Algebra - Rings and fields

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: nstejas@gmail.com Mobile: +91 9845410469

5. Divisibilty, assciates and units

  • Definition 5.1: Let $D$ be an integral domain. Let $a\neq 0$ and $b \in D$, we say that $a|b$ (read as $a$ divides $b$) in $D$ if there exists an element $k$ in $D$ such that $b=ka$.

  • Definition 5.2: Let $D$ be an integral domain. Let $a\neq 0$ and $b\neq 0$ in $D$, if $a|b$ and $b|a$ then $a$ and $b$ are called associates.

  • Definition 5.3: Let $D$ be an integral domain. Let $a\neq 0$ in $D$, if $a|1$ then $a$ is said to be unit in $D$.

  • Definition 5.4: A non zero element $a$ of an integral domain $D$ is called a prime element or irreducible element if it is not a unit and its only divisors are units and associates of $a$.

  • Theorems on associates and units

  • Theorem 5.1: The relation "$a$ is an associate of $b$" is an equivalence relation in an integral domain.
    Proof: Let $D$ be an integral domain.
    Since $a|a$ and $a|a$ in $D$ $\Rightarrow$ $a$ and $a$ are associates in $D$.
    $\therefore$ The relation is reflexive.
    Let $a$ and $b$ be associates in $D$ then $a|b$ and $b|a$ in $D$ $\Rightarrow$ $b|a$ and $a|b$ in $D$
    $\Rightarrow$ $b$ and $a$ are associates in $D$
    $\therefore$ The relation is symmetric.
    Let $a$ and $b$ be associates in $D$ then $a|b$ and $b|a$ in $D$ and Let $b$ and $c$ be associates in $D$ then $b|c$ and $c|b$ in $D$
    $a|b$ and $b|c$ $\Rightarrow$ $a|c$
    $c|b$ and $b|a$ $\Rightarrow$ $c|a$
    $\Rightarrow$ $a$ and $c$ are associates in $D$
    $\therefore$ The relation is transitive. $\blacksquare$

  • Theorem 5.2: If $a$ is an associate of $a'$, $b$ is an associate of $b'$, then
    (1.) $a+b$ is an associate of $a'+b'$.
    (2.) $ab$ is an associate of $a'b'$.
    Proof: Given, $a$ is an associate of $a'$ $\Rightarrow$ $a|a'$ and $a'|a$
    $b$ is an associate of $b'$ $\Rightarrow$ $b|b'$ and $b'|b$
    (1.) To prove: $a+b$ is an associate of $a'+b'$.
    $a|a'$ and $b|b'$ $\Rightarrow$ $a+b|a'+b'$
    $a'|a$ and $b'|b$ $\Rightarrow$ $a'+b'|a+b$
    $a+b$ and $a'+b'$ are associates.
    (2.) To prove: $ab$ is an associate of $a'b'$.
    $a|a'$ and $b|b'$ $\Rightarrow$ $ab|a'b'$
    $a'|a$ and $b'|b$ $\Rightarrow$ $a'b'|ab$
    $ab$ and $a'b'$ are associates. $\blacksquare$

  • Theorem 5.3: An element of $a$ in $\mathbb{Z}_n$ is a unit in $\mathbb{Z}_n$ if $(a,n)=1$. And all other non zero, non units are zero divisors.
    Proof: Let $a$ is a unit in $\mathbb{Z}_n$
    $\Rightarrow$ $a|1$
    $\Rightarrow$ $\exists$ $b$ in $\mathbb{Z}_n$ such that, $1=ab$
    $\Rightarrow$ $1=ab$
    $\Rightarrow$ $1-ab=0$
    $\Rightarrow$ $1-ab=tn$ for some $t=1,2,3,...$
    $\Rightarrow$ $1=ab+tn$
    $\therefore$ $(a,n)=1$
    Let $a\neq 0$ and $a$ be a non unit in $\mathbb{Z}_n$. Then $(a,n)\neq 1$
    So $(a,n)=d$, for some integer $d$ $\Rightarrow$ $d|a$ and $d|n$
    $\Rightarrow$ $\exists$ $s,t$ non zero integers such that $a=sd$ and $n=td$
    $\Rightarrow$ $at=ns$
    $\Rightarrow$ $at=0$
    $\Rightarrow$ $a$ is a zero divisor in $\mathbb{Z}_n$ $\blacksquare$

