $p_i$ and $q_j$ are irreducible, then $m=n$ and there is a permutation $\\sigma\\in S_n$ with $p_i$ and $q_{\\sigma(i)}$ associates for all $i$.<\/li>\n<\/ol>\n

    Proposition 1.9.1<\/strong> Let $R$ be a domain in which every $r\\in R$, neither $0$ nor a unit, is a product of irreducibles. Then $R$ is a UFD if and only if $(p)$ is a prime ideal in $R$ for every irreducible element $p\\in R$.<\/p>\n

    Proof.<\/p>\n

      \n
    • \n

      ($\\Rightarrow$).<\/p>\n

      Let $p\\in R$ be an irreducible element.<\/p>\n

      $\\forall ab\\in (p)$, assume that $a\\notin (p)$.<\/p>\n

      For $ab\\in (p)$, $\\exists r\\in R$ with $ab=rp$.<\/p>\n

      Then let $a=p_1\\cdots p_n,b=q_1\\cdots q_m,$ and $r=r_1\\cdots r_t$ with $p$’s, $q$’s, and $r$’s are irreducible.<\/p>\n

      $p_1\\cdots p_nq_1\\cdots q_m=r_1\\cdots r_tp$, then $\\exists j$ such that $q_j$ and $p$ associates.<\/p>\n

      Therefore $b\\in (p)$, and $(p)$ is a prime ideal.<\/p>\n<\/li>\n

    • \n

      ($\\Leftarrow$).<\/p>\n

      Let $a\\in R$ with $a=p_1\\cdots p_n=q_1\\cdots q_m$ in which $p$’s and $q$’s are irreducible.<\/p>\n

      By induction on $M=\\max \\{m,n\\}$.<\/p>\n

      If $M=1$, we have $p_1=q_1$.<\/p>\n

      For inductive step, because $q_1\\cdots q_m\\in(p_n)$, $\\exists j,q_j\\in (p_n)$; that is $q_j$ and $p_n$ are associates.<\/p>\n

      By reindexing, let $q_m=up_n$, where $u$ is a unit.<\/p>\n

      Then by cancelling,we have $p_1\\cdots p_{n-1}=q_1\\cdots (uq_{m-1})$. <\/p>\n

      Since $q_{m-1}u$ is irreducible, the inductive hypothesis gives $n-1=m-1$ (hence, $m=k$) and, after reindexing, $p_i$ and $q_i$ are associates for all $i$. $\\square$<\/p>\n<\/li>\n<\/ul>\n

      Lemma 1.9.1<\/strong><\/p>\n

        \n
      1. \n

        If $R$ is a commutative ring and
        \n$$I_1\\subset I_2\\subset \\cdots\\subset I_n\\subset I_{n+1}\\subset \\cdots
        \n$$
        \nis a ascending chain of ideals in $R$, then $J=\\bigcup_{n\\ge 1} I_n$ is an ideal in $R$.<\/p>\n<\/li>\n

      2. \n

        If $R$ is a PID, then it has no infinite strictly ascending chain of ideals
        \n$$I_1 \\subsetneq I_2 \\subsetneq \\cdots \\subsetneq I_{n}\\subsetneq I_{n+1}\\subsetneq \\cdots.
        \n$$<\/p>\n<\/li>\n

      3. \n

        If $R$ is a PID and $r\\in R$ is neither $0$ or a unit, then $r$ is a product of irreducibles.<\/p>\n<\/li>\n<\/ol>\n

        Proof.<\/p>\n

          \n
        1. \n
            \n
          • \n

            $\\forall x,y\\in \\bigcup_{n\\ge 1} I_n$, $\\exists m,k\\in \\mathbb N_+$ with $x\\in I_m,y\\in I_k$.<\/p>\n

            Let $M=\\max\\{m,k\\}$. Then $x,y\\in I_M$.<\/p>\n

            We have $x-y\\in I_M\\subset \\bigcup_{n\\ge 1} I_n$.<\/p>\n<\/li>\n

          • \n

            $\\forall x\\in \\bigcup_{n\\ge 1} I_n,\\exists m\\in \\mathbb N_+$ with $x\\in I_m$.<\/p>\n

