Integral domain - Embedding

Question

Let $R$ be an integral domain and the homomorphism \begin{align} \phi\colon \mathbb{Z} &\rightarrow R \\ n &\mapsto n \cdot 1_R \end{align}

What does it mean that if $\ker \phi =\{0\}$ then $\phi$ is an embedding ($\operatorname{char}R=0$) and if $\ker \phi \neq \{0\}$ then $\phi$ is not an embedding?

Also why does it stand that if $\operatorname{char}R =p$ ($p$: prime) then there is an embedding \begin{align} \widetilde{\phi}: \mathbb{Z}_p &\hookrightarrow R \\ \overline{a} &\mapsto a \cdot 1_R \end{align} ?

Answer 1

Embedding here means injective. For the second part, it is just the map induced by $\phi$ on the quotient (by the universal property of the quotient ring), which is injective as you mod out by the kernel.

Answer 2

If $\ker\phi \neq {0}$, then there exists $m\in\mathbb Z$ such that $\stackrel{(m)}{1_R+\cdots + 1_R} = 0$. A bit of reasoning shows that the smallest such $m$ must be prime. Call it $p$, and we now have $\ker\phi = p\mathbb Z$ (because we chose the smallest $p$, and $\mathbb Z$ is PID). Therefore, we have $$\mathbb Z/\ker\phi = \mathbb Z/p\mathbb Z \simeq \mathrm{Im}\phi$$

Which is given by $\tilde a\mapsto \stackrel{(a)}{1_R+\cdots + 1_R}$ (easy to verify). Compose with the inclusion $$\mathrm{Im}\phi\hookrightarrow R \atop x\mapsto x$$

to get an injective homomorphism, otherwise known as an embedding: $$\widetilde{\phi}: \mathbb{Z}_p \hookrightarrow R$$