{"id":403,"date":"2023-06-20T16:39:12","date_gmt":"2023-06-20T08:39:12","guid":{"rendered":"https:\/\/qwq.cafe\/?p=403"},"modified":"2023-06-20T16:39:12","modified_gmt":"2023-06-20T08:39:12","slug":"abstract-algebra-1-4-from-arithmetic-to-polynomials","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=403","title":{"rendered":"Abstract Algebra \u2013 1.4 From Arithmetic to Polynomials"},"content":{"rendered":"

Theorem 1.4.1 (Division Algorithm)<\/strong> If $K$ is a field and $f(x),g(x)\\in K[x]$ with $f\\ne 0$, then there are unique polynomials $q(x),r(X)\\in K[x]$ with
\n$$g(x)=q(x)f(x)+r(x),
\n$$
\nwhere either $r=0$ or $\\deg r(x)<\\deg f(x)$.<\/p>\n

Definition 1.4.1<\/strong> If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then the polynomials $q(x)$ and $r(x)$ occurring in the Division Algorithm are called the quotient<\/strong> and the remainder<\/strong> after dividing $g$ by $f$.<\/p>\n

Corollary 1.4.1<\/strong> Let $R$ be a commutative ring, and let $f(x)\\in R[x]$ be a monic polynomial. If $g(x)\\in R[x]$, then there exist $q(x),r(x)\\in R[x]$ with
\n$$g(x)=q(x)f(x)+r(x),
\n$$
\nwhere either $r=0$ or $\\deg r(x)<\\deg f(x)$.<\/p>\n

Theorem 1.4.2<\/strong> If $K$ is a field, then every ideal $I$ in $K[x]$ is a principal ideal; that is, there is $d\\in I$ with $I=(d)$, then $d$ can be chosen to be a monic polynomial. Moreover, if $I\\ne (0)$, then $d$ can be chosen to be a monic polynomial.<\/p>\n

Definition 1.4.2<\/strong> If $f(x)\\in R[x]$, where $R$ is a ring, then a root<\/strong> of $f(x)$ in $R$ is an element $a\\in R$ with $f(a)=0$.<\/p>\n

Lemma 1.4.1<\/strong> Let $f(x)\\in K[x]$, where $K$ is a field, and let $u\\in K$. Then there is $q(x)\\in K[x]$ with
\n$$f(x)=q(x)(x-u)+f(u).
\n$$
\nProposition 1.4.1<\/strong> If $f(x)\\in K[x]$, where $K$ is a field, then $a$ is a root of $f(x)$ in $K$ if and only if $x-a\\mid f(x)$.<\/p>\n

Proposition 1.4.2<\/strong> Let $K$ be a field and let $f(x)\\in K[x]$. If $f$ has degree $n$, then $f$ has at most $n$ roots in $K$.<\/p>\n

Corollary 1.4.2<\/strong> Every $n$th root of unity in $\\mathbb C$ is equal to
\n$$e^{2\\pi ik\/n}=\\cos(2\\pi k\/n)+i\\sin(2\\pi k\/n),
\n$$
\nwhere $k=0,1,2,\\dots,n-1$.<\/p>\n

Corollary 1.4.3<\/strong> Let $K$ be an infinite field and let $f(x)$ and $g(x)$ be polynomials in $K[x]$. If $f$ and $g$ determine the same polynomial function (that is, $f(a)=g(a)$ for all $a\\in K$), then $f=g$.<\/p>\n

Proof.<\/p>\n

Let $h(x)=f(x)-g(x)$. Then if $h\\ne 0$, $h$ has some degree.<\/p>\n

However, the roots of $h$ are too much, which contradicts the Proposition 1.4.2. $\\square$<\/p>\n

Corollary 1.4.4<\/strong> Let $K$ be a (possibly finite) field, let $f(x),g(x)\\in K[x]$, and let $\\deg f(x)\\le\\deg g(x)=n$. If $f(a)=g(a)$ for $n+1$ elements $a\\in K$, then $f=g$.<\/p>\n

Proof.<\/p>\n

Let $h(x)=f(x)-g(x)$. Then if $h\\ne 0$, $\\deg h(x)\\le n$.<\/p>\n

That $h(x)$ have $n+1$ roots contradicts the Proposition 1.4.2. $\\square$<\/p>\n

Proposition 1.4.3<\/strong> Let $f(X),g(X)\\in K[X]=K[x_1,\\dots,x_n]$, where $K$ is an infinite field.<\/p>\n