            $\\forall r\\in R,rx\\in I_m\\subset \\bigcup_{n\\ge 1} I_n$ for $I_m$ is an ideal.<\/p>\n<\/li>\n<\/ul>\n

            Therefore $\\bigcup_{n\\ge 1} I_n$ is an ideal.<\/p>\n<\/li>\n

          • \n

            Assume that $I_1\\subset I_2\\subset \\cdots\\subset I_n\\subset I_{n+1}\\subset \\cdots$ is a ascending chain of ideals in $R$.<\/p>\n

            For $R$ is a PID, there is $p\\in R$ with $\\bigcup_{n\\ge 1} I_n=(p)$.<\/p>\n

            Then we have $p\\in \\bigcup_{n\\ge 1} I_n$; that is $\\exists m$ with $p\\in I_m$.<\/p>\n

            Therefore $(p)\\subset I_m\\subset (p)\\Rightarrow I_m=(p)$, so does $I_{m+1}$.<\/p>\n<\/li>\n

          • \n

            A divisor $r$ of an element $a$ is called a proper divisor<\/strong> of $a$ if $r$ is neither a unit nor an associate of $a$.<\/p>\n

            If $r$ is a divisor of $a$, then $(a)\\subset (r)$; if $r$ is a proper divisor, then $(a)\\subsetneq (r)$, for f the inequality if not strict, then $(a)=(r)$, and this force $a$ and $r$ to be associates.<\/p>\n

            Call a nonzero non-unit $a\\in R$ good<\/em>, if it is a product of irreducibles; call it bad<\/em> otherwise.<\/p>\n

            If $a$ is bad, it is not irreducible, and so $a=rs$, where $r$ and $s$ are proper divisors. But the product of good elements is good, and so at least one of the factors, say $r$, is bad. We can get $(a)\\subsetneq (r)$. It follows, by induction, that there exists a sequence $a_1=1,a_2=r,a_3,\\dots,a_n,\\dots$ of bad elements with each $a_{n+1}$ a proper divisor of $a_n$, and this sequence yields a strictly ascending chain
            \n$$ (a_1)\\subsetneq (a_2)\\subsetneq\\cdots\\subsetneq(a_n)\\subsetneq(a_{n+1})\\subsetneq\\cdots,
            \n$$
            \ncontradicting (2). $\\square$<\/p>\n<\/li>\n<\/ol>\n

            Theorem 1.9.1<\/strong> Every PID is a UFD.<\/p>\n

            Proposition 1.9.2<\/strong> If $R$ is a UFD, then a $\\gcd(a_1,\\dots,a_n)$ of any finite set of elements $a_1,\\dots,a_n$ in $R$ exists.<\/p>\n

            Proof.<\/p>\n

            We prove first that a gcd of two elements $a$ and $b$ exists. There are distinct irreducibles $p_1,\\dots,p_t$ with
            \n$$a={p_1}^{e_1}p_2^{e_2}\\cdots p_t^{e_t}\\text{ and }b=p_1^{f_1}p_2^{f_2}\\cdots p_t^{f_t},
            \n$$
            \nwhere $e_i\\ge 0$ and $f_i\\ge 0$ for all $i$. It is easy to see that if $c\\mid a$, then the factorization of $c$ into irreducibles is $c=wp_1^{g_1}\\cdots p_t^{g_t}$, where $0\\le g_i\\le e_i$ for all $i$ and $w$ is a unit. Thus, $c$ is a common divisor of $a$ and $b$ if $g_i<m_i$ for all $i$, where $m_i=\\min\\{e_i,f_i\\}$.<\/p>\n

            It is now clear that $p_1^{m_1}p_2^{m_2}\\cdots p_t^{m_t}$ is a gcd of $a$ and $b$.<\/p>\n