Problems 5
  1. Show that $3+\sqrt{5}$ and $1-\sqrt{5}$ are associates in $\mathbb{Z}[\sqrt{5}]$.
    Solution: Consider, $\frac{3+\sqrt{5}}{1-\sqrt{5}}$
    $=\frac{3+\sqrt{5}}{1-\sqrt{5}} \times \frac{1+\sqrt{5}}{1+\sqrt{5}}$
    $=\frac{(3+\sqrt{5})(1+\sqrt{5})}{(1)^2-(\sqrt{5})^2}$
    $=\frac{3+3\sqrt{5}+\sqrt{5}+5}{1-5}$
    $=\frac{8+4\sqrt{5}}{-4}$
    $=-2-\sqrt{5}$ $\in \mathbb{Z}[\sqrt{5}]$
    So, $1-\sqrt{5}|3+\sqrt{5}$
    Now consider $\frac{1-\sqrt{5}}{3+\sqrt{5}}$
    $=\frac{1-\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}$
    $=\frac{3-\sqrt{5}-3\sqrt{5}+5}{(3)^2-(\sqrt{5})^2}$
    $=\frac{8-4\sqrt{5}}{4}$
    $=2-\sqrt{5} \in \mathbb{Z}[\sqrt{2}]$
    So, $3+\sqrt{5}|1-\sqrt{5}$ in $\mathbb{Z}[\sqrt{5}]$
    $\Rightarrow$ $1-\sqrt{5}$ and 3$+\sqrt{5}$ are associates in $\mathbb{Z}[\sqrt{5}]$ $\spadesuit$

  2. Show that $1-\sqrt{5}$ and $1+\sqrt{5}$ are associates in $\mathbb{Z}[\sqrt{5}]$
    Solution: Consider $\frac{1-\sqrt{5}}{1+\sqrt{5}}$
    $=\frac{1-\sqrt{5}}{1+\sqrt{5}}\times \frac{1-\sqrt{5}}{1-\sqrt{5}}$
    $=\frac{(1-\sqrt{5})^2}{1-5}$
    $=\frac{1+5-2\sqrt{5}}{-4}$
    $=\frac{-3}{2}+\frac{1}{2}\sqrt{5} \notin \mathbb{Z}[\sqrt{5}]$
    So $1-\sqrt{5}$ and $1+\sqrt{5}$ are not associates in $\mathbb{Z}[\sqrt{5}]$ $\spadesuit$

  3. Show that $9+4\sqrt{5}$ is a unit in $\mathbb{Z}[\sqrt{5}]$
    Solution: Consider, $\frac{1}{9+4\sqrt{5}}$
    $=\frac{1}{9+4\sqrt{5}} \times \frac{9-4\sqrt{5}}{9-4\sqrt{5}}$
    $=\frac{9+4\sqrt{5}}{(9)^2-(4\sqrt{5})^2}$
    $=\frac{9+4\sqrt{5}}{81-(16 \times 5)}$
    $=\frac{9+4\sqrt{5}}{1}$
    $=9+4\sqrt{5} \in \mathbb{Z}[\sqrt{5}]$
    So $9+4\sqrt{5}|1$
    $\Rightarrow$ $9+4\sqrt{5} $ is a unit in $\mathbb{Z}[\sqrt{5}]$ $\spadesuit$