    \n
  1. If $f(X)$ is nonzero, then there are $a_1,\\dots,a_n\\in K$ with $f(a_1,\\dots,a_n)\\ne 0$.<\/li>\n
  2. If $f(a_1,\\dots,a_n)=g(a_1,\\dots,a_n)$ for all $(a_1,\\dots,a_n)\\in K^n$, then $f=g$.<\/li>\n<\/ol>\n

    Theorem 1.4.3<\/strong> Let $K$ be a field. If $G$ is a finite subgroup of the multiplicative group $K^\\times$, then $G$ is cyclic. In particular, if $K$ is finite, then $K^\\times$ is cyclic.<\/p>\n

    Proof.<\/p>\n

    $G$ is an abelian group and $x^m-1$ have less than $m$ roots in $K$ (also in $G$).<\/p>\n

    For \u547d\u98981.1.7 in \u62bd\u8c61\u4ee3\u6570\u7b14\u8bb0<\/a>, $G$ is cyclic. $\\square$<\/p>\n

    Definition 1.4.3<\/strong> If $K$ is a finite field, a generator of the cyclic group $K^\\times$ is called a primitive element<\/strong> of $K$.<\/p>\n

    Definition 1.4.4<\/strong> If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then a common divisor<\/strong> is a polynomial $c(x)\\in K[x]$ with $c\\mid f$ and $c\\mid g$. If $f$ and $g$ in $K[x]$ are not both $0$, define their greatest common divisor<\/strong> abbreviated $\\gcd$, to be the monic common divisor having largest degree. If $f=0=g$, define $\\gcd(f,g)=0$.<\/p>\n

    Theorem 1.4.4<\/strong> If $K$ is a field and $f(x),g(x)\\in K[x]$, then their gcd $d(x)$ is a linear combination of $f$ and $g$; that is there are $s(x),t(x)\\in K[x]$ with
    \n$$d=sf+tg.
    \n$$
    \nProof.<\/p>\n

    The set $(f,g)$ of all linear combinations of $f$ and $g$ is an ideal in $K[x]$.<\/p>\n

    Since every ideal in $K[x]$ is a principle ideal, there is $d(x)\\in K[x]$ such that $(d)=(f,g)$.<\/p>\n

    Then there are $s(x),t(x)\\in K[x]$ with $d=sf+tg$.<\/p>\n

    Because $f(x),g(x)\\in (d)$, $d$ is a common divisor of $f$ and $g$.<\/p>\n

    If $q(x)$ is a common divisor of $f$ and $g$, then there are $f_1(x),g_1(x)\\in K[x]$ with $f=f_1q,g=g_1q$.<\/p>\n

    $d=sf+tg=(sf_1+tg_1)q$ shows that $q$ is a divisor of $d$, and $\\deg q(x)\\le\\deg d(x)$.<\/p>\n

    Therefore, $d$ is the gcd of $f$ and $g$. $\\square$<\/p>\n

    Corollary 1.4.5<\/strong> Let $K$ be a field and let $f(x),g(x)\\in K[x]$.<\/p>\n

      \n
    1. A monic common divisor $d(x)$ is the gcd if and only if $d$ is divisible by every common divisor; that is, if $h(x)$ is a common divisor, then $h\\mid d$.<\/li>\n
    2. $f$ and $g$ have a unique gcd.<\/li>\n<\/ol>\n

      Definition 1.4.5<\/strong> An element $p$ in a domain $R$ is irreducible<\/strong> if $p$ is neither $0$ nor a unit and, in every factorization $p=uv$ in $R$, either $u$ or $v$ is a unit.<\/p>\n

      For example, a prime $p\\in \\mathbb Z$ is irreducible element, as is $-p$.<\/p>\n

      Proposition 1.4.4<\/strong> If $K$ is a field, then a polynomial $p(x)\\in K[x]$ is irreducible if and only if $\\deg(p)=n\\ge 1$ and there is no factorization in $K[x]$ for the form $p(x)=g(x)h(x)$ in which both factors have degree smaller than $n$.<\/p>\n

      Corollary 1.4.6<\/strong> Let $K$ be a field and let $f(x)\\in K[x]$ be a quadratic or cubic polynomial. Then $f$ is irreducible in $K[x]$ if and only if $f$ has no roots in $K$.<\/p>\n

