Let $R$ be an integral domain and consider the set $\mathbb{Z1}$ of all integral multiples of the identity element: $$\mathbb{Z1}=\{n1:n\in\mathbb{Z}\}.$$ It is required to prove that $\mathbb{Z1}$ is a field if and only if $R$ has positive characteristic.

The following is my attempt.

$(\leftarrow)$ Suppose $R$ has positive characteristic $p$. Then $p$ is prime as the characteristic of an integral domain is ether $0$ or a prime number. But $\mathbb{Z1}$ is a subdomain of $R$ since $R$ is an integral domain. Let $n1$ be a nonzero element in $\mathbb{Z1}$. Then $p1\in\mathbb{Z1}$ and since $\operatorname{gcd}(n,p)=1$ we have $1=na+pb=(na+pb)1=n(a1)+b(p1)=n(a1)=(n1)(a1)$ for some $a,b\in\mathbb{Z}$. Thus $n1$ is invertible and hence $\mathbb{Z1}$ is a field.

When proving the forward direction I was stuck. The idea I had was that if $\mathbb{Z1}$ is a field (and thus a subdomain of $R$) then $\operatorname{Char} \mathbb{Z1}=\operatorname{Char}R$ due to the fact that characteristic of an integral domain is either $0$ or a prime number. I cannot find a way to proceed. Could someone please help? Thanks.


1 Answer 1



$\mathbf Z_1$ is a homomorphic image of $\mathbf Z$, hence a quotient $\mathbf Z/n\mathbf Z$ of $\mathbf Z$, and an integral domain.

  • $\begingroup$ $\phi:\mathbf {Z}\to \mathbf{R}$ given by $\phi(n)=n.1$ is a homomorphism and $\mathbf{Z}/\operatorname{ker} \phi \cong \mathbf{Z_1}$. Since $\operatorname{ker}\phi$ is an ideal of $\mathbf{Z}$ $\operatorname{ker}\phi=(n)$. Since $\mathbf{R}$ is an integral domain $n$ is prime or $0$. If $n=0$ then $\mathbf{Z_1}\cong \mathbf{Z}$ which is impossible as $\mathbf{Z}$ is not a field. So $\mathbf{Z_1}\cong \mathbf{Z_n}$ and hence $\mathbf{Z}$ is finite and $\operatorname{Char}\mathbf{Z_1}>0$. Is this argument alright? $\endgroup$
    – Janitha357
    Jul 1, 2017 at 10:04
  • $\begingroup$ Almost. Just don't use $n$ for both defining $\phi$ an the generator of $\ker \phi$. $\endgroup$
    – Bernard
    Jul 1, 2017 at 10:24

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