$\operatorname{char}R=0 \implies\mathbb{Q} \hookrightarrow R$ Let $R$ be any field, then:
$$\operatorname{char}R=0 \implies \mathbb{Q}  \hookrightarrow R$$
Proof: 
We know that $\mathbb{Q} = Q(\mathbb{Z})=\{[(x,y)]\subseteq\mathbb{Z}\times \mathbb{Z^*}:(x,y) \in \mathbb{Z}\times \mathbb{Z^*}\}$
We consider $φ:Q(\mathbb{Z})\longrightarrow R, \ [(r,s)]\longmapsto rs^{-1}$ and it's easy to prove that $φ$ is homomorphism, well ordered, and $\operatorname{ker}φ=\{[(x,y)]: φ[(x,y)]=0_R\}=\{0_{Q(\mathbb{Z})}\}$. So $φ$ is monomorphism.
Is it right? And if it is, why must $\operatorname{char}R=0$ necessary?
 A: Suppose char$\,R=p>0\;,\;\;(p\;$ a prime), then
$$\phi(p)=p\cdot\phi(1)=0\implies \phi\;\;\text{isn't injective}$$
even assuming you don't divide by zero because of some miracle.
A: The map $\chi\colon\mathbb{Z}\to R$ that maps $n$ to $n1$ (the multiple of the identity element $1\in R$ by the integer $n$) is a ring homomorphism. It is injective if and only if $\operatorname{char}R=0$. In this case this map $\chi$ can be extended to $\varphi\colon \mathbb{Q}\to R$ by
$$
\varphi\left(\frac{m}{n}\right)=\chi(m)\chi(n)^{-1}
$$
(here we use injectivity, so $n\ne0$ ensures $\chi(n)\ne0$). The map $\varphi$ is a ring homomorphism and it is injective, since it maps $1\in\mathbb{Q}$ to $1\in R$ and $\mathbb{Q}$ is a field, so its only ideals are $\{0\}$ and $\mathbb{Q}$.
If $R$ has characteristic $p>0$, then there is no ring homomorphism $f\colon \mathbb{Q}\to R$ such that $f(1)=1$. The reason is very simple:
$$
f(p)=pf(1)=0
$$
so $p\in\ker f$, but since $\ker f\ne\mathbb{Q}$, it follows $\ker f=\{0\}$: contradiction.
