{"id":401,"date":"2023-06-20T16:36:48","date_gmt":"2023-06-20T08:36:48","guid":{"rendered":"https:\/\/qwq.cafe\/?p=401"},"modified":"2023-06-20T16:37:49","modified_gmt":"2023-06-20T08:37:49","slug":"abstract-algebra-1-3-quotient-rings","status":"publish","type":"post","link":"https:\/\/qwq.cafe\/?p=401","title":{"rendered":"Abstract Algebra \u2013 1.3 Quotient Rings"},"content":{"rendered":"

Definition 1.3.1<\/strong> Let $I$ be an ideal in a commutative ring $R$. If $a\\in R$, then the coset<\/strong> $a+I$ is the subset
\n$$a+I=\\{a+i;i\\in I\\}.
\n$$
\nThe coset $a+I$ is often called $a\\bmod I$. The family of all cosets is denoted by $R\/I$:
\n$$R\/I=\\{a+I;a\\in R\\}
\n$$
\nThe relation $\\equiv$ on $R$, defined by $a\\equiv b$ if $a-b\\in I$, is called congruence mod $I$<\/strong>; it is an equivalence relation on $R$, and its equivalence classes are the cosets.<\/p>\n

Proposition 1.3.1<\/strong> Let $I$ be an ideal in a commutative ring $R$. If $a,b\\in R$, then $a+I=b+I$ if and only if $a-b\\in I$. In particular, $a+I=I$ if and only if $a\\in I$.<\/p>\n

Definition 1.3.2<\/strong> Let $R$ be a commutative ring and $I$ be an ideal in $R$. Define addition $+:R\/I\\times R\/I\\to R\/I$ by
\n$$+:(a+I,b+I)\\mapsto a+b+I,
\n$$
\nand multiplication $\\cdot:R\/I\\times R\/I\\to R\/I $ by
\n$$\\cdot:(a+I,b+I)\\mapsto ab+I
\n$$
\nIt’s easy to prove addition and multiplication $R\/I\\times R\/I\\to R\/I$ are well-defined.<\/p>\n

Theorem 1.3.1<\/strong> If $I$ is an ideal in a commutative ring $R$, then $R\/I$ is a commutative ring.<\/p>\n

Definition 1.3.2<\/strong> The commutative ring $R\/I$ equipped with addition and multiplication is called quotient ring<\/strong> of $R$ modulo $I$; it is usually pronounced $R\\bmod I$.<\/p>\n

Definition 1.3.3<\/strong> Let $I$ be an ideal in a commutative ring $R$. The natural function<\/strong> is the function $\\pi:R\\to R\/I,a\\mapsto a+I$.<\/p>\n

Proposition 1.3.2<\/strong> If $I$ is an ideal in a commutative ring $R$, then the natural map $\\pi: R\\to R\/I$ is a surjective homomorphism and $\\ker \\pi = I$.<\/p>\n

Corollary 1.3.1<\/strong> Given an ideal $I$ in a commutative ring $R$, there exists a commutative ring $A$ and a (surjective) homomorphism $\\varphi:R\\to A$ with $I=\\ker \\varphi$.<\/p>\n

If we set $A=R\/I$, then the natural map $\\pi:R\\to R\/I$ is a homomorphism with $I=\\ker \\pi$.<\/p>\n

Theorem 1.3.2 (First Isomorphism Theorem)<\/strong> Let $R$ and $A$ be commutative rings. If $\\varphi:R\\to A$ is a homomorphism, then $\\ker \\varphi$ is an ideal in $R$, and $\\operatorname{im}\\varphi$ is a subring of $A$, and
\n$$R\/\\ker\\varphi\\cong \\operatorname{im}\\varphi
\n$$
\nLet $I=\\ker \\varphi$.<\/p>\n

In the commutative diagram below, $\\pi:R\\to R\/I$ is the natural map, $i:\\operatorname{im}\\varphi\\to A$ is the inclusion, and define $\\tilde \\varphi:R\/I\\to A,a+I\\mapsto \\varphi(a)$ ($a+I=b+I\\Leftrightarrow b-a\\in I\\Leftrightarrow \\varphi(a)=\\varphi(a)+\\varphi(b-a)=\\varphi(b)$ show that $\\tilde\\varphi$ is well-defined).
\n$$% https:\/\/darknmt.github.io\/res\/xypic-editor\/#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
\n\\xymatrix{
\nR \\ar@{->}[d]_{\\pi} \\ar@{->}[r]^{\\varphi} & A \\\\
\nR\/I \\ar@{->}[r]_{\\tilde\\varphi} & \\operatorname{im} \\varphi \\ar@{->}[u]_{i}
\n}
\n$$
\nDefinition 1.3.4<\/strong> If $K$ is a field, the intersection of all the subfields of $K$ is called the prime field<\/strong> of $K$.<\/p>\n

For example, the prime field of $\\mathbb C$ is $\\mathbb Q$, because every subfield of $\\mathbb C$ contains $\\mathbb Q$.<\/p>\n

Definition 1.3.5<\/strong> If $X$ is a subset of a field, define $\\left<X\\right>$, the subfield of generated by $X$<\/strong>, to be the intersection of all the subfields containing $X$.<\/p>\n

Proposition 1.3.3<\/strong> Let $K$ be a field with identity $\\mathscr l$, and let $\\chi:\\mathbb Z\\to K,n\\mapsto n\\mathscr l$ be a homomorphism.<\/p>\n

    \n
  1. Either $\\operatorname{im}\\chi\\cong \\mathbb Z$ or $\\operatorname{im}\\chi \\cong \\mathbb Z_p$ for some prime $p$.<\/li>\n
  2. The prime field of $K$ is isomorphism to $\\mathbb Q$ or to $\\mathbb Z_p$ for some prime $p$.<\/li>\n<\/ol>\n