            More generally, if $a_i=u_ip_1^{e_{i1}}\\dots p_t^{e_{it}}$, where $e_{ij}\\ge 0$ and $i=1,\\dots,n$ and $u_i$ are units, then
            \n$$d=p_1^{\\mu_1}\\cdots p_t^{\\mu_t}
            \n$$
            \nis a gcd of $a_1,\\dots,a_n$, where $\\mu_j=\\min\\{e_{1j},\\dots,e_{nj}\\}$. $\\square$<\/p>\n

            Proposition 1.9.3<\/strong> Let $R$ be a UFD, and let $a_1,\\dots,a_n$ in $R$. An lcm of $a_1,\\dots, a_n$ exists, and
            \n$$a_1\\cdots a_n=\\gcd(a_1,\\dots,a_n)\\operatorname{lcm}(a_1,\\dots,a_n).
            \n$$
            \nDefinition 1.9.2<\/strong> A polynomial $f(x)=a_nx^n+\\cdots+a_1x+a_0\\in R[x]$, where $R$ is a UFD, is called primitive<\/strong> if its coefficients are relatively prime; that is, the only common divisors of $a_n,\\dots,a_1,a_0$ are units.<\/p>\n

            We can see that every monic polynomial is primitive. Observe that if $f(x)$ is not primitive, then there exists an irreducible $q\\in R$ that divides each of its coefficients: if the gcd is a non-unit $d$, then take for $q$ any irreducible factor of $d$.<\/p>\n

            Lemma 1.9.2 (Gauss)<\/strong> If $R$ is a UFD and $f(x),g(x)\\in R[x]$ are both primitive, then their product $fg$ is also primitive.<\/p>\n

            Proof.<\/p>\n

            If $fg$ is not primitive, there is an irreducible $p\\in R$ which divides all its coefficients.<\/p>\n

            Let $P=(p)$ and let $\\pi: R\\to R\/P$ be the natural map $a\\mapsto a+P$.<\/p>\n

            By replacing each coefficient $c$ of a polynomial by $\\pi(c)$, we have the homomorphism $\\tilde\\pi:R[x]\\to (R\/P)[x]$.<\/p>\n

            Now $\\tilde\\pi(fg)=0$ in $(R\/P)[x]$. Since $P$ is a prime ideal, both $R\/P$ and $(R\/P)[x]$ are domains.<\/p>\n

            But neither $\\tilde\\pi(f)$ nor $\\tilde\\pi(g)$ is $0$ in $(R\/P)[x]$, because $f$ and $g$ are primitive, and this contradicts $(R\/P)[x]$ being a domain. $\\square$<\/p>\n

            Proposition 1.9.4<\/strong> Let $R$ be a UFD and let $Q=\\operatorname{Frac}(R)$ be its fraction field. Each nonzero $a\/b\\in Q$ has an expression in lowest terms; that is, $a$ and $b$ are relatively prime.<\/p>\n

            Proof.<\/p>\n

            Since $R$ is a UFD, there is a gcd of $a$ and $b$ in $R$. Let $d$ be a gcd of $a$ and $b$.<\/p>\n

            Then there are $u,v\\in R$ with $a=ud,b=vd$ and $u,v$ are relatively prime.<\/p>\n

            Therefore $u\/v\\in Q$ is the expression in lowest term of $a\/b$. $\\square$<\/p>\n

            Proposition 1.9.5<\/strong> Let $R$ be a UFD. If $a,b,c\\in R$ and $a$ and $b$ are relatively prime. Then $a\\mid bc$ implies $a\\mid c$.<\/p>\n

            Proof.<\/p>\n

            Since $R$ is a UFD, there are irreducibles $p_1,\\dots,p_t$ with $a=p_1\\cdots p_t$.<\/p>\n

            We will show $c=up_1\\cdots p_t$ for some $u\\in R$.<\/p>\n

            Let’s prove by induction on $t$.<\/p>\n

            When $t=1$, since $a=p_1$ is an irreducible and $p_1\\nmid b$, $p_1\\mid c$.<\/p>\n

