I saw a statement: every field of characteristic 0, with its underlying additive group structure, is divisible.
I met with troubles in checking the above statement.
Suppose $\operatorname{char}(F)=0$, for each $g\in F$ and for each $n$, how to find $h\in F$ such that $nh=g$
The solution is:
$$h=(n 1)^{-1}\cdot g$$
Here "$1$" denotes the multiplicative identity of $F$, "$nx$" means "add $x$ to itself $n$ times" (as usual) and "$\cdot$" denotes the field multiplication. Note that when $\operatorname{char}(F)=0$ then $n 1$ is invertible in $F$ when $n$ is a positive natural.
The rest follows from the observation that for $x\in F$ and $n\in\mathbb{N}$ we have $nx=(n1)\cdot x$. And thus
$$nh=(n1)\cdot h=(n1)\cdot (n1)^{-1}\cdot g=g$$