# Characteristic of an Integral Domain is 0 or a Prime Number

## Problem 228

Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.

## Definition of the characteristic of a ring.

The characteristic of a commutative ring $R$ with $1$ is defined as follows.
Let us define the map $\phi: \Z \to R$ by sending $n \in \Z$ to
$\phi(n)= \begin{cases} \underbrace{1+\cdots+1}_{n\text{ times}} \text{ if } n>0\\ 0 \text{ if } n=0\\ – (\underbrace{1+\cdots+1}_{-n\text{ times}}) \text{ if } n<0. \end{cases}$ Then this map $\phi$ is a ring homomorphism and we define the characteristic $c$ of $R$ to be the integer $c$ such that
$\ker(\phi)=(c).$ (Note that the kernel of $\phi$ is an ideal in $\Z$, and $Z$ is a principal ideal domain (PID), thus such an integer $c$ exists.)

## Proof.

Let us now prove the problem.
Let $c$ be the characteristic of an integral domain $R$.

Then by the first isomorphism theorem with the ring homomorphism $\phi: \Z\to R$ as above, we have an injective homomorphism
\begin{align*}
\Zmod{c}=\Z/\ker(\phi) \to R.
\end{align*}
Since $R$ is an integral domain, $\Zmod{c}$ is also an integral domain.

This yields that $c\Z$ is a prime ideal of $\Z$.
Therefore $c=0$ or $c$ is a prime number.

##### 5 is Prime But 7 is Not Prime in the Ring $\Z[\sqrt{2}]$
In the ring $\Z[\sqrt{2}]=\{a+\sqrt{2}b \mid a, b \in \Z\},$ show that $5$ is a prime element but $7$ is not...