Answer to Question #178768 in Abstract Algebra for Zainab

Question

Suppose that R is a ring with identity such that char R=n>0 .if n is not prime, show that char R=n>0.if n is not prime, show that R has divisors of zero

Answer 1

Let's not assume R has a unit and is not the zero ring. Consider

"I={n\u2208Z:nr=0\\text{" for some } r\u2208R,r\u22600}."

If $n∈I _1\text{ and } m∈Z$ , then obviously $m,n∈I$ . If $m,n∈I$ and $mr=0, ns=0$ , with $r≠0$ and $s≠0$ , then

$(m+n)(rs)=(mr)s+(ns)r=0$

and rs≠0. Therefore m+n∈I and we have proved that I is an ideal of Z, because obviously $0∈I$ , as R≠{0}. Also $1∉I.$

Thus $I=kZ$ for a unique $k≥0$ . If $k=ab$ with $1<a<k \text{ and }1<b<k$ , then ar≠0 and br≠0 for any $r∈R.$ Let r≠0 with $kr=0$ : then ar≠0 and br≠0 because $a,b∉I,$ but

$(ar)(br)=(ab)r2=(kr)r=0$

which is a contradiction. Thus either k=0 or k is a prime.

In the first case nr≠0 for every $n∈Z, n≠0$ , and every $r∈R, r≠0$ , so that every nonzero element of R has infinite order.

Suppose k is prime; we want to show that $kr=0$ , for every $r∈R$ . Assume the contrary and let $k_r0=0,\text{ with } r_0≠0, \text{ and } k_r≠0: then$

$0≠r_0(k_r)=(k_r0)_r=0$