  4. Show that $2+\sqrt{2}$ is not a unit in $\mathbb{Z}[\sqrt{2}]$
    Solution: Consider, $\frac{1}{2+\sqrt{2}}$
    $=\frac{1}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
    $=\frac{2-\sqrt{2}}{(2)^2-(\sqrt{2})^2}$
    $=\frac{2-\sqrt{2}}{4-2}$
    $=\frac{2-\sqrt{2}}{2}$
    $=1-\frac{1}{2}\sqrt{2}$ $\notin$ $\mathbb{Z}[\sqrt{2}]$
    $2+\sqrt{2} \nmid 1$ in $\mathbb{Z}[\sqrt{2}]$
    $\Rightarrow$ $2+\sqrt{2}$ is not a unit in $\mathbb{Z}[\sqrt{2}]$ $\spadesuit$

  5. Show that $2+\sqrt{5}$ and $2-\sqrt{5}$ are units in $\mathbb{Z}[\sqrt{7}]$
    Solution: Consider $\frac{1}{2+\sqrt{5}}$
    $=\frac{1}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}}$
    $=\frac{2-\sqrt{5}}{4-5}$
    $=-2+\sqrt{5} \in \mathbb{Z}[\sqrt{5}]$
    Hence $2+\sqrt{5}|1$ and so $2+\sqrt{5}$ is a unit in $\mathbb{Z}[\sqrt{5}]$
    Now consider $\frac{1}{2-\sqrt{5}}$
    $=\frac{1}{2-\sqrt{5}}\times \frac{2+\sqrt{5}}{2+\sqrt{5}}$
    $=\frac{2+\sqrt{5}}{4-5}$
    $=-2-\sqrt{5} \in \mathbb{Z}[\sqrt{5}]$
    Hence $2-\sqrt{5}|1$ and so $2-\sqrt{5}$ is a unit in $\mathbb{Z}[\sqrt{5}]$ $\spadesuit$

  6. Find all the units and zero divisors of $\mathbb{Z}_8$
    Solution: $\mathbb{Z}_8=\{0,1,2,3,4,5,6,7 \} $
    Units of $\mathbb{Z}_8$ are $1,3,5,7$
    Zero divisors of $\mathbb{Z}_8$ are $2,4,6$ $\spadesuit$

  7. Find all the units and zero divisors of $\mathbb{Z}_{12}$
    Solution: $\mathbb{Z}_{12}=\{0,1,2,3,4,5,6,7,8,9,10,11 \} $
    Units of $\mathbb{Z}_{12}$ are $1,5,7,11$
    Zero divisors of $\mathbb{Z}_{12}$ are $2,3,4,6,8,9,10$ $\spadesuit$

  8. Find all the units and zero divisors of $\mathbb{Z}_{7}$
    Solution: $\mathbb{Z}_7=\{0,1,2,3,4,5,6 \} $
    Units of $\mathbb{Z}_7$ are $1,2,3,4,5,6$
    No zero divisors in $\mathbb{Z}_7$ $\spadesuit$

