Field with characteristic zero is vector space over $\mathbb{Q}$

Question

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?

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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?

Answer 1

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 $S$.