Field with characteristic zero is vector space over $\mathbb{Q}$ Here’s my problem:

Prove that a field R of characteristic $0$ is a vector space over $\mathbb{Q}$.

I am unsure how to proceed here, as checking the vector space axioms seems wrong. 
Here are my thoughts:
Since $\operatorname{char}(R) =0$, $\mathbb{Z}$ is isomorphic to some subdomain of $R$. Now $\mathbb{Z}$ is itself not a field but has field of fractions $\mathbb{Q}$... 
Can anyone help me out here?

Edit:
Based on the comments I received:
$\mathbb{Z}$ is isomorphic to some subdomain of $R$. Extend this to a field of fractions inside $R$. This extension will be isomorphic to $\mathbb{Q}$ as this is the field of fractions of $\mathbb{Z}$. Then verifying the vector space axioms (is this necessary or is there  shortcut?) gives the desired result. 
Is this correct?
 A: Hint: Because $R$ is a field of characteristic $0$, every nonzero element of $\Bbb{Z}\subset R$ is invertible.
Once you have shown that $\Bbb{Q}\subset R$ all the vector space axioms are easily verified because $R$ is a field that contains $\Bbb{Q}$ as a subfield. Note that half of the axioms are already satisfied a priori because $R$ is a field. It might even be worth proving that in general:

If $R$ is a field and $S\subset R$ is a subfield then $R$ is a vector space over $S$.

A: Consider $Z \subseteq R$ defined by
$$Z= \{\dots,-1-1,-1,0,1, 1+1, \dots\}$$
and put $$Q := \{ab^{-1}: a,b \in Z, b \neq 0\}$$
Then $\mathbb{Q}\cong Q$. Let $f:  \mathbb{Q} \to Q$ be this isomorphism. We can view $R$ as a $Q$-vector space by letting $Q$ act on $R$ by ring multiplication. 
Then $$q. r = \phi(q)r; \quad q \in \mathbb{Q}, r \in R$$
defines a $\mathbb{Q}$-scalar multiplication on $R$ and we get that $R$ is a $\mathbb{Q}$-vector space.
So basically the idea is to find an isomorphic copy of $\mathbb{Q}$ inside $R$ and use this to get the vector space structure you like.
