Multiplication in $\mathbb{Z}$ corresponds to multiplication in rings?

Question

So I was trying to prove that the characteristic of an integral domain is either $0$ or prime. I got stuck, so I searched for a proof and I came across the following proof online

Now I almost want to accept this proof, except for the following (possibly silly and pedantic) issue. In the above proof we know that $n_0 \in \mathbb{N} \subseteq \mathbb{Z}$, so since $n$ is not prime, we factorize $n = m \cdot k$ (where $\cdot$ represents multiplication on the integers in the ring $(\mathbb{Z}, +, \cdot)$). Now in the above proof the following is asserted $$n(1_D) = \underbrace{1_D + \dots + 1_D}_{n \text{ times}} =0_D \implies m\cdot k(1_D) = \underbrace{\left(1_D + \dots + 1_D\right)}_{m \text{ times}} \ \bullet \underbrace{\left(1_D + \dots + 1_D\right)}_{k \text{ times}}$$

Now seemingly it seems that multiplication of $m$ and $k$ in $\mathbb{Z}$ is inducing (ring) multiplication of elements in $D$ (when I thought we'd only end up with addition of the $1_D$'s $mk$ times).

Is there a reason why this happens?

Answer 1

This happens simply because by induction and definition of $k\cdot 1_D$, one may prove $(n\cdot 1_D)\times (m\cdot 1_D) = nm\cdot 1_D$.

The proof by induction on, say, $m$ is easy : $(n\cdot 1_D)((m+1)\cdot 1_D) = (n\cdot 1_D)(m\cdot 1_D + 1_D) =...$

Answer 2

As @Bernard pointed out in the comment above, for any commutative ring $R$ the map $\varphi : \mathbb{Z} \to R$ defined by $$\varphi(n) = n\cdot 1_R = \underbrace{1_R + \dots + 1_R}_{n \text{ times}}$$ is a ring homomorphism. Thus in my problem above where we have as our ring the integral domain $(D, +, \bullet)$ since we can factorize $n = m\cdot k$ in the ring $(\mathbb{Z}, +, \cdot)$ then we have \begin{align*} mn \cdot 1_R &= \varphi(m \cdot n) \\ &= \varphi(m) \bullet \varphi(n) \\ &= (m\cdot 1_R) \bullet (n\cdot 1_R) \\ &= \underbrace{\left(1_D + \dots + 1_D\right)}_{m \text{ times}} \ \bullet \underbrace{\left(1_D + \dots + 1_D\right)}_{k \text{ times}} \end{align*}