  9. Consider an integral domain $\mathbb{Z}(\sqrt{5})$, define norm of an element $N$ in $\mathbb{Z}$ and prove that

    (1.) $N(xy)=N(x)N(y)$ $\forall$ $x,y \in \mathbb{Z}(\sqrt{5}) $.
    (2.) If $x$ is a unit in $\mathbb{Z}(\sqrt{5})$ $\Rightarrow$ $N(x)=\pm 1$.
    (3.) If $N(x)$ is prime, then $x$ is prime.
    (4.) $4+\sqrt{5}$ is a prime in $\mathbb{Z}(\sqrt{5})$.
    Solution: $\mathbb{Z}[\sqrt{5}]=$ $\{a+b\sqrt{5}/ a,b \in \mathbb{Z} \}$ Let $x,y \in $ $\mathbb{Z}[\sqrt{5}]$ then $x=a_1+b_1\sqrt{5}$ and $y=a_2+b_2\sqrt{5}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
    We define a norm, $N$ of an element $x=a_1+b_1\sqrt{5}$ in $\mathbb{Z}[\sqrt{5}]$ to be
    $N(x)=(a_1+b_1\sqrt{5})(a_1-b_1\sqrt{5})=(a_1)^2-(b_1\sqrt{5})^2=a_1^2-5b_1^2$
    i.e., $N(x)=N(a_1+b_1\sqrt{5})=a_1^2-5b_1^2$
    Similarly $N(y)=N(a_2+b_2\sqrt{5})=a_2^2-5b_2^2$
    (1.) To prove: $N(xy)=N(x)N(y)$ $\forall$ $x,y \in \mathbb{Z}(\sqrt{5}) $.
    Consider $N(xy)$
    $=N[(a_1+b_1\sqrt{5})(a_2+b_2\sqrt{5})]$
    $=N[(a_1a_2+5b_1b_2)+(a_1b_2+a_2b_1)\sqrt{5}]$
    $=(a_1a_2+5b_1b_2)^2-5(a_1b_2+a_2b_1)^2$
    $=a_1^2a_2^2+25b_1^2b_2^2+10a_1a_2b_1b_2-5(a_1^2b_2^2+a_2^2b_1^2+2a_1a_2b_1b_2)$
    $=a_1^2a_2^2+25b_1^2b_2^2+10a_1a_2b_1b_2-5a_1^2a_2^2-5a_2^2b_1^2-10a_1a_2b_1b_2$
    $=a_1^2a_2^2-5a_1^2b_2^2+25b_1^2b_2^2-5a_2^2b_1^2$ $.......... (1)$
    Considr $N(x)N(y)$
    $=N(a_1+b_1\sqrt{5})N(a_2+b_2\sqrt{5})$
    $=(a_1^2-5b_1^2)(a_2^2-5b_2^2)$
    $=a_1^2a_2^2-5b_1^2a_2^2-5a_1^2b_2^2+25b_1^2b_2^2$ $.......... (2)$
    From, $(1)$ and $(2)$ we have $N(xy)=N(x)N(y)$
    (2.) To prove: If $x$ is a unit in $\mathbb{Z}(\sqrt{5})$ $\Rightarrow$ $N(x)=\pm 1$
    Let $x$ be a unit in $\mathbb{Z}(\sqrt{5})$
    $\Rightarrow$ $x|1$ in $\mathbb{Z}(\sqrt{5})$
    $\Rightarrow$ $\exists$ $y$ in $\mathbb{Z}(\sqrt{5})$ such that $xy=1$
    $\Rightarrow$ $N(xy)=N(1)$
    $\Rightarrow$ $N(x)N(y)=1$
    $\Rightarrow$ $N(y)=\frac{1}{N(x)}$
    $\Rightarrow$ $N(x)|1$ $\Rightarrow$ $N(x)=\pm 1$ [$\because$ $N(x)$ is an integer]
    (3.) To prove: If $N(x)$ is prime, then $x$ is prime.
    Let $N(x)$ be prime and Suppose $x$ is not a prime. Then $x=n_1n_2$ with $n_1<n$ and $n_2<n$
    $\Rightarrow$ $N(x)=N(n_1n_2)$
    $\Rightarrow$ $N(x)=N(n_1)N(n_2)$
    $\Rightarrow$ $N(x)$ is not a prime, which is a contradiction to the fact that $N(x)$ is prime.
    $\therefore$ $x$ must be prime.
    (4.) To prove: $4+\sqrt{5}$ is a prime in $\mathbb{Z}(\sqrt{5})$.
    $N(4+\sqrt{5})=4^2-5(1)^2=11$, which is a prime
    $\therefore$ $4+\sqrt{5}$ is prime in $\mathbb{Z}(\sqrt{5})$ $\spadesuit$

