Exercise 3.13

Prove that the characteristic of an integral domain is a prime integer.

Answer

Let $R$ be an integral domain with characteristic $n$. First we will prove that $n\ne 1$. Suppose $n=1$, then it follows that $1=0$, but this is a contradiction since an integral domain is nontrivial. Thus, $n\ne 1$. If $n=0$, then $n$ is prime. So, consider the case $n>1$. Suppose $n|ab$. Then, $ab=nk$ for some integer $k$. In the ring $R$, this equation becomes, $(ab)1_R=n(k{1}_R)=0$ since the characteristic is $n$. Now, we have $(a{1}_R)(b{1}_R)=0$. Since $R$ is an integral domain, we have either $a{1}_R=0$ or $b{1}_R=0$. Thus, by the characteristic divisibility lemma, we have $n|a$ or $n|b$. So, $n$ is prime. $■$