Algebra - Rings and fields

2. Characteristic of a ring

• Definition 2.1:The characteristic of a ring $R$ is a smallest positve integer $n$ such that $na=0$ $\forall$ $a \in R$.
i.e., $na=\smash{\underbrace{a+a+...+a}_{n\hspace{0.1cm}times}}=0$ $\forall$ $a \in R$ denoted by $ch(R)=n$.

• Examples: $ch(\mathbb{Z})=0$, $ch(\mathbb{Q})=0$, $ch(\mathbb{R})=0$, $ch(\mathbb{C})=0$, $ch(\mathbb{Z}_n)=n$

• Theorem 2.1: The characteristic of an integral domain is either $0$ or prime number.
Proof: Let $D$ be an integral domain. To prove $ch(D)=0$ or $ch(D)=n$ where $n$ is a prime number.
If $ch(D)=0$ then nothing to prove.
Let $ch(D)=n$ where $n$ is a smallest positive integer such that $na=0$ $\forall$ $a \in D$.
Suppose $n$ is a prime. Then we can have $n=n_1n_2$ where $n_1\neq 1$, $n_2\neq 1$ and $n_1<n$, $n_2<n$
Let $a \in D$ $\Rightarrow$ $a^2 \in D$. Then $na^2=0$
$\Rightarrow$ $(n_1n_2)a^2=0$
$\Rightarrow$ $\smash{\underbrace{a^2+a^2+...+a^2}_\text{$n_1n_2$times}}=0$

$\Rightarrow$ $\smash{\underbrace{(a+a+...+a)}_\text{$n_1$times}}$ $\smash{\underbrace{(a+a+...+a)}_\text{$n_2$times}}$ $=0$

$\Rightarrow$ $(n_1a)(n_2a)=0$
$\Rightarrow$ $n_1a=0$ or $n_2a=0$ which is a contradiction to the fact $ch(D)=n$ [$\because$ $n_1<n$ and $n_2<n$] Hence $n$ must be prime. $\blacksquare$

• Definiton 2.2: An element $a$ of a ring $R$ is said to nilpotent if $\exists$ a positive integer $n$ such that $a^n=0$.

• Definition 2.3: An element $a$ of a ring $R$ is said to be idempotent if $a^2=a$.

• Definition 2.4: A ring $R$ is said to be boolean ring if $a^2=a$ $\forall$ $a \in R$.

• Problem 2.1: In a ring $R$, if $(ab)^2=a^2b^2$ $\forall$ $a,b \in R$. Then show that $R$ is commutative.
Solution: Given $(ab)^2=a^2b^2$ $\forall$ $a,b \in R$
$\Rightarrow$ $(ab)(ab)=(aa)(bb)$
$\Rightarrow$ $a[b(ab)]=a[a(bb)]$
$\Rightarrow$ $b(ab)=a(bb)$ [by left cancellation law in ring $R$]
$\Rightarrow$ $(ba)b=(ab)b$ [by associative law in $R$]
$\Rightarrow$ $ba=ab$ [by right cancellation in $R$]
$\therefore$ $ab=ba$ $\forall$ $a,b \in R$ $\spadesuit$

• Problem 2.2: If $a^2=a$ $\forall$ $a \in R$, then show that
(1.) $a+a=0$
(2.) $a+b=0$ $\Rightarrow$ $a=b$
(3.) $R$ is commutative
Solution: (1.) Given, $a^2=a$ $\forall$ $a \in R$ $\Rightarrow$ $a+a \in R$
So $(a+a)^2=a+a$
$\Rightarrow$ $(a+a)(a+a)=a+a$
$\Rightarrow$ $a^2+a^2+a^2+a^2=a+a$ [by distributive law in ring $R$]
$\Rightarrow$ $a+a+a+a=a+a$ [$\because$ Given, $a^2=a$ $\forall$ $a \in R$]
$\Rightarrow$ $a+a+a=a$ [by left cancellation in $R$]
$\Rightarrow$ $a+a+a=a+0$
$\Rightarrow$ $a+a=0$ [by left cancellation in $R$]
(2.) Given $a+b=0$ and from (1.) we have $a+a=0$
$\Rightarrow$ $a+b=a+a$
$\Rightarrow$ $b=a$ [by right cancellation in $R$]
(3.) Given $a^2=a$ $\forall$ $a \in R$
Let $a,b \in R$ $\Rightarrow$ $a^2=a$ and $b^2=b$ also $a,b \in R$ $\Rightarrow$ $ab \in R$
So we have $(ab)^2=ab$
$\Rightarrow$ $(ab)^2=a^2b^2$
$\Rightarrow$ $(ab)(ab)=(aa)(bb)$
$\Rightarrow$ $a[b(ab)]=a[a(bb)]$
$\Rightarrow$ $b(ab)=a(bb)$
$\Rightarrow$ $(ba)b=(ab)b$ $\Rightarrow$ $ba=ab$ $\forall$ $a,b \in R$
$\therefore$ $R$ is commutative. $\spadesuit$

• Problem 2.3: Find all the idempotent elements of $(\mathbb{Z}_6,\oplus_6,\otimes_6)$
Solution: $\mathbb{Z}_6$ $=\{0,1,2,3,4,5\}$
$0^2=0\otimes_60=0$
$1^2=1\otimes_61=1$
$2^2=2\otimes_62=4$
$3^2=3\otimes_63=3$
$4^2=4\otimes_64=4$
$5^2=5\otimes_65=1$
$\therefore$ $0,1,3,4$ are idempotent elements in $\mathbb{Z}_6$. $\spadesuit$

• Exercise 2.1: Prove that a skew field has no zero divisors.

• Exercise 2.2: Find all the idempotents and nilpotents of $(\mathbb{Z}_4,\oplus_4,\otimes_4)$.

• Exercise 2.3: Show that the only idempotents of an integral domain is $0$ and $1$.

This website is only viewable in landscape mode.