  10. Consider an integral domain $\mathbb{Z}[i]$, define norm of an element $N$ in $\mathbb{Z}[i]$. Prove that
    (1.) $N(xy)=N(x)N(y)$ $\forall$ $x.y \in \mathbb{Z}[i] $
    (2.) If $N(x)$ is a zero $\Rightarrow$ $x=0$
    (3.) If $x$ is a unit in $\mathbb{Z}[i]$ $\Rightarrow$ $N(x)= 1$
    (4.) Find all the units $\mathbb{Z}[i]$
    Solution: $\mathbb{Z}[i]=$ $\{a+ib/ a,b \in \mathbb{Z} \}$
    Let $x,y \in \mathbb{Z}[i]$ then $x=a_1+ib_1$ and $y=a_2+ib_2$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
    We define a norm $N$ of an element $x=a_1+ib_1$ in $\mathbb{Z}[i]$ to be
    $N(x)=N(a_1+ib_1)=(a_1+ib_1)(a_1-ib_1)=a_1^2-(ib_1)^2=a_1^2+b_1^2$
    i.e., $N(x)=N(a_1+ib_1)=a_1^2+b_1^2$
    Similarly $N(y)=N(a_2+ib_2)=a_2^2+ib_2^2$
    (1.) To prove: $N(xy)=N(x)N(y)$ $\forall$ $x,y \in \mathbb{Z} $.
    Consider $N(xy)$
    $=N[(a_1+ib_1)(a_2+ib_2)]$
    $=N[a_1a_2+ib_1a_2+ia_1b_2+i^2b_1b_2]$
    $=N[(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)]$
    $=(a_1a_2-b_1b_2)^2+(a_1b_2+a_2b_1)^2$
    $=a_1^2a_2^2+b_1^2b_2^2-2a_1a_2b_1b_2+a_1^2b_2^2+a_2^2b_1^2+2a_1a_2b_1b_2$
    $=a_1^2a_2^2+b_1^2b_2^2+a_1^2b_2^2+a_2^2b_1^2$
    $=a_1^2a_2^2+a_1^2b_2^2+b_1^2b_2^2+a_2^2b_1^2$
    $=a_1^2(a_2^2+b_2^2)+b_1^2(a_2^2+b_2^2)$
    $=(a_2^2+b_2^2)(a_1^2+a_2^2)$
    $=N(y)N(x)=N(x)N(y)$
    $\therefore$ $N(xy)=N(x)N(y)$
    (2.) To prove: If $N(x)$ is a zero $\Rightarrow$ $x=0$
    Let $N(a_1+ib_1)=0$
    $\Rightarrow$ $a_1^2+b_1^2=0$
    $\Rightarrow$ $a_1^2=0$ and $b_1^2=0$
    $\Rightarrow$ $a_1=0$ and $b_1=0$ $\Rightarrow$ $x=a_1+ib_1=0+i0=0$
    $\therefore$ $x=0$
    (3.) If $x$ is a unit in $\mathbb{Z}[i]$ $\Rightarrow$ $N(x)= 1$
    Let $x$ be a unit in $\mathbb{Z}[i]$
    $\Rightarrow$ $x|1$
    $\Rightarrow$ $\exists$ $y$ in $\mathbb{Z}[i]$ such that $1=xy$
    $\Rightarrow$ $N(1)=N(x)N(y)$
    $\Rightarrow$ $1=N(x)N(y)$
    $\Rightarrow$ $N(x)|1$ $\Rightarrow$ $N(x)=\pm 1$
    $\Rightarrow$ $N(x)=1$
    (3.) Find all the units $\mathbb{Z}[i]$
    The units of $\mathbb{Z}[i]$ are $-1,1,i,-i$ $\spadesuit$

Exercise 5
  1. Show that $3+2\sqrt{5}$ and $3-2\sqrt{5}$ are associates in $\mathbb{Z}[\sqrt{5}]$

  2. Show that $3+\sqrt{2}$ and $5+4\sqrt{2}$ are associates in $\mathbb{Z}[\sqrt{2}]$

  3. Show that $3+\sqrt{5}$ is not a unit in $\mathbb{Z}[\sqrt{5}]$

  4. Show that $5+\sqrt{7}$ is not a unit in $\mathbb{Z}[\sqrt{7}]$

  5. Find all the units and zero divisors of $\mathbb{Z}_{15}$

  6. Find all the units and zero divisors of $\mathbb{Z}_{17}$

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