            Otherwise, $(p_t)$ is a prime ideal in $R$, then $c\\in (p_t)$ for $bc\\in (p_t)$ and $b\\notin (p_t)$.<\/p>\n

            Therefore, $c=c'p_t$ with some $c'\\in R$. By cancelling, we have $p_1\\cdots p_{t-1}\\mid bc$.<\/p>\n

            By inductive suppose we have $c'=up_1\\cdots p_{t-1}$. Therefore $c=c'p_t=up_1\\cdots p_{t}=ua$. $\\square$<\/p>\n

            Lemma 1.9.3<\/strong> Let $R$ be a UFD, let $Q=\\operatorname{Frac}(R)$, and let $f(x)\\in Q[x]$ be nonzero.<\/p>\n

              \n
            1. \n

              There is a factorization
              \n$$
              \nf(x)=c(f)f^*(x),
              \n$$
              \nwhere $c(f)\\in Q$ and $f^*\\in R[x]$ is primitive. This factorization is unique in the sense that if $f(x)=qg^*(x)$, where $q\\in Q$ and $g^*\\in R[x]$ is primitive, then there is a unit $w\\in R$ with $q=wc(f)$ and $f^*=wg^*$.<\/p>\n<\/li>\n

            2. \n

              If $f(x),g(x)\\in R[x]$, then $c(fg)$ and $c(f)c(g)$ are associates in $R$ and $(fg)^*$ and $f^*g^*$ are associates in $R[x]$.<\/p>\n<\/li>\n

            3. \n

              Let $f(x)\\in Q[x]$ have a factorization $f=qg^*$, where $q\\in Q$ and $g^*\\in R[x]$ is primitive. Then $f\\in R[x]$ if and only if $q\\in R$.<\/p>\n<\/li>\n

            4. \n

              Let $g^*,f\\in R[x]$. If $g^*$ is primitive and $g^*\\mid bf$, where $b\\in R$ and $b\\ne 0$, then $g^*\\mid f$.<\/p>\n<\/li>\n<\/ol>\n

              Proof.<\/p>\n

                \n
              1. \n

                Clearing denominators, there is $b\\in R$ with $bf\\in R[x]$. If $d$ is the gcd of the coefficients of $bf$, then $f^*(x)=(b\/d)f\\in R[x]$ is a primitive polynomial. If we define $c(f)=d\/b$, then $f=c(f)f^*$.<\/p>\n

                To prove uniqueness, suppose that $c(f)f^*=f=qg^*$, where $c(f),q\\in Q$ and $f^*(x),g^*(x)\\in R[x]$ are primitive. Proposition 1.9.4 allows us to write $q\/c(f)$ in lowest terms: $q\/c(f)=u\/v$, where $u$ and $v$ are relatively prime elements of $R$.<\/p>\n

                The equation $vf^*(x)=ug^*(x)$ holds in $R[x]$. Since $u$ and $v$ are relatively prime, Proposition 1.9.5 says that $v$ is a common divisor of all the coefficients of $g^*$. But $g$ is primitive, and so $v$ is a unit.<\/p>\n

                A similar argument shows that $u$ is a unit. Therefore, $q\/c(f)=u\/v$ is a unit in $R$, call it $w$; we have $q=wc(f)$ and $f^*=wg^*$.<\/p>\n<\/li>\n

              2. \n

                There are two factorizations of $f(x)g(x)$ in $R[x]$:
                \n$$\\begin{align}
                \nfg&=c(fg)(fg)^*,\\\\
                \nfg&=c(f)f^*c(g)g^*=c(f)c(g)f^*g^*.
                \n\\end{align}
                \n$$
                \nSince the product of primitive polynomials is primitive, each of these is a factorization as in (1); the uniqueness assertion there says that $c(fg)$ is an associate of $c(f)c(g)$ and $(fg)^*$ is an associate of $f^*g^*$.<\/p>\n<\/li>\n

