## Algebra - Rings and fields

### 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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