Question #178768

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


1
Expert's answer
2021-04-15T17:24:01-0400

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


I=nZ:nr=0" for some rR,r0.I={n∈Z:nr=0\text{" for some } r∈R,r≠0}.


If nI1 and mZn∈I _1\text{ and } m∈Z , then obviously m,nIm,n∈I . If m,nIm,n∈I and mr=0,ns=0mr=0, ns=0 , with r0r≠0 and s0s≠0 , then


(m+n)(rs)=(mr)s+(ns)r=0(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 0I0∈I , as R≠{0}. Also 1I.1∉I.


Thus I=kZI=kZ for a unique k0k≥0 . If k=abk=ab with 1<a<k and 1<b<k1<a<k \text{ and }1<b<k , then ar≠0 and br≠0 for any rR.r∈R. Let r≠0 with kr=0kr=0 : then ar≠0 and br≠0 because a,bI,a,b∉I, but


(ar)(br)=(ab)r2=(kr)r=0(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 nZ,n0n∈Z, n≠0 , and every rR,r0r∈R, r≠0 , so that every nonzero element of R has infinite order.



Suppose k is prime; we want to show that kr=0kr=0 , for every rRr∈R . Assume the contrary and let kr0=0, with r00, and kr0:thenk_r0=0,\text{ with } r_0≠0, \text{ and } k_r≠0: then


0r0(kr)=(kr0)r=00≠r_0(k_r)=(k_r0)_r=0


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