              3. \n

                If $q\\in R$, then it is obvious that $f=qg^*\\in R[x]$. Conversely, if $f(x)\\in R[x]$, then there is no need to clear denominators, and so $c(f)=d\\in R$, where $d$ is the gcd of the coefficients of $f(x)$. Thus $f=df^*$. By uniqueness, there is a unit $w\\in R$ with $q=wd\\in R$.<\/p>\n<\/li>\n

              4. \n

                Since $bf=hg^*$, we have $bc(f)f^*=c(h)h^*g^*=c(h)(hg)^*$. By uniqueness, $f^*$, $(hg)^*$ and $h^*g^*$ are associates, and so $g^*\\mid f^*$, But $f=c(f)f^*$, and so $g^*\\mid f$.<\/p>\n<\/li>\n<\/ol>\n

                Definition 1.9.3<\/strong> Let $R$ be a UFD with $Q=\\operatorname{Frac}(R)$. If $f(x)\\in Q[x]$, there is a factorization $f=c(f)f^*$, where $c(f)\\in Q$ and $f^*\\in R[x]$ is primitive. We call $c(f)$ the content<\/strong> of $f$ and $f^*$ the associated primitive polynomial<\/strong>. <\/p>\n

                Corollary 1.9.1<\/strong> Let $k$ be a field, and let
                \n$$f(x,y)=y^n+\\frac{g_{n-1}(x)}{h_{n-1}(x)}y^{n-1}+\\cdots+\\frac{g_0(x)}{h_0(x)}\\in k(x)[y],
                \n$$
                \nwhere each $g_i\/h_i$ is in lowest terms. If $f^*(x,y)\\in k[x][y]$ is the associated primitive polynomial of $f$, then
                \n$$\\max_i\\{\\deg(g_i),\\deg(h_i)\\}\\le \\deg_x(f^*)\\text{ and }n=\\deg_y(f^*),
                \n$$
                \nwhere $\\deg_x(f^*)$ (or $\\deg_y(f^*)$) is the highest power of $x$ (or $y$) occurring in $f^*$.<\/p>\n

                Proof.<\/p>\n

                Let $l$ be the lcm of $h_0,\\dots,h_{n-1}\\in k[x]$. Then, the associated primitive polynomial is
                \n$$f^*(x,y)=lf(x,y)=ly^n+l\\frac{g_{n-1}}{h_{n-1}}+\\cdots+l\\frac{g_0}{h_0}\\in k[x,y].
                \n$$
                \nSince $l$ is the lcm, there are $u_i\\in k[x]$ with $l=u_ih_i$ for all $i$. Hence, each coefficient $c(g_i\/h_i)=u_ig_i\\in k[x]$. If $m=\\deg_x(f^*)$, then
                \n$$m=\\max\\{\\deg(l),\\deg(c(g_i\/h_i))\\}=\\max\\{\\deg(l),\\deg(u_ig_i)\\},
                \n$$
                \nfor $l$ is a coefficient of $f^*$. Now $h_i\\mid l$ for all $i$, so that $\\deg(h_i)\\le \\deg(l)\\le m$.<\/p>\n

                Also $\\deg(g_i)\\le \\deg(u_ig_i)\\le m$. We conclude that $\\max_i\\{\\deg(g_i),\\deg(h_i)\\}\\le m=\\deg_x(f^*)$. $\\square$<\/p>\n

                Theorem 1.9.1 (Gauss)<\/strong> If $R$ is a UFD, then $R[x]$ is also a UFD.<\/p>\n

                Proof.<\/p>\n

                  \n
                • \n

                  First, let’s prove that every $f(x)\\in R[x]$, neither zero nor a unit, is a product of irreducibles by induction.<\/p>\n

                  The base step $\\deg(f)=0$ is true, because $f$ is a constant, hence lies in $R$, and hence is a product of irreducibles (for $R$ is a UFD).<\/p>\n

                  For the inductive step $\\deg(f)>0$, we have $f=c(f)f^*$, where $c(f)\\in R$ and $f^*(x)$ is primitive.<\/p>\n

                  Now $c(f)$ is either a unit or a product of irreducibles, by the base step.<\/p>\n

