Characteristic of an integral domain must be either $0$ or a prime number.

Hint $\ $ Whenever you have problems understanding such an abstract statement you should look at concrete instances. For example $\,\rm \Bbb Z\ mod\ 3\ $ has characteristic $3$ because $ 3n := n + n + n \equiv 0\ $ for $ n\equiv 0,1,2.\:$ See how you argument breaks down in this simple concrete case, then generalize.

Note that $\,m\cdot a\,$ does not denote an element obtained by applying the ring multiplication to two elements of the ring. Rather, the $m$'th multiple $\,m\cdot a\,$ is the additive analog of the $m$'th power $a^m.$ In the first case we add $m$ copies of $a$ to obtain $\,m\cdot a\,$ and the second we multiply them to get $\,a^m.\,$ They are both well-defined operations in any ring.

One may rigorously define such operations by recursion, viz.

$$\begin{eqnarray} 0\cdot a \,&=&\, 0 \\ (1+n)\cdot a\,&=&\, a + n\cdot a\end{eqnarray}$$

You are correct - $m$ is not an element of the integral domain D. $m$ is a natural number.

But then of course a fair question to consider is how $m \cdot a$ is defined. Somewhere in your text it should define $m \cdot a$ in which $m$ is a natural number and $a$ is an element of an integral domain D as $a$ added to itself $m$ times.

Consider the integral domain $\mathbb{Z}_5$ with addition and multiplication modulo $5$. $\mathbb{Z}_5$ is finite characteristic because each element may be added to itself a number of times (modulo $5$) to reach $0$.

• $0 \equiv 0$ (mod $5$)
• $1+1+1+1+1 = 5 \equiv 0$ (mod $5$)
• $2+2+2+2+2 = 10 \equiv 0$ (mod $5$)
• $3+3+3+3+3 = 15 \equiv 0$ (mod $5$)
• $4+4+4+4+4 = 20 \equiv 0$ (mod $5$)

In a ring $R$ we define $p*x= {(1_R+1_R+1_R+\cdots+1_R)}x= \sum\limits_{i=1}^p x $, so even if $p= 1_R+1_R+1_R+\cdots+1_R=0 $ as a ring element, it is not necessarily $0$ in $\Bbb Z$.

$ \mathbb F_p$ (the integers modulo $p$ a prime, see here) is an integral domain with characteristic $p$. If $R$ was a ring with characteristic $mn$ then $m \ne 0$ and $ n \ne 0$ but $mn$=0, so $R$ could not be an integral domain.

Note in $\Bbb F_p$ the equivalence class of an integer $n \equiv 0 \mod p$ if and only if $n=pm$ for some $m \in \Bbb Z$, now if $xy \equiv 0 \mod p $ then $p|xy$. By primality of $p$ then $p|x$ or $ p|y$. So either $x$ or $y \equiv 0 \mod p$.

The integer is not an element of $D$. You have the canonical $\mathbb{Z}$-action ($0_\mathbb{Z}a = 0_D$, and $(k+1)a = ka + a$) on the abelian group $(D,+)$, and for that $\mathbb{Z}$-action, you have $ma = 0$ for all $a \in D$.