Algebra - Rings and fields

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: Mobile: +91 9845410469

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: $\mathbb{Z}_6$ $=\{0,1,2,3,4,5\}$
    $\therefore$ $0,1,3,4$ 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 $1$.

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