Question

The following definition is needed for Exercises 41-43. Let R be a ring with identity. If there is a smallest positive integer n such that$n{I}_R=0_R$ , then R is said to have a characteristic n. If no such n exists, R is said to have a characteristic zero.

43. Let R be a ring with an identity of characteristic,$n>0$

1. Prove that$na=0_R$ for every $a\in R$.
2. If $R$is an integral domain, prove that n is prime.

Short answer

1. It is proved that$na=0_R$ for every $a\in R$.
2. It is proved that$n$ is prime.

Step-by-step solution

Show that na=0R for every a∈R 

Calculate$na$ as follows:

$\begin{array}{rcl}na& =& \sum _{i=1}^{n}a\\ & =& \sum _{i=1}^n{1}_{R}a\\ & =& \left(\underset{i=1}{\overset{n}{\sum 1_R}}\right)a\\ & =& \left(n{1}_R\right)a\\ & =& 0_{R}a\\ & =& 0_R\end{array}$

Hence, it is proved that$na=0_R$ for every$a\in R$ .

Show that n is prime

b)

Assume that $n$is composite, for $n=rs$with both$r$ and$s$ greater than 1.

It is observed that$r,s<n$ .From the minimality of the characteristic, $r{1}_R\ne 0_R$and$s{1}_R\ne 0_R$ . However,

$\begin{array}{rcl}\left(r{1}_R\right)\left(s{1}_R\right)& =& rs{1}_R\\ & =& n{1}_R\\ & =& 0_R\end{array}$

In other words,$r_1$ and$s_1$ are both zero-divisors which contradicts the fact that $R$is an integral.

Hence, it is proved that$n$ is prime.