      Theorem 1.4.5 (Gauss’s Lemma)<\/strong> Let $f(x)\\in \\mathbb Z[x]$. If $f(x)=G(x)H(x)$ in $\\mathbb Q[x]$, where $\\deg(G),\\deg(H)<\\deg(f)$, then $f(x)=g(x)h(h)$ in $\\mathbb Z[x]$, where $\\deg(g)=\\deg(G)$ and $\\deg(h)=\\deg(H)$.<\/p>\n

      Proof.<\/p>\n

      Clearing denominators, there are positive integers $n',n''$ such that $g(x)=n'G(x)$ and $h(x)=n''H(x)$.<\/p>\n

      Setting $n=n'n''$, we have
      \n$$nf(x)=n'G(x)n''H(x)=g(x)h(x)\\text{ in }\\mathbb Z[x].
      \n$$
      \nIf $p$ is a prime divisor of $n$, consider the map $\\mathbb Z[x]\\to \\mathbb Z_p[x]$, denoted by $g\\to \\overline g$, which reduces all coefficients mod $p$. The equation becomes
      \n$$0=\\overline g(x)\\overline h(x).
      \n$$
      \nBut $\\mathbb Z_p[x]$ is a domain, and so at least one of these factors is $0$. <\/p>\n

      Let’s assume that $\\overline g(x)=0$; that is all the coefficients of $g(x)$ are multiples of $p$.<\/p>\n

      Therefore, we may write $g(x)=pg'(x)$, where all the coefficients of $g'(x)$ lie in $\\mathbb Z$. If $n=pm$, then
      \n$$pmf(x)=pg'(x)h(x)\\text{ in } \\mathbb Z.
      \n$$
      \nCancel $p$, and continue canceling primes until we reach a factorization $f(x)=g^*(x)h^*(x)$ in $\\mathbb Z[x]$. $\\square$<\/p>\n

      The contrapositive version of Gauss’s Lemma is more convenient to use. If $f(x)\\in \\mathbb Z[x]$ has no factorization in $\\mathbb Z[x]$ as a product of two polynomials, each having degree smaller than $\\deg(f)$, then $f$ is irreducible in $\\mathbb Q[x]$.<\/p>\n

      Lemma 1.4.2<\/strong> Let $K$ be a field, let $p(x),f(x)\\in K[x]$, and let $d(x)=\\gcd(p,f)$. If $p$ is a monic irreducible polynomial, then
      \n$$d(x)=\\begin{cases}
      \n1,&p\\nmid f,\\\\
      \np(x),&p\\mid f.
      \n\\end{cases}
      \n$$<\/p>\n

      Proof. Since $d\\mid p$, we have $d=1$ or $d=p$. $\\square$<\/p>\n

      Theorem 1.4.6 (Euclid’s Lemma)<\/strong> Let $K$ be a field and let $f(x),g(x)\\in K[x]$. If $p(x)$ is an irreducible polynomial in $K[x]$, and $p\\mid fg$, then either
      \n$$p\\mid f\\text{ or }p\\mid g.
      \n$$
      \nMore generally, if $p\\mid f_1(x)\\cdots f_n(x)$, then $p\\mid f_i$ for some $i$.<\/p>\n

      Proof.<\/p>\n

      Assume that $p\\mid fg$ but that $p\\nmid f$.<\/p>\n

      Since $p$ is irreducible, $\\gcd(p,f)=1$, and so $1=sp+sf$ for some polynomials $s$ and $t$.<\/p>\n

      Therefore $g=spg+sfg$. Because $p\\mid fg$, $fg=hp$ for some polynomial $h$.<\/p>\n

      Then $g=spg+shp=(sg+sh)p$, so $p\\mid g$. $\\square$<\/p>\n

      Definition 1.4.6<\/strong> Two polynomials $f(x),g(x)\\in K[x]$, where $K$ is a field, are called relatively prime<\/strong> if their gcd is $1$.<\/p>\n

      Corollary 1.4.7<\/strong> Let $f(x),g(x),h(x)\\in K[x]$, where $K$ is a field, and let $h$ and $f$ be relatively prime. If $h\\mid fg$, then $h\\mid g$.<\/p>\n

      Definition 1.4.7<\/strong> If $K$ is a field, then a retaional function $f(x)\/g(x)\\in K(x)$ is in lowest terms<\/strong> if $f(x)$ and $g(x)$ are relatively prime.<\/p>\n