                  If $f^*$ is irreducible, we are done. Otherwise, $f^*=gh$, where neither $g$ nor $h$ is a unit.<\/p>\n

                  Since $f^*$ is primitive, however, neither $g$ nor $h$ is a constant; therefore, each of these has degree less than $\\deg(f^*)=\\deg(f)$.<\/p>\n

                  By indective hypothesis, each of $g$ and $h$ is a product of irreducibles. Therefore, $f$ is a product of irreducibles.<\/p>\n<\/li>\n

                • \n

                  Then, let’s prove $R[x]$ is a UFD by Proposition 1.9.1; that is let $p$ be an irreducible, then $(p)$ is a prime ideal in $R[x]$.<\/p>\n

                  Let $fg\\in (p)$.<\/p>\n

                    \n
                  • Suppose that $\\deg(p)=0$.<\/li>\n<\/ul>\n

                    Now $f=c(f)f^*(x),g=c(g)g^*(x)$, where $f^*,g^*$ are primitive and $c(f),c(g)\\in R$.<\/p>\n

                    Since $p\\mid fg$, we have $p\\mid c(f)c(g)f^*g^*$. Write $f^*g^*=\\sum_i a_ix^i$, where $a_i\\in R$, so that $p\\mid c(f)g(f)a_i$ in $R$ for all $i$.<\/p>\n

                    Now $f^*g^*$ is primitive, so there is some $i$ with $p\\nmid a_i$ in $R$. For $R$ is a UFD, $p\\mid c(f)$ or $p\\mid c(g)$.<\/p>\n

                    Thus, $p\\mid f$ or $p\\mid g$.<\/p>\n

                      \n
                    • Suppose that $\\deg(p)>0$ and $p\\nmid f$.<\/li>\n<\/ul>\n

                      Let $(p,f)=\\{s(x)p(x)+t(x)f(x);s(x),t(x)\\in R[x\\}$; then $(p,f)$ is an ideal in $R[x]$ containing $p$ and $f$.<\/p>\n

                      Choose $m(x)\\in (p,f)$ of minimal degree. If $Q=\\operatorname{Frac}(R)$ is the fraction field of $R$, then the division algorithm in $Q[x]$ gives polynomials $q(x),r(x)\\in R[x]$ with
                      \n$$
                      \nf=mq’+r’,
                      \n$$
                      \nwhere either $r'=0$ or $\\deg(r')<\\deg(m)$. Clearing denominators, there is a constant $b\\in R$ and polynomials $q(x),r(x)\\in R[x]$ with
                      \n$$
                      \nbf=qm+r,
                      \n$$
                      \nwhere $r=0$ or $\\deg(r)<\\deg(m)$.<\/p>\n

                      Since $m\\in (p,f)$, there are polynomials $s(x),t(x)\\in R[x]$ with $n=sp+tf$; hence $r=bf-qm\\in (q,f)$. Since $m$ has minimal degree in $(p,f)$, we must have $r=0$; that is $bf=mq$, and so $bf=c(m)m^*q$. But $m^*$ is primitive, and $m^*\\mid bf$, so that $m^*\\mid f$ for Lemma 1.9.3(4).<\/p>\n

                      A similar argument, replacing $f$ by $p$, we have $m^*\\mid p$. Since $p$ is irreducible, we have $m^*$ is a unit or $m^*$ and $p$ are associates. If $m^*$ were an associate of $p$, then $p\\mid f$ contrary to out assumption that $p\\nmid f$. Thus $m^*$ must be a unit. Therefore, $m\\in R$.<\/p>\n

                      We have $mg=(sf+pt)g=sfg+ptg$, so that $p\\mid mg$. Since $p$ is an irreducible, we have $p$ is primitive. By Lemma 1.9.3, we have $p\\mid g$. $\\square$<\/p>\n<\/li>\n<\/ul>\n

                      Corollary 1.9.2<\/strong> If $k$ is a field, then $k[x_1,\\dots,x_n]$ is a UFD.<\/p>\n