    Proof.<\/p>\n

      \n
    1. \n

      Since every ideal in $\\mathbb Z$ is principal, $\\ker \\chi = (m)$ for some integer $m\\ge 0$.<\/p>\n

      The First Isomorphism Theorem gives $\\operatorname{im}\\chi\\cong \\mathbb Z\/(m)$.<\/p>\n

      If $m=0$, $\\operatorname{im}\\chi\\cong\\mathbb Z\/(0)\\cong\\mathbb Z$.<\/p>\n

      If $m\\ne 0$, $\\operatorname{im}\\chi\\cong\\mathbb Z\/(m)=\\mathbb Z_m$. And then we will prove $m$ is a prime.<\/p>\n

      Let’s suppose $m$ isn’t a prime. Then there are two integers $a,b>0$ such that $ab=m$.<\/p>\n

      Therefore $\\chi(a)\\chi(b)=\\chi(ab)=0$, which contradicts the fact that $K$ is a domain.<\/p>\n<\/li>\n

    2. \n

      If $\\operatorname{im} \\chi\\cong \\mathbb Z$. There is a field $Q\\cong \\operatorname{Frac}(\\mathbb Z)=\\mathbb Q$ with $\\operatorname{im}\\chi\\subset Q\\subset K$. Now $Q$ is the prime field of $K$, for it is the subfield generated by $\\mathscr l$.<\/p>\n

      In case $\\operatorname{im}\\chi\\cong \\mathbb Z_p$, then $\\operatorname{im}\\chi$ must be the prime field of $K$, for it is a field which is obviously the subfield generated by $\\mathscr l$. $\\square$<\/p>\n<\/li>\n<\/ol>\n

      Definition 1.3.6<\/strong> A field $K$ has characteristic<\/strong> $0$ if its prime field is isomorphic to $\\mathbb Q$; it has characteristic<\/strong> $p$ if its prime field is isomorphic to $\\mathbb Z_p$ for some prime $p$.<\/p>\n

      Proposition 1.3.4<\/strong> If $K$ is a finite field, then $|K|=p^n$ for some prime $p$ and some $n\\ge 1$.<\/p>\n

      Proof.<\/p>\n

      The prime field of $K$ is isomorphic to $\\mathbb Z_p$ for some prime $p$.<\/p>\n

      We can regard $K$ as a vector space over $\\mathbb Z_p$. As $K$ is finite, it is obviously finite-dimensional.<\/p>\n

      If $\\operatorname{dim}_{\\mathbb Z_p}(K)=n$, then $|K|=p^n$. $\\square$<\/p>\n

      Proposition 1.3.5<\/strong> Let $I$ be an ideal in a commutative ring $R$. If $J$ is an ideal in $R$ containing $I$, define the subset $J\/I$ of $R\/I$ by
      \n$$J\/I=\\{a+I;a\\in I\\}.
      \n$$<\/p>\n

        \n
      1. $\\pi^{-1}(J\/I)=J$, where $\\pi:R\\to R\/I$ is the natural map.<\/li>\n
      2. $J\/I$ is an ideal in $R\/I$.<\/li>\n
      3. If $I\\subset J\\subset J'$ are ideals in $R$, $J\/I\\subset J'\/I$. Moreover, if $J\\ne J'$, then $J\/I\\ne J'\/I$.<\/li>\n<\/ol>\n

        Proof.<\/p>\n

          \n
        1. \n
            \n
          • \n

            First, let’s prove $\\pi^{-1}(J\/I)\\subset J$.<\/p>\n

            $\\forall a\\in \\pi^{-1}(J\/I),\\pi(a)=a+I\\in J\/I$.<\/p>\n

            Then there is $b\\in J,b+I=a+I\\Rightarrow a-b\\in I\\subset J$.<\/p>\n

            Therefore $a\\in J$.<\/p>\n<\/li>\n

          • \n

            Second, let’s prove $J\\subset \\pi^{-1}(J\/I)$.<\/p>\n

            $\\forall a\\in J,\\pi(a)=a+I\\in J\/I\\Rightarrow a\\in \\pi^{-1}(J\/I)$.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

          • \n

            $\\forall a+I\\in J\/I,\\forall b+I\\in R\/I,(a+I)(b+I)=ab+I$.<\/p>\n

            Since $ab\\in J\\Rightarrow ab+I\\in J\/I$, $J\/I$ is an ideal.<\/p>\n<\/li>\n

          • \n

            $J\\subset J'\\Rightarrow J\/I=\\pi(J)\\subset \\pi(J')=J'\/I$.<\/p>\n

            Then $J=\\pi^{-1}(J\/I)\\ne \\pi^{-1}(J'\/I)=J'\\Rightarrow J\/I\\ne J'\/I$. $\\square$<\/p>\n<\/li>\n<\/ol>\n

            Proposition 1.3.6<\/strong> A commutative ring $R$ is a field if and only if $R$ only have trivial ideals, $(0)$ and $R$.<\/p>\n

            Proof.<\/p>\n

              \n
            • \n

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

              Let $I$ be an ideal of $R$, with $I\\ne (0)$.<\/p>\n

              Then there is $a\\in I\\textbackslash \\{0\\}$ with $a$ is a unit.<\/p>\n

              Because $\\forall b\\in R,b=(ba^{-1})a\\in I$, $I=R$.<\/p>\n<\/li>\n

            • \n

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

              If $R$ isn’t a field, then there is $a\\in R$ with $a$ isn’t a unit.<\/p>\n

              Then $(a)$ is an ideal with $1\\notin (a)$, which lead to a contradiction. $\\square$<\/p>\n<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

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