      Theorem 1.4.7 (Euclidean Algorithms)<\/strong> If $K$ is a field $f(x),g(x)\\in K[x]$, then there are algorithms for computing $\\gcd(f,g)$, as well as for finding a pair of polynomials $s(x)$ and $t(x)$ with
      \n$$\\gcd(f,g)=sf+tg.
      \n$$
      \nCorollary 1.4.8<\/strong> Let $k$ be a subfield of a field $K$, so that $k[x]$ is a subring of $K[x]$. If $f(x),g(x)\\in k[x]$, then their gcd in $k[x]$ is equal to their gcd in $K[x]$.<\/p>\n

      Corollary 1.4.9<\/strong> If $f(x),g(x)\\in \\mathbb R[x]$ have no common root in $\\mathbb C$, then $f,g$ are relatively prime in $\\mathbb R[x]$.<\/p>\n

      Theorem 1.4.8 (Unique Factorization)<\/strong> If $K$ is a field, then every polynomial $f(x)\\in K[x]$ of degree $\\ge1$ is a product of a nonzero constant and monic irreducibles. Moreover, if $f(x)$ has two such factorizations,
      \n$$f(x)=ap_1(x)\\cdots p_m(x) \\text{ and } f(x)=bq_1(x)\\cdots q_n(x)
      \n$$
      \nthat is, $a$ and $b$ are nonzero constants and the $p$’s and $q$’s are monic irreducibles, then $a=b,m=n$ and the $q$’s may be reindexed so that $q_i=p_i$ for all $i$.<\/p>\n

      Definition 1.4.8<\/strong> Let $f(x)\\in K[x]$, where $K$ is a field. A prime factorization<\/strong> of $f(x)$ is
      \n$$f(x)=ap_1(x)^{e_1}\\cdots p_m(x)^{e_m},
      \n$$
      \nwhere $a$ is a nonzero constant, the $p_i$ are distinct monic irreducible polynomials, and $e_i\\ge 0$ for all $i$.<\/p>\n

      Let $K$ be a field, and assume that there are $a,r_1,\\dots,r_n\\in K$ with
      \n$$f(x)=a\\prod_{i=1}^n(x-r_i);
      \n$$
      \nwe say that $f$ splits<\/strong> over $k$. If $r_1,\\dots,r_s$, where $s\\le n$, are the distinct roots of $f(x)$, then a prime factorization of $f(x)$ is
      \n$$f(x)=a(x-r_1)^{e_1}(x-r_2)^{e_2}\\cdots(x-r_s)^{e_s}.
      \n$$
      \nWe call $e_j$ the multiplicity<\/strong> of the root $r_j$.<\/p>\n

      Definition 1.4.9<\/strong> If $f$ and $g$ are elements in a commutative ring $R$, then a common multiple<\/strong> is an element $m\\in R$ with $f\\mid m$ and $g\\mid m$. If $f$ and $g$ in $R$ are not both $0$, define their last common multiple<\/strong>, abbreviated $\\operatorname{lcm}(f,g)$, to be a common mutiple $c$ of them with $c\\mid m$ for every common multiple $m$. If $f=0=g$, define their $\\operatorname{lcm}(f,g)=0$, If $R=K[x]$, we require lcm’s to be monic.<\/p>\n

      Proposition 1.4.5<\/strong> If $K$ is a field and $f(x),g(x)\\in K[x]$ have prime factorizations $f(x)=a_0p_1^{a_1}\\cdots p_n^{a_n}$ and $g(x)=b_0p_1^{b_1}\\cdots p_n^{b_n}$ in $K[x]$, then<\/p>\n

        \n
      1. \n

        $f\\mid g$ if and only if $a_i\\le b_i,\\forall i$.<\/p>\n<\/li>\n

      2. \n

        If $m_i=\\min\\{a_i,b_i\\}$ and $M_i=\\max\\{a_i,b_i\\}$, then
        \n$$
        \n\\gcd(f,g)=p_1^{m_1}\\cdots p_n^{m_n}\\text{ and }\\operatorname{lcm}(f,g)=p_1^{M_1}\\cdots p_n^{M_n}.
        \n$$<\/p>\n<\/li>\n<\/ol>\n

        Corollary 1.4.10<\/strong> If $K$ is a field and $f(x),g(x)\\in K[x]$ are monic polynomials, then
        \n$$\\gcd(f,g)\\operatorname{lcm}(f,g)=fg.
        \n$$<\/p>\n","protected":false},"excerpt":{"rendered":"

        Theorem 1.4.1 (Division Algorithm) If $K$ is a field an […]<\/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-403","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\/403","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=403"}],"version-history":[{"count":1,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/403\/revisions"}],"predecessor-version":[{"id":404,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/403\/revisions\/404"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=403"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=403"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=403"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}