                      Corollary 1.9.3<\/strong> If $k$ is a field, then $p=p(x_1,\\dots,x_n)\\in k[x_1,\\dots,x_n]$ is irreducible if and only if $(p)$ is a prime ideal in $k[x_1,\\dots,x_n]$.<\/p>\n

                      Corollary 1.9.4 (Gauss’s Lemma)<\/strong> Let $R$ be a UFD, let $Q=\\operatorname{Frac}(R)$, and let $f(x)\\in R[x]$. If $f=GH$ in $Q[x]$, then there is a factorization
                      \n$$f=gh \\text{ in } R[x],
                      \n$$
                      \nwhere $\\deg(g)=\\deg(G)$ and $\\deg(h)=\\deg(H)$; in fact, $G$ is a constant multiple of $g$ and $H$ is a constant multiple of $h$. Therefore, if $f$ does not factor into polynomials of smaller degree in $R[x]$, then $f$ is irreducible in $Q[x]$.<\/p>\n

                      Proof.<\/p>\n

                      By Lemma 1.9.3(1), the factorization $f=GH$ in $Q[x]$ gives $q,q'\\in Q$ with
                      \n$$f=qG^*q'H^*\\text{ in } Q[x],
                      \n$$
                      \nwhere $G^*,H^*\\in R[x]$ are primitive; thus, $G^*H^*$ is primitive.<\/p>\n

                      Since $f\\in R[x]$, by Lemma 1.9.3, the equation $f=(qq')(G^*H^*)$ shows that $qq'\\in R$.<\/p>\n

                      Therefore, a factorization of $f$ in $R[x]$ is $f=(qq'G^*)H^*$. $\\square$<\/p>\n

                      Proposition 1.9.6<\/strong> Let $k$ be a field, and view $f(x_1,\\dots,x_n)\\in k[x_1,\\dots,x_n]$ as a polynomial in $R[x_n]$, where $R=k[x_1,\\dots,x_{n-1}]$:
                      \n$$f(x_n)=a_0(x_1,\\dots,x_{n-1})+a_1(x_1,\\dots,x_{n-1})x_n+\\cdots+a_m(x_1,\\dots,x_{n-1})x_n^m.
                      \n$$
                      \nIf $f(x_n)$ is primitive and cannot be factored into two polynomials of lower degree in $R[x_n]$, then $f(x_1,\\dots,x_n)$ is irreducible in $k[x_1,\\dots,x_n]$.<\/p>\n

                      Proof.<\/p>\n

                      Suppose that $f(x_n)=g(x_n)h(x_n)$ in $R[x_n]$; by hypothesis, the degree of $g$ and $h$ in $x_n$ cannot both be less than $\\deg(f)$; say, $\\deg(g)=0$.<\/p>\n

                      It follows, because $f$ is primitive, that $g$ is a unit in $k[x_1,\\dots,x_{n-1}]$. Therefore, $f(x_1,\\dots,x_n)$ is irreducible in $R[x_n]=k[x_1,\\dots,x_n]$. $\\square$<\/p>\n

                      Corollary 1.9.5<\/strong> If $k$ is a field and $g(x_1,\\dots,x_n),h(x_1,\\dots,x_n)\\in k[x_1,\\dots,x_n]$ are relatively prime, then $f(x_1,\\dots,x_n,y)=yg(x_1,\\dots,x_n)+h(x_1,\\dots,x_n)$ is irreducible in $k[x_1,\\dots,x_n,y]$.<\/p>\n","protected":false},"excerpt":{"rendered":"

                      Definition 1.9.1 A domain $R$ is a unique factorization […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12,6],"tags":[11,8],"class_list":["post-413","post","type-post","status-publish","format-standard","hentry","category-abstract-algebra","category-6","tag-11","tag-8"],"_links":{"self":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/413","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=413"}],"version-history":[{"count":2,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/413\/revisions"}],"predecessor-version":[{"id":419,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/413\/revisions\/419"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=413"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=413"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}